[{"Name":"Introduction to DC Circuits","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Current","Duration":"8m 18s","ChapterTopicVideoID":21479,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21479.jpeg","UploadDate":"2020-04-21T13:47:39.3400000","DurationForVideoObject":"PT8M18S","Description":null,"MetaTitle":"Current: Video + Workbook | Proprep","MetaDescription":"DC Circuits - Introduction to DC Circuits. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/dc-circuits/introduction-to-dc-circuits/vid22296","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.355","Text":"Hello, in this video,"},{"Start":"00:02.355 ","End":"00:05.730","Text":"we\u0027re going to be discussing what current is."},{"Start":"00:05.730 ","End":"00:10.815","Text":"If we have some circuit where we have an electronic components,"},{"Start":"00:10.815 ","End":"00:18.225","Text":"so the current is the amount of positive charge which passes in the unit of time."},{"Start":"00:18.225 ","End":"00:24.944","Text":"How many positive charges move through my electronic component or through any point,"},{"Start":"00:24.944 ","End":"00:31.035","Text":"some random point on my circuit in every unit of time."},{"Start":"00:31.035 ","End":"00:33.270","Text":"Now in mathematical terms,"},{"Start":"00:33.270 ","End":"00:39.230","Text":"our current is represented by the letter I and is equals to dq,"},{"Start":"00:39.230 ","End":"00:46.200","Text":"which is the amount of positive charge passing divided by dt per unit of time."},{"Start":"00:46.200 ","End":"00:48.195","Text":"This is our equation."},{"Start":"00:48.195 ","End":"00:53.080","Text":"Now notice that when we\u0027re dealing with the same wire,"},{"Start":"00:53.080 ","End":"00:56.545","Text":"our wire isn\u0027t splitting."},{"Start":"00:56.545 ","End":"01:00.110","Text":"If we measure our current over here,"},{"Start":"01:00.110 ","End":"01:02.345","Text":"and we measure our current over here,"},{"Start":"01:02.345 ","End":"01:05.375","Text":"and over here or anywhere else,"},{"Start":"01:05.375 ","End":"01:08.005","Text":"our current is going to be equal."},{"Start":"01:08.005 ","End":"01:10.145","Text":"That\u0027s because if we look at this point,"},{"Start":"01:10.145 ","End":"01:14.420","Text":"we have some I passing through and we\u0027re assuming that in"},{"Start":"01:14.420 ","End":"01:20.865","Text":"the wire our charges can\u0027t start clumping up and gets stuck."},{"Start":"01:20.865 ","End":"01:23.700","Text":"I currents carry on moving,"},{"Start":"01:23.700 ","End":"01:28.340","Text":"and so over here the same amount of current is moving as well."},{"Start":"01:28.340 ","End":"01:31.540","Text":"Unless another wire is added over here,"},{"Start":"01:31.540 ","End":"01:33.775","Text":"and then we have a parallel electric circuit,"},{"Start":"01:33.775 ","End":"01:35.530","Text":"which we\u0027ll speak about later."},{"Start":"01:35.530 ","End":"01:39.355","Text":"If we\u0027re just dealing with 1 wire with components attached,"},{"Start":"01:39.355 ","End":"01:43.090","Text":"the current along this 1 wire is going"},{"Start":"01:43.090 ","End":"01:47.200","Text":"to be equal at all the different points in the wire."},{"Start":"01:47.200 ","End":"01:52.135","Text":"This comes from the idea of conservation of charge,"},{"Start":"01:52.135 ","End":"01:56.170","Text":"which states exactly what we\u0027ve been speaking about up until now."},{"Start":"01:56.170 ","End":"01:58.365","Text":"Now, the units for current,"},{"Start":"01:58.365 ","End":"02:00.580","Text":"I call the ampere."},{"Start":"02:01.100 ","End":"02:09.485","Text":"Let\u0027s write this and that is equal to C for column or charge per second,"},{"Start":"02:09.485 ","End":"02:10.970","Text":"which we can see from here,"},{"Start":"02:10.970 ","End":"02:15.004","Text":"dq is given by charge which is in coulombs,"},{"Start":"02:15.004 ","End":"02:18.805","Text":"by dt, which is given in seconds."},{"Start":"02:18.805 ","End":"02:21.770","Text":"Up until now we\u0027ve said that current is the amount of"},{"Start":"02:21.770 ","End":"02:25.055","Text":"positive charge that\u0027s going through a circuit."},{"Start":"02:25.055 ","End":"02:27.830","Text":"But what if we have negative charge?"},{"Start":"02:27.830 ","End":"02:33.155","Text":"Let\u0027s say that we have a negative charge going in this direction."},{"Start":"02:33.155 ","End":"02:36.440","Text":"That\u0027s the exact same thing as saying that we have"},{"Start":"02:36.440 ","End":"02:41.155","Text":"a positive charge going in the opposite direction."},{"Start":"02:41.155 ","End":"02:46.550","Text":"Then if we want to see what the total amount of charges that\u0027s gone past here,"},{"Start":"02:46.550 ","End":"02:50.155","Text":"I count how many positive charges have gone"},{"Start":"02:50.155 ","End":"02:55.640","Text":"past and then I subtract the amount of negative charges that have gone past."},{"Start":"02:55.640 ","End":"03:00.860","Text":"If I have 7 positive charges going in this direction and 1 negative charge,"},{"Start":"03:00.860 ","End":"03:04.535","Text":"which means that I have a charge going in the opposite direction."},{"Start":"03:04.535 ","End":"03:14.375","Text":"Then in total, I\u0027ll have had 7 minus 1 which is 6 positive charges going past this point."},{"Start":"03:14.375 ","End":"03:17.930","Text":"Now something that is important to note is"},{"Start":"03:17.930 ","End":"03:22.565","Text":"that even though we have the movement of charge within our wire,"},{"Start":"03:22.565 ","End":"03:26.540","Text":"it doesn\u0027t mean that there\u0027s a net total charge."},{"Start":"03:26.540 ","End":"03:33.215","Text":"The net charge of every point in the wire is going to be equal to 0."},{"Start":"03:33.215 ","End":"03:40.180","Text":"Let\u0027s imagine that we\u0027re zooming in on this section of the wire."},{"Start":"03:40.180 ","End":"03:42.725","Text":"Let\u0027s draw it in black,"},{"Start":"03:42.725 ","End":"03:48.045","Text":"it\u0027s like some tube or some cylinder."},{"Start":"03:48.045 ","End":"03:51.020","Text":"Inside there\u0027s lots of different charges."},{"Start":"03:51.020 ","End":"03:55.465","Text":"There\u0027s pluses, there\u0027s minuses and lots and lots and lots inside."},{"Start":"03:55.465 ","End":"04:00.245","Text":"How do we get current if the net charge is equal to 0,"},{"Start":"04:00.245 ","End":"04:08.150","Text":"so that\u0027s because we\u0027re going to our have positive charges moving in 1 direction."},{"Start":"04:08.150 ","End":"04:10.360","Text":"Let\u0027s say this direction,"},{"Start":"04:10.360 ","End":"04:16.565","Text":"and our negative charges are going to be moving in the opposite direction."},{"Start":"04:16.565 ","End":"04:21.170","Text":"We can see that this movement of charges is causing a current."},{"Start":"04:21.170 ","End":"04:24.920","Text":"However, if I stop my time,"},{"Start":"04:24.920 ","End":"04:29.495","Text":"I pause the time and I take a look at the center over here,"},{"Start":"04:29.495 ","End":"04:32.390","Text":"and I count how many charges I have."},{"Start":"04:32.390 ","End":"04:36.200","Text":"I\u0027ll see that the amount of positive and negative charges"},{"Start":"04:36.200 ","End":"04:40.265","Text":"that I have are still going to cancel out,"},{"Start":"04:40.265 ","End":"04:47.345","Text":"which means that the charge of this section of wire is still going to be equal to 0."},{"Start":"04:47.345 ","End":"04:49.250","Text":"Even though we can clearly see that"},{"Start":"04:49.250 ","End":"04:53.195","Text":"my negative charges are moving in 1 direction and my positive and the other,"},{"Start":"04:53.195 ","End":"04:56.230","Text":"which means that I have current."},{"Start":"04:56.230 ","End":"05:00.125","Text":"Now, we\u0027ve spoken about a series circuit"},{"Start":"05:00.125 ","End":"05:04.355","Text":"where we just have 1 wire going around in some loop."},{"Start":"05:04.355 ","End":"05:10.895","Text":"We spoke about that means that the current at every single point is going to be the same."},{"Start":"05:10.895 ","End":"05:16.745","Text":"However, what happens if we add some wire which splits our circuit?"},{"Start":"05:16.745 ","End":"05:20.075","Text":"Then we\u0027re dealing with a parallel circuit."},{"Start":"05:20.075 ","End":"05:23.335","Text":"Let\u0027s see what happens to the current in that case."},{"Start":"05:23.335 ","End":"05:26.030","Text":"Here we have a section of a circuit where we have"},{"Start":"05:26.030 ","End":"05:30.740","Text":"some current I_1 going through this wire."},{"Start":"05:30.740 ","End":"05:33.200","Text":"Suddenly our wire splits,"},{"Start":"05:33.200 ","End":"05:39.915","Text":"and we have a branch going this way where some current I_2 going through it,"},{"Start":"05:39.915 ","End":"05:46.525","Text":"and the branch going this way with some current I_3 going through it."},{"Start":"05:46.525 ","End":"05:51.385","Text":"Now, because we know about the idea of conservation of charge,"},{"Start":"05:51.385 ","End":"05:56.240","Text":"which means that charge is never destroyed, it\u0027s always conserved."},{"Start":"05:56.240 ","End":"06:01.450","Text":"If we look again at the different points on this wire over here,"},{"Start":"06:01.450 ","End":"06:04.075","Text":"this first wire before the splits."},{"Start":"06:04.075 ","End":"06:09.985","Text":"Our current over here is still going to be equal at each 1 of these points."},{"Start":"06:09.985 ","End":"06:16.120","Text":"However, if we then go after this junction over here,"},{"Start":"06:16.120 ","End":"06:18.730","Text":"our current is going to change,"},{"Start":"06:18.730 ","End":"06:21.220","Text":"and our I_2 is not going to be equal to"},{"Start":"06:21.220 ","End":"06:25.670","Text":"our I_1 and I_3 is not going to be equal to our I_1."},{"Start":"06:25.670 ","End":"06:28.650","Text":"We can see then again that through this wire,"},{"Start":"06:28.650 ","End":"06:33.350","Text":"our current at these separate points is going to be equal to I_2."},{"Start":"06:33.350 ","End":"06:36.260","Text":"At this wire, our current at"},{"Start":"06:36.260 ","End":"06:39.830","Text":"these different points along the wire is going to be equal to I_3."},{"Start":"06:39.830 ","End":"06:42.815","Text":"Due to the conservation of charge,"},{"Start":"06:42.815 ","End":"06:50.740","Text":"we can say that our current I_1 is equal to each of the currents that it splits into."},{"Start":"06:50.740 ","End":"06:52.785","Text":"The addition of all of them together,"},{"Start":"06:52.785 ","End":"06:58.245","Text":"so I_1 is going to be equal to I_2 plus I_3."},{"Start":"06:58.245 ","End":"07:02.135","Text":"Or another way that\u0027s common to write it is that I_1"},{"Start":"07:02.135 ","End":"07:09.730","Text":"minus I_2 minus I_3 is equal to 0."},{"Start":"07:09.730 ","End":"07:12.260","Text":"Both of these are the exact same."},{"Start":"07:12.260 ","End":"07:15.770","Text":"Notice we\u0027ve just moved these to the other side of the equals sign."},{"Start":"07:15.770 ","End":"07:17.750","Text":"This must always be true,"},{"Start":"07:17.750 ","End":"07:23.255","Text":"if there was also more splits into I_4 and I_5 and whatever it might be,"},{"Start":"07:23.255 ","End":"07:25.590","Text":"all of the subsequent splits,"},{"Start":"07:25.590 ","End":"07:28.295","Text":"so I_2, 3, 4, and 5, etc."},{"Start":"07:28.295 ","End":"07:30.260","Text":"When we add all of them up,"},{"Start":"07:30.260 ","End":"07:33.820","Text":"they will be equal to our original current I_1."},{"Start":"07:33.820 ","End":"07:40.850","Text":"A handy way of remembering this is that the amount of charge that goes into a junction,"},{"Start":"07:40.850 ","End":"07:44.630","Text":"where this is our junction over here in gray,"},{"Start":"07:44.630 ","End":"07:47.570","Text":"the connecting point between all of the wires."},{"Start":"07:47.570 ","End":"07:53.515","Text":"The charge into the junction is equal to the charge out of the junction."},{"Start":"07:53.515 ","End":"07:57.320","Text":"As we set our I_1 is equal to the amount of"},{"Start":"07:57.320 ","End":"08:01.460","Text":"charge per unit time which enters the junction."},{"Start":"08:01.460 ","End":"08:08.690","Text":"Which means that the amount of charge per unit time which exits in this side"},{"Start":"08:08.690 ","End":"08:11.855","Text":"plus the amount of charge per unit time which"},{"Start":"08:11.855 ","End":"08:16.400","Text":"exits on this side is going to be equal to what went in."},{"Start":"08:16.400 ","End":"08:19.500","Text":"That\u0027s the end of this lesson."}],"ID":22296},{"Watched":false,"Name":"Ohms Law And Resistors","Duration":"15m 2s","ChapterTopicVideoID":21480,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"Hello. In this lesson,"},{"Start":"00:01.845 ","End":"00:05.130","Text":"we\u0027re going to be speaking about Ohm\u0027s law and how we can"},{"Start":"00:05.130 ","End":"00:08.909","Text":"connect resistors in series and in parallel."},{"Start":"00:08.909 ","End":"00:14.400","Text":"Here we can see that we have some battery which has a voltage."},{"Start":"00:14.400 ","End":"00:22.170","Text":"Now, the voltage inside the battery causes or induces electric field in the circuit."},{"Start":"00:22.170 ","End":"00:27.930","Text":"Now, the electric field produces a force on all of the charges inside the wire,"},{"Start":"00:27.930 ","End":"00:32.175","Text":"or inside the circuit and therefore moving the charges."},{"Start":"00:32.175 ","End":"00:38.785","Text":"Then this movement of charges causes a current to flow through the circuit."},{"Start":"00:38.785 ","End":"00:41.150","Text":"That is what our voltage does."},{"Start":"00:41.150 ","End":"00:44.990","Text":"Now here we have a resistor in this circuit."},{"Start":"00:44.990 ","End":"00:50.090","Text":"Now, the resistor is denoted by this triangular type squiggly line."},{"Start":"00:50.090 ","End":"00:53.810","Text":"The resistor is some kind of electrical components"},{"Start":"00:53.810 ","End":"00:59.179","Text":"which causes resistance in the circuit."},{"Start":"00:59.179 ","End":"01:03.050","Text":"Resistors are generally used to reduce current flow,"},{"Start":"01:03.050 ","End":"01:04.610","Text":"but there\u0027s also many,"},{"Start":"01:04.610 ","End":"01:07.750","Text":"many other uses which we won\u0027t go into now."},{"Start":"01:07.750 ","End":"01:09.200","Text":"That\u0027s a resistor."},{"Start":"01:09.200 ","End":"01:14.270","Text":"Now, generally, we\u0027re going to be using a resistor which is an ideal resistor."},{"Start":"01:14.270 ","End":"01:15.949","Text":"Now, what\u0027s an ideal resistor?"},{"Start":"01:15.949 ","End":"01:20.105","Text":"An ideal of resistor abides by Ohm\u0027s law."},{"Start":"01:20.105 ","End":"01:25.130","Text":"Ohm\u0027s law says that the relationship between voltage and"},{"Start":"01:25.130 ","End":"01:30.140","Text":"current is linear and depends on our resistance."},{"Start":"01:30.140 ","End":"01:33.605","Text":"Ohm\u0027s law is given by V=IR,"},{"Start":"01:33.605 ","End":"01:38.360","Text":"where V is the voltage over the resistor."},{"Start":"01:38.360 ","End":"01:42.620","Text":"If I were to add in over here some kind"},{"Start":"01:42.620 ","End":"01:46.610","Text":"of voltmeter to measure the voltage across my resistor,"},{"Start":"01:46.610 ","End":"01:49.640","Text":"so my voltage across my resistor would be equal"},{"Start":"01:49.640 ","End":"01:53.150","Text":"to the current running through the circuit, and of course,"},{"Start":"01:53.150 ","End":"01:55.100","Text":"because this is a circuit in series,"},{"Start":"01:55.100 ","End":"01:58.340","Text":"so the current at every single point in the circuit is going to be equal,"},{"Start":"01:58.340 ","End":"02:03.080","Text":"multiplied by the value for the resistance of this resistor."},{"Start":"02:03.080 ","End":"02:05.180","Text":"This is Ohm\u0027s law."},{"Start":"02:05.180 ","End":"02:12.095","Text":"Now, let\u0027s speak about how we can add more resistors into a circuit,"},{"Start":"02:12.095 ","End":"02:13.955","Text":"either in series or in parallel,"},{"Start":"02:13.955 ","End":"02:17.800","Text":"and how we find the total resistance of the circuit."},{"Start":"02:17.800 ","End":"02:21.709","Text":"Let\u0027s deal with the case of resistors connected in series."},{"Start":"02:21.709 ","End":"02:24.635","Text":"When we have our resistors connected in series,"},{"Start":"02:24.635 ","End":"02:29.225","Text":"that means that we\u0027re just taking this exact same circuit,"},{"Start":"02:29.225 ","End":"02:34.880","Text":"for instance, and we\u0027re just adding in resistors around this circuit."},{"Start":"02:34.880 ","End":"02:38.795","Text":"We\u0027re not adding other wires and other circuits to this,"},{"Start":"02:38.795 ","End":"02:42.955","Text":"we\u0027re just adding in resistors wherever we want in this 1 loop."},{"Start":"02:42.955 ","End":"02:46.520","Text":"When we\u0027re dealing with a series circuit,"},{"Start":"02:46.520 ","End":"02:49.130","Text":"so we know that the current at"},{"Start":"02:49.130 ","End":"02:53.900","Text":"every single point inside the series circuit is going to be equal."},{"Start":"02:53.900 ","End":"02:58.140","Text":"Let\u0027s say we have this resistor which is R_1,"},{"Start":"02:58.140 ","End":"03:01.035","Text":"and this resistor which is R_2."},{"Start":"03:01.035 ","End":"03:03.230","Text":"Here we\u0027ll have, let\u0027s say,"},{"Start":"03:03.230 ","End":"03:05.495","Text":"that this is called I_0,"},{"Start":"03:05.495 ","End":"03:08.945","Text":"which means that here we\u0027re also going to have I_0,"},{"Start":"03:08.945 ","End":"03:10.745","Text":"and here also I_0."},{"Start":"03:10.745 ","End":"03:12.665","Text":"What I can do, therefore,"},{"Start":"03:12.665 ","End":"03:14.495","Text":"if I\u0027m in series,"},{"Start":"03:14.495 ","End":"03:18.410","Text":"that means that my current at every single point is equal."},{"Start":"03:18.410 ","End":"03:23.645","Text":"Therefore, I can redraw this circuit as this,"},{"Start":"03:23.645 ","End":"03:29.165","Text":"where this resistor is called R_T or R total."},{"Start":"03:29.165 ","End":"03:37.598","Text":"I can consider this circuit with 2 resistors as a similar circuit with just 1 resistor,"},{"Start":"03:37.598 ","End":"03:39.530","Text":"with resistance R total."},{"Start":"03:39.530 ","End":"03:43.535","Text":"Now the value of this R total is going to be equal to"},{"Start":"03:43.535 ","End":"03:48.805","Text":"my resistance R_1 plus my resistance R_2."},{"Start":"03:48.805 ","End":"03:51.320","Text":"This is because in series,"},{"Start":"03:51.320 ","End":"03:55.719","Text":"my current is constant along the circuit."},{"Start":"03:55.719 ","End":"04:00.870","Text":"This is the total resistance for these 2,"},{"Start":"04:00.870 ","End":"04:06.200","Text":"and I can just work out all my equations for this exact same circuit,"},{"Start":"04:06.200 ","End":"04:08.390","Text":"but just using my R_T."},{"Start":"04:08.390 ","End":"04:12.040","Text":"Now, what happens if I want to measure the voltage?"},{"Start":"04:12.040 ","End":"04:15.590","Text":"If I want to measure the voltage across my resistor,"},{"Start":"04:15.590 ","End":"04:18.590","Text":"so I have to add a voltmeter across it."},{"Start":"04:18.590 ","End":"04:21.530","Text":"What is my voltage going to be equal to?"},{"Start":"04:21.530 ","End":"04:23.930","Text":"My total voltage, V_T,"},{"Start":"04:23.930 ","End":"04:31.325","Text":"is going to be equal to the voltage across my R_1 plus the voltage across my R_2."},{"Start":"04:31.325 ","End":"04:33.350","Text":"That will be this voltage over here."},{"Start":"04:33.350 ","End":"04:35.195","Text":"If we look what\u0027s happening,"},{"Start":"04:35.195 ","End":"04:41.200","Text":"it\u0027s as if I\u0027m connecting a voltmeter simply across these 2 resistors,"},{"Start":"04:41.200 ","End":"04:44.120","Text":"so I\u0027m not putting a voltmeter across each resistor,"},{"Start":"04:44.120 ","End":"04:46.415","Text":"I\u0027m putting in across the 2 resistors,"},{"Start":"04:46.415 ","End":"04:48.350","Text":"which is the exact same thing here and here."},{"Start":"04:48.350 ","End":"04:50.855","Text":"The last thing that we need to speak about is,"},{"Start":"04:50.855 ","End":"04:53.135","Text":"of course, our current."},{"Start":"04:53.135 ","End":"04:57.905","Text":"If in the original circuit I had I_0 or I naught,"},{"Start":"04:57.905 ","End":"05:01.894","Text":"here I\u0027m also going to have I naught and at every single point,"},{"Start":"05:01.894 ","End":"05:03.620","Text":"because we\u0027re still in series,"},{"Start":"05:03.620 ","End":"05:06.760","Text":"my current is going to be equal."},{"Start":"05:06.760 ","End":"05:13.180","Text":"We can just say that our I is going to be equal to I naught the whole time."},{"Start":"05:13.180 ","End":"05:18.065","Text":"Now, let\u0027s make a general case for what our R total is going to be."},{"Start":"05:18.065 ","End":"05:21.680","Text":"When we\u0027re dealing with resistors connected in series,"},{"Start":"05:21.680 ","End":"05:25.475","Text":"so we can say that our R total is equal to"},{"Start":"05:25.475 ","End":"05:32.845","Text":"the sum on i up until n of R_i."},{"Start":"05:32.845 ","End":"05:34.590","Text":"Here, specifically,"},{"Start":"05:34.590 ","End":"05:35.670","Text":"we had 2 resistors."},{"Start":"05:35.670 ","End":"05:37.470","Text":"We have R_1 plus R_2."},{"Start":"05:37.470 ","End":"05:40.475","Text":"But if we had 3 or 4 and so on and so forth,"},{"Start":"05:40.475 ","End":"05:42.600","Text":"we\u0027ll just have R_1 plus R_2,"},{"Start":"05:42.600 ","End":"05:45.500","Text":"plus R_3, plus R_4 plus R_5 and so on."},{"Start":"05:45.500 ","End":"05:47.210","Text":"Then in this case,"},{"Start":"05:47.210 ","End":"05:55.625","Text":"our V total would also simply be equal to the sum from i=1 until n(V_i)."},{"Start":"05:55.625 ","End":"05:58.855","Text":"Here we had V_1 plus V_2."},{"Start":"05:58.855 ","End":"06:03.930","Text":"We\u0027ll have V_1 plus V_2 plus V_3 plus V_4 and so on and so forth."},{"Start":"06:03.930 ","End":"06:09.909","Text":"Of course, our I is simply going to remain the same when we\u0027re in series."},{"Start":"06:09.909 ","End":"06:13.160","Text":"These are the important equations to take away when"},{"Start":"06:13.160 ","End":"06:16.490","Text":"we\u0027re dealing with resistors connected in series."},{"Start":"06:16.490 ","End":"06:19.700","Text":"Just let\u0027s see where these equations come about."},{"Start":"06:19.700 ","End":"06:22.404","Text":"Let\u0027s go back to this example."},{"Start":"06:22.404 ","End":"06:29.180","Text":"Let\u0027s say that we\u0027re measuring right now our voltage across Resistor 1."},{"Start":"06:29.180 ","End":"06:32.765","Text":"We add a voltmeter across R_1."},{"Start":"06:32.765 ","End":"06:41.130","Text":"Our voltage through Ohm\u0027s law is going to be equal to I naught or I multiplied by R_1."},{"Start":"06:41.130 ","End":"06:43.527","Text":"Our voltage across Resistor 2,"},{"Start":"06:43.527 ","End":"06:47.180","Text":"so we connect a voltmeter across Resistor 2,"},{"Start":"06:47.180 ","End":"06:52.610","Text":"will be equal to the same I naught multiplied by R_2 because it\u0027s the same current."},{"Start":"06:52.610 ","End":"06:59.200","Text":"Then we can say that our V total is going to be equal to our V_1 plus our V_2,"},{"Start":"06:59.200 ","End":"07:03.300","Text":"which is equal to our IR_1,"},{"Start":"07:03.300 ","End":"07:04.740","Text":"it doesn\u0027t matter the I naught, here,"},{"Start":"07:04.740 ","End":"07:06.870","Text":"we\u0027re just a variable,"},{"Start":"07:06.870 ","End":"07:11.840","Text":"plus IR_2, which is then we"},{"Start":"07:11.840 ","End":"07:17.820","Text":"can take out the common factor of I multiplied by R_1 plus R_2,"},{"Start":"07:18.230 ","End":"07:24.575","Text":"which is simply equal to IR total."},{"Start":"07:24.575 ","End":"07:28.315","Text":"Remember that R_T was R_1 plus R_2."},{"Start":"07:28.315 ","End":"07:32.320","Text":"Then we get that our V total is simply equal to the current"},{"Start":"07:32.320 ","End":"07:36.685","Text":"in the circuit multiplied by the total resistance of the circuit."},{"Start":"07:36.685 ","End":"07:40.869","Text":"Now let\u0027s talk about resistors connected in parallel."},{"Start":"07:40.869 ","End":"07:46.675","Text":"Here we have 2 resistors which are connected in parallel. What does that mean?"},{"Start":"07:46.675 ","End":"07:51.160","Text":"We have some kind of circuit with a voltage and current running through it"},{"Start":"07:51.160 ","End":"07:57.445","Text":"and then our wire splits into 2 different branches."},{"Start":"07:57.445 ","End":"08:01.980","Text":"Then some of the current will go through this resistor and some of"},{"Start":"08:01.980 ","End":"08:03.990","Text":"the current will go through this resistor and"},{"Start":"08:03.990 ","End":"08:06.105","Text":"then will meet again at the junction and carry on."},{"Start":"08:06.105 ","End":"08:09.525","Text":"Now here, when we\u0027re dealing with a connection in parallel,"},{"Start":"08:09.525 ","End":"08:15.490","Text":"our voltage at every point in the wire is going to be equal."},{"Start":"08:15.490 ","End":"08:18.969","Text":"Not current this time, our voltage."},{"Start":"08:18.969 ","End":"08:24.110","Text":"Our voltage here is going to be equal to our voltage here,"},{"Start":"08:24.110 ","End":"08:26.995","Text":"which is going to be equal to our voltage here,"},{"Start":"08:26.995 ","End":"08:28.525","Text":"and here, and here."},{"Start":"08:28.525 ","End":"08:33.715","Text":"Our voltage everywhere in the circuit is going to be equal."},{"Start":"08:33.715 ","End":"08:38.830","Text":"Let\u0027s say that V is equal to V naught throughout the entire circuit."},{"Start":"08:38.830 ","End":"08:41.125","Text":"Just like when we were speaking about the series,"},{"Start":"08:41.125 ","End":"08:46.870","Text":"I can say that this resistor and this resistor together can"},{"Start":"08:46.870 ","End":"08:53.545","Text":"behave like 1 resistor in a circuit with a total resistance of R_T, R total."},{"Start":"08:53.545 ","End":"08:55.104","Text":"When we\u0027re in parallel,"},{"Start":"08:55.104 ","End":"08:58.900","Text":"the voltage in the entire circuit is constant,"},{"Start":"08:58.900 ","End":"09:02.275","Text":"and therefore our R total,"},{"Start":"09:02.275 ","End":"09:07.810","Text":"the equation in order to find it is going to be equal to 1 divided by R total is equal"},{"Start":"09:07.810 ","End":"09:14.065","Text":"to 1 divided by R_1 plus 1 divided by R_2."},{"Start":"09:14.065 ","End":"09:18.130","Text":"Then we can just do 1 divided by all of this on"},{"Start":"09:18.130 ","End":"09:22.750","Text":"the right-hand side in order to get our value for R total."},{"Start":"09:22.750 ","End":"09:24.505","Text":"Now with our voltage,"},{"Start":"09:24.505 ","End":"09:31.765","Text":"if I connect some kind of voltmeter going like this over my R_2,"},{"Start":"09:31.765 ","End":"09:39.789","Text":"imagine there\u0027s a voltmeter or a voltmeter going like this over my R_1 over here,"},{"Start":"09:39.789 ","End":"09:46.075","Text":"my voltage on this side is going to be equal to my voltage on this side."},{"Start":"09:46.075 ","End":"09:50.740","Text":"Then we can write that my V total is going"},{"Start":"09:50.740 ","End":"09:55.375","Text":"to be equal to my V_1 which is my voltage over R_1,"},{"Start":"09:55.375 ","End":"09:57.490","Text":"which is equal to my V_2,"},{"Start":"09:57.490 ","End":"10:00.400","Text":"which is also my voltage over R_2."},{"Start":"10:00.400 ","End":"10:03.980","Text":"Now, what about my current?"},{"Start":"10:03.980 ","End":"10:07.380","Text":"As we saw in our previous lesson,"},{"Start":"10:07.380 ","End":"10:11.474","Text":"we saw that when our current is flowing through 1 circuit,"},{"Start":"10:11.474 ","End":"10:12.705","Text":"it remains the same."},{"Start":"10:12.705 ","End":"10:18.185","Text":"However, when it reaches a junction such as this one over here,"},{"Start":"10:18.185 ","End":"10:21.610","Text":"here\u0027s our junction and here we have another junction,"},{"Start":"10:21.610 ","End":"10:30.100","Text":"our current will either split like it is here or it will join back together like here."},{"Start":"10:30.100 ","End":"10:37.300","Text":"What we can see is if we have some kind of total current running through our circuit,"},{"Start":"10:37.300 ","End":"10:39.630","Text":"once it hits the junction,"},{"Start":"10:39.630 ","End":"10:43.060","Text":"we know that our current over here is going to be equal to,"},{"Start":"10:43.060 ","End":"10:44.320","Text":"let\u0027s say I_1,"},{"Start":"10:44.320 ","End":"10:48.145","Text":"and our current on this branch is going to be equal to I_2."},{"Start":"10:48.145 ","End":"10:52.585","Text":"Then the current will flow and then once our currents of"},{"Start":"10:52.585 ","End":"10:57.016","Text":"I_1 and I_2 connect back at this junction,"},{"Start":"10:57.016 ","End":"11:00.565","Text":"we\u0027ll have, again going out through here, our I total."},{"Start":"11:00.565 ","End":"11:06.584","Text":"In other words, our total current across our 2 resistors,"},{"Start":"11:06.584 ","End":"11:08.170","Text":"because the current splits,"},{"Start":"11:08.170 ","End":"11:11.170","Text":"is going to be equal to our I_1,"},{"Start":"11:11.170 ","End":"11:13.555","Text":"which is the current through this branch,"},{"Start":"11:13.555 ","End":"11:15.790","Text":"plus our I_2,"},{"Start":"11:15.790 ","End":"11:18.655","Text":"which is the current through this branch."},{"Start":"11:18.655 ","End":"11:22.450","Text":"We can see that our equations here are exactly the opposite of what"},{"Start":"11:22.450 ","End":"11:26.425","Text":"we saw when our resistors are connected in series."},{"Start":"11:26.425 ","End":"11:30.490","Text":"Now, let\u0027s write a general rule for R total."},{"Start":"11:30.490 ","End":"11:37.240","Text":"We can say that 1 divided by our R total is going to be equal to"},{"Start":"11:37.240 ","End":"11:45.355","Text":"the sum from I is equal to 1 until n of 1 divided by R_i."},{"Start":"11:45.355 ","End":"11:49.059","Text":"If we had in here another resistor,"},{"Start":"11:49.059 ","End":"11:50.575","Text":"let\u0027s say R_3,"},{"Start":"11:50.575 ","End":"11:54.520","Text":"so 1 divided by R total would be equal to 1"},{"Start":"11:54.520 ","End":"11:58.585","Text":"divided by R_1 plus 1 divided by R_2 plus 1 divided by R_3."},{"Start":"11:58.585 ","End":"12:00.490","Text":"Then for our current,"},{"Start":"12:00.490 ","End":"12:03.400","Text":"our I total is going to be equal,"},{"Start":"12:03.400 ","End":"12:07.450","Text":"so here we would have a current of I_3 running through."},{"Start":"12:07.450 ","End":"12:13.990","Text":"Our current is equal to the sum from I is equal to 1 up until n I_1."},{"Start":"12:13.990 ","End":"12:20.575","Text":"Here, our I total would be equal to I_1 plus I_2 plus I_3,"},{"Start":"12:20.575 ","End":"12:26.755","Text":"and of course, our V total is simply equal to our V naught,"},{"Start":"12:26.755 ","End":"12:28.420","Text":"what we had over here."},{"Start":"12:28.420 ","End":"12:36.295","Text":"We can just even write our voltage is simply constant and that\u0027s it."},{"Start":"12:36.295 ","End":"12:40.480","Text":"Now let\u0027s see where these equations come from."},{"Start":"12:40.480 ","End":"12:43.180","Text":"Let\u0027s write this over here."},{"Start":"12:43.180 ","End":"12:48.910","Text":"We know that the voltage across my R_1, so V_1,"},{"Start":"12:48.910 ","End":"12:52.360","Text":"is going to be equal to my current running through it which"},{"Start":"12:52.360 ","End":"12:56.350","Text":"is equal to I_1 multiplied by my resistance there."},{"Start":"12:56.350 ","End":"12:59.590","Text":"My voltage across my second resistor is going"},{"Start":"12:59.590 ","End":"13:02.620","Text":"to be the current running through that resistor because we\u0027re in parallel,"},{"Start":"13:02.620 ","End":"13:04.315","Text":"so it\u0027s a different current,"},{"Start":"13:04.315 ","End":"13:07.630","Text":"multiplied by that resistance."},{"Start":"13:07.630 ","End":"13:13.720","Text":"Now as we know, my total current here is"},{"Start":"13:13.720 ","End":"13:17.050","Text":"specifically here because I only have 2 resistors is going to be"},{"Start":"13:17.050 ","End":"13:20.845","Text":"equal to my I_1 plus my I_2."},{"Start":"13:20.845 ","End":"13:22.210","Text":"From Ohm\u0027s law,"},{"Start":"13:22.210 ","End":"13:24.880","Text":"I know that V is equal to IR,"},{"Start":"13:24.880 ","End":"13:31.900","Text":"or I can say that V divided by I is equal to my current."},{"Start":"13:31.900 ","End":"13:35.005","Text":"Then I can do for everything,"},{"Start":"13:35.005 ","End":"13:41.035","Text":"V_1 divided by R_1 is equal to I_1,"},{"Start":"13:41.035 ","End":"13:47.605","Text":"and V_2 divided by R_2 is equal to I_2."},{"Start":"13:47.605 ","End":"13:50.650","Text":"Then I know that my I total,"},{"Start":"13:50.650 ","End":"13:57.130","Text":"which is equal to V total divided by R total,"},{"Start":"13:57.130 ","End":"14:00.085","Text":"from my same Ohm\u0027s law,"},{"Start":"14:00.085 ","End":"14:03.775","Text":"is going to be equal to I_1 plus I_2."},{"Start":"14:03.775 ","End":"14:13.195","Text":"My I_1 is equal to V_1 over R_1 and my I_2 is equal to V_2 over R_2."},{"Start":"14:13.195 ","End":"14:18.115","Text":"However, we know that my voltage throughout the circuit is a constant,"},{"Start":"14:18.115 ","End":"14:19.705","Text":"it\u0027s the same value."},{"Start":"14:19.705 ","End":"14:25.105","Text":"That means that my V total is equal to my V_1 which is equal to my V_2."},{"Start":"14:25.105 ","End":"14:32.410","Text":"Then I can say that my V total divided by R total is equal"},{"Start":"14:32.410 ","End":"14:41.095","Text":"to V total divided by R_1 plus V total divided by R_2."},{"Start":"14:41.095 ","End":"14:45.475","Text":"Then I can divide both sides by V total."},{"Start":"14:45.475 ","End":"14:49.750","Text":"I\u0027ll get 1 divided by R_T is equal to 1 divided by R_1"},{"Start":"14:49.750 ","End":"14:54.905","Text":"plus 1 divided by R_2 and that\u0027s exactly what we got over here."},{"Start":"14:54.905 ","End":"14:59.420","Text":"Of course, we would get the exact same equation if we had more resistors."},{"Start":"14:59.420 ","End":"15:02.610","Text":"Okay, that\u0027s the end of this lesson."}],"ID":22297},{"Watched":false,"Name":"Exercise 1","Duration":"15m 18s","ChapterTopicVideoID":21481,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"Hello. In this lesson,"},{"Start":"00:01.740 ","End":"00:04.320","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.320 ","End":"00:08.400","Text":"We are given the following resistances and voltages."},{"Start":"00:08.400 ","End":"00:12.280","Text":"The resistance of R_1 is 2 ohms,"},{"Start":"00:12.280 ","End":"00:14.070","Text":"of R_2 is 3 ohms,"},{"Start":"00:14.070 ","End":"00:16.755","Text":"and of R_3 is 5 ohms."},{"Start":"00:16.755 ","End":"00:19.395","Text":"The voltage of the battery,"},{"Start":"00:19.395 ","End":"00:22.740","Text":"V_0, is equal to 31 volts."},{"Start":"00:22.740 ","End":"00:28.600","Text":"Question number 1 is to calculate the total resistance of the circuit."},{"Start":"00:29.510 ","End":"00:36.540","Text":"The first thing that we want to do is we want to convert the resistance of R_1 and"},{"Start":"00:36.540 ","End":"00:44.045","Text":"R_2 into the resistance as if this is just one resistor instead of 2 resistors."},{"Start":"00:44.045 ","End":"00:49.730","Text":"What we can see is that R_1 and R_2 are connected in parallel."},{"Start":"00:49.730 ","End":"00:55.760","Text":"The definition for a resistor to be connected in parallel is that"},{"Start":"00:55.760 ","End":"00:58.760","Text":"the voltage going through resistor number"},{"Start":"00:58.760 ","End":"01:03.245","Text":"2 has to be equal to the voltage going through resistor number 1,"},{"Start":"01:03.245 ","End":"01:07.980","Text":"or the voltage between the 2 resistors has to be the same."},{"Start":"01:09.200 ","End":"01:12.630","Text":"Let\u0027s see why that happens."},{"Start":"01:12.630 ","End":"01:16.910","Text":"First of all, voltage is in fact the potential difference."},{"Start":"01:16.910 ","End":"01:21.770","Text":"If I were to look at the potential over here,"},{"Start":"01:21.770 ","End":"01:24.010","Text":"over resistor number 2,"},{"Start":"01:24.010 ","End":"01:27.260","Text":"I would look at the potential over here and at the potential over"},{"Start":"01:27.260 ","End":"01:31.660","Text":"here and I\u0027ll find the potential difference between these two points."},{"Start":"01:31.660 ","End":"01:40.820","Text":"This point over here is connected by an ideal wire to this point over here."},{"Start":"01:40.820 ","End":"01:44.495","Text":"That means because it\u0027s connected by an ideal wire,"},{"Start":"01:44.495 ","End":"01:46.610","Text":"that the potential over here,"},{"Start":"01:46.610 ","End":"01:50.120","Text":"Phi_1, has to be equal to the potential over here."},{"Start":"01:50.120 ","End":"01:53.825","Text":"We\u0027ll call this also Phi_1."},{"Start":"01:53.825 ","End":"01:59.060","Text":"Then similarly, this point over here is"},{"Start":"01:59.060 ","End":"02:04.625","Text":"connected by an ideal wire to this point over here."},{"Start":"02:04.625 ","End":"02:10.310","Text":"Which means that the voltage at these two points has to be also the same."},{"Start":"02:10.310 ","End":"02:15.050","Text":"If I call the potential over here Phi_2,"},{"Start":"02:15.050 ","End":"02:19.370","Text":"so then the potential over here has to be the exact same thing."},{"Start":"02:19.370 ","End":"02:22.700","Text":"The potential over here will also be Phi_2."},{"Start":"02:22.700 ","End":"02:27.650","Text":"Then what I can see is that the potential difference between these two points,"},{"Start":"02:27.650 ","End":"02:28.880","Text":"or in other words,"},{"Start":"02:28.880 ","End":"02:34.770","Text":"the voltage over each one of these resistors is the exact same."},{"Start":"02:35.450 ","End":"02:39.710","Text":"The potential difference over both resistors over here,"},{"Start":"02:39.710 ","End":"02:42.290","Text":"we can call that, let\u0027s say V_2,"},{"Start":"02:42.290 ","End":"02:43.790","Text":"which we know is equal to V_1,"},{"Start":"02:43.790 ","End":"02:50.775","Text":"so that\u0027s just equal to Phi_1 minus Phi_2."},{"Start":"02:50.775 ","End":"02:53.975","Text":"This is of course, also equal to V_1."},{"Start":"02:53.975 ","End":"02:56.870","Text":"That is why the potential difference or"},{"Start":"02:56.870 ","End":"03:01.710","Text":"the voltage on each of these resistors is the same."},{"Start":"03:02.510 ","End":"03:05.850","Text":"If you ever see two resistors,"},{"Start":"03:05.850 ","End":"03:11.030","Text":"that a point just before the resistor is connected to"},{"Start":"03:11.030 ","End":"03:16.415","Text":"a point just before the other resistor and the same on the other side,"},{"Start":"03:16.415 ","End":"03:22.410","Text":"then you can say that these two resistors are connected in parallel."},{"Start":"03:23.150 ","End":"03:29.270","Text":"Now what we\u0027re going to do is we\u0027re going to see what"},{"Start":"03:29.270 ","End":"03:35.165","Text":"the total voltage of this over here is."},{"Start":"03:35.165 ","End":"03:40.115","Text":"Because we already saw that R_1 and R_2 are connected in parallel,"},{"Start":"03:40.115 ","End":"03:47.195","Text":"we know that the equation to find the total resistance is 1 divided by R_1,2"},{"Start":"03:47.195 ","End":"03:54.600","Text":"is equal to 1 divided by R_1 plus 1 divided by R_2."},{"Start":"03:54.600 ","End":"04:00.555","Text":"Of course, if there were 3 resistors over here connected in parallel,"},{"Start":"04:00.555 ","End":"04:02.790","Text":"then the total resistance,"},{"Start":"04:02.790 ","End":"04:06.990","Text":"1 divided by R_1,2,3 would be 1 divided by"},{"Start":"04:06.990 ","End":"04:13.090","Text":"R_1 plus 1 divided by R_2 plus 1 divided by R_3, and so on."},{"Start":"04:13.400 ","End":"04:18.180","Text":"Now, we can do the algebra."},{"Start":"04:18.180 ","End":"04:24.544","Text":"This is simply equal to, common denominators."},{"Start":"04:24.544 ","End":"04:34.735","Text":"R_2 plus R_1 divided by R_1 multiplied by R_2."},{"Start":"04:34.735 ","End":"04:41.780","Text":"Then we want to find the reciprocal because we want R_1,2 and not 1 divided by that."},{"Start":"04:41.780 ","End":"04:48.599","Text":"Then what we\u0027ll have is that R_1,2 is equal to R_1"},{"Start":"04:48.599 ","End":"04:56.085","Text":"multiplied by R_2 divided by R_1 plus R_2."},{"Start":"04:56.085 ","End":"04:59.085","Text":"Now we can plug in our values."},{"Start":"04:59.085 ","End":"05:03.150","Text":"R_1 is equal to 2 ohms multiplied by R_2,"},{"Start":"05:03.150 ","End":"05:04.635","Text":"which is 3 ohms,"},{"Start":"05:04.635 ","End":"05:07.170","Text":"divided by R_1 plus R_2,"},{"Start":"05:07.170 ","End":"05:09.120","Text":"so 2 plus 3."},{"Start":"05:09.120 ","End":"05:18.585","Text":"What we have is that R_1,2 is equal to 6 divided by 5 ohms."},{"Start":"05:18.585 ","End":"05:23.960","Text":"We can say that this blue box over here,"},{"Start":"05:23.960 ","End":"05:27.875","Text":"we can imagine that we have a resistor attached to"},{"Start":"05:27.875 ","End":"05:35.610","Text":"R_3 with a resistance of 6 divided by 5 ohms."},{"Start":"05:35.610 ","End":"05:39.740","Text":"Now we want to find the total resistance of the circuit."},{"Start":"05:39.740 ","End":"05:44.315","Text":"Now we can imagine that this is just one resistor"},{"Start":"05:44.315 ","End":"05:49.705","Text":"connected in series to this resistor R_3."},{"Start":"05:49.705 ","End":"05:54.770","Text":"How do we know that these 2 resistors are connected in series?"},{"Start":"05:54.770 ","End":"05:59.135","Text":"They\u0027re connected via the same wire,"},{"Start":"05:59.135 ","End":"06:04.060","Text":"which means that the same current flows through both of them."},{"Start":"06:04.060 ","End":"06:09.124","Text":"Now, the equation for 2 resistors in series,"},{"Start":"06:09.124 ","End":"06:14.140","Text":"and in this case, the total resistance of the circuit."},{"Start":"06:14.140 ","End":"06:15.735","Text":"We just add them up."},{"Start":"06:15.735 ","End":"06:21.975","Text":"We have the resistor 1,2 plus our R_3."},{"Start":"06:21.975 ","End":"06:28.965","Text":"R_1,2 is 6 over 5 ohms plus R_3,"},{"Start":"06:28.965 ","End":"06:31.645","Text":"which is 5 ohms."},{"Start":"06:31.645 ","End":"06:40.680","Text":"Then we can see that this is simply going to be equal to 31 divided by 5 ohms."},{"Start":"06:41.600 ","End":"06:44.505","Text":"That\u0027s the answer to question number 1."},{"Start":"06:44.505 ","End":"06:46.950","Text":"Now let\u0027s answer question number 2."},{"Start":"06:46.950 ","End":"06:50.480","Text":"Question number 2 is to calculate the current passing through"},{"Start":"06:50.480 ","End":"06:56.850","Text":"the battery and to calculate the current and voltage on each resistor."},{"Start":"06:57.500 ","End":"07:02.570","Text":"In order to calculate the current passing through the battery,"},{"Start":"07:02.570 ","End":"07:08.835","Text":"I can look at this circuit as a battery connected to one resistor."},{"Start":"07:08.835 ","End":"07:11.714","Text":"The one resistor of this resistance,"},{"Start":"07:11.714 ","End":"07:13.430","Text":"that\u0027s our R total,"},{"Start":"07:13.430 ","End":"07:16.590","Text":"which we found in the previous question."},{"Start":"07:17.060 ","End":"07:19.715","Text":"I can use Ohm\u0027s law,"},{"Start":"07:19.715 ","End":"07:21.620","Text":"and Ohm\u0027s law, as we know,"},{"Start":"07:21.620 ","End":"07:23.599","Text":"V is equal to IR."},{"Start":"07:23.599 ","End":"07:27.320","Text":"I have that V_0 is equal to"},{"Start":"07:27.320 ","End":"07:31.505","Text":"the total current passing through the battery"},{"Start":"07:31.505 ","End":"07:37.130","Text":"multiplied by the total resistance of the circuit."},{"Start":"07:37.130 ","End":"07:39.155","Text":"I\u0027m trying to find the current."},{"Start":"07:39.155 ","End":"07:40.460","Text":"This, I don\u0027t know."},{"Start":"07:40.460 ","End":"07:47.320","Text":"I know that my V_0 is equal to 31 volts."},{"Start":"07:47.600 ","End":"07:51.635","Text":"I have 31 volts,"},{"Start":"07:51.635 ","End":"07:54.710","Text":"and then I\u0027ll divide it by the total resistance."},{"Start":"07:54.710 ","End":"07:58.850","Text":"Divided by 31, divided by 5."},{"Start":"07:58.850 ","End":"08:06.323","Text":"This will give me the total current passing through the battery."},{"Start":"08:06.323 ","End":"08:12.130","Text":"Then we can see that our total current flowing"},{"Start":"08:12.130 ","End":"08:19.210","Text":"through the battery is equal to 5 amps, or 5 amperes."},{"Start":"08:19.210 ","End":"08:22.870","Text":"This is the current passing through the battery and now let\u0027s"},{"Start":"08:22.870 ","End":"08:27.410","Text":"calculate the current and voltage on each of the resistors."},{"Start":"08:28.890 ","End":"08:35.245","Text":"In order to find the total resistance of the circuit,"},{"Start":"08:35.245 ","End":"08:45.280","Text":"we made this R_1 and R_2 into one resistor that was connected in series to R_3."},{"Start":"08:45.280 ","End":"08:48.069","Text":"If something is connected in series,"},{"Start":"08:48.069 ","End":"08:49.795","Text":"we know that by definition,"},{"Start":"08:49.795 ","End":"08:56.270","Text":"that means that the same current is passing through the two resistors."},{"Start":"08:57.210 ","End":"09:04.750","Text":"In other words, we can say that the total current passing through"},{"Start":"09:04.750 ","End":"09:13.270","Text":"is equal to the current passing through resistor 1,2,"},{"Start":"09:13.270 ","End":"09:15.550","Text":"so this total resistor,"},{"Start":"09:15.550 ","End":"09:22.750","Text":"and it\u0027s equal to the current passing through resistor number 3."},{"Start":"09:22.750 ","End":"09:28.810","Text":"That is simply equal to 5 amps as we saw before."},{"Start":"09:30.710 ","End":"09:38.965","Text":"Now let\u0027s calculate the voltage over R_3. What does that mean?"},{"Start":"09:38.965 ","End":"09:43.885","Text":"That means that we take this point over here and this point over here,"},{"Start":"09:43.885 ","End":"09:51.470","Text":"and all we\u0027re doing is measuring the voltage over here, V_3."},{"Start":"09:52.140 ","End":"09:56.140","Text":"V_3, again, using Ohm\u0027s law,"},{"Start":"09:56.140 ","End":"10:04.270","Text":"is equal to the current over resistor 3 multiplied by the resistance of resistance 3."},{"Start":"10:04.270 ","End":"10:06.805","Text":"The current over resistor 3,"},{"Start":"10:06.805 ","End":"10:09.745","Text":"we already saw, is equal to 5 amps,"},{"Start":"10:09.745 ","End":"10:16.750","Text":"and the resistance of resistor number 3 is equal to 5 ohms."},{"Start":"10:16.750 ","End":"10:24.920","Text":"Then we\u0027ll get that the voltage over resistance number 3 is equal to 25 volts."},{"Start":"10:26.370 ","End":"10:32.270","Text":"Now let\u0027s find the voltage on resistor R_1,2."},{"Start":"10:32.270 ","End":"10:36.625","Text":"That means we\u0027re finding the voltage between these two points."},{"Start":"10:36.625 ","End":"10:39.335","Text":"As if this is just one resistor."},{"Start":"10:39.335 ","End":"10:46.240","Text":"It\u0027s as if we were to put a voltmeter over here. This is V_1,2."},{"Start":"10:46.740 ","End":"10:49.090","Text":"Again, using Ohm\u0027s law,"},{"Start":"10:49.090 ","End":"10:54.130","Text":"we have V_1,2 which is equal to the current"},{"Start":"10:54.130 ","End":"11:00.790","Text":"on that resistor multiplied by the resistance of that resistor."},{"Start":"11:00.790 ","End":"11:02.440","Text":"The current, as we saw,"},{"Start":"11:02.440 ","End":"11:11.845","Text":"is equal to 5 amps and the resistance is equal to over here. Where was it?"},{"Start":"11:11.845 ","End":"11:19.540","Text":"Here, 6 divided by 5 ohms."},{"Start":"11:19.540 ","End":"11:24.680","Text":"Then what we get is that this is equal to 6 volts."},{"Start":"11:25.230 ","End":"11:32.260","Text":"That\u0027s one way to calculate the voltage over our resistor R_1,2."},{"Start":"11:32.260 ","End":"11:37.390","Text":"Another way to calculate it is we know that V_0 is equal to 31 volts,"},{"Start":"11:37.390 ","End":"11:40.810","Text":"so 31 volts, and then as it passes through the resistor,"},{"Start":"11:40.810 ","End":"11:43.015","Text":"the resistance goes down,"},{"Start":"11:43.015 ","End":"11:46.270","Text":"and then when we get to here,"},{"Start":"11:46.270 ","End":"11:52.390","Text":"we know that the resistance over R_3 is equal to 25 volts."},{"Start":"11:52.390 ","End":"11:56.905","Text":"If we have 31 minus 25,"},{"Start":"11:56.905 ","End":"11:58.735","Text":"we\u0027re left with 6,"},{"Start":"11:58.735 ","End":"12:07.100","Text":"so 6 volts over resistor R_1,2 which is exactly what we got from this calculation."},{"Start":"12:07.860 ","End":"12:18.625","Text":"This we could also write as saying that V_1,2 is equal to V_0 minus V_3,"},{"Start":"12:18.625 ","End":"12:21.805","Text":"and then we also would get 6 volts."},{"Start":"12:21.805 ","End":"12:24.985","Text":"That\u0027s just another way to calculate it once we\u0027ve already"},{"Start":"12:24.985 ","End":"12:28.880","Text":"calculated the voltage over R_3."},{"Start":"12:29.100 ","End":"12:31.150","Text":"Earlier in the lesson,"},{"Start":"12:31.150 ","End":"12:36.190","Text":"we saw that when two resistors are connected in parallel,"},{"Start":"12:36.190 ","End":"12:42.190","Text":"the voltage over resistor number 1 will be equal to the voltage over resistor number 2."},{"Start":"12:42.190 ","End":"12:49.010","Text":"That\u0027s the definition of being connected in parallel."},{"Start":"12:49.830 ","End":"13:00.925","Text":"Therefore, we know that the voltage V_1,2 is equal to the voltage on resistor 1,"},{"Start":"13:00.925 ","End":"13:05.450","Text":"and that is equal to the voltage on resistor 2."},{"Start":"13:05.870 ","End":"13:09.030","Text":"The rule is, in parallel,"},{"Start":"13:09.030 ","End":"13:15.264","Text":"the voltages are equal and are equal to the total voltage."},{"Start":"13:15.264 ","End":"13:24.100","Text":"In series, the currents are equal and equal to the total current."},{"Start":"13:25.250 ","End":"13:30.610","Text":"This is of course equal to 6 volts."},{"Start":"13:30.690 ","End":"13:34.075","Text":"Now, again, via Ohm\u0027s law,"},{"Start":"13:34.075 ","End":"13:40.585","Text":"I\u0027m going to calculate the current on each resistor, R_1 and R_2."},{"Start":"13:40.585 ","End":"13:46.990","Text":"The current on resistor R_1 is via Ohm\u0027s law,"},{"Start":"13:46.990 ","End":"13:53.230","Text":"just equal to the voltage on the resistor 1 divided by the resistance of resistor 1."},{"Start":"13:53.230 ","End":"14:01.120","Text":"The voltage V_1 we saw is equal to 6 volts divided by the resistance R_1."},{"Start":"14:01.120 ","End":"14:05.150","Text":"Let\u0027s scroll up, which is equal to 2 ohms."},{"Start":"14:06.000 ","End":"14:14.140","Text":"We get that this is equal to 3 amps and the current on"},{"Start":"14:14.140 ","End":"14:17.290","Text":"resistor number 2 is equal to the voltage over"},{"Start":"14:17.290 ","End":"14:21.880","Text":"resistor number 2 divided by the resistance of resistor number 2,"},{"Start":"14:21.880 ","End":"14:25.600","Text":"so the voltage on resistor number 2 is also equal to"},{"Start":"14:25.600 ","End":"14:31.015","Text":"6 volts divided by the resistance of resistor number 2,"},{"Start":"14:31.015 ","End":"14:34.120","Text":"which is 3 ohms."},{"Start":"14:34.120 ","End":"14:41.210","Text":"Here, we have 3 ohms and then this is simply equal to 2 amps."},{"Start":"14:42.030 ","End":"14:46.990","Text":"Now, if we add on I_1 to I_2,"},{"Start":"14:46.990 ","End":"14:50.320","Text":"so that\u0027s 3 plus 2,"},{"Start":"14:50.320 ","End":"14:52.495","Text":"which is equal to 5 amps,"},{"Start":"14:52.495 ","End":"14:56.740","Text":"which is exactly the same current passing through I_3,"},{"Start":"14:56.740 ","End":"15:04.240","Text":"which is what we expected because these two resistors are in series with this resistor,"},{"Start":"15:04.240 ","End":"15:11.830","Text":"and that\u0027s also equal to the total current passing through the circuit."},{"Start":"15:11.830 ","End":"15:15.310","Text":"Just as we would have expected."},{"Start":"15:15.310 ","End":"15:18.440","Text":"That\u0027s the end of this lesson."}],"ID":22298},{"Watched":false,"Name":"Exercise 2","Duration":"10m 26s","ChapterTopicVideoID":21310,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello. In this question,"},{"Start":"00:01.980 ","End":"00:06.780","Text":"we\u0027re being asked to calculate the total resistance of the following circuit."},{"Start":"00:06.780 ","End":"00:11.280","Text":"Here we have 2 connections that could go on to"},{"Start":"00:11.280 ","End":"00:15.945","Text":"connect onto another circuit or a battery. It doesn\u0027t really matter."},{"Start":"00:15.945 ","End":"00:19.890","Text":"But we have this area which has lots of different resistors"},{"Start":"00:19.890 ","End":"00:24.615","Text":"and we want to find what the equivalent resistance to this is."},{"Start":"00:24.615 ","End":"00:29.850","Text":"The first thing that we can do is we can start from one of these connecting points."},{"Start":"00:29.850 ","End":"00:31.455","Text":"Let\u0027s start from here,"},{"Start":"00:31.455 ","End":"00:34.020","Text":"and let\u0027s follow the wire along."},{"Start":"00:34.020 ","End":"00:38.025","Text":"We can see that we start from there and then we go this way"},{"Start":"00:38.025 ","End":"00:43.265","Text":"and I have a 3 ohm resistor over here."},{"Start":"00:43.265 ","End":"00:49.475","Text":"Then I carry on with the wire and then I see that my wire splits in to 3."},{"Start":"00:49.475 ","End":"00:51.365","Text":"Here\u0027s 1 split,"},{"Start":"00:51.365 ","End":"00:55.735","Text":"2 splits and 3 splits."},{"Start":"00:55.735 ","End":"00:58.845","Text":"Now let\u0027s go to each split."},{"Start":"00:58.845 ","End":"01:00.705","Text":"Let\u0027s go to this split."},{"Start":"01:00.705 ","End":"01:03.900","Text":"I follow this wire,"},{"Start":"01:03.900 ","End":"01:10.635","Text":"and I get to my 2 ohm resistor up until this point. Let\u0027s stop here."},{"Start":"01:10.635 ","End":"01:18.225","Text":"Then my middle split so I follow that and they get to a 4 ohm resistor."},{"Start":"01:18.225 ","End":"01:23.760","Text":"Then after that, we\u0027ll go back to it later and then this split over here,"},{"Start":"01:23.760 ","End":"01:29.625","Text":"I follow it and I have a 3 ohm resistor."},{"Start":"01:29.625 ","End":"01:32.885","Text":"We\u0027ve reached this point over here."},{"Start":"01:32.885 ","End":"01:37.280","Text":"Let\u0027s carry on. Let\u0027s go back to my first split."},{"Start":"01:37.280 ","End":"01:41.765","Text":"Let\u0027s just actually label our splits so this was my first split,"},{"Start":"01:41.765 ","End":"01:43.195","Text":"my second split,"},{"Start":"01:43.195 ","End":"01:45.045","Text":"and my third split."},{"Start":"01:45.045 ","End":"01:46.830","Text":"Let\u0027s go back to my first."},{"Start":"01:46.830 ","End":"01:53.240","Text":"I got to my 2 ohm resistor and now I can see that I have another 2 ohm resistor."},{"Start":"01:53.240 ","End":"01:59.440","Text":"Let\u0027s draw that in as well, 2 ohm resistor."},{"Start":"01:59.450 ","End":"02:03.425","Text":"Now that\u0027s done. Let\u0027s go back to my second split."},{"Start":"02:03.425 ","End":"02:06.425","Text":"I\u0027ve passed my 4 ohm resistor,"},{"Start":"02:06.425 ","End":"02:11.600","Text":"and then I can see it carries on to this same point over here."},{"Start":"02:11.600 ","End":"02:14.075","Text":"Let\u0027s draw this in gray."},{"Start":"02:14.075 ","End":"02:15.830","Text":"So this same point over here,"},{"Start":"02:15.830 ","End":"02:21.800","Text":"which means that my 4 ohm resistor has to connect to my 2 ohm resistor."},{"Start":"02:21.800 ","End":"02:24.050","Text":"Let\u0027s draw that."},{"Start":"02:24.050 ","End":"02:28.495","Text":"Great. Now they\u0027ve both connected."},{"Start":"02:28.495 ","End":"02:33.115","Text":"Now after this junction over here in gray,"},{"Start":"02:33.115 ","End":"02:37.240","Text":"I go onto my next resistor. What is that?"},{"Start":"02:37.240 ","End":"02:43.270","Text":"That is going to be this 6 ohm resistor over here."},{"Start":"02:43.270 ","End":"02:46.810","Text":"Great. I\u0027ve drawn my 6 ohm resistor,"},{"Start":"02:46.810 ","End":"02:51.999","Text":"and I\u0027m carrying on along this wire after my 6 ohm resistor."},{"Start":"02:51.999 ","End":"02:56.965","Text":"Now I\u0027m reaching this junction over here,"},{"Start":"02:56.965 ","End":"03:00.610","Text":"which is connected to my third split,"},{"Start":"03:00.610 ","End":"03:03.095","Text":"where I have my 3 ohm resistor."},{"Start":"03:03.095 ","End":"03:13.695","Text":"Now I\u0027m going to connect up my 6 ohm resistor with my third split, 3 ohm resistor."},{"Start":"03:13.695 ","End":"03:17.480","Text":"Now I\u0027m at that junction and now I\u0027m going to carry on along"},{"Start":"03:17.480 ","End":"03:23.525","Text":"the wire where I have another 3 ohm resistor."},{"Start":"03:23.525 ","End":"03:32.665","Text":"Then I carry on and I reach my second connecting area."},{"Start":"03:32.665 ","End":"03:36.380","Text":"What we can see here is that the second that you are given"},{"Start":"03:36.380 ","End":"03:39.845","Text":"some complicated circuit like so,"},{"Start":"03:39.845 ","End":"03:44.120","Text":"you have to go from one of the wires, you start,"},{"Start":"03:44.120 ","End":"03:49.350","Text":"and then you just follow through along each one of the wires."},{"Start":"03:49.350 ","End":"03:54.390","Text":"If we call this junction number 1 between our 2 ohm and 6 ohm"},{"Start":"03:54.390 ","End":"03:59.700","Text":"so that is going to be our 4 ohms,"},{"Start":"03:59.700 ","End":"04:02.060","Text":"so that\u0027s this junction over here."},{"Start":"04:02.060 ","End":"04:03.635","Text":"This is junction number 1,"},{"Start":"04:03.635 ","End":"04:10.140","Text":"we can see and then our junction number 2 is when our 6 ohm and"},{"Start":"04:10.140 ","End":"04:16.940","Text":"3 ohm have joined over here and then they both are connected to this 3 ohm."},{"Start":"04:16.940 ","End":"04:20.895","Text":"This is our junction number 2."},{"Start":"04:20.895 ","End":"04:23.075","Text":"Now that I have this circuit,"},{"Start":"04:23.075 ","End":"04:26.630","Text":"it\u0027s a lot easier for me to look at what is going on"},{"Start":"04:26.630 ","End":"04:30.395","Text":"here and start calculating. Let\u0027s begin."},{"Start":"04:30.395 ","End":"04:34.825","Text":"First of all, we have these 2 resistors,"},{"Start":"04:34.825 ","End":"04:37.475","Text":"which as we can see, they\u0027re in series."},{"Start":"04:37.475 ","End":"04:40.490","Text":"The wire between these 2 resistors doesn\u0027t split,"},{"Start":"04:40.490 ","End":"04:45.740","Text":"which means that our resistors are in series so my resistance of these 2 resistors"},{"Start":"04:45.740 ","End":"04:52.065","Text":"together is simply the resistance of this resistor plus the resistance of this."},{"Start":"04:52.065 ","End":"04:57.810","Text":"That\u0027s going to be equal to 2 plus 2 which is equal to 4."},{"Start":"04:57.810 ","End":"05:04.610","Text":"Now I can cross out this section and I can just pretend that I"},{"Start":"05:04.610 ","End":"05:11.255","Text":"have 1 resistor going on this side that has a resistance of 4 ohms."},{"Start":"05:11.255 ","End":"05:15.490","Text":"This green line goes instead of where my x is."},{"Start":"05:15.490 ","End":"05:25.785","Text":"Now what we want to do is we want to calculate the equivalent resistance due to these 2."},{"Start":"05:25.785 ","End":"05:27.860","Text":"My forum here, in my forum here,"},{"Start":"05:27.860 ","End":"05:30.565","Text":"resistors connected in parallel."},{"Start":"05:30.565 ","End":"05:34.025","Text":"Let\u0027s draw this out again."},{"Start":"05:34.025 ","End":"05:36.005","Text":"I have a resistor here,"},{"Start":"05:36.005 ","End":"05:38.810","Text":"and then this splits into 2,"},{"Start":"05:38.810 ","End":"05:41.675","Text":"so let\u0027s draw this a bit more clearly."},{"Start":"05:41.675 ","End":"05:47.645","Text":"Then I have these 2 resistors which are in parallel,"},{"Start":"05:47.645 ","End":"05:52.578","Text":"which is 4 ohm and 4 ohms."},{"Start":"05:52.578 ","End":"05:59.295","Text":"There\u0027s another resistor over here 6 ohm,"},{"Start":"05:59.295 ","End":"06:04.560","Text":"and this connects to a 3 ohm over"},{"Start":"06:04.560 ","End":"06:12.345","Text":"here and then I have another 3 ohm and here is also 3 ohms."},{"Start":"06:12.345 ","End":"06:15.150","Text":"This is what I have right now."},{"Start":"06:15.150 ","End":"06:20.720","Text":"Now, we want to find the equivalent resistance to this."},{"Start":"06:20.720 ","End":"06:23.240","Text":"I can say 1 divided by my R,"},{"Start":"06:23.240 ","End":"06:25.880","Text":"so they\u0027re just different R totals."},{"Start":"06:25.880 ","End":"06:33.120","Text":"Now, this is R total for this parallel circuit over here, subcircuit."},{"Start":"06:33.120 ","End":"06:36.155","Text":"That\u0027s going to be equal to 1 divided by"},{"Start":"06:36.155 ","End":"06:40.730","Text":"this resistance plus 1 divided by this resistance,"},{"Start":"06:40.730 ","End":"06:43.740","Text":"which is equal to 2 over 4,"},{"Start":"06:43.740 ","End":"06:47.655","Text":"which is a 1/2 and then we have to find the reciprocal."},{"Start":"06:47.655 ","End":"06:51.690","Text":"We do 1 divided by both sides and then we\u0027ll get"},{"Start":"06:51.690 ","End":"06:56.970","Text":"that R total over here is equal to 2 ohms."},{"Start":"06:56.970 ","End":"07:01.730","Text":"That means that we can cross off all of this and we can"},{"Start":"07:01.730 ","End":"07:07.600","Text":"say that it\u0027s simply equal to 1 resistor of 2 ohms."},{"Start":"07:07.600 ","End":"07:10.970","Text":"Now let\u0027s substitute in this 2 ohm resistor in"},{"Start":"07:10.970 ","End":"07:14.635","Text":"place of this parallel thing going on over here."},{"Start":"07:14.635 ","End":"07:19.200","Text":"Here\u0027s our 2 ohm from the parallel subcircuit,"},{"Start":"07:19.200 ","End":"07:25.695","Text":"and now what I want to do is I want to find the resistance of these 2 resistors,"},{"Start":"07:25.695 ","End":"07:30.170","Text":"because we can see that these 2 resistors are in series with one another."},{"Start":"07:30.170 ","End":"07:33.005","Text":"I can\u0027t start working out their resistance with this yet,"},{"Start":"07:33.005 ","End":"07:36.260","Text":"because first I need to find the total resistance on"},{"Start":"07:36.260 ","End":"07:41.410","Text":"this branch of my now new sub parallel circuit."},{"Start":"07:41.410 ","End":"07:45.260","Text":"As we know, my resistance over here,"},{"Start":"07:45.260 ","End":"07:46.475","Text":"because it\u0027s in series,"},{"Start":"07:46.475 ","End":"07:50.215","Text":"is going to be 2 plus 6,"},{"Start":"07:50.215 ","End":"07:52.275","Text":"which is equal to 8."},{"Start":"07:52.275 ","End":"08:00.225","Text":"Now I can switch these 2 resistors with 1 resistor of 8 ohms. Let\u0027s do that."},{"Start":"08:00.225 ","End":"08:04.610","Text":"Now I can see that my 8 ohms over here is connected in"},{"Start":"08:04.610 ","End":"08:09.905","Text":"parallel to this 3 ohm resistor over here."},{"Start":"08:09.905 ","End":"08:20.760","Text":"Now I\u0027m trying to find the total resistance of this sub parallel circuit."},{"Start":"08:20.760 ","End":"08:24.000","Text":"1 over R total,"},{"Start":"08:24.000 ","End":"08:28.520","Text":"in order to find R total when we\u0027re dealing with a parallel circuit is going to"},{"Start":"08:28.520 ","End":"08:33.375","Text":"be 1 divided by 3 plus 1 divided by 8."},{"Start":"08:33.375 ","End":"08:36.890","Text":"Then once we use cross multiplication to add up"},{"Start":"08:36.890 ","End":"08:42.755","Text":"these 2 fractions and then do 1 divided by it to find the reciprocal,"},{"Start":"08:42.755 ","End":"08:46.145","Text":"because we have 1 divided by R_T and we want R_T."},{"Start":"08:46.145 ","End":"08:53.730","Text":"Our R total is going to be equal to 24 divided by 11."},{"Start":"08:53.870 ","End":"09:00.500","Text":"Now the resistance of this is 24 divided by 11 ohms."},{"Start":"09:00.500 ","End":"09:03.115","Text":"Let\u0027s now substitute that in."},{"Start":"09:03.115 ","End":"09:07.505","Text":"Now we\u0027re left with this circuit over here,"},{"Start":"09:07.505 ","End":"09:10.910","Text":"which we can see we have 2 points and we have"},{"Start":"09:10.910 ","End":"09:17.185","Text":"3 resistors connected in series because the wire doesn\u0027t split."},{"Start":"09:17.185 ","End":"09:20.070","Text":"We\u0027re not splitting into 3 different things."},{"Start":"09:20.070 ","End":"09:22.200","Text":"It\u0027s just 1 wire, so it\u0027s in series."},{"Start":"09:22.200 ","End":"09:26.945","Text":"Now we can say that our total resistance is equal to"},{"Start":"09:26.945 ","End":"09:33.720","Text":"3 plus 24 divided by 11 plus 3,"},{"Start":"09:33.720 ","End":"09:36.240","Text":"which is equal to,"},{"Start":"09:36.240 ","End":"09:41.060","Text":"multiply everything by 11."},{"Start":"09:41.060 ","End":"09:47.645","Text":"Then 33 plus 33 is 66 plus 24 divided by 11."},{"Start":"09:47.645 ","End":"09:53.645","Text":"That is just equal to 90 divided by 11."},{"Start":"09:53.645 ","End":"09:58.170","Text":"That\u0027s that over here. Now we can substitute that in."},{"Start":"09:58.170 ","End":"10:02.900","Text":"Now we can see that we can replace this entire circuit,"},{"Start":"10:02.900 ","End":"10:05.090","Text":"which was really complicated,"},{"Start":"10:05.090 ","End":"10:10.404","Text":"with just 1 single resistor between these 2 points."},{"Start":"10:10.404 ","End":"10:16.190","Text":"With a resistance value of 90 divided by 11 ohms."},{"Start":"10:16.190 ","End":"10:21.260","Text":"Now we can see that this can just become this."},{"Start":"10:21.260 ","End":"10:24.665","Text":"This is the equivalent resistance of this circuit."},{"Start":"10:24.665 ","End":"10:27.870","Text":"That\u0027s the end of this lesson."}],"ID":21390},{"Watched":false,"Name":"Exercise 3","Duration":"8m 52s","ChapterTopicVideoID":21482,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this question,"},{"Start":"00:01.920 ","End":"00:06.900","Text":"we\u0027re being asked to calculate the total resistance of the following circuit."},{"Start":"00:06.900 ","End":"00:14.685","Text":"Between these two connecting points that can connect to another circuit or to a battery,"},{"Start":"00:14.685 ","End":"00:18.495","Text":"we have all of these resistors and some ladder circuit."},{"Start":"00:18.495 ","End":"00:24.510","Text":"What we want to do is we want to find the equivalent resistance for this circuit."},{"Start":"00:24.510 ","End":"00:27.335","Text":"When dealing with a ladder circuit,"},{"Start":"00:27.335 ","End":"00:32.150","Text":"the trick is to deal with each rung individually."},{"Start":"00:32.150 ","End":"00:35.435","Text":"What we\u0027re going to do is first,"},{"Start":"00:35.435 ","End":"00:39.545","Text":"we\u0027re going to work out the resistance of all of this,"},{"Start":"00:39.545 ","End":"00:41.195","Text":"this rung in the ladder,"},{"Start":"00:41.195 ","End":"00:47.540","Text":"and then we\u0027ll switch out these three resistors with the equivalent resistor for that,"},{"Start":"00:47.540 ","End":"00:55.800","Text":"and then we\u0027ll work out the equivalent resistor for this and this,"},{"Start":"00:55.800 ","End":"00:58.505","Text":"1-ohm resistor joined in parallel,"},{"Start":"00:58.505 ","End":"01:03.560","Text":"and so on and so forth until we reach the bottom rung."},{"Start":"01:03.560 ","End":"01:09.855","Text":"Let\u0027s call this bubble that we\u0027ll say that this is equal to R_1."},{"Start":"01:09.855 ","End":"01:12.240","Text":"Let\u0027s see what R_1 is equal to."},{"Start":"01:12.240 ","End":"01:16.865","Text":"We can see that these three resistors are in series with one another."},{"Start":"01:16.865 ","End":"01:18.430","Text":"The wire doesn\u0027t split,"},{"Start":"01:18.430 ","End":"01:20.370","Text":"they come one after another."},{"Start":"01:20.370 ","End":"01:24.530","Text":"In series, our total resistance is going to be"},{"Start":"01:24.530 ","End":"01:31.965","Text":"equal to 2+1+2=5 ohms."},{"Start":"01:31.965 ","End":"01:35.085","Text":"So our R_1 is equal to 5 ohms."},{"Start":"01:35.085 ","End":"01:38.780","Text":"Now what we can do is we can replace this,"},{"Start":"01:38.780 ","End":"01:40.780","Text":"so let\u0027s redraw it."},{"Start":"01:40.780 ","End":"01:45.285","Text":"I\u0027ve just drawn this circuit again."},{"Start":"01:45.285 ","End":"01:52.395","Text":"Now we\u0027ve found the equivalent resistance of these three resistors."},{"Start":"01:52.395 ","End":"01:59.909","Text":"What I can do is I can just replace them by my equivalent resistor."},{"Start":"01:59.909 ","End":"02:04.650","Text":"Let\u0027s rub all of these out."},{"Start":"02:04.650 ","End":"02:06.870","Text":"Let\u0027s replace it with the green."},{"Start":"02:06.870 ","End":"02:13.600","Text":"This is equal to one resistor with a resistance of 5 ohms."},{"Start":"02:13.600 ","End":"02:18.185","Text":"Now what we\u0027re doing is we\u0027re now"},{"Start":"02:18.185 ","End":"02:23.015","Text":"working out the equivalent resistance of these two together."},{"Start":"02:23.015 ","End":"02:25.795","Text":"Let\u0027s call this R_2."},{"Start":"02:25.795 ","End":"02:30.920","Text":"We can see that our 1-ohm resistor and our 5-ohm resistor are"},{"Start":"02:30.920 ","End":"02:36.160","Text":"connected in parallel because we can see that our wire splits,"},{"Start":"02:36.160 ","End":"02:37.610","Text":"so our current splits,"},{"Start":"02:37.610 ","End":"02:42.020","Text":"but we have the same exact voltage on each branch."},{"Start":"02:42.020 ","End":"02:45.110","Text":"This is parallel. Our R_2,"},{"Start":"02:45.110 ","End":"02:50.100","Text":"so as we know it\u0027s 1 over R_2 is going to be 1/1+1/5."},{"Start":"02:53.150 ","End":"02:58.955","Text":"Now, when we use cross-multiplication to find the reciprocals so that we get R_2,"},{"Start":"02:58.955 ","End":"03:04.960","Text":"we\u0027ll get that R_2 is equal to 5 divided by 6 ohms."},{"Start":"03:04.960 ","End":"03:09.320","Text":"Now we can replace this 1 ohm connected in parallel with"},{"Start":"03:09.320 ","End":"03:16.000","Text":"5 ohms as one resistor with a value of 5/6 ohms."},{"Start":"03:16.000 ","End":"03:18.735","Text":"Now we\u0027re left with this circuit."},{"Start":"03:18.735 ","End":"03:25.910","Text":"Now what we want to do is we can see that these three resistors are in series."},{"Start":"03:25.910 ","End":"03:31.330","Text":"Now what we want to do is we want to work out the resistance of this,"},{"Start":"03:31.330 ","End":"03:33.420","Text":"let\u0027s call this R_3."},{"Start":"03:33.420 ","End":"03:35.490","Text":"We can see that they\u0027re all in series,"},{"Start":"03:35.490 ","End":"03:37.280","Text":"we have the same current running through,"},{"Start":"03:37.280 ","End":"03:38.720","Text":"our wire doesn\u0027t split,"},{"Start":"03:38.720 ","End":"03:42.110","Text":"so we can say that our R_3 is equal to."},{"Start":"03:42.110 ","End":"03:44.940","Text":"In series, it\u0027s just 2+5/6+2,"},{"Start":"03:47.000 ","End":"03:49.670","Text":"which is equal to,"},{"Start":"03:49.670 ","End":"03:51.770","Text":"once we add up the fractions,"},{"Start":"03:51.770 ","End":"03:55.665","Text":"29 divided by 6 ohms."},{"Start":"03:55.665 ","End":"04:03.030","Text":"All of this can be replaced by one resistor of 29/6 ohms."},{"Start":"04:03.030 ","End":"04:06.030","Text":"Let\u0027s substitute that in."},{"Start":"04:06.030 ","End":"04:08.415","Text":"We can just draw that over here,"},{"Start":"04:08.415 ","End":"04:12.700","Text":"29 divided by 6 ohms."},{"Start":"04:13.070 ","End":"04:16.830","Text":"Now what we want to do is we want to work out"},{"Start":"04:16.830 ","End":"04:22.340","Text":"the equivalent resistance of these two resistors together."},{"Start":"04:22.340 ","End":"04:24.455","Text":"Now we can see that our wire splits,"},{"Start":"04:24.455 ","End":"04:26.509","Text":"which means that our current splits,"},{"Start":"04:26.509 ","End":"04:30.850","Text":"but our voltage across each wire is going to be the same."},{"Start":"04:30.850 ","End":"04:34.685","Text":"Right now we\u0027re connected in parallel."},{"Start":"04:34.685 ","End":"04:40.230","Text":"Let\u0027s call this R_4."},{"Start":"04:40.230 ","End":"04:41.600","Text":"When we\u0027re connected in parallel,"},{"Start":"04:41.600 ","End":"04:51.560","Text":"we know that 1 divided by R_4 is equal to 1 divided by 1 plus 1 divided by 29/6,"},{"Start":"04:51.560 ","End":"04:53.880","Text":"which is simply 6/29."},{"Start":"04:54.380 ","End":"05:00.680","Text":"Then, once we add up these fractions and find the reciprocal,"},{"Start":"05:00.680 ","End":"05:02.015","Text":"because we want R_4,"},{"Start":"05:02.015 ","End":"05:03.950","Text":"not 1 divided by R_4,"},{"Start":"05:03.950 ","End":"05:11.960","Text":"we\u0027ll get that our R_4 is equal to 29 divided by 35 ohms."},{"Start":"05:11.960 ","End":"05:14.285","Text":"That\u0027s all of this over here,"},{"Start":"05:14.285 ","End":"05:20.090","Text":"so we can simply replace all of this, and this,"},{"Start":"05:20.090 ","End":"05:29.100","Text":"of course, with one resistor equal to 29 divided by 35 ohms."},{"Start":"05:29.100 ","End":"05:36.360","Text":"Now again, we want to add up these three resistors."},{"Start":"05:36.360 ","End":"05:38.815","Text":"Let\u0027s call this R_5."},{"Start":"05:38.815 ","End":"05:40.265","Text":"Now as we can see,"},{"Start":"05:40.265 ","End":"05:41.380","Text":"our wire doesn\u0027t split,"},{"Start":"05:41.380 ","End":"05:43.175","Text":"it\u0027s the same current running through."},{"Start":"05:43.175 ","End":"05:47.465","Text":"These three resistors are in series."},{"Start":"05:47.465 ","End":"05:49.170","Text":"In series, we just add them,"},{"Start":"05:49.170 ","End":"05:56.240","Text":"so we have 2+29 divided by 35+2."},{"Start":"05:56.240 ","End":"06:03.465","Text":"That is equal to 169 divided by 35 ohms."},{"Start":"06:03.465 ","End":"06:07.515","Text":"Now again, we can rub out all of this,"},{"Start":"06:07.515 ","End":"06:13.740","Text":"and we can add in our equivalent resistance to those three,"},{"Start":"06:13.740 ","End":"06:19.570","Text":"which is equal to 169 divided by 35 ohms."},{"Start":"06:20.060 ","End":"06:22.785","Text":"Again, this is very repetitive,"},{"Start":"06:22.785 ","End":"06:26.435","Text":"now we\u0027re trying to find our resistance of these two."},{"Start":"06:26.435 ","End":"06:28.730","Text":"Let\u0027s call this R_5."},{"Start":"06:28.730 ","End":"06:32.000","Text":"We can see again that they\u0027re connected in parallel."},{"Start":"06:32.000 ","End":"06:33.765","Text":"We\u0027ll go over it again."},{"Start":"06:33.765 ","End":"06:41.989","Text":"We have 1 divided by R_5 is equal to 1 divided by 1 plus 1 divided by 169 divided by 35,"},{"Start":"06:41.989 ","End":"06:45.410","Text":"which is 35 divided by 169."},{"Start":"06:45.410 ","End":"06:51.050","Text":"Then once we add all of our fractions and find the reciprocal,"},{"Start":"06:51.050 ","End":"06:55.035","Text":"because we want R_5 and not 1 divided by R_5,"},{"Start":"06:55.035 ","End":"07:04.990","Text":"we\u0027ll have that this is equal to 169 divided by 204 ohms."},{"Start":"07:04.990 ","End":"07:15.810","Text":"Now again, we can replace these two resistors simply with 1 equivalent resistor,"},{"Start":"07:15.810 ","End":"07:22.015","Text":"which is equal to 169 divided by 204 ohms."},{"Start":"07:22.015 ","End":"07:26.720","Text":"Now finally, we have our final thing."},{"Start":"07:26.720 ","End":"07:29.105","Text":"Let\u0027s call this R_6."},{"Start":"07:29.105 ","End":"07:33.330","Text":"This is our final rung in our ladder."},{"Start":"07:33.330 ","End":"07:38.180","Text":"Remember, we started off from this and we\u0027ve worked our way down to just this."},{"Start":"07:38.180 ","End":"07:43.370","Text":"Again, all these three are connected in series,"},{"Start":"07:43.370 ","End":"07:44.780","Text":"which means that we can just add them."},{"Start":"07:44.780 ","End":"07:50.160","Text":"We have 2 plus 169 divided by 204 plus 2,"},{"Start":"07:51.700 ","End":"07:56.340","Text":"let\u0027s cancel this out."},{"Start":"07:56.340 ","End":"08:04.440","Text":"Let\u0027s just say that it\u0027s equal to 169 divided by 204 plus 4 ohms."},{"Start":"08:04.440 ","End":"08:08.555","Text":"Now our final stage,"},{"Start":"08:08.555 ","End":"08:10.580","Text":"we can cross out all of this,"},{"Start":"08:10.580 ","End":"08:13.910","Text":"and we can just put in this place one resistor,"},{"Start":"08:13.910 ","End":"08:19.895","Text":"which is equal to 169 divided by 204 plus 4,"},{"Start":"08:19.895 ","End":"08:24.450","Text":"you can just add this up in your calculator, ohms."},{"Start":"08:24.450 ","End":"08:27.380","Text":"Now we can see that the total resistance of"},{"Start":"08:27.380 ","End":"08:32.145","Text":"the following circuit is equal to this over here."},{"Start":"08:32.145 ","End":"08:37.715","Text":"I could replace all of this whole mess that we have over here,"},{"Start":"08:37.715 ","End":"08:42.850","Text":"with simply one resistor with a resistance of this,"},{"Start":"08:42.850 ","End":"08:49.080","Text":"and I will be keeping the physics in my question exactly the same."},{"Start":"08:49.080 ","End":"08:52.480","Text":"That\u0027s the end of this lesson."}],"ID":22299},{"Watched":false,"Name":"Exercise 4","Duration":"4m 53s","ChapterTopicVideoID":21483,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.425","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.425 ","End":"00:09.570","Text":"Find as many ways to connect 3 identical resistors of resistance"},{"Start":"00:09.570 ","End":"00:15.570","Text":"R. Then we\u0027re being told to calculate the total resistance of these circuits."},{"Start":"00:15.570 ","End":"00:21.885","Text":"The fast way to connect these 3 resistors is in series."},{"Start":"00:21.885 ","End":"00:24.810","Text":"We have 1 resistor,"},{"Start":"00:24.810 ","End":"00:27.375","Text":"another resistor connected to that,"},{"Start":"00:27.375 ","End":"00:31.720","Text":"and another resistor connected to that."},{"Start":"00:32.090 ","End":"00:35.900","Text":"We\u0027ll calculate the total resistance afterward."},{"Start":"00:35.900 ","End":"00:44.450","Text":"Our next option is to have 2 resistors in parallel and the 3rd in series with them."},{"Start":"00:44.450 ","End":"00:52.950","Text":"Then we would have something like so and then the 3rd resistor over"},{"Start":"00:52.950 ","End":"00:57.590","Text":"here and the 3rd and"},{"Start":"00:57.590 ","End":"01:05.375","Text":"final option is if we were to have 3 resistors in parallel,"},{"Start":"01:05.375 ","End":"01:14.905","Text":"we would have something like so connected, like so."},{"Start":"01:14.905 ","End":"01:18.250","Text":"If the resistors weren\u0027t identical and each 1 had"},{"Start":"01:18.250 ","End":"01:23.320","Text":"a different resistance then I could switch up other things over here."},{"Start":"01:23.320 ","End":"01:29.140","Text":"Like I could have this one in parallel over here and then I would have more options,"},{"Start":"01:29.140 ","End":"01:31.555","Text":"but because they\u0027re identical,"},{"Start":"01:31.555 ","End":"01:34.670","Text":"these are my only options."},{"Start":"01:35.180 ","End":"01:37.455","Text":"We have options 1,"},{"Start":"01:37.455 ","End":"01:39.675","Text":"2, and 3."},{"Start":"01:39.675 ","End":"01:45.265","Text":"Let\u0027s first answer the total resistance for option 1,"},{"Start":"01:45.265 ","End":"01:48.850","Text":"we can see all the resistors are in series, therefore,"},{"Start":"01:48.850 ","End":"01:54.350","Text":"our total is R1 plus R2 plus R3."},{"Start":"01:54.620 ","End":"01:59.930","Text":"Like so, and because each resistor is identical,"},{"Start":"01:59.930 ","End":"02:06.600","Text":"so we have R plus R plus R so we have that this is equal to 3R."},{"Start":"02:07.820 ","End":"02:11.715","Text":"Then for option number 2,"},{"Start":"02:11.715 ","End":"02:19.240","Text":"so here we have 2 resistors in parallel and then connected to the 3rd in series."},{"Start":"02:19.240 ","End":"02:21.870","Text":"This, I have to split up into 2 sections."},{"Start":"02:21.870 ","End":"02:28.040","Text":"First, I\u0027m going to calculate the total resistance of the 2 resistors in parallel."},{"Start":"02:28.040 ","End":"02:31.355","Text":"Let\u0027s call this, R total parallel."},{"Start":"02:31.355 ","End":"02:35.810","Text":"We know that 1 divided by R total parallel is equal to 1"},{"Start":"02:35.810 ","End":"02:40.310","Text":"divided by the first resistor plus 1 divided by"},{"Start":"02:40.310 ","End":"02:49.365","Text":"the 2nd resistor and so we know that this is going to be equal too,"},{"Start":"02:49.365 ","End":"02:55.100","Text":"once we also take the reciprocal so R total parallel is going to be equal"},{"Start":"02:55.100 ","End":"03:03.190","Text":"to R1 multiplied by R2 divided by R1 plus R2."},{"Start":"03:03.190 ","End":"03:07.655","Text":"I worked out the common denominator and then took the reciprocal."},{"Start":"03:07.655 ","End":"03:11.660","Text":"Here we have R multiplied by R,"},{"Start":"03:11.660 ","End":"03:17.610","Text":"which is R^2 divided by R plus R, which is 2R."},{"Start":"03:17.610 ","End":"03:22.320","Text":"Then this cancels out and this is equal to 1.5R."},{"Start":"03:22.850 ","End":"03:27.440","Text":"Now, in order to find the total resistance,"},{"Start":"03:27.440 ","End":"03:32.065","Text":"so I have this resistor in parallel,"},{"Start":"03:32.065 ","End":"03:37.550","Text":"which is R total parallel connected in series to this resistor."},{"Start":"03:37.550 ","End":"03:46.215","Text":"All I\u0027m going to do is I\u0027m going to add R total parallel plus my R3,"},{"Start":"03:46.215 ","End":"03:51.240","Text":"and that is going to be equal to 1.5R plus R3 is"},{"Start":"03:51.240 ","End":"03:57.130","Text":"equal to R so this is equal to 3 divided by 2R."},{"Start":"03:58.970 ","End":"04:03.030","Text":"Then for number 3,"},{"Start":"04:03.030 ","End":"04:09.615","Text":"we have 3 resistors connected in parallel,"},{"Start":"04:09.615 ","End":"04:12.240","Text":"again, we\u0027re going to use this equation,"},{"Start":"04:12.240 ","End":"04:14.460","Text":"so we have 1 over, this time it\u0027s,"},{"Start":"04:14.460 ","End":"04:19.050","Text":"R total because they\u0027re all in parallel and this is equal to 1 divided by"},{"Start":"04:19.050 ","End":"04:24.930","Text":"R1 plus 1 divided by R2 plus 1 divided by R3."},{"Start":"04:24.930 ","End":"04:27.720","Text":"Now, as we know, R1, 2,"},{"Start":"04:27.720 ","End":"04:36.810","Text":"and 3 are equal to R so we have that this is equal simply to 3"},{"Start":"04:36.810 ","End":"04:41.075","Text":"divided by R and then we take the reciprocal in order to get"},{"Start":"04:41.075 ","End":"04:48.210","Text":"R total and we get that R total is equal to R divided by 3."},{"Start":"04:49.310 ","End":"04:53.830","Text":"That is the onset to this question."}],"ID":22300},{"Watched":false,"Name":"Exercise 5","Duration":"18m 41s","ChapterTopicVideoID":21484,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.935","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.935 ","End":"00:10.410","Text":"We\u0027re being asked to calculate the current and voltage in each resistor."},{"Start":"00:10.410 ","End":"00:13.020","Text":"Here we have a circuit where we have a battery of"},{"Start":"00:13.020 ","End":"00:17.670","Text":"37 volts and connected we have 2 resistors,"},{"Start":"00:17.670 ","End":"00:20.430","Text":"each of 1 ohm, 2 resistors,"},{"Start":"00:20.430 ","End":"00:21.615","Text":"each of 2 ohms,"},{"Start":"00:21.615 ","End":"00:23.670","Text":"and another 2 resistors,"},{"Start":"00:23.670 ","End":"00:26.140","Text":"each of 3 ohms."},{"Start":"00:26.960 ","End":"00:32.085","Text":"Let\u0027s begin. The first thing that we can see is that"},{"Start":"00:32.085 ","End":"00:38.235","Text":"our 2 resistors of 1 ohm are connected in parallel to one another."},{"Start":"00:38.235 ","End":"00:42.980","Text":"We can see that because if we look at the potential over here,"},{"Start":"00:42.980 ","End":"00:47.525","Text":"then it\u0027s connected via wire to this point over here,"},{"Start":"00:47.525 ","End":"00:49.940","Text":"which means that the potential at these 2 points is the"},{"Start":"00:49.940 ","End":"00:53.420","Text":"same and similarly with these 2 points,"},{"Start":"00:53.420 ","End":"00:56.135","Text":"the potential over here"},{"Start":"00:56.135 ","End":"01:01.050","Text":"is the same as the potential over here because they\u0027re connected by the same wire."},{"Start":"01:01.760 ","End":"01:06.110","Text":"That means that the voltage over these 2 resistors is the"},{"Start":"01:06.110 ","End":"01:11.375","Text":"same and that means that these resistors are connected indeed in parallel."},{"Start":"01:11.375 ","End":"01:21.980","Text":"Then we can say that the total resistance of these 2 resistors is 1 divided by R_T,"},{"Start":"01:21.980 ","End":"01:28.770","Text":"which is 1 divided by R_1 plus 1 divided by R_2."},{"Start":"01:28.770 ","End":"01:31.109","Text":"If there were third and fourth resistors,"},{"Start":"01:31.109 ","End":"01:36.990","Text":"then we would add plus 1 divided by R_3 plus 1 divided by R_4, and so on."},{"Start":"01:36.990 ","End":"01:41.615","Text":"Once we find the common denominator and take the reciprocal,"},{"Start":"01:41.615 ","End":"01:47.450","Text":"we get that our total for 2 resistors in parallel is equal to"},{"Start":"01:47.450 ","End":"01:57.555","Text":"R_1 multiplied by R_2 divided by R_1 plus R_2."},{"Start":"01:57.555 ","End":"02:01.370","Text":"The resistance for Resistor number 1 is"},{"Start":"02:01.370 ","End":"02:05.300","Text":"1 ohm multiplied by the resistance for Resistor number 2,"},{"Start":"02:05.300 ","End":"02:14.080","Text":"which is another 1 ohms divided by 1 ohm plus 1 ohm."},{"Start":"02:14.080 ","End":"02:22.830","Text":"Then we can see that here we\u0027ll have 1 ohm^2 divided by 2 ohms."},{"Start":"02:22.830 ","End":"02:28.460","Text":"Then we can cancel out one of this and therefore we get that the total resistance"},{"Start":"02:28.460 ","End":"02:35.430","Text":"of these 2 resistors in parallel is 1/2 ohm."},{"Start":"02:36.590 ","End":"02:44.150","Text":"Now we can consider these 2 resistors as just one resistor with the following resistance."},{"Start":"02:44.150 ","End":"02:48.390","Text":"Our circuit diagram will look like so."},{"Start":"02:48.980 ","End":"02:56.630","Text":"I just switched these 2 resistors with 1 resistor of value of 1/2 ohm."},{"Start":"02:56.630 ","End":"03:03.530","Text":"Now my next step is to find the total resistance due to this resistor,"},{"Start":"03:03.530 ","End":"03:06.770","Text":"the blue resistor that I just calculated,"},{"Start":"03:06.770 ","End":"03:09.755","Text":"and this resistor of 2 ohms."},{"Start":"03:09.755 ","End":"03:17.940","Text":"These 2 resistors have the same current passing through them because they\u0027re in series."},{"Start":"03:17.940 ","End":"03:23.585","Text":"That means that in order to find the total resistance,"},{"Start":"03:23.585 ","End":"03:26.570","Text":"I\u0027m just going to add on the 2 resistors."},{"Start":"03:26.570 ","End":"03:29.023","Text":"I have my R_1,"},{"Start":"03:29.023 ","End":"03:33.480","Text":"2 plus my R_3."},{"Start":"03:33.480 ","End":"03:39.090","Text":"That\u0027s equal to 1 divided by 2 ohm plus 2 ohms,"},{"Start":"03:39.090 ","End":"03:46.350","Text":"which is equal to 5 divided by 2 ohms."},{"Start":"03:46.350 ","End":"03:51.000","Text":"Now I can replace this diagram with the following."},{"Start":"03:51.430 ","End":"03:53.780","Text":"This is our new diagram."},{"Start":"03:53.780 ","End":"04:01.700","Text":"I\u0027ve just replaced these 2 resistors with this resistor and now we can see that"},{"Start":"04:01.700 ","End":"04:04.025","Text":"the potential on this point is"},{"Start":"04:04.025 ","End":"04:06.800","Text":"equal to the potential at this point because they\u0027re both connected with"},{"Start":"04:06.800 ","End":"04:09.290","Text":"the same wire and the same thing for"},{"Start":"04:09.290 ","End":"04:12.830","Text":"these 2 points over here because they\u0027re connected by the same wire."},{"Start":"04:12.830 ","End":"04:20.038","Text":"Therefore, these 2 resistors are connected in parallel."},{"Start":"04:20.038 ","End":"04:25.160","Text":"Our R Total is going to be equal to, again,"},{"Start":"04:25.160 ","End":"04:30.485","Text":"let\u0027s call this our R_1,"},{"Start":"04:30.485 ","End":"04:32.765","Text":"2, 3 over here,"},{"Start":"04:32.765 ","End":"04:34.700","Text":"multiplied by our R4 divided by R_1,"},{"Start":"04:34.700 ","End":"04:36.750","Text":"2, 3 plus R4."},{"Start":"04:42.410 ","End":"04:46.335","Text":"Here we have our R_1, 2, 3,"},{"Start":"04:46.335 ","End":"04:51.795","Text":"which is 5 over 2 ohms multiplied by this,"},{"Start":"04:51.795 ","End":"04:56.290","Text":"which is 2 ohms divided by 5 over 2 ohms plus 2 ohms."},{"Start":"05:01.330 ","End":"05:04.625","Text":"This 2 and this 2 can cancel out,"},{"Start":"05:04.625 ","End":"05:08.495","Text":"this ohm can cancel out with these 2."},{"Start":"05:08.495 ","End":"05:11.450","Text":"Then what we have"},{"Start":"05:11.450 ","End":"05:20.785","Text":"is 5 divided by 9 over 2,"},{"Start":"05:20.785 ","End":"05:24.045","Text":"when we add these 2 ohms,"},{"Start":"05:24.045 ","End":"05:29.985","Text":"which is simply equal to 10 divided by 9 ohms."},{"Start":"05:29.985 ","End":"05:36.550","Text":"Now I\u0027m going to replace these 2 resistors with the following diagram."},{"Start":"05:37.520 ","End":"05:44.250","Text":"Now we\u0027ve replaced these 2 resistors by this resistor and now we can"},{"Start":"05:44.250 ","End":"05:50.330","Text":"see that these 2 resistors are connected by the same string or the same wire,"},{"Start":"05:50.330 ","End":"05:53.820","Text":"so they\u0027re connected in series."},{"Start":"05:53.820 ","End":"05:57.800","Text":"That means that they have the same current running through them"},{"Start":"05:57.800 ","End":"06:03.215","Text":"so the total resistance is just the sum of the 2."},{"Start":"06:03.215 ","End":"06:13.200","Text":"We have 10 divided by 9 ohms plus 3 ohms and this is just going"},{"Start":"06:13.200 ","End":"06:16.018","Text":"to be equal to 37"},{"Start":"06:16.018 ","End":"06:21.261","Text":"divided"},{"Start":"06:21.261 ","End":"06:29.490","Text":"by 9 ohms."},{"Start":"06:29.490 ","End":"06:32.140","Text":"Now I can replace these 2."},{"Start":"06:32.630 ","End":"06:37.445","Text":"I just drew it on the same diagram to save time and then again,"},{"Start":"06:37.445 ","End":"06:41.090","Text":"this point is connected to this point via the same string"},{"Start":"06:41.090 ","End":"06:44.750","Text":"and this point is connected to this point via the same string,"},{"Start":"06:44.750 ","End":"06:46.790","Text":"via the same wire, which means that"},{"Start":"06:46.790 ","End":"06:49.700","Text":"the potential difference across each resistor is the same,"},{"Start":"06:49.700 ","End":"06:55.150","Text":"which means that these 2 resistors are connected in parallel."},{"Start":"06:55.150 ","End":"06:57.780","Text":"Again, we can write this out."},{"Start":"06:57.780 ","End":"07:05.880","Text":"We have 37 divided by 9 ohms multiplied by 3 ohms"},{"Start":"07:05.880 ","End":"07:08.864","Text":"divided by 37 divided by"},{"Start":"07:08.864 ","End":"07:17.320","Text":"9 ohms plus 3 ohms so this ohm can cancel out with these 2."},{"Start":"07:17.960 ","End":"07:23.470","Text":"Then we can cancel this 3 with this 9 over here."},{"Start":"07:25.460 ","End":"07:30.470","Text":"Then when we add these 2 numbers,"},{"Start":"07:30.470 ","End":"07:32.270","Text":"so on the denominator,"},{"Start":"07:32.270 ","End":"07:35.840","Text":"we\u0027re going to have 64 divided by"},{"Start":"07:35.840 ","End":"07:45.720","Text":"9 and then in the numerator we have 37 divided by 3."},{"Start":"07:47.180 ","End":"07:51.500","Text":"Then once you play around with this fraction,"},{"Start":"07:51.500 ","End":"07:54.680","Text":"what we\u0027ll get is 111 divided by 64"},{"Start":"07:54.680 ","End":"08:03.280","Text":"and of course here we have ohms and here as well."},{"Start":"08:03.740 ","End":"08:14.145","Text":"Now we can replace these 2 diagrams with this 1 resistor,"},{"Start":"08:14.145 ","End":"08:24.050","Text":"which is equal to 111 divided by 64 ohms."},{"Start":"08:24.050 ","End":"08:26.795","Text":"The original question was,"},{"Start":"08:26.795 ","End":"08:28.550","Text":"if we scroll up,"},{"Start":"08:28.550 ","End":"08:33.720","Text":"to calculate the current and voltage on each resistor."},{"Start":"08:34.240 ","End":"08:37.550","Text":"Let\u0027s begin doing that."},{"Start":"08:37.550 ","End":"08:39.760","Text":"We\u0027ll scroll down over here."},{"Start":"08:39.760 ","End":"08:48.060","Text":"First of all, we see that we have a voltage of 37 volts coming from the battery and"},{"Start":"08:48.060 ","End":"08:56.870","Text":"we can see that it\u0027s hitting this resistor or these resistors which are in parallel."},{"Start":"08:56.870 ","End":"09:00.110","Text":"From Ohm\u0027s law,"},{"Start":"09:00.110 ","End":"09:09.710","Text":"we know that V=I_R."},{"Start":"09:09.710 ","End":"09:12.875","Text":"In order to find the current,"},{"Start":"09:12.875 ","End":"09:20.390","Text":"so that is equal to V divided by R. We have V voltage,"},{"Start":"09:20.390 ","End":"09:25.520","Text":"which is 37 volts divided by the resistance."},{"Start":"09:25.520 ","End":"09:34.889","Text":"The voltage is hitting both this resistor and this resistor."},{"Start":"09:34.889 ","End":"09:37.920","Text":"Let\u0027s first find for this resistor,"},{"Start":"09:37.920 ","End":"09:38.940","Text":"the 3 ohm resistor,"},{"Start":"09:38.940 ","End":"09:42.480","Text":"so 37 divided by 3 ohms,"},{"Start":"09:42.480 ","End":"09:47.098","Text":"which is simply equal to 37 divided"},{"Start":"09:47.098 ","End":"09:52.765","Text":"3 amps or Amperes because that\u0027s the unit for current."},{"Start":"09:52.765 ","End":"09:58.140","Text":"Let\u0027s write that in our original diagram."},{"Start":"09:58.280 ","End":"10:00.920","Text":"We can write it in red."},{"Start":"10:00.920 ","End":"10:08.210","Text":"We have a 3 ohm resistor and then we have 37 volts going"},{"Start":"10:08.210 ","End":"10:16.590","Text":"over it and we have 37 divided by 3 amps going over it."},{"Start":"10:17.590 ","End":"10:24.440","Text":"Then we had these 2 resistors in the stage before that,"},{"Start":"10:24.440 ","End":"10:27.415","Text":"what we calculated over here."},{"Start":"10:27.415 ","End":"10:35.330","Text":"First of all,"},{"Start":"10:35.330 ","End":"10:42.155","Text":"the voltage going through this point and this point and this point and this point"},{"Start":"10:42.155 ","End":"10:49.280","Text":"is the same and these 2 resistors together were equal to this resistor."},{"Start":"10:49.280 ","End":"10:51.845","Text":"We calculated that over here."},{"Start":"10:51.845 ","End":"10:55.430","Text":"In order to calculate the current going"},{"Start":"10:55.430 ","End":"10:59.690","Text":"through these 2 resistors of this total resistance,"},{"Start":"10:59.690 ","End":"11:03.860","Text":"so we have that I is equal to the voltage,"},{"Start":"11:03.860 ","End":"11:08.840","Text":"so it\u0027s the same voltage because these 2 are in parallel."},{"Start":"11:08.840 ","End":"11:14.460","Text":"We have 37 divided by the total resistance of this,"},{"Start":"11:14.460 ","End":"11:20.445","Text":"which is 37 divided by 9."},{"Start":"11:20.445 ","End":"11:26.940","Text":"This is simply equal to 9 amps."},{"Start":"11:26.940 ","End":"11:31.995","Text":"We know that the same current runs through both of these."},{"Start":"11:31.995 ","End":"11:35.430","Text":"However, a different voltage runs through the 2."},{"Start":"11:35.430 ","End":"11:42.735","Text":"Now we\u0027re going to use Ohm\u0027s law to find the voltage over this resistor of 3 ohms."},{"Start":"11:42.735 ","End":"11:44.700","Text":"I made a mistake before I wrote 2,"},{"Start":"11:44.700 ","End":"11:46.365","Text":"but it\u0027s actually 3 ohms."},{"Start":"11:46.365 ","End":"11:50.070","Text":"The voltage is equal to the current,"},{"Start":"11:50.070 ","End":"11:53.160","Text":"which is 9 amps,"},{"Start":"11:53.160 ","End":"11:55.440","Text":"multiplied by the resistance,"},{"Start":"11:55.440 ","End":"11:59.070","Text":"which is 3 ohms."},{"Start":"11:59.070 ","End":"12:05.505","Text":"We get a voltage of 27 volts for this over here."},{"Start":"12:05.505 ","End":"12:07.200","Text":"The 3 on the corner."},{"Start":"12:07.200 ","End":"12:11.655","Text":"Let\u0027s go back to our original diagram."},{"Start":"12:11.655 ","End":"12:18.735","Text":"Here we have a resistance of 3 ohms and a current of 9 amps,"},{"Start":"12:18.735 ","End":"12:24.040","Text":"and a voltage of 27 volts."},{"Start":"12:24.980 ","End":"12:31.875","Text":"Now we want to find the voltage over this resistor."},{"Start":"12:31.875 ","End":"12:34.200","Text":"There are 2 ways to do this."},{"Start":"12:34.200 ","End":"12:36.045","Text":"Either we can say,"},{"Start":"12:36.045 ","End":"12:39.960","Text":"let\u0027s call this V_2 and this V_1."},{"Start":"12:39.960 ","End":"12:46.545","Text":"We can say that V_1 is equal to V total minus V_2,"},{"Start":"12:46.545 ","End":"12:53.565","Text":"which is equal to 37 volts minus 27 volts,"},{"Start":"12:53.565 ","End":"12:55.680","Text":"which is equal to 10 volts,"},{"Start":"12:55.680 ","End":"13:00.700","Text":"or the other way that we can do this is by Ohm\u0027s law."},{"Start":"13:01.010 ","End":"13:05.355","Text":"V_1 is equal to the current,"},{"Start":"13:05.355 ","End":"13:10.635","Text":"which is 9 amps multiplied by the resistance,"},{"Start":"13:10.635 ","End":"13:14.985","Text":"which is 10 divided by 9 ohms."},{"Start":"13:14.985 ","End":"13:20.410","Text":"Then again, we get 10 amps either way."},{"Start":"13:21.380 ","End":"13:26.355","Text":"Sorry, of course, this is in volts, not in amps."},{"Start":"13:26.355 ","End":"13:36.340","Text":"Then I can go back over here and so on all of this,"},{"Start":"13:37.340 ","End":"13:41.250","Text":"I have a voltage of 10 volts."},{"Start":"13:41.250 ","End":"13:44.795","Text":"Let\u0027s write this over here,"},{"Start":"13:44.795 ","End":"13:49.835","Text":"10 volts over here and over here,"},{"Start":"13:49.835 ","End":"13:51.320","Text":"I also have 10 volts,"},{"Start":"13:51.320 ","End":"13:54.570","Text":"we\u0027ll carry on to these in a minute."},{"Start":"13:54.650 ","End":"13:57.720","Text":"Then let\u0027s find the current."},{"Start":"13:57.720 ","End":"14:04.710","Text":"From Ohm\u0027s law we know that V is equal to I_R."},{"Start":"14:04.710 ","End":"14:10.635","Text":"Therefore, what we\u0027ll get is that our current is equal to the voltage,"},{"Start":"14:10.635 ","End":"14:16.365","Text":"which is 10 volts divided by the resistance."},{"Start":"14:16.365 ","End":"14:22.455","Text":"Again, we\u0027re looking at this resistor."},{"Start":"14:22.455 ","End":"14:25.215","Text":"It has a resistance of 2 ohms,"},{"Start":"14:25.215 ","End":"14:28.230","Text":"so divided by 2 ohms."},{"Start":"14:28.230 ","End":"14:33.645","Text":"We get that the current passing through this is equal to 5 amps."},{"Start":"14:33.645 ","End":"14:40.710","Text":"Let\u0027s scroll back up and then the current here is 5 amps."},{"Start":"14:40.710 ","End":"14:46.410","Text":"Now we\u0027re looking at all of these resistors,"},{"Start":"14:46.410 ","End":"14:51.435","Text":"which is just the same as this resist over here."},{"Start":"14:51.435 ","End":"14:56.250","Text":"We\u0027re going to carry on the calculation just over here looking here."},{"Start":"14:56.250 ","End":"15:03.720","Text":"We already know that the voltage over here,"},{"Start":"15:03.720 ","End":"15:07.185","Text":"as we remember, was 10 volts."},{"Start":"15:07.185 ","End":"15:13.620","Text":"That means that we have the same voltage over this red resistor over here."},{"Start":"15:13.620 ","End":"15:15.734","Text":"In order to find the current,"},{"Start":"15:15.734 ","End":"15:19.590","Text":"current is equal to voltage divided by resistance."},{"Start":"15:19.590 ","End":"15:22.950","Text":"We have 10 volts divided by the resistance,"},{"Start":"15:22.950 ","End":"15:27.285","Text":"which is 5 over 2 ohms."},{"Start":"15:27.285 ","End":"15:32.500","Text":"What we\u0027re going to get is that this is equal to 4 amps."},{"Start":"15:32.840 ","End":"15:40.720","Text":"The current flowing through this resistor over here is 4 amps."},{"Start":"15:41.120 ","End":"15:44.370","Text":"Now we go one stage back,"},{"Start":"15:44.370 ","End":"15:47.190","Text":"and we remember that this red resistor is made"},{"Start":"15:47.190 ","End":"15:51.630","Text":"up of this blue resistor and this 2 ohm resistor."},{"Start":"15:51.630 ","End":"15:56.910","Text":"If the red resistor has a current of 4 amps,"},{"Start":"15:56.910 ","End":"16:00.555","Text":"that means that the same current is flowing through"},{"Start":"16:00.555 ","End":"16:05.070","Text":"the blue resistor and through the 2 ohm resistor."},{"Start":"16:05.070 ","End":"16:09.119","Text":"Therefore, we know that on the 2 ohm resistor,"},{"Start":"16:09.119 ","End":"16:13.935","Text":"we have a current of 4 amps flowing through it."},{"Start":"16:13.935 ","End":"16:15.495","Text":"Let\u0027s write this here."},{"Start":"16:15.495 ","End":"16:18.940","Text":"We have a current of 4 amps."},{"Start":"16:18.940 ","End":"16:22.610","Text":"Now if we want to know the voltage on this resistor,"},{"Start":"16:22.610 ","End":"16:25.080","Text":"so through Ohm\u0027s law,"},{"Start":"16:25.080 ","End":"16:26.775","Text":"let\u0027s just go down to here,"},{"Start":"16:26.775 ","End":"16:33.105","Text":"we know that V is equal to I_R and our current is 4,"},{"Start":"16:33.105 ","End":"16:36.515","Text":"and our resistance is 2."},{"Start":"16:36.515 ","End":"16:39.875","Text":"We get that this is equal to 8 volts."},{"Start":"16:39.875 ","End":"16:47.010","Text":"We can scroll up here and write that on this resistor we have 8 volts."},{"Start":"16:47.010 ","End":"16:52.140","Text":"We saw that our 4 amps was working over here."},{"Start":"16:52.140 ","End":"16:55.110","Text":"Over here we have 4 amps."},{"Start":"16:55.110 ","End":"16:59.250","Text":"Also over here on this blue resistor,"},{"Start":"16:59.250 ","End":"17:04.500","Text":"which in reality is made up of these 2 resistors of 1 ohm each."},{"Start":"17:04.500 ","End":"17:10.290","Text":"In order to know the voltage across this blue resistor,"},{"Start":"17:10.290 ","End":"17:12.930","Text":"we can simply write 10,"},{"Start":"17:12.930 ","End":"17:20.445","Text":"because we know that 10 volts is flowing all through here from this stage down here."},{"Start":"17:20.445 ","End":"17:24.270","Text":"Then what we can write is 10 minus the voltage flowing"},{"Start":"17:24.270 ","End":"17:28.530","Text":"through the 2 ohm resistor, which is 8."},{"Start":"17:28.530 ","End":"17:37.500","Text":"Then we have that this is equal to 2 volts going over this blue resistor, or of course,"},{"Start":"17:37.500 ","End":"17:39.720","Text":"we could have done V is equal to I_R,"},{"Start":"17:39.720 ","End":"17:47.130","Text":"so I is 4 amps and R is 1/2 an ohm and we would have gotten 2 volts."},{"Start":"17:47.130 ","End":"17:49.830","Text":"Two volts is over the blue resistor."},{"Start":"17:49.830 ","End":"17:53.160","Text":"Of course, the blue resistor is made out of these 2 resistors,"},{"Start":"17:53.160 ","End":"17:54.915","Text":"which are in parallel,"},{"Start":"17:54.915 ","End":"17:59.160","Text":"which means that each resistor has the same voltage across it,"},{"Start":"17:59.160 ","End":"18:05.895","Text":"so this has a voltage of 2 volts and also this one is 2 volts."},{"Start":"18:05.895 ","End":"18:09.030","Text":"Then in order to know the current,"},{"Start":"18:09.030 ","End":"18:11.850","Text":"so V over R is equal to I,"},{"Start":"18:11.850 ","End":"18:18.660","Text":"so a voltage is 2 and our resistance is 1."},{"Start":"18:18.660 ","End":"18:25.785","Text":"Therefore we get a current of 2 amps over each one of these resistors."},{"Start":"18:25.785 ","End":"18:29.830","Text":"We can just write this in like so."},{"Start":"18:30.110 ","End":"18:32.415","Text":"That\u0027s the end of the lesson."},{"Start":"18:32.415 ","End":"18:40.869","Text":"We found the voltage and the current in each one of the resistors in this diagram."}],"ID":22301},{"Watched":false,"Name":"Open And Short Circuits","Duration":"9m 34s","ChapterTopicVideoID":21476,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:06.960","Text":"we\u0027re going to be speaking about open circuits and short-circuits."},{"Start":"00:06.960 ","End":"00:11.055","Text":"Let\u0027s imagine that we have here some circuit,"},{"Start":"00:11.055 ","End":"00:19.515","Text":"and it has at least 1 electrical components over here, like so."},{"Start":"00:19.515 ","End":"00:25.155","Text":"Now we can see that we have our current running through our circuit."},{"Start":"00:25.155 ","End":"00:29.819","Text":"Now, if we were to cut our string at any point,"},{"Start":"00:29.819 ","End":"00:32.145","Text":"so let\u0027s say here."},{"Start":"00:32.145 ","End":"00:36.465","Text":"Due to both of our wires not connecting over here,"},{"Start":"00:36.465 ","End":"00:42.045","Text":"the current flowing through the circuit will immediately stop every point,"},{"Start":"00:42.045 ","End":"00:46.135","Text":"because it won\u0027t be able to pass over here in this space."},{"Start":"00:46.135 ","End":"00:52.630","Text":"A lot of the times in your circuit diagrams you\u0027ll have something called a switch."},{"Start":"00:52.630 ","End":"00:57.320","Text":"Now, what the switch does is when it\u0027s open,"},{"Start":"00:57.320 ","End":"01:00.595","Text":"no current can flow through the circuit,"},{"Start":"01:00.595 ","End":"01:04.165","Text":"you can imagine it like at home with your light switch."},{"Start":"01:04.165 ","End":"01:06.565","Text":"When it\u0027s in some position,"},{"Start":"01:06.565 ","End":"01:08.960","Text":"it means that the circuit is open and your light has"},{"Start":"01:08.960 ","End":"01:11.615","Text":"switched off because no current is flowing."},{"Start":"01:11.615 ","End":"01:14.795","Text":"But then if you close the switch,"},{"Start":"01:14.795 ","End":"01:17.610","Text":"then or you switch on the light."},{"Start":"01:17.610 ","End":"01:21.770","Text":"When you switch on the light, you are closing the circuits and"},{"Start":"01:21.770 ","End":"01:26.905","Text":"then current can flow and your light bulb can illuminate your room."},{"Start":"01:26.905 ","End":"01:31.940","Text":"An open circuit is when a circuit is enclosed,"},{"Start":"01:31.940 ","End":"01:33.515","Text":"like in the case of a switch,"},{"Start":"01:33.515 ","End":"01:35.210","Text":"or that we cut the wires,"},{"Start":"01:35.210 ","End":"01:39.050","Text":"which means that there\u0027s no path for the current to flow through,"},{"Start":"01:39.050 ","End":"01:42.650","Text":"because the current can flow through the air,"},{"Start":"01:42.650 ","End":"01:45.770","Text":"it needs to pass through a wire or a conductor,"},{"Start":"01:45.770 ","End":"01:48.290","Text":"or some electrical component."},{"Start":"01:48.290 ","End":"01:53.060","Text":"Therefore, the current in the entire circuit stops."},{"Start":"01:53.060 ","End":"01:55.720","Text":"Now let\u0027s talk about short-circuits."},{"Start":"01:55.720 ","End":"01:57.885","Text":"In the short-circuit,"},{"Start":"01:57.885 ","End":"01:59.790","Text":"what I\u0027m doing is I\u0027m taking"},{"Start":"01:59.790 ","End":"02:05.120","Text":"some electrical component and then canceling out its resistance."},{"Start":"02:05.120 ","End":"02:07.010","Text":"If you can imagine,"},{"Start":"02:07.010 ","End":"02:11.600","Text":"let\u0027s say I have this component over here and I connect a wire from"},{"Start":"02:11.600 ","End":"02:17.915","Text":"here to here all the way around my electrical components."},{"Start":"02:17.915 ","End":"02:23.905","Text":"By doing this and canceling out the effect of this component."},{"Start":"02:23.905 ","End":"02:25.640","Text":"How does that happen?"},{"Start":"02:25.640 ","End":"02:31.835","Text":"My electric components will probably have a higher resistance than my wire."},{"Start":"02:31.835 ","End":"02:34.220","Text":"Every electric component is like that."},{"Start":"02:34.220 ","End":"02:41.965","Text":"My current in my circuit is going to prefer to go through paths with less resistance."},{"Start":"02:41.965 ","End":"02:46.310","Text":"The electric current will go like this,"},{"Start":"02:46.310 ","End":"02:48.540","Text":"and then it will reach the junction over here."},{"Start":"02:48.540 ","End":"02:54.890","Text":"It will prefer to go via this wire and come back at my agenda,"},{"Start":"02:54.890 ","End":"03:00.670","Text":"switches shots, and come back over here because it\u0027s the path with the least resistance."},{"Start":"03:00.670 ","End":"03:06.215","Text":"Then what we have is if this is some current."},{"Start":"03:06.215 ","End":"03:12.395","Text":"The current going through the component itself is going to be equal to 0."},{"Start":"03:12.395 ","End":"03:16.430","Text":"The current going through this wire is going to be equal"},{"Start":"03:16.430 ","End":"03:20.755","Text":"to the total current going through the circuit,"},{"Start":"03:20.755 ","End":"03:23.775","Text":"because no current will go through here."},{"Start":"03:23.775 ","End":"03:28.815","Text":"In that case, if here specifically we have 1 component."},{"Start":"03:28.815 ","End":"03:32.975","Text":"We\u0027ve short-circuited the 1 components and if we had more components,"},{"Start":"03:32.975 ","End":"03:35.825","Text":"if we did the exact same trick with this wire,"},{"Start":"03:35.825 ","End":"03:38.560","Text":"bypassing all of those components."},{"Start":"03:38.560 ","End":"03:40.730","Text":"We short-circuited all of them."},{"Start":"03:40.730 ","End":"03:48.185","Text":"What we actually have is that our battery or our voltage source is connected to itself."},{"Start":"03:48.185 ","End":"03:51.065","Text":"Now, of course, in our battery,"},{"Start":"03:51.065 ","End":"03:53.750","Text":"there\u0027s going to be some resistance,"},{"Start":"03:53.750 ","End":"03:55.745","Text":"but it\u0027s a very small amount."},{"Start":"03:55.745 ","End":"03:57.394","Text":"Also in the wire,"},{"Start":"03:57.394 ","End":"04:00.380","Text":"there\u0027s a very small amount of resistance as well."},{"Start":"04:00.380 ","End":"04:04.725","Text":"Because in every component there\u0027s going to be resistance,"},{"Start":"04:04.725 ","End":"04:08.270","Text":"however theoretically or compared to the voltage and"},{"Start":"04:08.270 ","End":"04:12.350","Text":"the currents that can run through the wires and the battery itself,"},{"Start":"04:12.350 ","End":"04:17.035","Text":"their existence is considered approaching 0."},{"Start":"04:17.035 ","End":"04:19.790","Text":"If we go back to our Ohm\u0027s law,"},{"Start":"04:19.790 ","End":"04:21.680","Text":"so we know that our I,"},{"Start":"04:21.680 ","End":"04:27.605","Text":"our current is equal to voltage divided by our resistance."},{"Start":"04:27.605 ","End":"04:30.930","Text":"Because we\u0027ve short-circuited all of I components,"},{"Start":"04:30.930 ","End":"04:36.465","Text":"that means that our resistance is approaching 0."},{"Start":"04:36.465 ","End":"04:40.910","Text":"If we have some number are voltage from a voltage source divided by 0,"},{"Start":"04:40.910 ","End":"04:46.660","Text":"that means that this over here is going to be approaching infinity."},{"Start":"04:46.660 ","End":"04:50.915","Text":"Then if we have this infinite value for currents,"},{"Start":"04:50.915 ","End":"04:55.570","Text":"that means that either our battery will just empty out,"},{"Start":"04:55.570 ","End":"04:58.560","Text":"and it will get ruined,"},{"Start":"04:58.560 ","End":"05:02.540","Text":"or if we\u0027re dealing with our electrical grids,"},{"Start":"05:02.540 ","End":"05:06.015","Text":"providing us electricity to our homes."},{"Start":"05:06.015 ","End":"05:09.530","Text":"If we have such a high current running through the grids,"},{"Start":"05:09.530 ","End":"05:16.427","Text":"it\u0027s very dangerous and it can also burn and ruin the electrical grids,"},{"Start":"05:16.427 ","End":"05:18.540","Text":"and cause power outages."},{"Start":"05:18.540 ","End":"05:24.155","Text":"We can see that a short circuit is when there is no resistance in the circuit,"},{"Start":"05:24.155 ","End":"05:26.959","Text":"leading to an infinite current,"},{"Start":"05:26.959 ","End":"05:31.585","Text":"which is also dangerous and can ruin the electrical components."},{"Start":"05:31.585 ","End":"05:34.475","Text":"Now a lot of the times we\u0027ll talk about"},{"Start":"05:34.475 ","End":"05:39.410","Text":"some circuit which has multiple electrical components."},{"Start":"05:39.410 ","End":"05:42.155","Text":"What I can do is I can short-circuit"},{"Start":"05:42.155 ","End":"05:45.615","Text":"just specific components as well. What does that mean?"},{"Start":"05:45.615 ","End":"05:48.915","Text":"Let\u0027s say, I have a circuit with 2 components,"},{"Start":"05:48.915 ","End":"05:54.015","Text":"here\u0027s 1 component and here\u0027s the other and here I\u0027ve short-circuited this component."},{"Start":"05:54.015 ","End":"05:57.075","Text":"Now that means that when I work out"},{"Start":"05:57.075 ","End":"06:00.830","Text":"whatever question they\u0027re asking me to solve for this circuit,"},{"Start":"06:00.830 ","End":"06:02.465","Text":"all I have to do,"},{"Start":"06:02.465 ","End":"06:07.610","Text":"is I have to consider it as if it\u0027s a circuit with just 1 component."},{"Start":"06:07.610 ","End":"06:09.170","Text":"With just this components,"},{"Start":"06:09.170 ","End":"06:13.465","Text":"and I ignore this component completely as if it\u0027s not connected at all."},{"Start":"06:13.465 ","End":"06:17.285","Text":"Of course, if I would short-circuit this component as well,"},{"Start":"06:17.285 ","End":"06:21.650","Text":"then again, we get this infinite currents and then we have a problem."},{"Start":"06:21.650 ","End":"06:24.965","Text":"In most of the questions that they\u0027re going to be asking you,"},{"Start":"06:24.965 ","End":"06:28.910","Text":"is going to be some circuit with multiple components and you\u0027ll"},{"Start":"06:28.910 ","End":"06:32.840","Text":"be asked to just short-circuit 1 of them or a few of them,"},{"Start":"06:32.840 ","End":"06:37.510","Text":"but always leaving at least 1 components in the circuit."},{"Start":"06:37.510 ","End":"06:42.430","Text":"That will be similar or analogous to"},{"Start":"06:42.430 ","End":"06:44.110","Text":"just a circuit with"},{"Start":"06:44.110 ","End":"06:47.855","Text":"just that 1 component and without the others that were short-circuited."},{"Start":"06:47.855 ","End":"06:50.110","Text":"How does a short-circuit happen in"},{"Start":"06:50.110 ","End":"06:53.830","Text":"our day-to-day lives if we say we have a short circuit at home."},{"Start":"06:53.830 ","End":"07:00.175","Text":"A lot of the time that\u0027s somewhere in our electricity system,"},{"Start":"07:00.175 ","End":"07:03.190","Text":"we have exposed wires,"},{"Start":"07:03.190 ","End":"07:07.000","Text":"so sections where the plastic covering on them has peeled off and"},{"Start":"07:07.000 ","End":"07:12.895","Text":"these exposed wires are in contact with each other forming a short-circuit."},{"Start":"07:12.895 ","End":"07:15.184","Text":"Now, let\u0027s just clear something up."},{"Start":"07:15.184 ","End":"07:20.155","Text":"A lot of the time at home something will go wrong with your electricity and you\u0027ll say,"},{"Start":"07:20.155 ","End":"07:22.505","Text":"I have a short circuit."},{"Start":"07:22.505 ","End":"07:25.140","Text":"What does that actually mean?"},{"Start":"07:25.140 ","End":"07:29.660","Text":"In your households electrical circuit system,"},{"Start":"07:29.660 ","End":"07:33.020","Text":"you\u0027ll have current flowing through all of your different components,"},{"Start":"07:33.020 ","End":"07:36.535","Text":"for your fridge and your lights and whatever it might be."},{"Start":"07:36.535 ","End":"07:41.379","Text":"What might happen is that either the bare wires are"},{"Start":"07:41.379 ","End":"07:45.665","Text":"touching 1 of the components bare wires,"},{"Start":"07:45.665 ","End":"07:50.690","Text":"and then it will short circuit at least 1 component or more."},{"Start":"07:50.690 ","End":"07:56.150","Text":"Then what will happen is that the current flowing through the entire circuit,"},{"Start":"07:56.150 ","End":"07:59.405","Text":"because now there\u0027s less resistance in the circuit because there\u0027s"},{"Start":"07:59.405 ","End":"08:01.880","Text":"1 or more components have been"},{"Start":"08:01.880 ","End":"08:04.835","Text":"eliminated from the circuit due to the short circuiting it,"},{"Start":"08:04.835 ","End":"08:08.970","Text":"due to the current being able to bypass the components."},{"Start":"08:08.970 ","End":"08:13.610","Text":"The current flowing through the entire circuit will be a lot higher."},{"Start":"08:13.610 ","End":"08:20.045","Text":"What the electrical companies do is they install fuses."},{"Start":"08:20.045 ","End":"08:22.385","Text":"Let\u0027s call this a fuse."},{"Start":"08:22.385 ","End":"08:27.440","Text":"They install a fuse into your circuitry at home."},{"Start":"08:27.440 ","End":"08:33.230","Text":"The fuse can notice when this high current is flowing through,"},{"Start":"08:33.230 ","End":"08:36.020","Text":"because we need that otherwise this high-current can"},{"Start":"08:36.020 ","End":"08:40.430","Text":"destroy your electrical machines at home,"},{"Start":"08:40.430 ","End":"08:42.440","Text":"such as your computers,"},{"Start":"08:42.440 ","End":"08:45.140","Text":"laptops, televisions, and so on."},{"Start":"08:45.140 ","End":"08:48.380","Text":"In order to protect all of that, there is a fuse,"},{"Start":"08:48.380 ","End":"08:52.565","Text":"which notices when the current is reaching dangerously high levels."},{"Start":"08:52.565 ","End":"08:54.890","Text":"Then when the current reaches the fuse,"},{"Start":"08:54.890 ","End":"08:58.544","Text":"it will burn the fuse or destroy the fuse"},{"Start":"08:58.544 ","End":"09:03.235","Text":"and then that means that we\u0027ll have an open circuit."},{"Start":"09:03.235 ","End":"09:06.320","Text":"You can imagine that there\u0027s some wire in"},{"Start":"09:06.320 ","End":"09:10.715","Text":"the fuse and then when the current flows through,"},{"Start":"09:10.715 ","End":"09:12.625","Text":"it will melt the wire,"},{"Start":"09:12.625 ","End":"09:14.920","Text":"or cut the wire, cut the connection here."},{"Start":"09:14.920 ","End":"09:19.580","Text":"Then they\u0027ll be a break in the wire which will lead to an open circuit."},{"Start":"09:19.580 ","End":"09:24.245","Text":"Which means that there\u0027s no path for this dangerously high current to flow,"},{"Start":"09:24.245 ","End":"09:27.440","Text":"which means that the current stops everywhere and then all of"},{"Start":"09:27.440 ","End":"09:31.910","Text":"your precious computers and so on are saved."},{"Start":"09:31.910 ","End":"09:34.860","Text":"That\u0027s the end of the lesson."}],"ID":22293},{"Watched":false,"Name":"Kirchhoffs Circuit Law","Duration":"23m 16s","ChapterTopicVideoID":21309,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"Hello. In this lesson,"},{"Start":"00:01.860 ","End":"00:05.565","Text":"we\u0027re going to be speaking about Kirchhoff\u0027s circuit law."},{"Start":"00:05.565 ","End":"00:12.795","Text":"Now, we\u0027ve seen in previous lessons that we can find our total resistance of a circuit,"},{"Start":"00:12.795 ","End":"00:16.965","Text":"when our resistors are connected in parallel,"},{"Start":"00:16.965 ","End":"00:19.560","Text":"such as over here or in"},{"Start":"00:19.560 ","End":"00:25.455","Text":"series and we had laws of how to do this and it was based on Ohm\u0027s law."},{"Start":"00:25.455 ","End":"00:28.590","Text":"Now, sometimes we\u0027ll have complex circuits,"},{"Start":"00:28.590 ","End":"00:33.525","Text":"where we can simply use Ohm\u0027s law in order to find the voltages,"},{"Start":"00:33.525 ","End":"00:38.202","Text":"and the currents circulating within the circuit."},{"Start":"00:38.202 ","End":"00:41.660","Text":"For that we use Kirchhoff\u0027s circuit law in order to"},{"Start":"00:41.660 ","End":"00:46.275","Text":"simplify the circuit and in order to find out the equations."},{"Start":"00:46.275 ","End":"00:50.130","Text":"That\u0027s what we\u0027re going to be discussing in this lesson."},{"Start":"00:51.260 ","End":"00:55.896","Text":"Let\u0027s look at this circuit as an example."},{"Start":"00:55.896 ","End":"01:00.710","Text":"What we can see is that we have a few places where we have"},{"Start":"01:00.710 ","End":"01:06.620","Text":"some kind of battery or something that has a voltage source."},{"Start":"01:06.620 ","End":"01:11.795","Text":"That means that we can\u0027t work out straight away via Ohm\u0027s law,"},{"Start":"01:11.795 ","End":"01:17.240","Text":"the total resistance of the circuit because we have multiple voltage sources."},{"Start":"01:17.240 ","End":"01:22.630","Text":"We\u0027re going to use Kirchhoff\u0027s circuit law in order to solve this."},{"Start":"01:23.750 ","End":"01:26.145","Text":"How are we going to do this?"},{"Start":"01:26.145 ","End":"01:32.000","Text":"Now the first thing that we\u0027re going to want to do is we\u0027re"},{"Start":"01:32.000 ","End":"01:39.130","Text":"going to define our current\u0027s direction for every point in our circuit."},{"Start":"01:40.790 ","End":"01:43.440","Text":"Here we have a voltage source,"},{"Start":"01:43.440 ","End":"01:50.075","Text":"so we can say that we have a current coming through here in this direction,"},{"Start":"01:50.075 ","End":"01:51.830","Text":"and let\u0027s call it I_1."},{"Start":"01:51.830 ","End":"01:53.750","Text":"Now it doesn\u0027t matter if our arrow is pointing in"},{"Start":"01:53.750 ","End":"01:56.405","Text":"this direction or in the opposite direction,"},{"Start":"01:56.405 ","End":"01:59.730","Text":"we will see it will sort itself out."},{"Start":"02:00.460 ","End":"02:03.860","Text":"Now, so long as we\u0027re dealing with the same wire,"},{"Start":"02:03.860 ","End":"02:09.015","Text":"the current flowing through the same wire is going to be the same."},{"Start":"02:09.015 ","End":"02:11.150","Text":"If we have I_1 over here,"},{"Start":"02:11.150 ","End":"02:15.605","Text":"so also on this side in the same direction we\u0027ll have I_1,"},{"Start":"02:15.605 ","End":"02:20.880","Text":"and also coming back up here we also have I_1."},{"Start":"02:21.260 ","End":"02:28.250","Text":"Now as we know every time our wire splits are we reach a junction, like over here."},{"Start":"02:28.250 ","End":"02:32.755","Text":"Our current is also going to split."},{"Start":"02:32.755 ","End":"02:39.500","Text":"Now we can say that in this direction we have some current I_2,"},{"Start":"02:39.500 ","End":"02:44.585","Text":"and again I\u0027ve just chosen an arbitrary direction but it doesn\u0027t really matter,"},{"Start":"02:44.585 ","End":"02:49.480","Text":"and over here we\u0027ll have I_3."},{"Start":"02:50.240 ","End":"02:55.350","Text":"Of course, when we\u0027re dealing with the same wire I current is going to be the same,"},{"Start":"02:55.350 ","End":"02:58.730","Text":"so that means along this same wire."},{"Start":"02:58.730 ","End":"03:08.540","Text":"We\u0027ll have over here current of I_2 and along here we\u0027ll have a current of I_3,"},{"Start":"03:08.540 ","End":"03:10.760","Text":"and up here as well,"},{"Start":"03:10.760 ","End":"03:13.710","Text":"we\u0027ll have our current I_3."},{"Start":"03:14.690 ","End":"03:22.895","Text":"Now we\u0027ve chosen our currents for our circuit and their direction."},{"Start":"03:22.895 ","End":"03:27.575","Text":"Now I\u0027m going to do the most important thing in Kirchhoff\u0027s circuit law,"},{"Start":"03:27.575 ","End":"03:33.710","Text":"and that\u0027s that I\u0027m going to write an equation for the voltages in the circuit,"},{"Start":"03:33.710 ","End":"03:37.740","Text":"and an equation for the currents in the circuit."},{"Start":"03:39.140 ","End":"03:42.500","Text":"The first thing that we\u0027re going to do is,"},{"Start":"03:42.500 ","End":"03:45.740","Text":"we\u0027re going to write our voltage equations."},{"Start":"03:45.740 ","End":"03:50.405","Text":"The sum of all the voltages in a closed circuit is equal to 0."},{"Start":"03:50.405 ","End":"03:55.133","Text":"If I choose for instance this point over here as point A"},{"Start":"03:55.133 ","End":"03:59.765","Text":"and then I go around some closed loop,"},{"Start":"03:59.765 ","End":"04:04.025","Text":"either like that or like so and the outer loop."},{"Start":"04:04.025 ","End":"04:07.025","Text":"Once I get back to my starting point,"},{"Start":"04:07.025 ","End":"04:08.675","Text":"my point A over here,"},{"Start":"04:08.675 ","End":"04:14.435","Text":"the sum of all of the voltages that I would have added up will be equal to 0."},{"Start":"04:14.435 ","End":"04:16.830","Text":"Let\u0027s see how we do that."},{"Start":"04:17.980 ","End":"04:22.880","Text":"Let\u0027s begin, I\u0027m starting at this point A over here,"},{"Start":"04:22.880 ","End":"04:26.585","Text":"and let\u0027s go anticlockwise."},{"Start":"04:26.585 ","End":"04:30.455","Text":"We can also go clockwise it doesn\u0027t really matter, let\u0027s go anticlockwise."},{"Start":"04:30.455 ","End":"04:35.135","Text":"I\u0027m going along and then I reach this voltage source of 3 volts."},{"Start":"04:35.135 ","End":"04:36.890","Text":"Now we can see over here,"},{"Start":"04:36.890 ","End":"04:42.365","Text":"this short or a low bar over here or a line,"},{"Start":"04:42.365 ","End":"04:44.730","Text":"means that it\u0027s a low voltage,"},{"Start":"04:44.730 ","End":"04:46.685","Text":"and this high line,"},{"Start":"04:46.685 ","End":"04:50.065","Text":"high bar, means that there\u0027s a higher voltage."},{"Start":"04:50.065 ","End":"04:53.945","Text":"Every time we go from a lower voltage to a higher voltage,"},{"Start":"04:53.945 ","End":"04:56.935","Text":"because we\u0027re going like so in this direction,"},{"Start":"04:56.935 ","End":"05:00.350","Text":"we\u0027re going from a low voltage to a higher voltage which means"},{"Start":"05:00.350 ","End":"05:03.985","Text":"that we\u0027re adding 3 volts on."},{"Start":"05:03.985 ","End":"05:10.300","Text":"Let\u0027s begin. I\u0027ve passed my voltage source from low to high I\u0027m adding on 3 volts."},{"Start":"05:10.300 ","End":"05:14.045","Text":"Then I\u0027m carrying on in the direction of the current,"},{"Start":"05:14.045 ","End":"05:18.210","Text":"and now I\u0027ve reached this resistor over here."},{"Start":"05:19.060 ","End":"05:23.420","Text":"Because my current is in this direction,"},{"Start":"05:23.420 ","End":"05:30.765","Text":"I can say that this is my positive side and this is my negative side of the resistor."},{"Start":"05:30.765 ","End":"05:33.980","Text":"Because as my current passes my resistor,"},{"Start":"05:33.980 ","End":"05:36.485","Text":"it experiences resistance,"},{"Start":"05:36.485 ","End":"05:39.620","Text":"so when it gets to this end,"},{"Start":"05:39.620 ","End":"05:43.160","Text":"we\u0027re going to have a lower value of voltage."},{"Start":"05:43.160 ","End":"05:49.140","Text":"Because we have to work harder to get through the resistor over here."},{"Start":"05:49.880 ","End":"05:53.810","Text":"If we\u0027re going with the direction of the current,"},{"Start":"05:53.810 ","End":"05:57.230","Text":"we\u0027re starting at a higher voltage and as we go through"},{"Start":"05:57.230 ","End":"06:00.545","Text":"the resistor in the direction of the current flow,"},{"Start":"06:00.545 ","End":"06:05.230","Text":"we\u0027re going to have a lower voltage at the other end of the resistor."},{"Start":"06:05.230 ","End":"06:13.440","Text":"That means that we can minus off the voltage that goes over the resistor."},{"Start":"06:13.440 ","End":"06:14.910","Text":"What\u0027s that going to be?"},{"Start":"06:14.910 ","End":"06:18.538","Text":"Through Ohm\u0027s law that\u0027s going to be equal to the current."},{"Start":"06:18.538 ","End":"06:21.665","Text":"Here we have a current of I_1 flowing through the resistor,"},{"Start":"06:21.665 ","End":"06:25.550","Text":"multiplied by the resistance of the resistor which here is given to"},{"Start":"06:25.550 ","End":"06:31.720","Text":"us as being equal to 2 Ohms."},{"Start":"06:31.720 ","End":"06:35.330","Text":"Now we\u0027ve passed the resistor and I keep going and I keep going,"},{"Start":"06:35.330 ","End":"06:37.405","Text":"and now I reach my junction."},{"Start":"06:37.405 ","End":"06:40.405","Text":"I can either turn right over here,"},{"Start":"06:40.405 ","End":"06:42.260","Text":"and go through the smaller loop,"},{"Start":"06:42.260 ","End":"06:46.235","Text":"or I can carry on straight and go over the larger loop."},{"Start":"06:46.235 ","End":"06:48.515","Text":"Right now, let\u0027s go through the smaller loop,"},{"Start":"06:48.515 ","End":"06:54.235","Text":"and then afterwards we\u0027ll see how to deal with the larger loop."},{"Start":"06:54.235 ","End":"06:57.450","Text":"I\u0027m at I_1, I\u0027ve reach the junction and I turn"},{"Start":"06:57.450 ","End":"07:01.005","Text":"right and and my current has now become I_2."},{"Start":"07:01.005 ","End":"07:03.525","Text":"Now I\u0027m carrying on and carrying on,"},{"Start":"07:03.525 ","End":"07:07.050","Text":"and I\u0027m reaching a voltage source."},{"Start":"07:07.050 ","End":"07:11.455","Text":"We can see that here I have a high bar or a high line,"},{"Start":"07:11.455 ","End":"07:15.755","Text":"that means I have a high voltage."},{"Start":"07:15.755 ","End":"07:19.600","Text":"Then I\u0027m going in this direction to the low bar or the low line,"},{"Start":"07:19.600 ","End":"07:22.140","Text":"which means that I have a low voltage."},{"Start":"07:22.140 ","End":"07:25.905","Text":"I\u0027m going from high voltage to a lower voltage."},{"Start":"07:25.905 ","End":"07:30.220","Text":"That means that I\u0027m minusing off my voltage over here,"},{"Start":"07:30.220 ","End":"07:33.460","Text":"and my voltage is given as 2 volts."},{"Start":"07:33.460 ","End":"07:36.410","Text":"Minus 2V."},{"Start":"07:37.950 ","End":"07:45.795","Text":"Now I carry on with my I_2 and I reach this resistor."},{"Start":"07:45.795 ","End":"07:48.535","Text":"Again, in the direction of the current,"},{"Start":"07:48.535 ","End":"07:52.580","Text":"this side is going to be positive and this side is going to be negative."},{"Start":"07:52.580 ","End":"07:57.140","Text":"Because if we\u0027re going through our resistor in the direction of the current,"},{"Start":"07:57.140 ","End":"07:59.120","Text":"our voltage is going to go down."},{"Start":"07:59.120 ","End":"08:02.315","Text":"We\u0027re going from a high voltage to a low voltage."},{"Start":"08:02.315 ","End":"08:08.445","Text":"Because our resistor reduces our value for the voltage."},{"Start":"08:08.445 ","End":"08:14.330","Text":"Again, we have minus the voltage over our resistor."},{"Start":"08:14.330 ","End":"08:17.960","Text":"Which is according to Ohm\u0027s law equal to I_2,"},{"Start":"08:17.960 ","End":"08:20.000","Text":"multiplied by the resistance."},{"Start":"08:20.000 ","End":"08:24.090","Text":"Which here is given as 1 Ohm."},{"Start":"08:24.090 ","End":"08:26.810","Text":"Then after that resistance,"},{"Start":"08:26.810 ","End":"08:32.110","Text":"I go back up and I reach my original starting point my point A,"},{"Start":"08:32.110 ","End":"08:35.990","Text":"which means that I\u0027ve finished a closed loop."},{"Start":"08:35.990 ","End":"08:41.945","Text":"Then, I know that the sum of all my voltages in a close circuit is equal to 0."},{"Start":"08:41.945 ","End":"08:45.268","Text":"All of this is equal to 0."},{"Start":"08:45.268 ","End":"08:50.950","Text":"Now I\u0027ve covered this small loop over here,"},{"Start":"08:50.950 ","End":"08:54.250","Text":"and now what I want to do is I want to cover"},{"Start":"08:54.250 ","End":"08:58.315","Text":"these components that I haven\u0027t gotten to yet,"},{"Start":"08:58.315 ","End":"09:01.570","Text":"that I didn\u0027t go past in this loop over here."},{"Start":"09:01.570 ","End":"09:05.155","Text":"It doesn\u0027t matter if I start from my original point A,"},{"Start":"09:05.155 ","End":"09:09.430","Text":"or if I start from a different point over here or over here, wherever it might be."},{"Start":"09:09.430 ","End":"09:16.400","Text":"The most important thing is that I go over these untouched electrical components."},{"Start":"09:17.370 ","End":"09:21.040","Text":"Let\u0027s define a new starting point and"},{"Start":"09:21.040 ","End":"09:24.915","Text":"let\u0027s say that we\u0027re starting at the junction over here,"},{"Start":"09:24.915 ","End":"09:27.720","Text":"or over here just after the junction."},{"Start":"09:27.720 ","End":"09:29.760","Text":"We\u0027ve defined this new point over here."},{"Start":"09:29.760 ","End":"09:33.190","Text":"This is our starting point, and let\u0027s call it B."},{"Start":"09:33.210 ","End":"09:38.470","Text":"Now, again, it doesn\u0027t matter if I go clockwise or anticlockwise,"},{"Start":"09:38.470 ","End":"09:40.585","Text":"but let\u0027s go clockwise."},{"Start":"09:40.585 ","End":"09:44.650","Text":"Now we\u0027re going to do exactly what we did over here."},{"Start":"09:44.650 ","End":"09:49.075","Text":"We\u0027re going with our I_2 and then we reach this voltage source."},{"Start":"09:49.075 ","End":"09:52.210","Text":"We can see that we\u0027re going from a high bar to a low bar,"},{"Start":"09:52.210 ","End":"09:54.760","Text":"so we\u0027re going from high voltage to a low voltage,"},{"Start":"09:54.760 ","End":"09:59.185","Text":"which means that we have to minus 2 volts over here."},{"Start":"09:59.185 ","End":"10:04.660","Text":"Then we\u0027re carrying on and we\u0027re getting to our resistor over here, again,"},{"Start":"10:04.660 ","End":"10:09.070","Text":"with our plus at the beginning of the resistor and our minus at the end of"},{"Start":"10:09.070 ","End":"10:14.455","Text":"the resistor if we\u0027re going in the positive direction of the current."},{"Start":"10:14.455 ","End":"10:20.200","Text":"That means, again, we\u0027re minusing the voltage over our resistor,"},{"Start":"10:20.200 ","End":"10:21.865","Text":"which is according to Ohm\u0027s Law,"},{"Start":"10:21.865 ","End":"10:25.660","Text":"equal to I_2 multiplied by the resistance,"},{"Start":"10:25.660 ","End":"10:28.750","Text":"which over here, is equal to 1 ohm."},{"Start":"10:28.750 ","End":"10:34.600","Text":"Then we\u0027re carrying on, and now we\u0027re not going to turn up here because we want to"},{"Start":"10:34.600 ","End":"10:41.095","Text":"complete a closed loop that incorporates these components that we haven\u0027t gone over yet."},{"Start":"10:41.095 ","End":"10:42.820","Text":"In our previous equation,"},{"Start":"10:42.820 ","End":"10:44.900","Text":"we went over these components,"},{"Start":"10:44.900 ","End":"10:46.600","Text":"so now we want to go over these."},{"Start":"10:46.600 ","End":"10:52.270","Text":"Now we\u0027re going to continue in this clockwise direction, downwards over here."},{"Start":"10:52.270 ","End":"10:54.369","Text":"We continue, we move down,"},{"Start":"10:54.369 ","End":"10:59.540","Text":"and then we get to this resistor over here."},{"Start":"11:01.050 ","End":"11:05.140","Text":"What we can see here is that the current flowing"},{"Start":"11:05.140 ","End":"11:08.950","Text":"through this resistor is I_3 first of all,"},{"Start":"11:08.950 ","End":"11:14.025","Text":"we\u0027ve left to the region of our I_2 current and it\u0027s I_3,"},{"Start":"11:14.025 ","End":"11:15.560","Text":"and as we can see,"},{"Start":"11:15.560 ","End":"11:20.220","Text":"we\u0027re currently traveling in this direction,"},{"Start":"11:20.220 ","End":"11:24.090","Text":"which means that we\u0027re going to be going through the resistor in"},{"Start":"11:24.090 ","End":"11:29.275","Text":"the direction negative to the direction of our current."},{"Start":"11:29.275 ","End":"11:31.840","Text":"That means that first of all,"},{"Start":"11:31.840 ","End":"11:35.320","Text":"I direction of current is in this upwards direction,"},{"Start":"11:35.320 ","End":"11:38.620","Text":"which means that this is the positive side and this is the negative side."},{"Start":"11:38.620 ","End":"11:42.445","Text":"But our direction of travel is according to this gray arrow,"},{"Start":"11:42.445 ","End":"11:46.285","Text":"which means that we\u0027re going from negative to positive."},{"Start":"11:46.285 ","End":"11:51.475","Text":"That means that because we\u0027re going against the direction of the current,"},{"Start":"11:51.475 ","End":"11:53.950","Text":"we\u0027re going to be this time adding in."},{"Start":"11:53.950 ","End":"11:55.330","Text":"According to Ohm\u0027s Law,"},{"Start":"11:55.330 ","End":"11:57.490","Text":"that\u0027s going to be equal to I_3,"},{"Start":"11:57.490 ","End":"11:59.275","Text":"the current over the resistor,"},{"Start":"11:59.275 ","End":"12:01.525","Text":"multiplied by the resistance,"},{"Start":"12:01.525 ","End":"12:04.225","Text":"which is 2 ohms."},{"Start":"12:04.225 ","End":"12:06.895","Text":"Then we\u0027re carrying on and we\u0027re carrying on."},{"Start":"12:06.895 ","End":"12:10.760","Text":"Now we\u0027ve gotten to a voltage source where here,"},{"Start":"12:10.760 ","End":"12:13.240","Text":"we have a low bar, and here, we have a high bar."},{"Start":"12:13.240 ","End":"12:18.650","Text":"We\u0027re going from low voltage to high voltage."},{"Start":"12:18.720 ","End":"12:24.535","Text":"That means that we\u0027re going to add 4 volts."},{"Start":"12:24.535 ","End":"12:28.120","Text":"Then we carry on and we get to another resistor."},{"Start":"12:28.120 ","End":"12:30.280","Text":"As we can see, we\u0027re still in the region of"},{"Start":"12:30.280 ","End":"12:34.630","Text":"our I_3 and we can see that our current is flowing in this direction,"},{"Start":"12:34.630 ","End":"12:39.055","Text":"which means that this is the positive and this is the negative side."},{"Start":"12:39.055 ","End":"12:41.755","Text":"But we\u0027re going according to this gray arrow,"},{"Start":"12:41.755 ","End":"12:43.660","Text":"which means that we\u0027re coming from this side,"},{"Start":"12:43.660 ","End":"12:45.970","Text":"from the negative to the positive,"},{"Start":"12:45.970 ","End":"12:49.943","Text":"which means that again, we\u0027re adding on."},{"Start":"12:49.943 ","End":"12:53.710","Text":"Then the voltage across the resistor is going to be"},{"Start":"12:53.710 ","End":"12:57.670","Text":"equal to I_3 multiplied by the resistance,"},{"Start":"12:57.670 ","End":"12:59.920","Text":"which is 1 ohm."},{"Start":"12:59.920 ","End":"13:03.840","Text":"Now we\u0027ve passed the resistor and then we\u0027re going up and"},{"Start":"13:03.840 ","End":"13:07.875","Text":"then we\u0027re meeting again at point B which was our starting point."},{"Start":"13:07.875 ","End":"13:12.810","Text":"That means we\u0027ve completed a full circuit or a full loop,"},{"Start":"13:12.810 ","End":"13:16.430","Text":"which means that the sum of all of these voltages is equal to 0."},{"Start":"13:16.430 ","End":"13:21.700","Text":"Now, what we have over here are 2 equations and we can see that"},{"Start":"13:21.700 ","End":"13:28.045","Text":"these 2 equations cover all of the components in my circuit over here."},{"Start":"13:28.045 ","End":"13:32.080","Text":"Now, if I had more loops such that I had"},{"Start":"13:32.080 ","End":"13:37.366","Text":"more components that weren\u0027t covered somewhere within these 2 equations,"},{"Start":"13:37.366 ","End":"13:40.270","Text":"then I would have to repeat this process of"},{"Start":"13:40.270 ","End":"13:43.870","Text":"going around from some starting point and a closed loop,"},{"Start":"13:43.870 ","End":"13:50.290","Text":"incorporating each time, more and more of those components but right now here,"},{"Start":"13:50.290 ","End":"13:53.770","Text":"we\u0027ve incorporated everything in just these 2 equations."},{"Start":"13:53.770 ","End":"13:57.020","Text":"Now we can move on to the next step."},{"Start":"13:57.390 ","End":"14:00.970","Text":"Now what I\u0027m going to do is I\u0027m going to do step 3,"},{"Start":"14:00.970 ","End":"14:05.140","Text":"which is to write out all of the current equations,"},{"Start":"14:05.140 ","End":"14:09.220","Text":"remembering that at each junction that we get to,"},{"Start":"14:09.220 ","End":"14:11.230","Text":"where our wire splits,"},{"Start":"14:11.230 ","End":"14:18.230","Text":"the entering current equals the current exiting."},{"Start":"14:18.450 ","End":"14:22.075","Text":"For example, here we have a junction."},{"Start":"14:22.075 ","End":"14:25.600","Text":"We can say that I current going in to the junction,"},{"Start":"14:25.600 ","End":"14:28.540","Text":"which is I_1, is equal to the current coming out,"},{"Start":"14:28.540 ","End":"14:30.955","Text":"which is I_2 plus I_3."},{"Start":"14:30.955 ","End":"14:35.755","Text":"This law obviously comes from the idea of conservation of charge."},{"Start":"14:35.755 ","End":"14:41.185","Text":"We have lots of little charges moving along the wire and when they reach the junction,"},{"Start":"14:41.185 ","End":"14:43.660","Text":"some of the charges will go down"},{"Start":"14:43.660 ","End":"14:47.095","Text":"this branch and some of the charges will go down this branch."},{"Start":"14:47.095 ","End":"14:52.390","Text":"But the total charges that have split from this junction"},{"Start":"14:52.390 ","End":"14:55.720","Text":"into one of the branches is equal to all of"},{"Start":"14:55.720 ","End":"14:59.900","Text":"the charges that were approaching the junction."},{"Start":"15:01.860 ","End":"15:04.885","Text":"Let\u0027s write out the current equations."},{"Start":"15:04.885 ","End":"15:09.625","Text":"I go along my circuit and I try and reach every single junction."},{"Start":"15:09.625 ","End":"15:13.075","Text":"The first junction that I\u0027m going to reach is this one over here."},{"Start":"15:13.075 ","End":"15:16.000","Text":"We can see that my current entering is equal to"},{"Start":"15:16.000 ","End":"15:19.930","Text":"I_1 and that is equal to the currents exiting."},{"Start":"15:19.930 ","End":"15:26.365","Text":"That is equal to I_2 plus I_3."},{"Start":"15:26.365 ","End":"15:30.640","Text":"Now I have to look for another junction somewhere in the circuit."},{"Start":"15:30.640 ","End":"15:34.870","Text":"My other junction is over here and I can see again that"},{"Start":"15:34.870 ","End":"15:38.500","Text":"my 2 currents going into the junction is equal to"},{"Start":"15:38.500 ","End":"15:43.150","Text":"I_2 and I_3 and my junction exiting is equal to I_1."},{"Start":"15:43.150 ","End":"15:46.540","Text":"So then I get I_2 plus I_3 is equal to I_1,"},{"Start":"15:46.540 ","End":"15:49.360","Text":"which is the same equation that we have over here."},{"Start":"15:49.360 ","End":"15:51.490","Text":"I don\u0027t have to write it out again,"},{"Start":"15:51.490 ","End":"15:58.009","Text":"I\u0027ve already written it and those are my current equations."},{"Start":"15:58.009 ","End":"16:00.745","Text":"Now we have 3 equations,"},{"Start":"16:00.745 ","End":"16:03.220","Text":"and we have 3 unknowns."},{"Start":"16:03.220 ","End":"16:04.840","Text":"Our unknowns are I_1,"},{"Start":"16:04.840 ","End":"16:06.670","Text":"I_2, and I_3."},{"Start":"16:06.670 ","End":"16:09.205","Text":"Everything else is a given."},{"Start":"16:09.205 ","End":"16:12.145","Text":"Now, if we have 3 equations and 3 unknowns,"},{"Start":"16:12.145 ","End":"16:15.385","Text":"then we know that we can solve to find out our unknowns."},{"Start":"16:15.385 ","End":"16:18.640","Text":"We can either solve it the usual way by substitution and"},{"Start":"16:18.640 ","End":"16:21.790","Text":"lots of algebra in order to find out what I_1,"},{"Start":"16:21.790 ","End":"16:23.800","Text":"I_2, and I_3 are equal to."},{"Start":"16:23.800 ","End":"16:27.370","Text":"Another way to solve this is by using a matrix."},{"Start":"16:27.370 ","End":"16:29.140","Text":"If you know how to do that,"},{"Start":"16:29.140 ","End":"16:31.990","Text":"we\u0027ll speak about it in future lessons but you can"},{"Start":"16:31.990 ","End":"16:35.470","Text":"also use in order to solve this, matrix."},{"Start":"16:35.470 ","End":"16:37.810","Text":"I\u0027m not going to do all of the algebra."},{"Start":"16:37.810 ","End":"16:41.245","Text":"Let\u0027s just write down the answers."},{"Start":"16:41.245 ","End":"16:44.215","Text":"After solving via algebra or a matrix,"},{"Start":"16:44.215 ","End":"16:50.474","Text":"we\u0027ll get that our I_1 is equal to 2 divided by 11 Amps,"},{"Start":"16:50.474 ","End":"16:57.396","Text":"our I_2 is equal to 7 divided by 11 Amps,"},{"Start":"16:57.396 ","End":"17:06.295","Text":"and our I_3 is equal to negative 5 divided by 11 Amps."},{"Start":"17:06.295 ","End":"17:13.405","Text":"Of course, I\u0027ve given the units of Amps or Amperes because we were using volts and Ohms,"},{"Start":"17:13.405 ","End":"17:15.820","Text":"which are units in MKS,"},{"Start":"17:15.820 ","End":"17:18.385","Text":"and so is our Amp."},{"Start":"17:18.385 ","End":"17:24.670","Text":"Now, we can see that our I_3 is equal to negative 5 divided by 11 Amps."},{"Start":"17:24.670 ","End":"17:28.250","Text":"What does this negative sign over here mean?"},{"Start":"17:28.590 ","End":"17:36.625","Text":"This minus simply means that the direction that we said that our I_3 is pointing in,"},{"Start":"17:36.625 ","End":"17:38.455","Text":"is in the opposite direction."},{"Start":"17:38.455 ","End":"17:40.690","Text":"If you remember at the beginning of the lesson I said it"},{"Start":"17:40.690 ","End":"17:42.970","Text":"doesn\u0027t really matter in which direction you"},{"Start":"17:42.970 ","End":"17:47.890","Text":"draw your arrows because at the end when you get your values for your currents,"},{"Start":"17:47.890 ","End":"17:51.070","Text":"if you have a positive number,"},{"Start":"17:51.070 ","End":"17:53.775","Text":"then that means you chose the right direction."},{"Start":"17:53.775 ","End":"17:55.225","Text":"If you have a negative number,"},{"Start":"17:55.225 ","End":"18:00.130","Text":"then you know that the arrow just needs to point in the other direction."},{"Start":"18:00.130 ","End":"18:04.180","Text":"Here that means that our I_3 is pointing"},{"Start":"18:04.180 ","End":"18:09.890","Text":"in this direction with the red arrows. That\u0027s it."},{"Start":"18:11.010 ","End":"18:14.770","Text":"Now we\u0027ve actually finished the lesson."},{"Start":"18:14.770 ","End":"18:20.125","Text":"If you don\u0027t want to see how I solve this algebraically,"},{"Start":"18:20.125 ","End":"18:24.280","Text":"so you can just click out of this video and go on to watch the next 1."},{"Start":"18:24.280 ","End":"18:29.500","Text":"If however you do want to see how I solve this algebraically to get my values for I_1,"},{"Start":"18:29.500 ","End":"18:32.340","Text":"I_2, and I_3, it\u0027s just going to be a bit of algebra,"},{"Start":"18:32.340 ","End":"18:34.990","Text":"then carry on watching."},{"Start":"18:36.080 ","End":"18:39.500","Text":"Let\u0027s do some algebra."},{"Start":"18:39.500 ","End":"18:42.430","Text":"Let\u0027s go on to this equation,"},{"Start":"18:42.430 ","End":"18:43.735","Text":"our first equation over here."},{"Start":"18:43.735 ","End":"18:47.020","Text":"We can see that we have 3 volts minus 2 volts."},{"Start":"18:47.020 ","End":"18:50.215","Text":"Let\u0027s just write out that this is equal to 1 volt."},{"Start":"18:50.215 ","End":"18:53.665","Text":"Then I\u0027m going to stop writing the units, volts and Ohms."},{"Start":"18:53.665 ","End":"18:55.510","Text":"We just have 1 volt."},{"Start":"18:55.510 ","End":"18:58.390","Text":"Now we have minus."},{"Start":"18:58.390 ","End":"19:03.070","Text":"We have 2 multiplied by our I_1."},{"Start":"19:03.070 ","End":"19:08.140","Text":"Now our I_1, is equal to I_2 plus I_3."},{"Start":"19:08.140 ","End":"19:11.620","Text":"Let\u0027s substitute in what our I_1 is equal to."},{"Start":"19:11.620 ","End":"19:16.765","Text":"We have I_2 plus I_3."},{"Start":"19:16.765 ","End":"19:21.400","Text":"Our negative 2 volts we dealt with over here."},{"Start":"19:21.400 ","End":"19:27.655","Text":"Then we have minus 1 times I_2 minus I_2,"},{"Start":"19:27.655 ","End":"19:30.685","Text":"and this is equal to 0."},{"Start":"19:30.685 ","End":"19:33.850","Text":"Now, if we open up these brackets,"},{"Start":"19:33.850 ","End":"19:39.250","Text":"and then I\u0027ll have negative 2I_2 and then another 2 negative I_2."},{"Start":"19:39.250 ","End":"19:42.235","Text":"Let\u0027s just rub this out and rewrite this."},{"Start":"19:42.235 ","End":"19:51.490","Text":"In the end I\u0027ll have 1 minus 3I_2 minus 2I_3."},{"Start":"19:51.490 ","End":"19:54.950","Text":"That is simply going to be equal to 0."},{"Start":"19:55.260 ","End":"19:57.880","Text":"Now for our second equation,"},{"Start":"19:57.880 ","End":"20:01.090","Text":"so we have negative 2 volts plus 4 volts."},{"Start":"20:01.090 ","End":"20:03.520","Text":"We have 2 volts again without the units."},{"Start":"20:03.520 ","End":"20:10.075","Text":"Then we have negative I_2. That\u0027s it."},{"Start":"20:10.075 ","End":"20:13.060","Text":"We have negative I_2."},{"Start":"20:13.060 ","End":"20:18.490","Text":"Then with our I_3 here we have 2_I3 and here we have another I_3."},{"Start":"20:18.490 ","End":"20:21.477","Text":"Then we have plus 3_I3,"},{"Start":"20:21.477 ","End":"20:24.800","Text":"and this is equal to 0."},{"Start":"20:26.100 ","End":"20:28.540","Text":"Now, what I want to do is,"},{"Start":"20:28.540 ","End":"20:32.095","Text":"I want to get rid of my I_2 over here."},{"Start":"20:32.095 ","End":"20:36.040","Text":"Here we can see I have negative 3I_2 and here I have 1I_2."},{"Start":"20:36.040 ","End":"20:40.195","Text":"Let\u0027s call this equation number 1 and this equation number 2."},{"Start":"20:40.195 ","End":"20:45.155","Text":"Now what I\u0027m going to do is I\u0027m going to do equation number 1"},{"Start":"20:45.155 ","End":"20:50.910","Text":"minus 3 times equation number 2."},{"Start":"20:50.910 ","End":"20:54.870","Text":"What we\u0027ll get is therefore we\u0027ll have"},{"Start":"20:54.870 ","End":"21:04.495","Text":"1 minus 3I_2 minus 2I_3 minus 3 times equation number 2,"},{"Start":"21:04.495 ","End":"21:10.960","Text":"which is 2 minus I_2 plus 3I_3."},{"Start":"21:10.960 ","End":"21:13.480","Text":"Then on our equals side,"},{"Start":"21:13.480 ","End":"21:18.530","Text":"we have 0 minus 3 times 0, which is 0."},{"Start":"21:18.990 ","End":"21:21.970","Text":"Now, what we can do is we can solve this."},{"Start":"21:21.970 ","End":"21:24.535","Text":"We have 1 minus 6."},{"Start":"21:24.535 ","End":"21:26.710","Text":"We have negative 5."},{"Start":"21:26.710 ","End":"21:28.765","Text":"Then our I_2 will cancel out."},{"Start":"21:28.765 ","End":"21:32.200","Text":"Negative 3I_2 plus 3I_2."},{"Start":"21:32.200 ","End":"21:34.240","Text":"These will cancel out."},{"Start":"21:34.240 ","End":"21:38.860","Text":"Then we have negative 2I_3, negative 9I_3."},{"Start":"21:38.860 ","End":"21:42.700","Text":"That is equal to negative 11I_3,"},{"Start":"21:42.700 ","End":"21:44.410","Text":"which is equal to 0."},{"Start":"21:44.410 ","End":"21:47.980","Text":"Therefore, we can isolate out our I_3,"},{"Start":"21:47.980 ","End":"21:53.215","Text":"which is equal to negative 5 divided by 11,"},{"Start":"21:53.215 ","End":"21:57.130","Text":"and I\u0027ll straightaway add in my units of the Amps."},{"Start":"21:57.130 ","End":"22:02.740","Text":"There we got that. Now we know what our I_3 is equal to."},{"Start":"22:02.740 ","End":"22:06.340","Text":"We can see that here we have that our I_2,"},{"Start":"22:06.340 ","End":"22:09.490","Text":"if we move this over to the other side of the equals sign."},{"Start":"22:09.490 ","End":"22:13.675","Text":"Our I_2 is equal to 2 plus 3I_3."},{"Start":"22:13.675 ","End":"22:20.710","Text":"Let\u0027s write that I_2 is equal to 2 plus 3I_3."},{"Start":"22:20.710 ","End":"22:26.030","Text":"Then we can substitute in what our I_3 is equal to."},{"Start":"22:26.100 ","End":"22:34.570","Text":"Then we\u0027ll get, so let\u0027s write that in 2 plus 3 multiplied by negative 5 divided by 11."},{"Start":"22:34.570 ","End":"22:41.440","Text":"Then once we solve that will get that that is equal to 7 divided by 11 Amps."},{"Start":"22:41.440 ","End":"22:46.855","Text":"Now we have to find our I_1."},{"Start":"22:46.855 ","End":"22:48.325","Text":"Let\u0027s just scroll down."},{"Start":"22:48.325 ","End":"22:52.225","Text":"We know that our I_1 from over here,"},{"Start":"22:52.225 ","End":"22:56.110","Text":"from our current equation is simply equal to I_2 plus I_3."},{"Start":"22:56.110 ","End":"23:01.900","Text":"I_2 is 7 divided by 11 plus I_3,"},{"Start":"23:01.900 ","End":"23:05.185","Text":"so negative 5 divided by 11."},{"Start":"23:05.185 ","End":"23:10.120","Text":"Then we\u0027ll get that, that is equal to 2 divided by 11 Amps."},{"Start":"23:10.120 ","End":"23:12.535","Text":"Which is what we got over there."},{"Start":"23:12.535 ","End":"23:15.590","Text":"That\u0027s the end of this lesson."}],"ID":21389},{"Watched":false,"Name":"Kirchhoffs Circuit Law Continued - Voltage Between Two points","Duration":"6m 30s","ChapterTopicVideoID":21477,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.135","Text":"Hello. We\u0027re back to an extension of our previous lesson."},{"Start":"00:06.135 ","End":"00:10.770","Text":"I know it was a long lesson and it\u0027s not done yet."},{"Start":"00:10.770 ","End":"00:14.475","Text":"As we learned how to use"},{"Start":"00:14.475 ","End":"00:19.245","Text":"our different laws in order to find the currents flowing through our circuit."},{"Start":"00:19.245 ","End":"00:23.040","Text":"Now, there\u0027s a question that they can ask to do with this,"},{"Start":"00:23.040 ","End":"00:29.620","Text":"and that is to find the voltage between 2 points."},{"Start":"00:30.290 ","End":"00:37.050","Text":"Let\u0027s say that we have our point A over here and that"},{"Start":"00:37.050 ","End":"00:45.230","Text":"our point B is like so and we\u0027re being asked to find the voltage between these 2 points."},{"Start":"00:45.230 ","End":"00:47.310","Text":"How are we going to do that?"},{"Start":"00:47.780 ","End":"00:50.268","Text":"How do we do this?"},{"Start":"00:50.268 ","End":"00:54.890","Text":"Now, the first thing that we have to do is like what we did over here,"},{"Start":"00:54.890 ","End":"00:58.190","Text":"we choose our direction of currents,"},{"Start":"00:58.190 ","End":"01:00.980","Text":"which we\u0027ve done, and then we write"},{"Start":"01:00.980 ","End":"01:07.270","Text":"our voltage equations but this time our voltage equations are gonna be different."},{"Start":"01:07.270 ","End":"01:11.925","Text":"Here, we did around a whole closed loop,"},{"Start":"01:11.925 ","End":"01:18.125","Text":"so if we started at A, then we did some kind of circle and we ended at A."},{"Start":"01:18.125 ","End":"01:21.444","Text":"Now, here, because we\u0027re trying to find the voltage between A and B,"},{"Start":"01:21.444 ","End":"01:23.390","Text":"we\u0027re starting at A,"},{"Start":"01:23.390 ","End":"01:24.665","Text":"we\u0027re going around,"},{"Start":"01:24.665 ","End":"01:31.360","Text":"and we\u0027re ending at B."},{"Start":"01:31.360 ","End":"01:35.705","Text":"The first thing that we need to realize is when answering these types of questions,"},{"Start":"01:35.705 ","End":"01:38.450","Text":"we do have to do the whole process that we did in"},{"Start":"01:38.450 ","End":"01:41.405","Text":"the last video in order to find out what"},{"Start":"01:41.405 ","End":"01:48.920","Text":"our current values are and we saw that our I_3 is,"},{"Start":"01:48.920 ","End":"01:51.370","Text":"in fact, pointing in the direction of the red arrow,"},{"Start":"01:51.370 ","End":"01:53.640","Text":"so let\u0027s just draw that in now,"},{"Start":"01:53.640 ","End":"02:03.680","Text":"so we saw that our I_3 is going like this."},{"Start":"02:03.680 ","End":"02:09.930","Text":"That means that here we also have to switch around our positives and negatives,"},{"Start":"02:09.930 ","End":"02:13.195","Text":"so here we will have a positive and here a negative,"},{"Start":"02:13.195 ","End":"02:22.097","Text":"and then here, we\u0027re also going to have a positive here and a negative here."},{"Start":"02:22.097 ","End":"02:26.270","Text":"So we fix this according to what we got in the previous step where I_3"},{"Start":"02:26.270 ","End":"02:31.150","Text":"is pointing in this direction because we had a negative."},{"Start":"02:31.150 ","End":"02:33.780","Text":"Originally, we said that our I_3 was pointing"},{"Start":"02:33.780 ","End":"02:38.780","Text":"upwards in the opposite direction to how the red arrows are now,"},{"Start":"02:38.780 ","End":"02:44.180","Text":"but then we got that our I_3 is equal to negative some kind of amp,"},{"Start":"02:44.180 ","End":"02:46.010","Text":"so we flipped it around,"},{"Start":"02:46.010 ","End":"02:49.325","Text":"which means that we also have to change"},{"Start":"02:49.325 ","End":"02:53.785","Text":"our plus and minus signs along each one of the resistors."},{"Start":"02:53.785 ","End":"02:56.570","Text":"Now, let\u0027s get back to it."},{"Start":"02:56.570 ","End":"03:04.860","Text":"What we\u0027re going to do is we\u0027re going to write our voltage between points A and B."},{"Start":"03:04.860 ","End":"03:06.585","Text":"That means that first of all,"},{"Start":"03:06.585 ","End":"03:09.035","Text":"we want to find our voltage at point A,"},{"Start":"03:09.035 ","End":"03:11.565","Text":"so let\u0027s just call it V_A."},{"Start":"03:11.565 ","End":"03:14.450","Text":"It\u0027s more correct to write that it\u0027s our potential at A,"},{"Start":"03:14.450 ","End":"03:16.250","Text":"but it doesn\u0027t really matter."},{"Start":"03:16.250 ","End":"03:18.385","Text":"We have our voltage at A,"},{"Start":"03:18.385 ","End":"03:20.190","Text":"and then let\u0027s go this way,"},{"Start":"03:20.190 ","End":"03:23.564","Text":"so now we\u0027re jumping over our battery."},{"Start":"03:23.564 ","End":"03:27.754","Text":"Now, we can see that we\u0027re going from a low voltage to a high voltage"},{"Start":"03:27.754 ","End":"03:33.490","Text":"so that means that we\u0027re going to be adding on 3 volts."},{"Start":"03:33.490 ","End":"03:35.792","Text":"I\u0027m not going to write in the units, just 3."},{"Start":"03:35.792 ","End":"03:37.175","Text":"Then we carry on,"},{"Start":"03:37.175 ","End":"03:39.425","Text":"and then we get to our resistor,"},{"Start":"03:39.425 ","End":"03:42.620","Text":"so we can see that we\u0027re going with the current flow"},{"Start":"03:42.620 ","End":"03:48.356","Text":"from a higher voltage to lower voltage because we\u0027re going through a resistor."},{"Start":"03:48.356 ","End":"03:54.620","Text":"So here we\u0027re going to minus our voltage across the resistor,"},{"Start":"03:54.620 ","End":"03:58.550","Text":"which is equal to I_1 multiplied by 2 ohms."},{"Start":"03:58.550 ","End":"04:01.640","Text":"Now, our I_1, we already worked that out,"},{"Start":"04:01.640 ","End":"04:07.530","Text":"so that is equal to 2 divided by 11 multiplied by 2 ohms, again,"},{"Start":"04:07.530 ","End":"04:10.275","Text":"I\u0027m not going to include in the units,"},{"Start":"04:10.275 ","End":"04:12.870","Text":"and then we carry on, we\u0027re heading to B,"},{"Start":"04:12.870 ","End":"04:14.640","Text":"so we\u0027re going to carry on."},{"Start":"04:14.640 ","End":"04:18.205","Text":"Now, notice we\u0027re in the region of our I_3,"},{"Start":"04:18.205 ","End":"04:22.580","Text":"and now, we\u0027re going against the current along our resistor,"},{"Start":"04:22.580 ","End":"04:24.950","Text":"so our voltage is getting bigger,"},{"Start":"04:24.950 ","End":"04:27.530","Text":"we\u0027re going from a minus to a positive,"},{"Start":"04:27.530 ","End":"04:32.675","Text":"so we\u0027re going to add on our voltage across the resistor."},{"Start":"04:32.675 ","End":"04:38.965","Text":"So that\u0027s going to be equal to I_3 multiplied by our resistance, which is1 ohm."},{"Start":"04:38.965 ","End":"04:45.136","Text":"Now, notice here we can change this to a positive because we flipped over our arrows."},{"Start":"04:45.136 ","End":"04:47.060","Text":"So if we did all of this equation again,"},{"Start":"04:47.060 ","End":"04:51.290","Text":"taking into account that our arrow was pointing in this clockwise direction,"},{"Start":"04:51.290 ","End":"04:58.178","Text":"so our I_3 would be a positive value because we changed this I_3 is a positive now."},{"Start":"04:58.178 ","End":"05:03.270","Text":"So we\u0027re adding on our I_3,"},{"Start":"05:03.270 ","End":"05:07.280","Text":"so 5 divided by 11 multiplied by the resistance,"},{"Start":"05:07.280 ","End":"05:10.440","Text":"which is multiplied by 1."},{"Start":"05:11.630 ","End":"05:14.770","Text":"After we\u0027ve passed this resistor,"},{"Start":"05:14.770 ","End":"05:17.170","Text":"we carry on and we reach our point B,"},{"Start":"05:17.170 ","End":"05:19.519","Text":"which is our final destination."},{"Start":"05:19.519 ","End":"05:23.800","Text":"We can say that this is equal to our voltage at point B,"},{"Start":"05:23.800 ","End":"05:25.900","Text":"or which is more correct to write,"},{"Start":"05:25.900 ","End":"05:32.570","Text":"our potential at point B but you can write voltage."},{"Start":"05:32.570 ","End":"05:37.110","Text":"Now, what I want to do is I want to find the voltage between A and B."},{"Start":"05:37.110 ","End":"05:40.420","Text":"We generally call this our V_AB,"},{"Start":"05:40.420 ","End":"05:43.225","Text":"so this is the voltage between A and B,"},{"Start":"05:43.225 ","End":"05:48.530","Text":"and it is equal to V_B minus V_A."},{"Start":"05:48.530 ","End":"05:53.695","Text":"All I have to do is I have to move over my V_A to the other side of the equal sign,"},{"Start":"05:53.695 ","End":"05:59.445","Text":"so I\u0027m going to minus V_A from both sides and now I have"},{"Start":"05:59.445 ","End":"06:05.420","Text":"my V_B minus V_A and we can see that that\u0027s just equal to all of this over here,"},{"Start":"06:05.420 ","End":"06:13.100","Text":"so that\u0027s equal to 3 minus 4 divided by 11 plus 5 divided by 11,"},{"Start":"06:13.100 ","End":"06:20.560","Text":"which is equal to 3 plus 1/11 volts."},{"Start":"06:20.560 ","End":"06:24.065","Text":"I\u0027m writing in the units for my voltage,"},{"Start":"06:24.065 ","End":"06:28.610","Text":"and this is my voltage between points A and B."},{"Start":"06:28.610 ","End":"06:31.530","Text":"That\u0027s the end of this lesson."}],"ID":22294},{"Watched":false,"Name":"Cramers Rule","Duration":"15m 2s","ChapterTopicVideoID":21478,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this lesson,"},{"Start":"00:02.145 ","End":"00:05.055","Text":"we\u0027re going to be learning about Cramer\u0027s rule."},{"Start":"00:05.055 ","End":"00:09.825","Text":"Now Cramer\u0027s rule, you might have learned in your linear algebra course,"},{"Start":"00:09.825 ","End":"00:13.740","Text":"and it\u0027s a formula that helps us solve or find"},{"Start":"00:13.740 ","End":"00:17.970","Text":"a solution for a system of linear equations."},{"Start":"00:17.970 ","End":"00:23.415","Text":"As long as we have the same amount of equations as we have unknowns,"},{"Start":"00:23.415 ","End":"00:30.520","Text":"we can solve these equations via matrix system and using Cramer\u0027s rule."},{"Start":"00:30.980 ","End":"00:36.595","Text":"Let\u0027s take this system of linear equations as an example."},{"Start":"00:36.595 ","End":"00:42.680","Text":"Imagine that we have a complex circuit and we use Kirchhoff\u0027s method"},{"Start":"00:42.680 ","End":"00:48.990","Text":"in order to find our equations for our currents and for our voltages,"},{"Start":"00:48.990 ","End":"00:51.815","Text":"and then we rearrange everything so that we get"},{"Start":"00:51.815 ","End":"00:55.520","Text":"our equations in terms of our unknown currents."},{"Start":"00:55.520 ","End":"00:59.550","Text":"What we\u0027re trying to find is what our currents I_1,"},{"Start":"00:59.550 ","End":"01:02.640","Text":"I_2, and I_3 are equal to."},{"Start":"01:02.640 ","End":"01:04.850","Text":"About 2 lessons ago,"},{"Start":"01:04.850 ","End":"01:08.735","Text":"we saw that we can do this via substitution in algebra."},{"Start":"01:08.735 ","End":"01:12.650","Text":"Now depending how many unknowns and how many equations we have,"},{"Start":"01:12.650 ","End":"01:14.720","Text":"this could take a lot of time,"},{"Start":"01:14.720 ","End":"01:16.310","Text":"and especially when you\u0027re in an exam,"},{"Start":"01:16.310 ","End":"01:19.250","Text":"you really don\u0027t want to be wasting precious time."},{"Start":"01:19.250 ","End":"01:25.505","Text":"A way to solve this relatively quickly and easily is by using Cramer\u0027s rule."},{"Start":"01:25.505 ","End":"01:27.230","Text":"Let\u0027s see how we do this."},{"Start":"01:27.230 ","End":"01:36.200","Text":"The important thing to note before we begin is that if we have 3 unknowns,"},{"Start":"01:36.200 ","End":"01:38.570","Text":"so here are our unknowns, I_1, I_2,"},{"Start":"01:38.570 ","End":"01:41.780","Text":"and I_3, that means that we have to have 3 equations."},{"Start":"01:41.780 ","End":"01:45.010","Text":"We have 3 equations, so we can begin."},{"Start":"01:45.010 ","End":"01:47.810","Text":"What we\u0027re going to do is we\u0027re going to"},{"Start":"01:47.810 ","End":"01:52.910","Text":"take all the sections of the equation that are before the equal sign,"},{"Start":"01:52.910 ","End":"01:57.300","Text":"and we\u0027re going to write that out in matrix format."},{"Start":"01:57.920 ","End":"02:01.050","Text":"We have 3 unknowns and 3 equations,"},{"Start":"02:01.050 ","End":"02:03.760","Text":"so our matrix is going to be 3-by-3,"},{"Start":"02:03.760 ","End":"02:09.825","Text":"and each row is going to represent 1 of our unknown."},{"Start":"02:09.825 ","End":"02:11.235","Text":"Our first row is I_1,"},{"Start":"02:11.235 ","End":"02:13.945","Text":"our next is I_2, and our third is I_3."},{"Start":"02:13.945 ","End":"02:18.845","Text":"Then in each row we write down the coefficient of that specific unknown."},{"Start":"02:18.845 ","End":"02:22.520","Text":"We have I_1 and the coefficient is 2,"},{"Start":"02:22.520 ","End":"02:25.640","Text":"then we have I_2 and the coefficient is 3,"},{"Start":"02:25.640 ","End":"02:28.585","Text":"and then I_3 and the coefficient is 2."},{"Start":"02:28.585 ","End":"02:30.925","Text":"Next equation, I_1,"},{"Start":"02:30.925 ","End":"02:33.635","Text":"so it goes here and the coefficient is 3."},{"Start":"02:33.635 ","End":"02:35.600","Text":"I_2, there\u0027s no I_2,"},{"Start":"02:35.600 ","End":"02:37.555","Text":"so the coefficient is 0."},{"Start":"02:37.555 ","End":"02:42.080","Text":"If 1 of the unknowns is not in the equation,"},{"Start":"02:42.080 ","End":"02:43.940","Text":"then its coefficient is equal to 0,"},{"Start":"02:43.940 ","End":"02:47.980","Text":"and then I_3 and its coefficient is negative 1."},{"Start":"02:47.980 ","End":"02:49.715","Text":"Then our third equation,"},{"Start":"02:49.715 ","End":"02:51.650","Text":"we have negative I_1,"},{"Start":"02:51.650 ","End":"02:56.850","Text":"so the coefficient is negative 1 and then 1 and 1."},{"Start":"02:58.010 ","End":"03:01.725","Text":"Now this is equal to,"},{"Start":"03:01.725 ","End":"03:05.690","Text":"and then we\u0027re going to write a vector for our solutions,"},{"Start":"03:05.690 ","End":"03:08.405","Text":"so the solution to our first equation,"},{"Start":"03:08.405 ","End":"03:10.820","Text":"so that\u0027s our first row is 5,"},{"Start":"03:10.820 ","End":"03:13.010","Text":"our solution to our second equation is 2,"},{"Start":"03:13.010 ","End":"03:15.020","Text":"and our third equation is 0."},{"Start":"03:15.020 ","End":"03:17.735","Text":"This is our solution vector."},{"Start":"03:17.735 ","End":"03:21.755","Text":"Now obviously, if these equations were written in a separate order,"},{"Start":"03:21.755 ","End":"03:25.340","Text":"doesn\u0027t matter so long as as you go across,"},{"Start":"03:25.340 ","End":"03:33.430","Text":"you have the coefficients which match our unknowns and matches the specific solution."},{"Start":"03:33.430 ","End":"03:37.355","Text":"If you had 2 3 2 over here corresponding to this equation,"},{"Start":"03:37.355 ","End":"03:40.420","Text":"you can write that the solution here is 0."},{"Start":"03:40.420 ","End":"03:42.690","Text":"Everything has to be in a row."},{"Start":"03:42.690 ","End":"03:45.450","Text":"Similarly, with your unknowns,"},{"Start":"03:45.450 ","End":"03:54.540","Text":"if this is your column for your I_1 variables,"},{"Start":"03:54.540 ","End":"03:58.655","Text":"all of the coefficients only for I_1 can go here,"},{"Start":"03:58.655 ","End":"04:01.460","Text":"and all the coefficients only for I_2 can go here,"},{"Start":"04:01.460 ","End":"04:03.035","Text":"and for I_3 can go here."},{"Start":"04:03.035 ","End":"04:04.475","Text":"You can mix and match,"},{"Start":"04:04.475 ","End":"04:06.925","Text":"that\u0027s what\u0027s very important here."},{"Start":"04:06.925 ","End":"04:11.990","Text":"Great. Now we put this system into a matrix format."},{"Start":"04:11.990 ","End":"04:17.190","Text":"How are we going to solve this to find our unknowns?"},{"Start":"04:18.410 ","End":"04:26.085","Text":"The first thing that we have to do in Cramer\u0027s rule is let\u0027s give this matrix a name."},{"Start":"04:26.085 ","End":"04:28.860","Text":"Let\u0027s say that this matrix is matrix M."},{"Start":"04:28.860 ","End":"04:32.180","Text":"The first thing that we have to do is we have to find"},{"Start":"04:32.180 ","End":"04:35.875","Text":"the determinant of our matrix M."},{"Start":"04:35.875 ","End":"04:39.590","Text":"Now I\u0027m just going to write out this matrix in the determinant fashion."},{"Start":"04:39.590 ","End":"04:42.320","Text":"The determinant, how we write it is instead of"},{"Start":"04:42.320 ","End":"04:46.630","Text":"having square or circular brackets to represent that this is a matrix,"},{"Start":"04:46.630 ","End":"04:51.040","Text":"we just do a long a straight line and then we plug in our values,"},{"Start":"04:51.040 ","End":"04:52.565","Text":"so 2 3 2,"},{"Start":"04:52.565 ","End":"04:54.495","Text":"3 0 negative 1,"},{"Start":"04:54.495 ","End":"04:57.100","Text":"negative 1 1 1."},{"Start":"04:57.100 ","End":"05:02.040","Text":"Now the determinant, I\u0027m assuming that most of you know how to do this,"},{"Start":"05:02.040 ","End":"05:03.515","Text":"I\u0027m not going to solve it right now."},{"Start":"05:03.515 ","End":"05:05.270","Text":"If you don\u0027t know how to do this,"},{"Start":"05:05.270 ","End":"05:09.125","Text":"I will solve the determinant at the end of the video, so keep watching."},{"Start":"05:09.125 ","End":"05:10.865","Text":"But in the meantime, just trust me,"},{"Start":"05:10.865 ","End":"05:16.220","Text":"the determinant for this matrix M over here is equal to 2."},{"Start":"05:17.960 ","End":"05:20.445","Text":"Up until now, that\u0027s great."},{"Start":"05:20.445 ","End":"05:24.740","Text":"Now we\u0027re going to do Cramer\u0027s rule,"},{"Start":"05:24.740 ","End":"05:26.285","Text":"which is super cool."},{"Start":"05:26.285 ","End":"05:34.490","Text":"That means that for each column that we have in our matrix M,"},{"Start":"05:34.490 ","End":"05:38.735","Text":"we\u0027re going to replace that column with"},{"Start":"05:38.735 ","End":"05:43.545","Text":"our vector of solutions and then we\u0027ll work out the determinant."},{"Start":"05:43.545 ","End":"05:46.515","Text":"Let\u0027s give our columns numbers."},{"Start":"05:46.515 ","End":"05:47.880","Text":"This is column 1,"},{"Start":"05:47.880 ","End":"05:50.445","Text":"column 2, and column 3."},{"Start":"05:50.445 ","End":"05:58.495","Text":"Now what we\u0027re going to do is we\u0027re going to find the determinant of M_1."},{"Start":"05:58.495 ","End":"06:00.305","Text":"What does this 1 mean?"},{"Start":"06:00.305 ","End":"06:03.320","Text":"It means that we\u0027re replacing our first column in"},{"Start":"06:03.320 ","End":"06:07.040","Text":"our matrix M with our vector of solutions,"},{"Start":"06:07.040 ","End":"06:09.350","Text":"and then we\u0027re working out the determinant."},{"Start":"06:09.350 ","End":"06:11.030","Text":"Let\u0027s see how that is."},{"Start":"06:11.030 ","End":"06:12.290","Text":"Again, our determinants,"},{"Start":"06:12.290 ","End":"06:14.330","Text":"so instead of 2 3 negative 1,"},{"Start":"06:14.330 ","End":"06:17.440","Text":"we\u0027re going to replace it with 5 2 0."},{"Start":"06:17.440 ","End":"06:22.600","Text":"In the first column we have 5 2 0 because this is det M_1,"},{"Start":"06:22.600 ","End":"06:25.370","Text":"and then the rest of the matrix looks identical,"},{"Start":"06:25.370 ","End":"06:33.650","Text":"so 3 0 1 going down and 2 negative 1 1 again over there."},{"Start":"06:33.650 ","End":"06:35.930","Text":"Now again, I\u0027m not going to solve the determinant,"},{"Start":"06:35.930 ","End":"06:38.060","Text":"but it equals 3."},{"Start":"06:38.060 ","End":"06:44.970","Text":"Now what we\u0027re going to do is we\u0027re going to work out det M_2."},{"Start":"06:45.650 ","End":"06:52.310","Text":"Det M_2 just means working out the determinant of this matrix M,"},{"Start":"06:52.310 ","End":"06:55.040","Text":"where I\u0027m substituting my second column,"},{"Start":"06:55.040 ","End":"06:58.685","Text":"column number 2, with my vector of solutions."},{"Start":"06:58.685 ","End":"07:01.865","Text":"Now my column number 1 goes back to what it was."},{"Start":"07:01.865 ","End":"07:05.000","Text":"Det M_1 is 1 thing, and now, det M_2,"},{"Start":"07:05.000 ","End":"07:08.815","Text":"I\u0027m just replacing the middle column or the second column."},{"Start":"07:08.815 ","End":"07:13.280","Text":"My first column is again 2 3 negative 1 going down."},{"Start":"07:13.280 ","End":"07:17.840","Text":"Then my second column is being replaced with my vector of solutions,"},{"Start":"07:17.840 ","End":"07:19.550","Text":"so 5 2 0,"},{"Start":"07:19.550 ","End":"07:21.874","Text":"and then my third column is exactly the same,"},{"Start":"07:21.874 ","End":"07:23.950","Text":"2 negative 1 1."},{"Start":"07:23.950 ","End":"07:32.656","Text":"Then the determinant for this matrix is going to be equal to negative 2."},{"Start":"07:32.656 ","End":"07:36.430","Text":"Now this is all I need"},{"Start":"07:36.430 ","End":"07:40.450","Text":"for the meantime in order to work out what my I_1 and I_2 is equal to,"},{"Start":"07:40.450 ","End":"07:45.940","Text":"and then subsequently I can just substitute it in and then I\u0027ll get my answer for I_3."},{"Start":"07:45.940 ","End":"07:53.455","Text":"I_1 is equal to the determinant of M_1."},{"Start":"07:53.455 ","End":"07:59.440","Text":"The determinant, when I substitute in my first row the vector of solutions"},{"Start":"07:59.440 ","End":"08:06.160","Text":"divided by the determinant of my original matrix,"},{"Start":"08:06.160 ","End":"08:09.850","Text":"which is unaltered with no substitutions inside."},{"Start":"08:09.850 ","End":"08:16.165","Text":"My det M_1 is equal to 3 and my det M is equal to 2."},{"Start":"08:16.165 ","End":"08:21.940","Text":"I_1 is equal to 3 divided by 2 Amps."},{"Start":"08:21.940 ","End":"08:28.180","Text":"Because my currents are in Amps and this is my voltage, 5 volts."},{"Start":"08:28.180 ","End":"08:31.495","Text":"Now let\u0027s do our I_2."},{"Start":"08:31.495 ","End":"08:37.015","Text":"You can guess it, the equation for I_2 is equal to det M_2,"},{"Start":"08:37.015 ","End":"08:43.045","Text":"the determinant when we replace the second column in our M matrix"},{"Start":"08:43.045 ","End":"08:50.485","Text":"with the vector of solutions divided by the determinant of the original matrix M,"},{"Start":"08:50.485 ","End":"08:55.330","Text":"which was unaltered and not substituted with the vector of solutions."},{"Start":"08:55.330 ","End":"09:01.585","Text":"Then we\u0027ll get that this is equal to negative 2 divided by 2,"},{"Start":"09:01.585 ","End":"09:05.830","Text":"which is equal to negative 1 Amps."},{"Start":"09:05.830 ","End":"09:07.945","Text":"For our I_3,"},{"Start":"09:07.945 ","End":"09:10.240","Text":"we could technically do the exact same thing."},{"Start":"09:10.240 ","End":"09:14.050","Text":"We could substitute in our third column with this vector of"},{"Start":"09:14.050 ","End":"09:19.270","Text":"solutions and then work out det M_3 and then divide that by 2,"},{"Start":"09:19.270 ","End":"09:21.190","Text":"by our det M. However,"},{"Start":"09:21.190 ","End":"09:25.480","Text":"to work out the determinant of a 3 by 3 matrix can take a little bit of time."},{"Start":"09:25.480 ","End":"09:31.780","Text":"Instead, we can just substitute this in to our equation over here."},{"Start":"09:31.780 ","End":"09:35.200","Text":"We can even substitute this in over here."},{"Start":"09:35.200 ","End":"09:37.810","Text":"We have that 3 I_1,"},{"Start":"09:37.810 ","End":"09:43.825","Text":"so that\u0027s 3 times 3 over 2 minus I_3 is equal to 2."},{"Start":"09:43.825 ","End":"09:47.830","Text":"We can substitute in our I_1 and I_2 into either equation."},{"Start":"09:47.830 ","End":"09:50.380","Text":"But here it looks like it might be easier."},{"Start":"09:50.380 ","End":"09:56.080","Text":"Then we\u0027re going to therefore get that I_3 is equal"},{"Start":"09:56.080 ","End":"10:01.810","Text":"to 9 divided by 2 minus 2,"},{"Start":"10:01.810 ","End":"10:05.080","Text":"which is equal to 5,"},{"Start":"10:05.080 ","End":"10:08.510","Text":"divided by 2 Amps."},{"Start":"10:10.020 ","End":"10:12.400","Text":"Now we\u0027ve solved everything."},{"Start":"10:12.400 ","End":"10:16.360","Text":"You can also again do det M_3 divided by"},{"Start":"10:16.360 ","End":"10:22.310","Text":"det M and see if you get this same answer for I_3."},{"Start":"10:24.360 ","End":"10:27.190","Text":"This is technically the end of the lesson."},{"Start":"10:27.190 ","End":"10:32.965","Text":"Now what I\u0027m going to do is I\u0027m going to solve det M and also det M_1."},{"Start":"10:32.965 ","End":"10:36.985","Text":"If you don\u0027t know how to find the determinant of a matrix, carry on watching."},{"Start":"10:36.985 ","End":"10:39.860","Text":"If you do, you can switch off now."},{"Start":"10:41.100 ","End":"10:46.270","Text":"What I\u0027ve done over here is I\u0027ve rewritten out my determinant for my matrix"},{"Start":"10:46.270 ","End":"10:51.175","Text":"M. This is my original matrix that we had at the beginning of the lesson,"},{"Start":"10:51.175 ","End":"10:54.670","Text":"and now let\u0027s see how we work out the determinant."},{"Start":"10:54.670 ","End":"11:02.845","Text":"First things first, let\u0027s look at this value over here too."},{"Start":"11:02.845 ","End":"11:05.395","Text":"Then I\u0027m going to write 2 times,"},{"Start":"11:05.395 ","End":"11:07.240","Text":"and then if I\u0027m looking at this value,"},{"Start":"11:07.240 ","End":"11:10.375","Text":"I have to cross off the row and column,"},{"Start":"11:10.375 ","End":"11:13.885","Text":"which are joining on to this value over here."},{"Start":"11:13.885 ","End":"11:22.315","Text":"I\u0027ll have 2 multiplied by the determinant of this 2 by 2 mini sub matrix over here."},{"Start":"11:22.315 ","End":"11:27.625","Text":"I have 0 times 1 minus,"},{"Start":"11:27.625 ","End":"11:30.400","Text":"because we go this times this,"},{"Start":"11:30.400 ","End":"11:31.600","Text":"and then we crossover,"},{"Start":"11:31.600 ","End":"11:37.970","Text":"so negative 1 multiplied by negative 1."},{"Start":"11:38.250 ","End":"11:41.725","Text":"Now just before we carry on,"},{"Start":"11:41.725 ","End":"11:45.775","Text":"we\u0027re moving along this row over here."},{"Start":"11:45.775 ","End":"11:48.895","Text":"When we circle our first number,"},{"Start":"11:48.895 ","End":"11:53.380","Text":"the sign before our first number is going to be a positive."},{"Start":"11:53.380 ","End":"11:56.290","Text":"Then when we move on to the next column,"},{"Start":"11:56.290 ","End":"11:58.855","Text":"our sign is going to be a negative."},{"Start":"11:58.855 ","End":"12:00.939","Text":"Our next column is a positive,"},{"Start":"12:00.939 ","End":"12:05.410","Text":"and we\u0027re going to carry on alternating between positives and negatives."},{"Start":"12:05.410 ","End":"12:07.105","Text":"Now we can carry on."},{"Start":"12:07.105 ","End":"12:08.530","Text":"I\u0027m going to rub out."},{"Start":"12:08.530 ","End":"12:14.500","Text":"Now we\u0027re moving on to I_3.. We just add I_3."},{"Start":"12:14.500 ","End":"12:15.820","Text":"Now let\u0027s do I_3."},{"Start":"12:15.820 ","End":"12:17.590","Text":"So let\u0027s circle this."},{"Start":"12:17.590 ","End":"12:21.040","Text":"So we have our 3 and we know that this is a negative."},{"Start":"12:21.040 ","End":"12:23.050","Text":"Here we write negative 3,"},{"Start":"12:23.050 ","End":"12:25.060","Text":"here we had positive 2,"},{"Start":"12:25.060 ","End":"12:27.800","Text":"and here we have negative 3."},{"Start":"12:28.380 ","End":"12:33.250","Text":"First of all, we\u0027re going to cross out"},{"Start":"12:33.250 ","End":"12:38.845","Text":"the row and the column which has joined onto this 3."},{"Start":"12:38.845 ","End":"12:40.435","Text":"Now we\u0027re going to find"},{"Start":"12:40.435 ","End":"12:47.420","Text":"the sub determinant or the determinant of this sub-matrix in here."},{"Start":"12:47.430 ","End":"12:51.955","Text":"Again, that\u0027s 3 times 1 because we\u0027re crossing over,"},{"Start":"12:51.955 ","End":"12:56.000","Text":"so 3 times 1 and then minus,"},{"Start":"12:56.000 ","End":"13:00.600","Text":"and then we\u0027re going to have negative 1 multiplied by negative 1,"},{"Start":"13:00.600 ","End":"13:03.960","Text":"so negative 1 times negative 1."},{"Start":"13:03.960 ","End":"13:08.790","Text":"Now we\u0027re going to move over to our next number."},{"Start":"13:08.790 ","End":"13:12.345","Text":"Let\u0027s rub out this."},{"Start":"13:12.345 ","End":"13:14.915","Text":"Here there was a 2."},{"Start":"13:14.915 ","End":"13:17.470","Text":"Now we moved over to here over 2."},{"Start":"13:17.470 ","End":"13:22.990","Text":"We\u0027re going to cross out the corresponding row and the corresponding column."},{"Start":"13:22.990 ","End":"13:25.705","Text":"Here we have a plus."},{"Start":"13:25.705 ","End":"13:31.870","Text":"We\u0027ve gone back to a plus 2 multiplied by the determinant of the sub-matrix."},{"Start":"13:31.870 ","End":"13:34.255","Text":"Here we have 3 times 1."},{"Start":"13:34.255 ","End":"13:43.555","Text":"We\u0027re multiplying across an x-ray minus and then we have negative 1 times 0."},{"Start":"13:43.555 ","End":"13:45.445","Text":"Now let\u0027s solve this."},{"Start":"13:45.445 ","End":"13:49.255","Text":"We have 1 times negative 1 is negative 1,"},{"Start":"13:49.255 ","End":"13:51.310","Text":"negative and a negative is positive,"},{"Start":"13:51.310 ","End":"13:54.700","Text":"1 multiplied by 2 is 2,"},{"Start":"13:54.700 ","End":"13:57.130","Text":"3 times 1 is 3,"},{"Start":"13:57.130 ","End":"13:59.980","Text":"and then we have negative 1 times negative 1,"},{"Start":"13:59.980 ","End":"14:03.190","Text":"which is 1 and then negative."},{"Start":"14:03.190 ","End":"14:06.010","Text":"We have 3 negative 1 which is 2."},{"Start":"14:06.010 ","End":"14:08.455","Text":"Then that\u0027s being multiplied by negative 3,"},{"Start":"14:08.455 ","End":"14:10.615","Text":"so we have negative 6."},{"Start":"14:10.615 ","End":"14:16.240","Text":"Then we have 3 times 1 minus negative 1 times 0, so that\u0027s 0."},{"Start":"14:16.240 ","End":"14:17.875","Text":"We have 2 times 3,"},{"Start":"14:17.875 ","End":"14:19.885","Text":"so plus 6,"},{"Start":"14:19.885 ","End":"14:22.240","Text":"which is equal to 2."},{"Start":"14:22.240 ","End":"14:24.610","Text":"Now, let\u0027s go back over here."},{"Start":"14:24.610 ","End":"14:26.905","Text":"Remember det M is equal to 2,"},{"Start":"14:26.905 ","End":"14:29.570","Text":"and that\u0027s what I said over here."},{"Start":"14:31.110 ","End":"14:38.575","Text":"That\u0027s how we work out the determinant of any matrix and especially a 3 by 3."},{"Start":"14:38.575 ","End":"14:41.470","Text":"Then we do the same thing for our det M_1."},{"Start":"14:41.470 ","End":"14:43.750","Text":"The only difference is for our M_1,"},{"Start":"14:43.750 ","End":"14:45.640","Text":"we replaced the first column with"},{"Start":"14:45.640 ","End":"14:52.255","Text":"our vectors of solutions or vector solutions, which is this."},{"Start":"14:52.255 ","End":"14:55.270","Text":"Then our det M_2, we replace column number 2"},{"Start":"14:55.270 ","End":"14:58.885","Text":"with our vector of solutions and so on and so forth."},{"Start":"14:58.885 ","End":"15:01.820","Text":"That\u0027s the end of this lesson."}],"ID":22295},{"Watched":false,"Name":"Mesh Current Method","Duration":"26m 6s","ChapterTopicVideoID":21485,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"Hello. In this lesson,"},{"Start":"00:01.965 ","End":"00:06.170","Text":"we\u0027re going to be speaking about another way of solving different currents."},{"Start":"00:06.170 ","End":"00:09.270","Text":"We\u0027re going to be speaking about the mesh current method."},{"Start":"00:09.270 ","End":"00:15.840","Text":"Now the mesh current method works on Kirchhoff\u0027s law and this is how we use it."},{"Start":"00:15.840 ","End":"00:20.130","Text":"Now, first of all, let\u0027s understand what a mesh is."},{"Start":"00:20.130 ","End":"00:22.365","Text":"It\u0027s like a loop."},{"Start":"00:22.365 ","End":"00:25.740","Text":"A loop is any closed path around a circuit."},{"Start":"00:25.740 ","End":"00:31.530","Text":"We can say that this, starting from this point over here,"},{"Start":"00:31.530 ","End":"00:34.250","Text":"and coming back is a loop."},{"Start":"00:34.250 ","End":"00:37.850","Text":"Similarly, starting from this point over here and going"},{"Start":"00:37.850 ","End":"00:41.705","Text":"around this mini circuit is also a loop."},{"Start":"00:41.705 ","End":"00:44.710","Text":"Also, let\u0027s say from this same point over here."},{"Start":"00:44.710 ","End":"00:53.075","Text":"This is also a loop. What is a mesh?"},{"Start":"00:53.075 ","End":"00:58.910","Text":"A mesh is a loop that contains no other loops."},{"Start":"00:58.910 ","End":"01:03.440","Text":"We can see that this loop over here is a mesh because"},{"Start":"01:03.440 ","End":"01:08.300","Text":"it just has itself as the loop and no subloops within it."},{"Start":"01:08.300 ","End":"01:14.015","Text":"Similarly, this is also a mesh for the exact same reason."},{"Start":"01:14.015 ","End":"01:20.540","Text":"However, this loop over here that\u0027s going on the outside along the outer loop,"},{"Start":"01:20.540 ","End":"01:23.300","Text":"we can see that it contains 2 loops."},{"Start":"01:23.300 ","End":"01:26.145","Text":"This one and this one."},{"Start":"01:26.145 ","End":"01:29.675","Text":"This outer loop is not a mesh,"},{"Start":"01:29.675 ","End":"01:32.820","Text":"but these 2 inners are meshes."},{"Start":"01:34.130 ","End":"01:37.055","Text":"Now that we know what a mesh is,"},{"Start":"01:37.055 ","End":"01:39.920","Text":"let\u0027s speak about how to use the mesh current method."},{"Start":"01:39.920 ","End":"01:45.080","Text":"Now, this method is super easy to use and it makes the use of"},{"Start":"01:45.080 ","End":"01:50.965","Text":"all of the complicated equations that you can get from complex circuits like so."},{"Start":"01:50.965 ","End":"01:54.150","Text":"It makes them much easier to solve."},{"Start":"01:54.150 ","End":"01:56.740","Text":"Let\u0027s see how we do this."},{"Start":"01:57.560 ","End":"02:01.540","Text":"Now, we\u0027re going to speak about the mesh current method,"},{"Start":"02:01.540 ","End":"02:03.295","Text":"and later in the lesson,"},{"Start":"02:03.295 ","End":"02:06.125","Text":"I\u0027ll explain when to use it."},{"Start":"02:06.125 ","End":"02:11.425","Text":"The first thing that we\u0027re going to do is we\u0027re going to look at our circuit over here,"},{"Start":"02:11.425 ","End":"02:16.060","Text":"and we\u0027re going to split it up into different meshes."},{"Start":"02:16.060 ","End":"02:19.825","Text":"By the way, this is the exact same circuit"},{"Start":"02:19.825 ","End":"02:24.199","Text":"that are used in order to explain about Kirchhoff\u0027s laws."},{"Start":"02:24.199 ","End":"02:29.605","Text":"I just took out the values for the resistances and put in"},{"Start":"02:29.605 ","End":"02:35.953","Text":"arbitrary symbols but it\u0027s the same thing and you can go back to the Kirchhoff lesson."},{"Start":"02:35.953 ","End":"02:43.760","Text":"Then you can see the connection in correlation between this method and that method."},{"Start":"02:44.510 ","End":"02:47.770","Text":"Let\u0027s begin. This,"},{"Start":"02:47.770 ","End":"02:49.145","Text":"as we already saw,"},{"Start":"02:49.145 ","End":"02:52.205","Text":"is going to be my first mesh."},{"Start":"02:52.205 ","End":"02:55.880","Text":"It\u0027s going in the clockwise direction."},{"Start":"02:55.880 ","End":"02:59.040","Text":"I can even highlight it."},{"Start":"02:59.040 ","End":"03:05.790","Text":"This mesh includes all of these components over here."},{"Start":"03:06.250 ","End":"03:11.450","Text":"Now, which is something different to Kirchhoff\u0027s law,"},{"Start":"03:11.450 ","End":"03:15.440","Text":"I\u0027m going to say that the current flowing through this entire mesh,"},{"Start":"03:15.440 ","End":"03:19.040","Text":"so this entire area highlighted with yellow."},{"Start":"03:19.040 ","End":"03:23.255","Text":"The current flowing through is I_1"},{"Start":"03:23.255 ","End":"03:29.100","Text":"and it goes through all of these components in this mesh."},{"Start":"03:30.380 ","End":"03:36.065","Text":"Next, if I chose my I_1 to be in the clockwise direction,"},{"Start":"03:36.065 ","End":"03:40.160","Text":"then my current in the second mesh,"},{"Start":"03:40.160 ","End":"03:44.900","Text":"so I_2 also has to be in the clockwise direction."},{"Start":"03:44.900 ","End":"03:49.875","Text":"They\u0027re either both clockwise or both anticlockwise. It\u0027s very important."},{"Start":"03:49.875 ","End":"03:55.179","Text":"Now my current flowing through my second mesh is equal to I_2,"},{"Start":"03:55.179 ","End":"03:58.940","Text":"and now let\u0027s highlight my second mesh."},{"Start":"03:58.940 ","End":"04:03.940","Text":"My second mesh is this highlighted in green."},{"Start":"04:03.940 ","End":"04:11.795","Text":"This current is flowing through all of the components that are in this mesh as well."},{"Start":"04:11.795 ","End":"04:16.765","Text":"As we can see, we have this voltage source and this resistor,"},{"Start":"04:16.765 ","End":"04:22.020","Text":"and they appear both in mesh number 1 and in mesh number 2."},{"Start":"04:22.020 ","End":"04:24.670","Text":"That\u0027s fine. That\u0027s good."},{"Start":"04:24.670 ","End":"04:29.900","Text":"Now, the first thing that we can see is that I only have 2 currents flowing."},{"Start":"04:29.900 ","End":"04:32.600","Text":"I have current I_1 and current I_2,"},{"Start":"04:32.600 ","End":"04:35.705","Text":"which means that I only have 2 unknowns,"},{"Start":"04:35.705 ","End":"04:40.730","Text":"which means that I\u0027m only going to need 2 equations."},{"Start":"04:41.700 ","End":"04:44.800","Text":"Now let\u0027s begin to solve this."},{"Start":"04:44.800 ","End":"04:50.635","Text":"We\u0027re just going to look at each mesh as if it\u0027s its own circuit."},{"Start":"04:50.635 ","End":"04:56.615","Text":"We\u0027re going to solve these individual circuits like we would using Kirchhoff\u0027s law."},{"Start":"04:56.615 ","End":"05:00.550","Text":"What I\u0027m going to do is I\u0027m going to choose some starting position."},{"Start":"05:00.550 ","End":"05:03.415","Text":"Let\u0027s say this red point over here."},{"Start":"05:03.415 ","End":"05:08.933","Text":"Then I\u0027m going to complete a full loop around."},{"Start":"05:08.933 ","End":"05:11.455","Text":"Starting at the red point and ending at the red point,"},{"Start":"05:11.455 ","End":"05:13.885","Text":"just in this yellow mesh."},{"Start":"05:13.885 ","End":"05:18.505","Text":"First of all, I have my I_1 going in this direction."},{"Start":"05:18.505 ","End":"05:23.760","Text":"Here I have my positive and here I have my negative."},{"Start":"05:23.760 ","End":"05:26.540","Text":"Because as my current goes through my I_1,"},{"Start":"05:26.540 ","End":"05:30.005","Text":"my voltage here is going to be higher than my voltage over here"},{"Start":"05:30.005 ","End":"05:33.710","Text":"because my resistor reduces the voltage."},{"Start":"05:33.710 ","End":"05:38.180","Text":"That means I\u0027ll have a negative because I\u0027m reducing the voltage."},{"Start":"05:38.180 ","End":"05:43.185","Text":"Negative I_1 multiplied by I_1."},{"Start":"05:43.185 ","End":"05:46.745","Text":"Through Ohm\u0027s law, V is equal to IR."},{"Start":"05:46.745 ","End":"05:50.560","Text":"Then I carry on and I get to this voltage source."},{"Start":"05:50.560 ","End":"05:56.645","Text":"Then I can see I\u0027m going from a high bar so a high voltage to a low bar, a low voltage."},{"Start":"05:56.645 ","End":"05:59.080","Text":"My voltage is being reduced."},{"Start":"05:59.080 ","End":"06:02.565","Text":"Then I have negative Epsilon 1."},{"Start":"06:02.565 ","End":"06:05.795","Text":"Now I\u0027m carrying on along the circuit,"},{"Start":"06:05.795 ","End":"06:09.890","Text":"now I reach this component over here, R2."},{"Start":"06:09.890 ","End":"06:11.585","Text":"Now as we can see,"},{"Start":"06:11.585 ","End":"06:17.065","Text":"I have my current I_1 flowing in this direction,"},{"Start":"06:17.065 ","End":"06:20.765","Text":"I_1 but then I also have my current I2,"},{"Start":"06:20.765 ","End":"06:23.000","Text":"which is also going clockwise."},{"Start":"06:23.000 ","End":"06:26.070","Text":"Which is flowing in this direction."},{"Start":"06:26.070 ","End":"06:28.475","Text":"Through this resistor I_2,"},{"Start":"06:28.475 ","End":"06:30.670","Text":"I have 2 currents flowing."},{"Start":"06:30.670 ","End":"06:35.660","Text":"My I_1 from my yellow mesh and my I_2 for my green mesh."},{"Start":"06:35.660 ","End":"06:39.305","Text":"I have to take in both currents into account."},{"Start":"06:39.305 ","End":"06:44.690","Text":"First of all, we know that through my I_1 loop,"},{"Start":"06:44.690 ","End":"06:47.090","Text":"if we\u0027re going in this direction,"},{"Start":"06:47.090 ","End":"06:50.390","Text":"this is my positive and this is my negative."},{"Start":"06:50.390 ","End":"06:54.425","Text":"Because when we\u0027re going with the direction of the current,"},{"Start":"06:54.425 ","End":"06:56.720","Text":"the entering terminal is a positive,"},{"Start":"06:56.720 ","End":"06:59.760","Text":"and the exiting terminal is a negative."},{"Start":"07:00.110 ","End":"07:04.175","Text":"This is going to be equal to negative."},{"Start":"07:04.175 ","End":"07:08.790","Text":"Then we have our current I_1."},{"Start":"07:08.790 ","End":"07:15.955","Text":"Then this is going to be multiplied by the resistance R_2, according to Ohm\u0027s Law."},{"Start":"07:15.955 ","End":"07:20.755","Text":"However, we also have this I_2 flowing through over here."},{"Start":"07:20.755 ","End":"07:22.690","Text":"Now our I_2, as we can see,"},{"Start":"07:22.690 ","End":"07:25.715","Text":"is in the opposite direction to our I_1,"},{"Start":"07:25.715 ","End":"07:30.795","Text":"so negative I_2, we write over here."},{"Start":"07:30.795 ","End":"07:35.920","Text":"The total current flowing through this resistor I_2,"},{"Start":"07:36.150 ","End":"07:38.635","Text":"because we\u0027re looking at this mesh,"},{"Start":"07:38.635 ","End":"07:45.440","Text":"it\u0027s going to be equal to I_1 minus I_2 multiplied by R_2."},{"Start":"07:46.920 ","End":"07:49.930","Text":"Now I\u0027m carrying on and I\u0027m carrying on,"},{"Start":"07:49.930 ","End":"07:51.760","Text":"and I get to this voltage source."},{"Start":"07:51.760 ","End":"07:53.170","Text":"Now my voltage source,"},{"Start":"07:53.170 ","End":"07:58.840","Text":"I know that it is just a source of voltage of Epsilon 2."},{"Start":"07:58.840 ","End":"08:03.010","Text":"I can see that I\u0027m going from a low voltage to a high voltage."},{"Start":"08:03.010 ","End":"08:07.735","Text":"I\u0027m adding on this voltage Epsilon 2,"},{"Start":"08:07.735 ","End":"08:09.775","Text":"and then I carry on along my loop,"},{"Start":"08:09.775 ","End":"08:12.430","Text":"and I get back to my starting points."},{"Start":"08:12.430 ","End":"08:14.425","Text":"We know that in a closed loop,"},{"Start":"08:14.425 ","End":"08:18.955","Text":"the sum of all of the voltages is going to be equal to 0."},{"Start":"08:18.955 ","End":"08:22.765","Text":"Now let\u0027s do the same thing for the second mesh,"},{"Start":"08:22.765 ","End":"08:24.295","Text":"so my green mesh."},{"Start":"08:24.295 ","End":"08:26.725","Text":"Let\u0027s choose a starting point."},{"Start":"08:26.725 ","End":"08:29.005","Text":"This time let\u0027s start over here,"},{"Start":"08:29.005 ","End":"08:31.345","Text":"and now let\u0027s write all of our equations."},{"Start":"08:31.345 ","End":"08:36.160","Text":"First of all, I\u0027m going through this voltage source of Epsilon 2,"},{"Start":"08:36.160 ","End":"08:40.615","Text":"and we can see that I\u0027m going from high voltage to a low voltage."},{"Start":"08:40.615 ","End":"08:46.165","Text":"That\u0027s going to mean that I\u0027m reducing Epsilon 2,"},{"Start":"08:46.165 ","End":"08:47.590","Text":"minus Epsilon 2,"},{"Start":"08:47.590 ","End":"08:51.070","Text":"and then I carry on and I reach this resistor R_2."},{"Start":"08:51.070 ","End":"08:53.994","Text":"Now, because I\u0027m looking at this mesh,"},{"Start":"08:53.994 ","End":"08:55.750","Text":"I\u0027m looking at this side,"},{"Start":"08:55.750 ","End":"08:58.825","Text":"which means that here in the direction of the current,"},{"Start":"08:58.825 ","End":"09:01.135","Text":"my I_2 is going in from here."},{"Start":"09:01.135 ","End":"09:04.630","Text":"So this is the positive terminal and my I_2 is exiting on"},{"Start":"09:04.630 ","End":"09:09.885","Text":"this side because we\u0027re in the direction of the current which is clockwise."},{"Start":"09:09.885 ","End":"09:13.430","Text":"So this is my negative terminal right now."},{"Start":"09:13.710 ","End":"09:23.590","Text":"That means that my total current is going to be I_2 in the positive direction minus,"},{"Start":"09:23.590 ","End":"09:25.345","Text":"I can see I have my I_1,"},{"Start":"09:25.345 ","End":"09:27.385","Text":"which is this red arrow over here,"},{"Start":"09:27.385 ","End":"09:29.659","Text":"pointing in the opposite direction."},{"Start":"09:29.659 ","End":"09:34.320","Text":"I\u0027m going to have that the current flowing through my I_2 resistor is"},{"Start":"09:34.320 ","End":"09:39.480","Text":"going to be equal to I_2 minus I_1 this time."},{"Start":"09:39.480 ","End":"09:41.205","Text":"Each time you look at a mesh,"},{"Start":"09:41.205 ","End":"09:43.675","Text":"you\u0027re looking at it individually,"},{"Start":"09:43.675 ","End":"09:48.715","Text":"and as if the current that you are looking at in that mesh is the positive direction,"},{"Start":"09:48.715 ","End":"09:52.060","Text":"and then anything else in the opposite direction is now a negative."},{"Start":"09:52.060 ","End":"09:57.625","Text":"I have I_2 minus I_1 this time multiplied by the resistor R_2."},{"Start":"09:57.625 ","End":"10:00.250","Text":"Of course, this is a minus because we\u0027re going"},{"Start":"10:00.250 ","End":"10:03.250","Text":"from the positive terminal to the negative terminal,"},{"Start":"10:03.250 ","End":"10:05.800","Text":"which means that we have a minus sign over here because"},{"Start":"10:05.800 ","End":"10:08.950","Text":"our voltage is being reduced across the resistor."},{"Start":"10:08.950 ","End":"10:13.075","Text":"Then we carry on along this mesh in the direction of the current."},{"Start":"10:13.075 ","End":"10:17.950","Text":"Again, our current entering is here positive and here negative, so again,"},{"Start":"10:17.950 ","End":"10:22.645","Text":"we\u0027ll have minus I_2 multiplied by R_3,"},{"Start":"10:22.645 ","End":"10:24.250","Text":"and then we carry on,"},{"Start":"10:24.250 ","End":"10:29.170","Text":"and then we get to our Epsilon 3 over here, our voltage source."},{"Start":"10:29.170 ","End":"10:32.440","Text":"We\u0027re going from a low voltage to high voltage,"},{"Start":"10:32.440 ","End":"10:34.870","Text":"which means that we\u0027re adding on our voltage,"},{"Start":"10:34.870 ","End":"10:37.195","Text":"so adding on Epsilon 3,"},{"Start":"10:37.195 ","End":"10:40.510","Text":"and then we carry on in the direction of the current."},{"Start":"10:40.510 ","End":"10:43.240","Text":"This is a positive and this is the negative terminal,"},{"Start":"10:43.240 ","End":"10:48.592","Text":"which means that we have negative I_2R_4."},{"Start":"10:48.592 ","End":"10:52.345","Text":"Then we carry on and we get back to our starting position,"},{"Start":"10:52.345 ","End":"10:57.200","Text":"which means that the sum of all of the voltages is equal to 0."},{"Start":"10:58.290 ","End":"11:01.225","Text":"Now we have 2 equations."},{"Start":"11:01.225 ","End":"11:02.410","Text":"We have our 2 unknowns,"},{"Start":"11:02.410 ","End":"11:04.465","Text":"which are I_1 and I_2,"},{"Start":"11:04.465 ","End":"11:06.625","Text":"and so now we can solve this."},{"Start":"11:06.625 ","End":"11:09.715","Text":"What we\u0027re going to do is we\u0027re going to move"},{"Start":"11:09.715 ","End":"11:13.240","Text":"our voltage sources to 1 side of the equation,"},{"Start":"11:13.240 ","End":"11:18.250","Text":"and then we\u0027re going to open up our brackets and put like terms together."},{"Start":"11:18.250 ","End":"11:22.240","Text":"Let\u0027s first deal with the first equation,"},{"Start":"11:22.240 ","End":"11:24.670","Text":"so that\u0027s this over here."},{"Start":"11:24.670 ","End":"11:27.610","Text":"Let\u0027s put our currents together."},{"Start":"11:27.610 ","End":"11:30.835","Text":"Here we have negative I_1,"},{"Start":"11:30.835 ","End":"11:34.075","Text":"and it\u0027s being multiplied over here by R_1,"},{"Start":"11:34.075 ","End":"11:38.260","Text":"and it\u0027s being multiplied here by R_2."},{"Start":"11:38.260 ","End":"11:40.315","Text":"So plus R_2."},{"Start":"11:40.315 ","End":"11:43.600","Text":"Notice that here we have negative I_1 times R_1,"},{"Start":"11:43.600 ","End":"11:46.450","Text":"and here we have negative I_1 times R_2,"},{"Start":"11:46.450 ","End":"11:49.960","Text":"so that\u0027s why there\u0027s a negative over here and a plus over here."},{"Start":"11:49.960 ","End":"11:53.150","Text":"Then we have our I_2."},{"Start":"11:55.170 ","End":"11:58.510","Text":"Negative times a negative is a positive,"},{"Start":"11:58.510 ","End":"12:02.500","Text":"so we have positive I_2 multiplied by R_2."},{"Start":"12:02.500 ","End":"12:04.840","Text":"That\u0027s it. That is equal to."},{"Start":"12:04.840 ","End":"12:08.740","Text":"Let\u0027s move our Epsilon 1 over to this side of the equation,"},{"Start":"12:08.740 ","End":"12:14.990","Text":"so that becomes plus Epsilon 1 and then minus Epsilon 2."},{"Start":"12:15.110 ","End":"12:20.790","Text":"Now let\u0027s do the exact same thing for equation Number 2."},{"Start":"12:20.790 ","End":"12:22.725","Text":"Equation Number 2,"},{"Start":"12:22.725 ","End":"12:24.450","Text":"we\u0027ll move our Epsilons,"},{"Start":"12:24.450 ","End":"12:29.080","Text":"so our voltage sources over to the other side, and then let\u0027s see."},{"Start":"12:29.190 ","End":"12:31.675","Text":"Let\u0027s start with our I_1."},{"Start":"12:31.675 ","End":"12:33.280","Text":"Negative times a negative,"},{"Start":"12:33.280 ","End":"12:35.935","Text":"so we have positive I_1."},{"Start":"12:35.935 ","End":"12:40.000","Text":"Then here it\u0027s multiplied by R_2 and by nothing else,"},{"Start":"12:40.000 ","End":"12:43.000","Text":"so we can just have I_1 multiplied by R_2."},{"Start":"12:43.000 ","End":"12:44.620","Text":"Then we have our I_2,"},{"Start":"12:44.620 ","End":"12:47.695","Text":"so we have negative I_2,"},{"Start":"12:47.695 ","End":"12:52.555","Text":"and then here it\u0027s being multiplied by R_2."},{"Start":"12:52.555 ","End":"12:57.610","Text":"Here it\u0027s being multiplied by R_3."},{"Start":"12:57.610 ","End":"12:59.920","Text":"Again, there\u0027s a negative sign here."},{"Start":"12:59.920 ","End":"13:02.740","Text":"Here, it\u0027s being multiplied by R_4."},{"Start":"13:02.740 ","End":"13:04.150","Text":"Again there\u0027s a negative sign,"},{"Start":"13:04.150 ","End":"13:06.940","Text":"so that means that here in the brackets,"},{"Start":"13:06.940 ","End":"13:10.885","Text":"we have to have positive R_4,"},{"Start":"13:10.885 ","End":"13:14.200","Text":"and then that is equal to."},{"Start":"13:14.200 ","End":"13:17.110","Text":"Here we\u0027ll move our Epsilon 2 to the other side,"},{"Start":"13:17.110 ","End":"13:22.730","Text":"so it becomes positive Epsilon 2, negative Epsilon 3."},{"Start":"13:24.060 ","End":"13:28.240","Text":"Now we have 2 equations with 2 unknowns,"},{"Start":"13:28.240 ","End":"13:30.535","Text":"our I_1 and our I_2,"},{"Start":"13:30.535 ","End":"13:32.290","Text":"and now we can solve this."},{"Start":"13:32.290 ","End":"13:39.595","Text":"Now I just want to show you an easy way of getting into this setup of our 2 equations,"},{"Start":"13:39.595 ","End":"13:42.220","Text":"because then we can get to this a lot quicker instead of"},{"Start":"13:42.220 ","End":"13:45.070","Text":"doing Kirchhoff\u0027s law and doing all of this,"},{"Start":"13:45.070 ","End":"13:49.375","Text":"we can just get straight to this over here."},{"Start":"13:49.375 ","End":"13:51.550","Text":"Let\u0027s see what\u0027s going on."},{"Start":"13:51.550 ","End":"13:55.465","Text":"Let\u0027s look at equation Number 1 for mesh Number 1."},{"Start":"13:55.465 ","End":"14:02.320","Text":"We can see that I have the current in this mesh multiplied by R_1 plus R_2."},{"Start":"14:02.320 ","End":"14:04.975","Text":"If we\u0027re looking at just this mesh,"},{"Start":"14:04.975 ","End":"14:08.260","Text":"we can see that we have 2 resistors and they\u0027re"},{"Start":"14:08.260 ","End":"14:12.175","Text":"both connected in series with each other in this mesh."},{"Start":"14:12.175 ","End":"14:15.910","Text":"The total resistance of resistors connected in"},{"Start":"14:15.910 ","End":"14:19.405","Text":"series is simply equal to the sum of the resistors,"},{"Start":"14:19.405 ","End":"14:21.790","Text":"so R_1 plus R_2."},{"Start":"14:21.790 ","End":"14:23.920","Text":"Then according to Ohm\u0027s law,"},{"Start":"14:23.920 ","End":"14:25.720","Text":"it has to be multiplied by I_1,"},{"Start":"14:25.720 ","End":"14:28.625","Text":"and because we\u0027re going in the direction of our current,"},{"Start":"14:28.625 ","End":"14:30.140","Text":"we have a minus,"},{"Start":"14:30.140 ","End":"14:33.020","Text":"because our resistors are reducing the current."},{"Start":"14:33.020 ","End":"14:41.995","Text":"Then we also noticed that our R_2 is located both in mesh Number 1 and in mesh Number 2."},{"Start":"14:41.995 ","End":"14:46.985","Text":"Because our I_2 is flowing in the opposite direction,"},{"Start":"14:46.985 ","End":"14:49.070","Text":"so we have a plus over here,"},{"Start":"14:49.070 ","End":"14:54.930","Text":"and then we have our current I_2 multiplied by the resistance."},{"Start":"14:54.930 ","End":"14:58.640","Text":"That it\u0027s going through. That is equal to,"},{"Start":"14:58.640 ","End":"15:05.105","Text":"here we have negative Epsilon 1 plus Epsilon 2,"},{"Start":"15:05.105 ","End":"15:07.130","Text":"because here we\u0027re going from high to low,"},{"Start":"15:07.130 ","End":"15:11.000","Text":"and here we\u0027re going from low to high in the clockwise direction."},{"Start":"15:11.000 ","End":"15:14.585","Text":"Here on the other side of the equal side is just minus that,"},{"Start":"15:14.585 ","End":"15:17.780","Text":"so instead of getting Epsilon 2 minus Epsilon 1,"},{"Start":"15:17.780 ","End":"15:20.895","Text":"we get Epsilon 1 minus Epsilon 2."},{"Start":"15:20.895 ","End":"15:23.230","Text":"Now in equation Number 2,"},{"Start":"15:23.230 ","End":"15:24.700","Text":"for mesh Number 2,"},{"Start":"15:24.700 ","End":"15:27.130","Text":"we have the exact same thing."},{"Start":"15:27.130 ","End":"15:32.500","Text":"We have our I_2 which is flowing through this mesh,"},{"Start":"15:32.500 ","End":"15:36.800","Text":"and then we can see that it\u0027s flowing through the resistor R_2, R_3, and R_4,"},{"Start":"15:36.800 ","End":"15:43.130","Text":"so all of these resistors are in series with each other within this mesh."},{"Start":"15:43.130 ","End":"15:48.020","Text":"The total resistance is just the sum of the resistances multiplied by the current."},{"Start":"15:48.020 ","End":"15:51.785","Text":"Of course, because we\u0027re going in the direction of our current I_2,"},{"Start":"15:51.785 ","End":"15:56.640","Text":"so these resistors reduce the voltage and that\u0027s why there\u0027s a negative over here."},{"Start":"15:56.640 ","End":"16:01.870","Text":"We also see that our R_2 is located between the 2 meshes."},{"Start":"16:01.870 ","End":"16:07.360","Text":"That means that we also have our I_1 flowing through our R_2."},{"Start":"16:07.360 ","End":"16:10.975","Text":"Because it\u0027s in the opposite direction to our I_2,"},{"Start":"16:10.975 ","End":"16:12.620","Text":"so if here we had a negative,"},{"Start":"16:12.620 ","End":"16:17.180","Text":"so here we\u0027ll have a positive I_1 multiplied by R_2."},{"Start":"16:17.180 ","End":"16:22.580","Text":"Then again, we have from the high voltage to a low voltage,"},{"Start":"16:22.580 ","End":"16:26.860","Text":"so we have negative Epsilon 2 plus Epsilon 3."},{"Start":"16:26.860 ","End":"16:32.345","Text":"Then we multiply that by minus 1 on this side of the equation."},{"Start":"16:32.345 ","End":"16:35.075","Text":"Instead of Epsilon 3 minus Epsilon 2,"},{"Start":"16:35.075 ","End":"16:39.106","Text":"we have Epsilon 2 minus Epsilon 3."},{"Start":"16:39.106 ","End":"16:47.335","Text":"Now due to everything being a multiple of negative 1."},{"Start":"16:47.335 ","End":"16:52.300","Text":"What we\u0027re going to do is we\u0027re going to multiply everything by negative"},{"Start":"16:52.300 ","End":"16:58.120","Text":"1 and then our voltage sources will make a bit more sense."},{"Start":"16:58.120 ","End":"17:00.445","Text":"If we\u0027re going in this direction,"},{"Start":"17:00.445 ","End":"17:03.835","Text":"we\u0027ll get minus Epsilon_1 plus Epsilon_2,"},{"Start":"17:03.835 ","End":"17:06.955","Text":"and then the other side will also sort it out."},{"Start":"17:06.955 ","End":"17:13.040","Text":"What we\u0027ll have is let\u0027s multiply equation number 1 by negative 1."},{"Start":"17:13.290 ","End":"17:19.930","Text":"R_1 plus R_2 multiplied by I_1"},{"Start":"17:19.930 ","End":"17:29.605","Text":"minus R_2 multiplied by I_2 and that will equal to Epsilon_2 minus Epsilon_1."},{"Start":"17:29.605 ","End":"17:32.380","Text":"Then for equation number 2,"},{"Start":"17:32.380 ","End":"17:35.305","Text":"we\u0027ll multiply that also by negative 1."},{"Start":"17:35.305 ","End":"17:39.700","Text":"We\u0027ll get negative R_2,"},{"Start":"17:39.700 ","End":"17:45.580","Text":"I_1 plus I_2 multiplied"},{"Start":"17:45.580 ","End":"17:51.820","Text":"by R_2 plus R_3 plus R_4,"},{"Start":"17:51.820 ","End":"17:56.755","Text":"and that will be equal to Epsilon_3 minus Epsilon_2."},{"Start":"17:56.755 ","End":"18:02.380","Text":"This is the type of equation that we want to get,"},{"Start":"18:02.380 ","End":"18:04.990","Text":"and you can go straight to this by"},{"Start":"18:04.990 ","End":"18:09.835","Text":"simply writing this and multiplying everything by negative 1,"},{"Start":"18:09.835 ","End":"18:14.140","Text":"or automatically doing that in your head in the same way"},{"Start":"18:14.140 ","End":"18:19.670","Text":"that we explained how we added in the resistors, etc."},{"Start":"18:19.890 ","End":"18:26.720","Text":"Then all we have to do is we have to solve this via algebra."},{"Start":"18:27.120 ","End":"18:29.350","Text":"We have 2 unknowns,"},{"Start":"18:29.350 ","End":"18:32.905","Text":"which are I_1 and I_2, and 2 equations."},{"Start":"18:32.905 ","End":"18:36.265","Text":"Now notice that in Kirchhoff\u0027s law,"},{"Start":"18:36.265 ","End":"18:39.670","Text":"we had an I_3 going along one of"},{"Start":"18:39.670 ","End":"18:45.025","Text":"the branches and we said that our I_1 was equal to I_2 plus I_3."},{"Start":"18:45.025 ","End":"18:49.525","Text":"So if we substitute it in instead of I_1 or I_2 plus I_3,"},{"Start":"18:49.525 ","End":"18:55.610","Text":"we would have the exact same equations as we had in our lesson for Kirchhoff\u0027s law."},{"Start":"18:55.890 ","End":"18:59.200","Text":"You\u0027ll see in future that it\u0027s very"},{"Start":"18:59.200 ","End":"19:02.500","Text":"easily and recommended to use the mesh current method,"},{"Start":"19:02.500 ","End":"19:06.880","Text":"because here we can see that we have 2 unknowns and 2 equations,"},{"Start":"19:06.880 ","End":"19:09.115","Text":"which is something that\u0027s very easy to solve."},{"Start":"19:09.115 ","End":"19:11.860","Text":"Whereas in Kirchhoff\u0027s rule,"},{"Start":"19:11.860 ","End":"19:16.720","Text":"the exact same circuit had 3 equations and 3 unknowns,"},{"Start":"19:16.720 ","End":"19:19.120","Text":"which makes it a little bit harder to solve."},{"Start":"19:19.120 ","End":"19:20.695","Text":"With Kirchhoff in general,"},{"Start":"19:20.695 ","End":"19:22.130","Text":"we can sometimes get,"},{"Start":"19:22.130 ","End":"19:23.760","Text":"without any problem,"},{"Start":"19:23.760 ","End":"19:25.920","Text":"6 equations with 6 unknowns."},{"Start":"19:25.920 ","End":"19:29.090","Text":"The mesh current method always will have"},{"Start":"19:29.090 ","End":"19:33.685","Text":"an easier set of equations to solve because they\u0027ll be less of them."},{"Start":"19:33.685 ","End":"19:40.030","Text":"Now you can solve this using Cramer\u0027s method or Cramer\u0027s rule,"},{"Start":"19:40.030 ","End":"19:42.580","Text":"which is what we learned in the previous lesson,"},{"Start":"19:42.580 ","End":"19:45.025","Text":"or by solving with algebra."},{"Start":"19:45.025 ","End":"19:46.690","Text":"If you know how to do this,"},{"Start":"19:46.690 ","End":"19:49.810","Text":"you can click out of this video and move on to the next one."},{"Start":"19:49.810 ","End":"19:51.475","Text":"If you want to stay,"},{"Start":"19:51.475 ","End":"19:57.220","Text":"I\u0027m going to substitute in the values for all of our components in this circuit"},{"Start":"19:57.220 ","End":"20:03.280","Text":"to the exact same values that we had in our Kirchhoff\u0027s law lesson."},{"Start":"20:03.280 ","End":"20:06.685","Text":"Then we\u0027re going to see that we get to the exact same answer."},{"Start":"20:06.685 ","End":"20:11.770","Text":"I\u0027m also going to use Cramer\u0027s rule in order to solve the equation."},{"Start":"20:11.770 ","End":"20:15.830","Text":"If you want to carry on and see how we do that you\u0027re welcome."},{"Start":"20:17.220 ","End":"20:22.000","Text":"Here we can see the values of all of our components"},{"Start":"20:22.000 ","End":"20:26.770","Text":"and now we can just plug them into these equations."},{"Start":"20:26.770 ","End":"20:30.850","Text":"I\u0027m going to rub out everything here to give us a little bit more space."},{"Start":"20:30.850 ","End":"20:33.610","Text":"We can just plug in our values,"},{"Start":"20:33.610 ","End":"20:35.080","Text":"for our first equation,"},{"Start":"20:35.080 ","End":"20:37.140","Text":"we have R_1 plus R_2,"},{"Start":"20:37.140 ","End":"20:40.567","Text":"so that\u0027s 2 Ohms plus 1 Ohm."},{"Start":"20:40.567 ","End":"20:49.405","Text":"So we\u0027ll have 3 Ohms multiplied by our I_1 minus I_2,"},{"Start":"20:49.405 ","End":"20:54.025","Text":"which is 1 Ohm multiplied by our I_2,"},{"Start":"20:54.025 ","End":"20:58.450","Text":"and that is equal to Epsilon_2 minus Epsilon_1."},{"Start":"20:58.450 ","End":"21:01.930","Text":"That\u0027s 2 volts minus 3 volts,"},{"Start":"21:01.930 ","End":"21:04.705","Text":"so that\u0027s negative 1 volt."},{"Start":"21:04.705 ","End":"21:06.310","Text":"Then for our second equation,"},{"Start":"21:06.310 ","End":"21:09.715","Text":"we have negative R_2 multiplied by I_1,"},{"Start":"21:09.715 ","End":"21:16.360","Text":"we have negative 1 Ohm multiplied by I_1 and then we"},{"Start":"21:16.360 ","End":"21:24.190","Text":"have plus R_2 plus R_3 plus R_4 multiplied by I_2."},{"Start":"21:24.190 ","End":"21:29.905","Text":"That\u0027s 1 plus 2 plus 1,"},{"Start":"21:29.905 ","End":"21:39.625","Text":"so that\u0027s 4 Ohms multiplied by I_2 and that is equal to Epsilon 3 minus Epsilon 2."},{"Start":"21:39.625 ","End":"21:42.955","Text":"So that\u0027s 4 volts minus 2 volts,"},{"Start":"21:42.955 ","End":"21:45.385","Text":"that is equal to 2 volts."},{"Start":"21:45.385 ","End":"21:49.720","Text":"Now, we have these 2 equations and we\u0027re going to use Cramer\u0027s rule,"},{"Start":"21:49.720 ","End":"21:53.695","Text":"which means that we place these equations into a matrix format."},{"Start":"21:53.695 ","End":"21:57.020","Text":"Let\u0027s draw this in,"},{"Start":"21:57.020 ","End":"22:03.025","Text":"our matrix is going to be a 2-by-2 matrix because we have 2 equations and 2 unknowns."},{"Start":"22:03.025 ","End":"22:07.310","Text":"Here we have 3 Ohms multiplied by I_1."},{"Start":"22:07.310 ","End":"22:09.535","Text":"We\u0027ll put in a 3 over here,"},{"Start":"22:09.535 ","End":"22:11.995","Text":"then for our I_2 column,"},{"Start":"22:11.995 ","End":"22:16.405","Text":"we have negative 1 Ohm multiplied by I_2."},{"Start":"22:16.405 ","End":"22:18.400","Text":"Now for our second equation,"},{"Start":"22:18.400 ","End":"22:20.005","Text":"so that\u0027s the second row,"},{"Start":"22:20.005 ","End":"22:23.860","Text":"we have negative 1 Ohm multiplied by I_1."},{"Start":"22:23.860 ","End":"22:30.055","Text":"So in the I_1 column we have negative 1 and then we have 4 Ohms multiplied by I_2,"},{"Start":"22:30.055 ","End":"22:33.550","Text":"so that\u0027s 4 in the I_2 column."},{"Start":"22:33.550 ","End":"22:37.495","Text":"Then this is equal to our vector of solutions,"},{"Start":"22:37.495 ","End":"22:40.640","Text":"which is simply equal to negative 1,2."},{"Start":"22:41.040 ","End":"22:45.295","Text":"Now what we want to do is we want to find the determinant,"},{"Start":"22:45.295 ","End":"22:48.700","Text":"let\u0027s call this matrix matrix M. Now what we want"},{"Start":"22:48.700 ","End":"22:52.495","Text":"to do is we want to find the determinant of matrix M,"},{"Start":"22:52.495 ","End":"22:54.460","Text":"that\u0027s going to be 3 times 4,"},{"Start":"22:54.460 ","End":"23:00.625","Text":"which is 12 minus negative 1 times negative 1 that\u0027s positive 1,"},{"Start":"23:00.625 ","End":"23:04.540","Text":"and then negative 1 so that\u0027s 11."},{"Start":"23:04.540 ","End":"23:10.675","Text":"Now what we want to do is we want to find det M_1."},{"Start":"23:10.675 ","End":"23:12.700","Text":"So det M_1,"},{"Start":"23:12.700 ","End":"23:17.215","Text":"we replace our first column with our vector of solutions."},{"Start":"23:17.215 ","End":"23:21.070","Text":"That will be negative 1,2,"},{"Start":"23:21.070 ","End":"23:26.275","Text":"and then we\u0027ll have negative 1 for the second column remains the exact same thing,"},{"Start":"23:26.275 ","End":"23:28.570","Text":"so this is matrix M_1."},{"Start":"23:28.570 ","End":"23:31.525","Text":"That\u0027s going to be equal to negative 1 times 4,"},{"Start":"23:31.525 ","End":"23:35.830","Text":"that\u0027s negative 4 minus 2 times negative 1."},{"Start":"23:35.830 ","End":"23:40.015","Text":"That\u0027s a negative 2 and a negative and a negative so plus 2,"},{"Start":"23:40.015 ","End":"23:44.200","Text":"our det M_1 is equal to negative 2."},{"Start":"23:44.200 ","End":"23:49.345","Text":"Now we want to find our det M_2,"},{"Start":"23:49.345 ","End":"23:50.920","Text":"we don\u0027t have to find our det M_2."},{"Start":"23:50.920 ","End":"23:54.310","Text":"We can simply substitute in our numbers into this equation,"},{"Start":"23:54.310 ","End":"23:56.335","Text":"find out what our I_2 is."},{"Start":"23:56.335 ","End":"23:59.890","Text":"Our det M_1, finds out what our I_1 is equal to and"},{"Start":"23:59.890 ","End":"24:03.790","Text":"our det M_2 will help us find out what our I_2 is equal to."},{"Start":"24:03.790 ","End":"24:06.460","Text":"We don\u0027t have to find our det M_2 because we only have"},{"Start":"24:06.460 ","End":"24:09.205","Text":"2 unknowns so we can just plug in the numbers,"},{"Start":"24:09.205 ","End":"24:11.860","Text":"but just for the sake of seeing how that\u0027s done."},{"Start":"24:11.860 ","End":"24:14.620","Text":"Our det M_2, the first column,"},{"Start":"24:14.620 ","End":"24:17.695","Text":"is going to be the same as the original M matrix,"},{"Start":"24:17.695 ","End":"24:21.640","Text":"and then we replace the second column with our vector of solutions,"},{"Start":"24:21.640 ","End":"24:23.900","Text":"so it\u0027ll be negative 1,2."},{"Start":"24:24.480 ","End":"24:27.580","Text":"This is our M_2 matrix,"},{"Start":"24:27.580 ","End":"24:30.280","Text":"the determinant is 3 times 2,"},{"Start":"24:30.280 ","End":"24:35.965","Text":"which is 6 minus negative 1 times negative 1 minus 1,"},{"Start":"24:35.965 ","End":"24:39.020","Text":"and that is going to be equal to 5."},{"Start":"24:40.080 ","End":"24:46.810","Text":"Now in order to find what our current I_1 is equal to,"},{"Start":"24:46.810 ","End":"24:56.260","Text":"it is equal to det M_1 for current I_1 divided by the determinant of"},{"Start":"24:56.260 ","End":"25:06.310","Text":"our original matrix M. That is simply going to be equal to negative 2 divided by 11 amps."},{"Start":"25:06.310 ","End":"25:09.520","Text":"Now our I_2 current,"},{"Start":"25:09.520 ","End":"25:15.190","Text":"we could plug in what our I_1 is and simply solve or we can just use this,"},{"Start":"25:15.190 ","End":"25:16.840","Text":"which is also very easy."},{"Start":"25:16.840 ","End":"25:24.985","Text":"It\u0027s going to be equal to your det M_2 divided by det of the original matrix."},{"Start":"25:24.985 ","End":"25:29.890","Text":"Det M_2 is equal to 5 divided by 11,"},{"Start":"25:29.890 ","End":"25:34.640","Text":"which is the determinant of the original matrix amps."},{"Start":"25:35.430 ","End":"25:40.360","Text":"These are my current I_1 and I_2 and of course,"},{"Start":"25:40.360 ","End":"25:42.655","Text":"if I want to find my current I_3,"},{"Start":"25:42.655 ","End":"25:48.620","Text":"which is the current flowing through the intersection between my 2 meshes,"},{"Start":"25:49.560 ","End":"25:54.835","Text":"I\u0027ll have to do either I_1 minus I_2;"},{"Start":"25:54.835 ","End":"25:57.115","Text":"if we\u0027re going in this direction,"},{"Start":"25:57.115 ","End":"25:59.830","Text":"or I_2 minus I_1,"},{"Start":"25:59.830 ","End":"26:03.235","Text":"if we\u0027re going in this direction."},{"Start":"26:03.235 ","End":"26:06.530","Text":"That\u0027s the end of this lesson."}],"ID":22302},{"Watched":false,"Name":"Exercise 6","Duration":"17m 24s","ChapterTopicVideoID":21486,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"Hello. In this lesson,"},{"Start":"00:02.100 ","End":"00:04.635","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.635 ","End":"00:07.890","Text":"Question Number 1 is to calculate the current through"},{"Start":"00:07.890 ","End":"00:13.290","Text":"each resistor and question Number 2 is to calculate the voltage V_AB,"},{"Start":"00:13.290 ","End":"00:19.180","Text":"where A is this point over here and B is this point over here."},{"Start":"00:19.340 ","End":"00:26.625","Text":"We\u0027ll start with Question Number 1 and we\u0027re going to be using the mesh current method."},{"Start":"00:26.625 ","End":"00:33.690","Text":"The first thing I\u0027ll do is I\u0027ll say that we have this loop over"},{"Start":"00:33.690 ","End":"00:42.150","Text":"here and it\u0027s in the clockwise direction and we have a current flowing through here, I_1."},{"Start":"00:42.650 ","End":"00:45.290","Text":"Then over here, again,"},{"Start":"00:45.290 ","End":"00:47.300","Text":"in the clockwise direction and we have"},{"Start":"00:47.300 ","End":"00:53.080","Text":"current I_2 and over here as well in the clockwise direction,"},{"Start":"00:53.080 ","End":"00:55.800","Text":"we have current I_3."},{"Start":"00:56.270 ","End":"01:01.845","Text":"Here, this is meant to be 4 volts, not 4 ohms."},{"Start":"01:01.845 ","End":"01:10.155","Text":"First of all, the total resistance in this loop is 5 ohms plus 10 ohms."},{"Start":"01:10.155 ","End":"01:16.170","Text":"We have 15 ohms multiplied by I_1."},{"Start":"01:16.170 ","End":"01:18.165","Text":"This is the voltage,"},{"Start":"01:18.165 ","End":"01:20.475","Text":"V is equal to I_1."},{"Start":"01:20.475 ","End":"01:27.030","Text":"Then we\u0027re going to subtract what we have,"},{"Start":"01:27.030 ","End":"01:30.740","Text":"if we have any resistors in the other loops."},{"Start":"01:30.740 ","End":"01:32.480","Text":"We can see that in I_2,"},{"Start":"01:32.480 ","End":"01:39.590","Text":"we don\u0027t have any resistor so we can say negative 0 multiplied by I_2 and minus."},{"Start":"01:39.590 ","End":"01:41.945","Text":"So we can see we have over here"},{"Start":"01:41.945 ","End":"01:48.900","Text":"this 10 ohm resistor in both the I_1 loop and in the I_3 loop."},{"Start":"01:48.900 ","End":"01:52.215","Text":"Because of the direction that the loops are going,"},{"Start":"01:52.215 ","End":"02:01.488","Text":"so we have to subtract 10 ohms multiplied by I_3."},{"Start":"02:01.488 ","End":"02:03.270","Text":"This is of course,"},{"Start":"02:03.270 ","End":"02:05.430","Text":"equal to the voltage."},{"Start":"02:05.430 ","End":"02:13.115","Text":"Of course, we know that here we\u0027re going from the positive to the negative."},{"Start":"02:13.115 ","End":"02:19.205","Text":"Here we have plus 2 because we\u0027re going in the positive direction."},{"Start":"02:19.205 ","End":"02:22.320","Text":"Here we\u0027re going from the long to the short so we have minus"},{"Start":"02:22.320 ","End":"02:26.460","Text":"4 and again from the long to the short so minus another 4,"},{"Start":"02:26.460 ","End":"02:32.895","Text":"so if 2 minus 4 minus 4 is negative 6 volts."},{"Start":"02:32.895 ","End":"02:39.325","Text":"Sorry, I meant this is the negative and this is the positive."},{"Start":"02:39.325 ","End":"02:43.265","Text":"Then here, this is the positive and the negative."},{"Start":"02:43.265 ","End":"02:47.750","Text":"Here we\u0027re adding 2 volts and here we\u0027re subtracting"},{"Start":"02:47.750 ","End":"02:53.370","Text":"4 and then here again, subtracting another 4."},{"Start":"02:53.780 ","End":"02:58.005","Text":"Now we\u0027re writing the equation for our I_2 loop."},{"Start":"02:58.005 ","End":"03:03.000","Text":"Our total resistance is 6 plus 2 which is 8."},{"Start":"03:03.000 ","End":"03:09.127","Text":"We have 8 ohms multiplied by I_2."},{"Start":"03:09.127 ","End":"03:14.300","Text":"Then we have no resistors that are overlapping with our I_1 loop,"},{"Start":"03:14.300 ","End":"03:19.239","Text":"so we have 0 multiplied by I_1 plus."},{"Start":"03:19.239 ","End":"03:23.300","Text":"Then I_2 ohm resistor is overlapping with our I_3 loop,"},{"Start":"03:23.300 ","End":"03:31.055","Text":"so we have negative 2 ohms multiplied by I_3 and this is equal 2."},{"Start":"03:31.055 ","End":"03:32.990","Text":"As we go this way,"},{"Start":"03:32.990 ","End":"03:37.295","Text":"so of course, this is positive and this is negative."},{"Start":"03:37.295 ","End":"03:45.845","Text":"We have negative 3 volts and then we\u0027re adding 4 volts over here,"},{"Start":"03:45.845 ","End":"03:48.620","Text":"because we\u0027re going from negative to positive,"},{"Start":"03:48.620 ","End":"03:53.040","Text":"so this is equal to 1 volt."},{"Start":"03:53.210 ","End":"03:55.890","Text":"Now for the I_3 loop."},{"Start":"03:55.890 ","End":"04:00.020","Text":"We have a total resistance of 10 plus 10 plus 2,"},{"Start":"04:00.020 ","End":"04:06.075","Text":"which is 22 ohms multiplied by I_3."},{"Start":"04:06.075 ","End":"04:11.490","Text":"Then we have a 10 ohm resistor overlapping with our I_1 loop"},{"Start":"04:11.490 ","End":"04:17.190","Text":"so we have negative 10 multiplied by I_1."},{"Start":"04:17.190 ","End":"04:22.020","Text":"Then we have our 2 ohm resistor overlapping with our I_2 loop"},{"Start":"04:22.020 ","End":"04:27.660","Text":"so negative 2 ohms multiplied by I_2."},{"Start":"04:27.660 ","End":"04:29.360","Text":"This is equal 2,"},{"Start":"04:29.360 ","End":"04:32.240","Text":"so we only have this battery over"},{"Start":"04:32.240 ","End":"04:35.705","Text":"here and we\u0027re going from the positive to the negative,"},{"Start":"04:35.705 ","End":"04:40.140","Text":"so we have that this is equal to negative 2 volts."},{"Start":"04:40.270 ","End":"04:44.135","Text":"Now we have our 3 equations and what are we\u0027re going to do,"},{"Start":"04:44.135 ","End":"04:47.037","Text":"is we\u0027re going to write this in matrix form,"},{"Start":"04:47.037 ","End":"04:48.905","Text":"so what does that mean?"},{"Start":"04:48.905 ","End":"04:55.055","Text":"We\u0027ll do 3 by 3 matrix and we just write it without the I\u0027s."},{"Start":"04:55.055 ","End":"04:57.470","Text":"Here we have 15,"},{"Start":"04:57.470 ","End":"05:01.835","Text":"0, and then negative 10."},{"Start":"05:01.835 ","End":"05:05.195","Text":"Then we have 0, 8,"},{"Start":"05:05.195 ","End":"05:09.705","Text":"and negative 2 and then we have negative 10,"},{"Start":"05:09.705 ","End":"05:13.870","Text":"negative 2 and then 22."},{"Start":"05:13.870 ","End":"05:19.490","Text":"Then this is simply equal to negative 6,"},{"Start":"05:19.490 ","End":"05:22.765","Text":"1, negative 2,"},{"Start":"05:22.765 ","End":"05:26.340","Text":"which is just the results over here."},{"Start":"05:26.860 ","End":"05:30.865","Text":"Now, all that\u0027s left to do is to solve the matrix."},{"Start":"05:30.865 ","End":"05:33.110","Text":"What we\u0027re going to do is we\u0027re going to be using"},{"Start":"05:33.110 ","End":"05:37.660","Text":"Cramer\u0027s rule like we saw 1 or 2 lessons ago."},{"Start":"05:37.660 ","End":"05:39.770","Text":"The first thing that we\u0027re going to do is we\u0027re going to"},{"Start":"05:39.770 ","End":"05:43.045","Text":"find the determinant of the matrix."},{"Start":"05:43.045 ","End":"05:47.720","Text":"I\u0027m not going to do the calculation for finding the determinant,"},{"Start":"05:47.720 ","End":"05:56.000","Text":"but the determinant of this matrix works out to be 1,780."},{"Start":"05:56.000 ","End":"05:57.710","Text":"Now what I want to do is,"},{"Start":"05:57.710 ","End":"06:01.200","Text":"I want to find I_1, my current I_1."},{"Start":"06:01.200 ","End":"06:04.610","Text":"What I\u0027m going to do, is I\u0027m going to switch out"},{"Start":"06:04.610 ","End":"06:12.303","Text":"this first column in the matrix with the solutions column."},{"Start":"06:12.303 ","End":"06:15.525","Text":"Then we\u0027ll find the determinant of that."},{"Start":"06:15.525 ","End":"06:22.730","Text":"Let\u0027s call this determinant for I_1 and this is equal 2,"},{"Start":"06:22.730 ","End":"06:26.615","Text":"so now we\u0027re just switching out the first column."},{"Start":"06:26.615 ","End":"06:29.528","Text":"We\u0027re going to have negative 6, 1,"},{"Start":"06:29.528 ","End":"06:32.640","Text":"2, that\u0027s the solution."},{"Start":"06:32.640 ","End":"06:35.690","Text":"Then we have the 2nd column which has 0, 8,"},{"Start":"06:35.690 ","End":"06:38.920","Text":"negative 2 and then negative 10,"},{"Start":"06:38.920 ","End":"06:41.988","Text":"negative 2, and 22."},{"Start":"06:41.988 ","End":"06:44.285","Text":"The determinant of this,"},{"Start":"06:44.285 ","End":"06:46.460","Text":"again, I\u0027ll just do it on the calculator,"},{"Start":"06:46.460 ","End":"06:52.080","Text":"is equal to negative 1,172."},{"Start":"06:53.030 ","End":"06:59.450","Text":"From here, I can say that my I_1 is equal to"},{"Start":"06:59.450 ","End":"07:06.098","Text":"my determinant for I_1 divided by my original determinant."},{"Start":"07:06.098 ","End":"07:13.280","Text":"That is simply going to be equal to negative 1,172 divided"},{"Start":"07:13.280 ","End":"07:22.650","Text":"by 1,780 and this is equal to negative 0.6584."},{"Start":"07:25.490 ","End":"07:28.640","Text":"Of course, the units are in amps,"},{"Start":"07:28.640 ","End":"07:32.780","Text":"so amperes and the minus sign just means that"},{"Start":"07:32.780 ","End":"07:34.910","Text":"the direction that we chose over"},{"Start":"07:34.910 ","End":"07:38.768","Text":"here is wrong and we just have to do the opposite direction."},{"Start":"07:38.768 ","End":"07:42.930","Text":"So I_1 is flowing in the anti-clockwise direction."},{"Start":"07:43.270 ","End":"07:48.070","Text":"Now, let\u0027s calculate I_2."},{"Start":"07:48.070 ","End":"07:50.150","Text":"What we\u0027re going to do is again,"},{"Start":"07:50.150 ","End":"07:53.495","Text":"Cramer\u0027s rule where we switch out this time the second column,"},{"Start":"07:53.495 ","End":"07:57.080","Text":"with the solutions column and work out the determinant."},{"Start":"07:57.080 ","End":"08:00.845","Text":"We\u0027re working out the determinant for I_2."},{"Start":"08:00.845 ","End":"08:02.555","Text":"This is equal 2,"},{"Start":"08:02.555 ","End":"08:07.505","Text":"so the 1st column is 15, 0, negative 10."},{"Start":"08:07.505 ","End":"08:09.560","Text":"The 2nd column is now the solution,"},{"Start":"08:09.560 ","End":"08:11.405","Text":"so negative 6, 1,"},{"Start":"08:11.405 ","End":"08:14.885","Text":"negative 2 and the 3rd column is negative 10,"},{"Start":"08:14.885 ","End":"08:21.360","Text":"negative 2, 22 and this is equal to 50."},{"Start":"08:21.360 ","End":"08:30.125","Text":"Therefore, I_2 is equal to the determinant for I_2 divided by the original determinant,"},{"Start":"08:30.125 ","End":"08:32.270","Text":"which is simply equal to 50,"},{"Start":"08:32.270 ","End":"08:36.230","Text":"divided by 1,780,"},{"Start":"08:36.230 ","End":"08:43.338","Text":"which is equal to 0.0281 amps."},{"Start":"08:43.338 ","End":"08:49.450","Text":"Here, we can see that the direction that we chose for I_2 in our diagram,"},{"Start":"08:49.450 ","End":"08:53.485","Text":"so the clockwise direction is correct."},{"Start":"08:53.485 ","End":"09:01.690","Text":"Then the determinant for I_3 where we switch out the third column is equal, too."},{"Start":"09:01.690 ","End":"09:03.415","Text":"So we have 15, 0,"},{"Start":"09:03.415 ","End":"09:05.780","Text":"negative 10, 0, 8,"},{"Start":"09:05.910 ","End":"09:09.820","Text":"negative 2, and then we switch out the third column with the solution,"},{"Start":"09:09.820 ","End":"09:11.180","Text":"so negative 6,"},{"Start":"09:11.180 ","End":"09:13.450","Text":"1, negative 2."},{"Start":"09:13.450 ","End":"09:17.605","Text":"Then we work out the determinant,"},{"Start":"09:17.605 ","End":"09:20.125","Text":"and then we do this,"},{"Start":"09:20.125 ","End":"09:25.910","Text":"and what we get is that I_3=0.1449amps."},{"Start":"09:28.470 ","End":"09:31.795","Text":"We repeat all these steps."},{"Start":"09:31.795 ","End":"09:41.480","Text":"Again, the positive coefficient means that this clockwise direction for I_3 is correct."},{"Start":"09:41.610 ","End":"09:46.780","Text":"Here I wrote out the currents that we calculated and now let\u0027s"},{"Start":"09:46.780 ","End":"09:51.760","Text":"say I wanted to calculate the current going through the 5-ohm resistor."},{"Start":"09:51.760 ","End":"09:54.715","Text":"That would just be equal to I_1."},{"Start":"09:54.715 ","End":"09:57.580","Text":"Because that\u0027s the only current flowing through"},{"Start":"09:57.580 ","End":"10:02.065","Text":"the 5-ohm resistor but let\u0027s say this 10-ohm resistor,"},{"Start":"10:02.065 ","End":"10:04.405","Text":"so let\u0027s take it over here."},{"Start":"10:04.405 ","End":"10:09.085","Text":"Here it overlaps with the I_1 and the I_3."},{"Start":"10:09.085 ","End":"10:14.005","Text":"Let\u0027s say that this is the positive direction of travel."},{"Start":"10:14.005 ","End":"10:20.980","Text":"What we\u0027re going to have is we\u0027re going to have that the current over here,"},{"Start":"10:20.980 ","End":"10:28.960","Text":"I in this resistor is going to be equal to I_3 minus I_1,"},{"Start":"10:28.960 ","End":"10:38.620","Text":"which is equal to 0.1449 amps minus I_1,"},{"Start":"10:38.620 ","End":"10:44.725","Text":"which is equal to negative 0.6584 amps."},{"Start":"10:44.725 ","End":"10:52.690","Text":"This is just going to be equal to 0.8033 and of course,"},{"Start":"10:52.690 ","End":"10:59.380","Text":"in amps and let\u0027s say this 2-ohm resistor over here."},{"Start":"10:59.380 ","End":"11:00.895","Text":"So the current over here,"},{"Start":"11:00.895 ","End":"11:03.115","Text":"if this is again the positive direction,"},{"Start":"11:03.115 ","End":"11:10.015","Text":"so we\u0027ll have I_3 minus I_2,"},{"Start":"11:10.015 ","End":"11:19.315","Text":"which is equal to 0.1449 minus 0.0281,"},{"Start":"11:19.315 ","End":"11:23.423","Text":"and that is simply going to be equal too."},{"Start":"11:23.423 ","End":"11:29.814","Text":"Then the current through the 5-ohm resistor if this is the positive direction."},{"Start":"11:29.814 ","End":"11:34.195","Text":"So here, the current will just be I_1 and then over here,"},{"Start":"11:34.195 ","End":"11:38.110","Text":"if this is the positive direction in the clockwise direction,"},{"Start":"11:38.110 ","End":"11:42.955","Text":"then the current through the 6-ohm resistor will just be I_2."},{"Start":"11:42.955 ","End":"11:48.730","Text":"If this is the positive direction in the direction of the I_3,"},{"Start":"11:48.730 ","End":"11:54.890","Text":"so this, the current through the 10-ohm resistor will just be equal to I_3."},{"Start":"11:55.980 ","End":"11:59.530","Text":"Now we\u0027ve answered question Number 1 with all"},{"Start":"11:59.530 ","End":"12:02.335","Text":"of the currents through the resistance and now"},{"Start":"12:02.335 ","End":"12:08.960","Text":"let\u0027s go on to question Number 2 to calculate the voltage between points A and B."},{"Start":"12:09.090 ","End":"12:15.985","Text":"The voltage V_AB is simply equal to the potential at the end."},{"Start":"12:15.985 ","End":"12:21.130","Text":"The potential at point B minus the potential at the start,"},{"Start":"12:21.130 ","End":"12:24.920","Text":"which is V at A."},{"Start":"12:25.560 ","End":"12:31.090","Text":"The first thing that I\u0027m going to do is I\u0027m going to see what the voltage"},{"Start":"12:31.090 ","End":"12:36.790","Text":"is at A. I start at A and I find some route,"},{"Start":"12:36.790 ","End":"12:42.230","Text":"any route that I want that goes from A and ends at B."},{"Start":"12:42.900 ","End":"12:46.015","Text":"Let\u0027s imagine that from point A,"},{"Start":"12:46.015 ","End":"12:51.175","Text":"I go all the way over here and I pass my 5-ohm resistor."},{"Start":"12:51.175 ","End":"12:54.939","Text":"If this is the direction of the current,"},{"Start":"12:54.939 ","End":"12:57.114","Text":"so as I pass my resistor,"},{"Start":"12:57.114 ","End":"13:00.550","Text":"my voltage is going to go down."},{"Start":"13:00.550 ","End":"13:07.580","Text":"I have a higher voltage before the resistor and a lower voltage after the resistor."},{"Start":"13:10.080 ","End":"13:14.860","Text":"That means that I\u0027m subtracting so I have"},{"Start":"13:14.860 ","End":"13:21.317","Text":"I_1 multiplied by the resistance which is 5 ohms."},{"Start":"13:21.317 ","End":"13:28.300","Text":"Then I pass this battery over here and I\u0027m going from positive to negative."},{"Start":"13:28.300 ","End":"13:33.080","Text":"Again, I\u0027m subtracting 4 volts."},{"Start":"13:33.510 ","End":"13:37.570","Text":"Then I move on to this section over here."},{"Start":"13:37.570 ","End":"13:42.625","Text":"Assuming that this is the positive direction, again,"},{"Start":"13:42.625 ","End":"13:46.000","Text":"I\u0027m going from a higher voltage before"},{"Start":"13:46.000 ","End":"13:50.275","Text":"the resistor to a lower voltage after the resistor."},{"Start":"13:50.275 ","End":"13:53.060","Text":"I\u0027m subtracting, therefore,"},{"Start":"13:53.060 ","End":"13:57.600","Text":"I_2 multiplied by 6 ohm,"},{"Start":"13:57.600 ","End":"14:00.975","Text":"and then I reach my point B."},{"Start":"14:00.975 ","End":"14:09.460","Text":"All of this has to be equal to my voltage at point B. I want V_AB,"},{"Start":"14:09.460 ","End":"14:12.760","Text":"which is equal to V_B minus V_A."},{"Start":"14:12.760 ","End":"14:17.020","Text":"All I\u0027m going to do is I\u0027m going to move by V_A to the other side."},{"Start":"14:17.020 ","End":"14:20.120","Text":"I\u0027m going to subtract V_A from both sides,"},{"Start":"14:20.120 ","End":"14:24.700","Text":"so then we\u0027ll have is V_B minus V_A,"},{"Start":"14:24.700 ","End":"14:29.140","Text":"which is simply equal to negative I_1 multiplied by"},{"Start":"14:29.140 ","End":"14:36.650","Text":"5 ohms minus 4 volts minus I_2 multiplied by 6 ohms."},{"Start":"14:37.800 ","End":"14:40.870","Text":"Let\u0027s write this out."},{"Start":"14:40.870 ","End":"14:45.718","Text":"This is equal to, so negative I_1."},{"Start":"14:45.718 ","End":"14:49.240","Text":"So that\u0027s just a negative and a negative is a positive."},{"Start":"14:49.240 ","End":"14:55.285","Text":"I have 0.6584 multiplied by 5"},{"Start":"14:55.285 ","End":"15:03.925","Text":"minus 4 minus I_2,"},{"Start":"15:03.925 ","End":"15:10.435","Text":"which is 0.0281 multiplied by 6."},{"Start":"15:10.435 ","End":"15:20.500","Text":"This is simply equal to negative 0.8766 volts,"},{"Start":"15:20.500 ","End":"15:23.630","Text":"and this is of course, equal to V_AB."},{"Start":"15:24.090 ","End":"15:29.560","Text":"Now notice that V_AB is negative because the voltage or"},{"Start":"15:29.560 ","End":"15:36.290","Text":"the potential at A is higher than the voltage or the potential at B."},{"Start":"15:36.480 ","End":"15:41.290","Text":"Just a note, I could have chosen a different route to get from A to"},{"Start":"15:41.290 ","End":"15:46.390","Text":"B so I could have gone down here and then like so."},{"Start":"15:46.390 ","End":"15:48.700","Text":"Notice if you do do that,"},{"Start":"15:48.700 ","End":"15:51.505","Text":"then you are going through this 10-ohm resistor,"},{"Start":"15:51.505 ","End":"15:57.070","Text":"which of course is overlapping with the current I_1 and the current I_3."},{"Start":"15:57.070 ","End":"15:59.410","Text":"You would have to say that the current,"},{"Start":"15:59.410 ","End":"16:03.295","Text":"when you multiply the resistance by the current,"},{"Start":"16:03.295 ","End":"16:08.545","Text":"it would be equal to I_3 minus I_1,"},{"Start":"16:08.545 ","End":"16:11.110","Text":"would it be equal to this current over here,"},{"Start":"16:11.110 ","End":"16:13.820","Text":"you would have to multiply that by 10."},{"Start":"16:14.280 ","End":"16:16.720","Text":"Let\u0027s imagine that I do that,"},{"Start":"16:16.720 ","End":"16:22.750","Text":"so I start with V_A and then I go down this battery over here."},{"Start":"16:22.750 ","End":"16:29.440","Text":"So negative 2 volts and then I go across this resistor over here."},{"Start":"16:29.440 ","End":"16:34.210","Text":"What I\u0027m going to have is negative the current flowing through this resistor,"},{"Start":"16:34.210 ","End":"16:35.755","Text":"which is this over here,"},{"Start":"16:35.755 ","End":"16:40.060","Text":"which is just equal to I_3 minus I_1."},{"Start":"16:40.060 ","End":"16:43.795","Text":"This is of course multiplied by 10 ohms."},{"Start":"16:43.795 ","End":"16:46.900","Text":"Then I go up this battery over here,"},{"Start":"16:46.900 ","End":"16:49.630","Text":"so I add on 4 volts,"},{"Start":"16:49.630 ","End":"16:52.128","Text":"because I\u0027m going from negative to positive."},{"Start":"16:52.128 ","End":"16:54.970","Text":"Then I go down this resistor again,"},{"Start":"16:54.970 ","End":"17:01.345","Text":"so negative I_2 multiplied by 6 ohms."},{"Start":"17:01.345 ","End":"17:06.590","Text":"This is of course equal to V_B because then I get to point B."},{"Start":"17:06.960 ","End":"17:11.500","Text":"If you solve this and you subtract V_A from both sides,"},{"Start":"17:11.500 ","End":"17:13.810","Text":"and you plug in all the numbers,"},{"Start":"17:13.810 ","End":"17:20.450","Text":"you\u0027ll see that V_AB will be the same answer as you\u0027ve got choosing this route."},{"Start":"17:20.450 ","End":"17:24.040","Text":"That\u0027s the end of this lesson."}],"ID":22303},{"Watched":false,"Name":"Terminal Voltage and Electromotive Force","Duration":"8m 5s","ChapterTopicVideoID":21487,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:08.460","Text":"we\u0027re going to be learning about non ideal voltage sources and the electromotive force."},{"Start":"00:08.460 ","End":"00:14.550","Text":"Up until now, we\u0027ve looked at electric circuits where we have"},{"Start":"00:14.550 ","End":"00:20.955","Text":"some battery and we looked at this as some ideal voltage source."},{"Start":"00:20.955 ","End":"00:24.975","Text":"Well, it\u0027s just providing us with some voltage."},{"Start":"00:24.975 ","End":"00:31.005","Text":"Then we have a circuit and we have some resistors."},{"Start":"00:31.005 ","End":"00:36.030","Text":"Let\u0027s say these are resistors and that was it."},{"Start":"00:36.030 ","End":"00:38.750","Text":"Then, of course, we could calculate"},{"Start":"00:38.750 ","End":"00:46.170","Text":"the total resistance of these 2 resistors and just call this R total."},{"Start":"00:46.730 ","End":"00:53.705","Text":"From now, we are going to consider these voltage sources as non-ideal."},{"Start":"00:53.705 ","End":"00:56.540","Text":"There is actually sadly no such thing as"},{"Start":"00:56.540 ","End":"01:00.845","Text":"an ideal voltage source that just provides some voltage."},{"Start":"01:00.845 ","End":"01:04.385","Text":"Every voltage source, let\u0027s say batteries,"},{"Start":"01:04.385 ","End":"01:08.820","Text":"have some internal resistance."},{"Start":"01:09.830 ","End":"01:13.725","Text":"Let\u0027s draw out our battery."},{"Start":"01:13.725 ","End":"01:18.655","Text":"We have some battery like so."},{"Start":"01:18.655 ","End":"01:24.170","Text":"It\u0027s very difficult to just measure out the resistance inside"},{"Start":"01:24.170 ","End":"01:26.480","Text":"the battery because there\u0027s lots and lots of"},{"Start":"01:26.480 ","End":"01:30.150","Text":"different components within the battery itself,"},{"Start":"01:30.150 ","End":"01:36.550","Text":"and I don\u0027t know how to calculate the total resistance."},{"Start":"01:36.550 ","End":"01:40.070","Text":"This is how we do it. Inside the battery,"},{"Start":"01:40.070 ","End":"01:44.750","Text":"we imagine that we have an ideal voltage source, again,"},{"Start":"01:44.750 ","End":"01:46.030","Text":"we label it with Epsilon,"},{"Start":"01:46.030 ","End":"01:47.690","Text":"so this is an ideal voltage source,"},{"Start":"01:47.690 ","End":"01:53.420","Text":"it just provides voltage and itself doesn\u0027t have resistance."},{"Start":"01:53.420 ","End":"01:56.555","Text":"Then connected to this,"},{"Start":"01:56.555 ","End":"02:00.305","Text":"we have some resistor."},{"Start":"02:00.305 ","End":"02:06.725","Text":"This is our internal resistance because the battery itself is a non-ideal voltage source."},{"Start":"02:06.725 ","End":"02:12.770","Text":"The resistance inside a non-ideal voltage source is written via"},{"Start":"02:12.770 ","End":"02:19.195","Text":"this r. Then the battery is connected to the rest of the circuit."},{"Start":"02:19.195 ","End":"02:23.314","Text":"From now on, whenever we\u0027re told that we have a battery,"},{"Start":"02:23.314 ","End":"02:28.029","Text":"so we know the battery is a non-ideal voltage source,"},{"Start":"02:28.029 ","End":"02:32.000","Text":"and so we look at a non-ideal voltage source as being made up of"},{"Start":"02:32.000 ","End":"02:37.090","Text":"an ideal voltage source and a small resistor."},{"Start":"02:37.090 ","End":"02:40.490","Text":"All voltage sources are non-ideal,"},{"Start":"02:40.490 ","End":"02:42.515","Text":"they have an internal resistance."},{"Start":"02:42.515 ","End":"02:44.854","Text":"From now on, we will regard"},{"Start":"02:44.854 ","End":"02:51.580","Text":"all non-ideal voltage sources as being made up of a theoretical ideal voltage source,"},{"Start":"02:51.580 ","End":"02:56.840","Text":"connected in series to a small internal resistor."},{"Start":"02:56.840 ","End":"02:59.795","Text":"Now, we\u0027re going to learn the new term,"},{"Start":"02:59.795 ","End":"03:02.330","Text":"electromotive force or EMF."},{"Start":"03:02.330 ","End":"03:04.400","Text":"But in order to understand this,"},{"Start":"03:04.400 ","End":"03:08.225","Text":"we\u0027re going to speak about the terminal voltage."},{"Start":"03:08.225 ","End":"03:10.352","Text":"What is the terminal voltage?"},{"Start":"03:10.352 ","End":"03:17.135","Text":"It is the voltage or the potential difference across the 2 terminals of a battery."},{"Start":"03:17.135 ","End":"03:24.460","Text":"If here we have our battery and we were to measure the voltage across it,"},{"Start":"03:24.460 ","End":"03:28.820","Text":"the voltage that we would measure across it is the terminal voltage."},{"Start":"03:28.820 ","End":"03:32.165","Text":"Why is the terminal voltage interesting to us?"},{"Start":"03:32.165 ","End":"03:34.040","Text":"Because this is 1,"},{"Start":"03:34.040 ","End":"03:36.515","Text":"the voltage measured across the battery,"},{"Start":"03:36.515 ","End":"03:41.525","Text":"which means that the terminal voltage is the voltage of the circuit."},{"Start":"03:41.525 ","End":"03:48.380","Text":"If I were to connect this circuit to some resistor,"},{"Start":"03:48.380 ","End":"03:55.535","Text":"so the voltage that will go through this resistor is the terminal voltage."},{"Start":"03:55.535 ","End":"03:59.119","Text":"This is calculated by taking into account"},{"Start":"03:59.119 ","End":"04:06.720","Text":"the ideal voltage source and the internal resistance of the battery."},{"Start":"04:06.850 ","End":"04:16.910","Text":"This ideal voltage source is measured or it gives out EMF electromotive force."},{"Start":"04:16.910 ","End":"04:22.240","Text":"You can remember electromotive force is EMF, that\u0027s the abbreviation."},{"Start":"04:22.240 ","End":"04:26.495","Text":"EMF is denoted by the Greek letter Epsilon,"},{"Start":"04:26.495 ","End":"04:31.865","Text":"and it is not to be confused with force in Newtons."},{"Start":"04:31.865 ","End":"04:36.515","Text":"Electromotive force is measured in volts,"},{"Start":"04:36.515 ","End":"04:37.925","Text":"so it\u0027s a voltage,"},{"Start":"04:37.925 ","End":"04:39.020","Text":"it\u0027s not a force."},{"Start":"04:39.020 ","End":"04:42.325","Text":"The name can be a little bit confusing."},{"Start":"04:42.325 ","End":"04:52.180","Text":"EMF is the potential difference or the voltage of a source when no current is flowing."},{"Start":"04:52.180 ","End":"04:58.610","Text":"If I were to take this battery and not connect it to anything,"},{"Start":"04:58.610 ","End":"05:00.733","Text":"leave it as an open circuit,"},{"Start":"05:00.733 ","End":"05:06.350","Text":"and just measure the voltage across the battery when it\u0027s not connected to anything."},{"Start":"05:06.350 ","End":"05:10.670","Text":"Number 1, I\u0027ll be measuring the terminal voltage that we already know,"},{"Start":"05:10.670 ","End":"05:16.040","Text":"and in this case the terminal voltage will be equal to the EMF."},{"Start":"05:16.040 ","End":"05:21.485","Text":"However, once I do connect this to some circuit,"},{"Start":"05:21.485 ","End":"05:23.510","Text":"so I close the circuit."},{"Start":"05:23.510 ","End":"05:26.630","Text":"When I now measure the terminal voltage,"},{"Start":"05:26.630 ","End":"05:30.635","Text":"I\u0027ll see that the terminal voltage isn\u0027t equal to the EMF,"},{"Start":"05:30.635 ","End":"05:32.885","Text":"because now when I have a closed circuit,"},{"Start":"05:32.885 ","End":"05:37.290","Text":"a current is flowing through the battery."},{"Start":"05:37.290 ","End":"05:41.060","Text":"Therefore, we\u0027re going to get a voltage drop over"},{"Start":"05:41.060 ","End":"05:45.050","Text":"the internal resistance inside the battery."},{"Start":"05:45.050 ","End":"05:51.750","Text":"Therefore, the terminal voltage will be less than the EMF."},{"Start":"05:52.850 ","End":"05:55.880","Text":"EMF is measured in volts,"},{"Start":"05:55.880 ","End":"05:57.890","Text":"and it is not a force."},{"Start":"05:57.890 ","End":"06:01.940","Text":"Don\u0027t get confused with the fact that it\u0027s called the electromotive force."},{"Start":"06:01.940 ","End":"06:07.685","Text":"It is the voltage of a source when no current is flowing through,"},{"Start":"06:07.685 ","End":"06:10.295","Text":"and it is, if you want to look at it,"},{"Start":"06:10.295 ","End":"06:12.940","Text":"it is the ideal voltage source."},{"Start":"06:12.940 ","End":"06:15.650","Text":"We said that if we have a source, a battery,"},{"Start":"06:15.650 ","End":"06:19.955","Text":"we know that the battery is a non-ideal voltage source."},{"Start":"06:19.955 ","End":"06:23.180","Text":"Inside we have this Epsilon over here, the EMF,"},{"Start":"06:23.180 ","End":"06:29.670","Text":"which represents the ideal voltage source plus the internal resistance."},{"Start":"06:29.670 ","End":"06:36.780","Text":"Now, let\u0027s give an equation which links our terminal voltage and our EMF."},{"Start":"06:36.780 ","End":"06:40.310","Text":"Our terminal voltage, which is the voltage across"},{"Start":"06:40.310 ","End":"06:45.090","Text":"the battery and the voltage that will go all through the circuit."},{"Start":"06:45.140 ","End":"06:48.000","Text":"Let\u0027s call this V_T,"},{"Start":"06:48.000 ","End":"06:52.365","Text":"V terminal or V term to make it clearer."},{"Start":"06:52.365 ","End":"06:57.550","Text":"This is equal to the maximum voltage that we can have here,"},{"Start":"06:57.550 ","End":"07:00.430","Text":"which is equal to the EMF."},{"Start":"07:00.430 ","End":"07:04.630","Text":"This is the EMF"},{"Start":"07:04.630 ","End":"07:11.590","Text":"minus the voltage drop that we get from this internal resistance over here."},{"Start":"07:11.590 ","End":"07:13.360","Text":"What is the voltage drop?"},{"Start":"07:13.360 ","End":"07:15.295","Text":"V is equal to IR."},{"Start":"07:15.295 ","End":"07:17.545","Text":"It\u0027s minus I,"},{"Start":"07:17.545 ","End":"07:20.380","Text":"the current, multiplied by the resistance,"},{"Start":"07:20.380 ","End":"07:27.310","Text":"r. This is the equation for our terminal voltage."},{"Start":"07:27.310 ","End":"07:33.080","Text":"As we can see, it is dependent on current."},{"Start":"07:33.080 ","End":"07:37.010","Text":"Now, the problem with this is that our current is, of course,"},{"Start":"07:37.010 ","End":"07:41.600","Text":"dependent on the circuit that we will connect our battery to."},{"Start":"07:41.600 ","End":"07:46.700","Text":"It depends on the amount of resistance that we have in the circuit."},{"Start":"07:46.700 ","End":"07:52.750","Text":"This means that we cannot predict our terminal voltage in advance."},{"Start":"07:52.750 ","End":"07:55.115","Text":"This is still something that we can work with,"},{"Start":"07:55.115 ","End":"07:57.860","Text":"but it\u0027s just something to be aware of."},{"Start":"07:57.860 ","End":"08:02.420","Text":"The terminal voltage is dependent on the current,"},{"Start":"08:02.420 ","End":"08:05.790","Text":"and that is the end of this lesson."}],"ID":22304},{"Watched":false,"Name":"Exercise 7","Duration":"5m 29s","ChapterTopicVideoID":21311,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Hello. In this lesson,"},{"Start":"00:01.800 ","End":"00:04.485","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.485 ","End":"00:10.350","Text":"The following circuit is comprised of a non-ideal battery and the 10 ohm resistor."},{"Start":"00:10.350 ","End":"00:13.740","Text":"The internal resistance of the battery is 1 ohm,"},{"Start":"00:13.740 ","End":"00:18.225","Text":"and the current of 2 amperes flows through the circuit."},{"Start":"00:18.225 ","End":"00:19.695","Text":"Question number 1 is,"},{"Start":"00:19.695 ","End":"00:22.440","Text":"what is the battery\u0027s EMF?"},{"Start":"00:22.440 ","End":"00:26.565","Text":"Once we\u0027re told that we have a non-ideal battery,"},{"Start":"00:26.565 ","End":"00:33.975","Text":"we can look at it as an ideal voltage source with an EMF over here of epsilon."},{"Start":"00:33.975 ","End":"00:42.820","Text":"It is connected in series to an internal resistor of internal resistance r,"},{"Start":"00:42.820 ","End":"00:45.020","Text":"which were specifically given over here,"},{"Start":"00:45.020 ","End":"00:48.270","Text":"is equal to 1 ohm."},{"Start":"00:49.070 ","End":"00:59.170","Text":"This is what is going on inside our non-ideal battery."},{"Start":"00:59.990 ","End":"01:06.200","Text":"The fact that all of this is inside the non-ideal battery doesn\u0027t really interest us."},{"Start":"01:06.200 ","End":"01:13.490","Text":"In actual fact, we\u0027re looking at an EMF connected to a resistor in series."},{"Start":"01:13.490 ","End":"01:19.930","Text":"All of this in series is connected to this 10 ohm resistor as well."},{"Start":"01:19.930 ","End":"01:23.770","Text":"What we\u0027re going to do is we\u0027re going to look at it as if there"},{"Start":"01:23.770 ","End":"01:27.975","Text":"isn\u0027t this battery over here, that casing."},{"Start":"01:27.975 ","End":"01:30.260","Text":"We just have 2 resistors."},{"Start":"01:30.260 ","End":"01:32.170","Text":"We\u0027re answering question number 1."},{"Start":"01:32.170 ","End":"01:37.430","Text":"What we\u0027re going to do is we\u0027re going to calculate the total resistance of this circuit."},{"Start":"01:37.430 ","End":"01:40.855","Text":"Seeing is the 2 resistors are connected in series."},{"Start":"01:40.855 ","End":"01:44.440","Text":"The total resistance is just the resistance of"},{"Start":"01:44.440 ","End":"01:49.090","Text":"the first resistor plus the resistance of the second resistor."},{"Start":"01:49.090 ","End":"01:53.810","Text":"The total resistance is just 11 ohms."},{"Start":"01:53.810 ","End":"01:57.160","Text":"We know that the voltage,"},{"Start":"01:57.160 ","End":"02:05.735","Text":"or in this case the EMF is simply equal to IR, as we know."},{"Start":"02:05.735 ","End":"02:11.720","Text":"It\u0027s equal to I multiplied by the total resistance."},{"Start":"02:11.720 ","End":"02:15.635","Text":"The current we were told is equal to 2 amps."},{"Start":"02:15.635 ","End":"02:18.589","Text":"Then this is multiplied by the total resistance,"},{"Start":"02:18.589 ","End":"02:20.960","Text":"which is 11 ohms."},{"Start":"02:20.960 ","End":"02:26.610","Text":"What we\u0027re going to get is an EMF of 22 volts."},{"Start":"02:26.610 ","End":"02:29.265","Text":"That\u0027s the answer to question number 1."},{"Start":"02:29.265 ","End":"02:31.710","Text":"Now let\u0027s look at question number 2."},{"Start":"02:31.710 ","End":"02:35.865","Text":"Question number 2 is what is the terminal voltage?"},{"Start":"02:35.865 ","End":"02:41.020","Text":"First of all, we know that when looking at the ideal battery,"},{"Start":"02:41.020 ","End":"02:45.885","Text":"it comprises of the EMF and this internal resistance."},{"Start":"02:45.885 ","End":"02:49.985","Text":"When we\u0027re calculating the terminal voltage,"},{"Start":"02:49.985 ","End":"02:56.730","Text":"we\u0027re looking at the voltage across the battery."},{"Start":"02:56.730 ","End":"03:02.495","Text":"That means that we\u0027re looking at the voltage across the EMF and the internal resistance."},{"Start":"03:02.495 ","End":"03:08.760","Text":"We\u0027re measuring the voltage between these 2 points, V terminal."},{"Start":"03:09.230 ","End":"03:20.209","Text":"We know that the equation for the terminal voltage is equal to the EMF,"},{"Start":"03:20.209 ","End":"03:25.705","Text":"which has the maximum voltage that we would have if no current was flowing through."},{"Start":"03:25.705 ","End":"03:29.090","Text":"If it was an open circuit but it is a closed circuit,"},{"Start":"03:29.090 ","End":"03:31.010","Text":"which means that current is flowing."},{"Start":"03:31.010 ","End":"03:36.425","Text":"It\u0027s equal to the EMF minus Ir,"},{"Start":"03:36.425 ","End":"03:40.640","Text":"minus the current multiplied by the internal resistance."},{"Start":"03:40.640 ","End":"03:48.210","Text":"Over here, we saw that the EMF was equal to 22 volts,"},{"Start":"03:48.210 ","End":"03:50.535","Text":"and then we subtract the current."},{"Start":"03:50.535 ","End":"03:55.775","Text":"The current is 2 amps multiplied by the internal resistance,"},{"Start":"03:55.775 ","End":"03:58.710","Text":"which is 1 ohm."},{"Start":"03:58.710 ","End":"04:06.070","Text":"We get that the terminal voltage is simply equal to 20 volts."},{"Start":"04:06.070 ","End":"04:10.640","Text":"Notice that the terminal voltage is also"},{"Start":"04:10.640 ","End":"04:15.275","Text":"the voltage that is felt over the other components in the circuit."},{"Start":"04:15.275 ","End":"04:19.590","Text":"Terminal voltage is the voltage that is felt throughout this circuit."},{"Start":"04:19.610 ","End":"04:25.325","Text":"Here\u0027s specifically, it was easier we could have calculated it either way."},{"Start":"04:25.325 ","End":"04:28.665","Text":"But sometimes if you have lots and lots of components in the circuit"},{"Start":"04:28.665 ","End":"04:32.360","Text":"it\u0027s easier to calculate it the way that we did it now,"},{"Start":"04:32.360 ","End":"04:39.900","Text":"however, if you look as we know, v=IR in general."},{"Start":"04:40.370 ","End":"04:46.325","Text":"Let\u0027s say voltage over resistor is equal to I the current which is 2 amps,"},{"Start":"04:46.325 ","End":"04:49.255","Text":"multiplied by the resistance of the resistor,"},{"Start":"04:49.255 ","End":"04:50.845","Text":"which has 10 ohms,"},{"Start":"04:50.845 ","End":"04:54.235","Text":"which is equal to 20 volts."},{"Start":"04:54.235 ","End":"04:56.490","Text":"This is also equal to V term."},{"Start":"04:56.490 ","End":"05:00.530","Text":"You can calculate the terminal voltage either way by calculating"},{"Start":"05:00.530 ","End":"05:04.410","Text":"the voltage over the battery like we did over here."},{"Start":"05:04.410 ","End":"05:12.215","Text":"We take into account the total resistance of the circuit in order to calculate the EMF,"},{"Start":"05:12.215 ","End":"05:15.199","Text":"and then we just use this simple equation."},{"Start":"05:15.199 ","End":"05:18.890","Text":"Or you can use the other way of"},{"Start":"05:18.890 ","End":"05:22.460","Text":"just finding the voltage over all of the other components."},{"Start":"05:22.460 ","End":"05:24.080","Text":"If there\u0027s lots of components,"},{"Start":"05:24.080 ","End":"05:27.430","Text":"then the way that we did it is easier."},{"Start":"05:27.430 ","End":"05:30.350","Text":"That\u0027s the end of this lesson."}],"ID":21391},{"Watched":false,"Name":"Exercise 8","Duration":"6m 57s","ChapterTopicVideoID":21488,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:03.990","Text":"we\u0027ll be answering the following question."},{"Start":"00:03.990 ","End":"00:07.845","Text":"A non-ideal battery is attached to a 10 ohm resistor."},{"Start":"00:07.845 ","End":"00:12.000","Text":"The current in the circuit is measured as being equal to 2 amps."},{"Start":"00:12.000 ","End":"00:13.980","Text":"The battery is then disconnected from"},{"Start":"00:13.980 ","End":"00:18.255","Text":"the 10 ohm resistor and is instead connected to a 6 ohm resistor."},{"Start":"00:18.255 ","End":"00:21.390","Text":"Now a current of 3 amps is measured."},{"Start":"00:21.390 ","End":"00:23.004","Text":"Question number 1 is,"},{"Start":"00:23.004 ","End":"00:27.060","Text":"\"What is the EMF and internal resistance of the battery?\""},{"Start":"00:27.060 ","End":"00:30.720","Text":"First of all, let\u0027s draw out our circuit."},{"Start":"00:30.720 ","End":"00:35.205","Text":"First of all, we have our battery which is made up of"},{"Start":"00:35.205 ","End":"00:42.465","Text":"our EMF connected in series to our internal resistance."},{"Start":"00:42.465 ","End":"00:47.480","Text":"This is what we have inside the battery,"},{"Start":"00:47.480 ","End":"00:50.145","Text":"so we can look at it as this,"},{"Start":"00:50.145 ","End":"00:57.550","Text":"and then this is attached to some resistor."},{"Start":"00:57.590 ","End":"01:02.715","Text":"The first resistor it\u0027s connected to is 10 ohms,"},{"Start":"01:02.715 ","End":"01:07.350","Text":"and then we switch out the 10 ohm resistor with our 6 ohm resistor."},{"Start":"01:07.350 ","End":"01:10.865","Text":"Specifically, when we\u0027re dealing with a 10 ohm resistor,"},{"Start":"01:10.865 ","End":"01:19.200","Text":"we\u0027re being told that the current flowing through the circuit is equal to 2 amps."},{"Start":"01:19.790 ","End":"01:25.145","Text":"Let\u0027s see what we can get out of this in the meantime."},{"Start":"01:25.145 ","End":"01:30.815","Text":"First of all, we know that the total resistance inside"},{"Start":"01:30.815 ","End":"01:36.500","Text":"this circuit is equal to the 2 resistors connected in series."},{"Start":"01:36.500 ","End":"01:42.995","Text":"It\u0027s equal to 10 ohms plus whatever our internal resistance,"},{"Start":"01:42.995 ","End":"01:44.785","Text":"r, is equal to."},{"Start":"01:44.785 ","End":"01:49.115","Text":"Then we know that our EMF is simply equal to"},{"Start":"01:49.115 ","End":"01:55.805","Text":"the current flowing through the circle multiplied by the total resistance."},{"Start":"01:55.805 ","End":"01:59.790","Text":"It\u0027s the current, which is 2 amps,"},{"Start":"01:59.790 ","End":"02:01.860","Text":"multiplied by the total resistance,"},{"Start":"02:01.860 ","End":"02:04.815","Text":"which is equal to 10 ohms plus"},{"Start":"02:04.815 ","End":"02:12.805","Text":"r. Now we have 2 unknowns."},{"Start":"02:12.805 ","End":"02:16.960","Text":"We have our EMF and our internal resistance."},{"Start":"02:16.960 ","End":"02:20.660","Text":"Now let\u0027s draw again the diagram,"},{"Start":"02:20.660 ","End":"02:30.530","Text":"but this time when we have the 6 ohm resistor connected to our battery."},{"Start":"02:30.530 ","End":"02:32.195","Text":"This is what it looks like."},{"Start":"02:32.195 ","End":"02:37.990","Text":"We have 6 ohms and a current which is equal to 3 amps."},{"Start":"02:37.990 ","End":"02:41.975","Text":"Here, we have our EMF and our internal resistance."},{"Start":"02:41.975 ","End":"02:51.545","Text":"Now our equations, so our R total is going to be equal to 6 ohms plus r,"},{"Start":"02:51.545 ","End":"02:53.875","Text":"and our EMF is,"},{"Start":"02:53.875 ","End":"02:57.160","Text":"again, equal in this case."},{"Start":"02:57.160 ","End":"03:01.785","Text":"This is number 1 and this is number 2."},{"Start":"03:01.785 ","End":"03:05.205","Text":"This is case number 1,"},{"Start":"03:05.205 ","End":"03:08.775","Text":"this is case number 2 over here."},{"Start":"03:08.775 ","End":"03:12.680","Text":"It\u0027s equal to I multiplied by the total resistance,"},{"Start":"03:12.680 ","End":"03:17.280","Text":"which in this case is equal to I which"},{"Start":"03:17.280 ","End":"03:21.710","Text":"is equal to 3 amps multiplied by the total resistance,"},{"Start":"03:21.710 ","End":"03:26.270","Text":"which is equal to 6 ohms plus r. Now we know that"},{"Start":"03:26.270 ","End":"03:32.005","Text":"the EMF in both of these cases is equal because that\u0027s a constant."},{"Start":"03:32.005 ","End":"03:35.190","Text":"Therefore, we can equate the 2."},{"Start":"03:35.190 ","End":"03:43.890","Text":"We can therefore say that 3 multiplied by 6 plus r is equal"},{"Start":"03:43.890 ","End":"03:52.910","Text":"to 2 multiplied by 10 plus r. What we can do is we can open up the brackets,"},{"Start":"03:52.910 ","End":"04:02.235","Text":"so we have 18 plus 3r is equal to 20 plus 2r."},{"Start":"04:02.235 ","End":"04:09.875","Text":"Then we can subtract 18 from both sides and subtract 2r from both sides."},{"Start":"04:09.875 ","End":"04:17.640","Text":"Therefore, we\u0027ll get that r is simply equal to 2 ohms."},{"Start":"04:17.640 ","End":"04:20.473","Text":"This is the internal resistance."},{"Start":"04:20.473 ","End":"04:23.450","Text":"Therefore, our EMF is equal 2,"},{"Start":"04:23.450 ","End":"04:26.985","Text":"so we can use either 1 of these equations."},{"Start":"04:26.985 ","End":"04:28.545","Text":"Let\u0027s just use this one."},{"Start":"04:28.545 ","End":"04:36.400","Text":"Our EMF is equal to 2 amps multiplied by 10 plus our internal resistance, which is 2."},{"Start":"04:36.400 ","End":"04:45.410","Text":"2 times 12 is simply equal to 24 volts. That\u0027s it."},{"Start":"04:45.410 ","End":"04:47.750","Text":"This is the answer to question number 1."},{"Start":"04:47.750 ","End":"04:52.925","Text":"This is our EMF and this is our internal resistance."},{"Start":"04:52.925 ","End":"04:58.185","Text":"Now, let\u0027s go on to question number 2."},{"Start":"04:58.185 ","End":"05:02.270","Text":"Let\u0027s just show that this is still question number 1."},{"Start":"05:02.270 ","End":"05:07.160","Text":"Question number 2 is to calculate the terminal voltage in each case."},{"Start":"05:07.160 ","End":"05:14.379","Text":"Of course, the terminal voltage is the voltage across the 2 terminals of the battery."},{"Start":"05:14.379 ","End":"05:22.000","Text":"Here, V_T is for V terminal and that\u0027s what this is equal to."},{"Start":"05:22.520 ","End":"05:27.380","Text":"We\u0027re looking for the terminal voltage in each case."},{"Start":"05:27.380 ","End":"05:32.570","Text":"We know that the equation for the terminal voltage is equal to the EMF"},{"Start":"05:32.570 ","End":"05:39.835","Text":"minus the current flowing through the circuit multiplied by the internal resistance."},{"Start":"05:39.835 ","End":"05:44.145","Text":"Let\u0027s look at case number 1,"},{"Start":"05:44.145 ","End":"05:47.850","Text":"where we have the 10 ohm resistor in."},{"Start":"05:47.850 ","End":"05:57.000","Text":"In this case, our V terminal for the 10 ohm resistor, our EMF,"},{"Start":"05:57.000 ","End":"06:02.544","Text":"which we calculated is equal to 24 volts minus the current,"},{"Start":"06:02.544 ","End":"06:08.530","Text":"the current in this case was 2 amps multiplied by our internal resistance,"},{"Start":"06:08.530 ","End":"06:11.295","Text":"which is 2 ohms."},{"Start":"06:11.295 ","End":"06:14.265","Text":"Then we have 2 times 2 is 4,"},{"Start":"06:14.265 ","End":"06:19.150","Text":"24 minus 4 is equal to 20 volts."},{"Start":"06:19.150 ","End":"06:22.185","Text":"Then let\u0027s look at Case 2."},{"Start":"06:22.185 ","End":"06:25.645","Text":"Our V terminal, when we have a 6 ohm resistor,"},{"Start":"06:25.645 ","End":"06:28.270","Text":"is equal to our EMF,"},{"Start":"06:28.270 ","End":"06:32.560","Text":"which is 24 volts minus the current flowing through,"},{"Start":"06:32.560 ","End":"06:34.771","Text":"so in the 6 ohm case,"},{"Start":"06:34.771 ","End":"06:41.090","Text":"we have 3 amps flowing through multiplied by the internal resistance, which is 2."},{"Start":"06:41.090 ","End":"06:43.345","Text":"3 times 2 is 6,"},{"Start":"06:43.345 ","End":"06:48.270","Text":"24 minus 6 is equal to 18 volts."},{"Start":"06:48.270 ","End":"06:52.700","Text":"This is the answer to question number 2,"},{"Start":"06:52.700 ","End":"06:55.205","Text":"the terminal voltage in each case,"},{"Start":"06:55.205 ","End":"06:58.290","Text":"and that is the end of this lesson."}],"ID":22305},{"Watched":false,"Name":"Method for Measuring EMF","Duration":"2m 33s","ChapterTopicVideoID":21489,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"Hello. In this lesson,"},{"Start":"00:01.860 ","End":"00:07.935","Text":"we\u0027re going to be learning about a method for measuring EMF so up till now,"},{"Start":"00:07.935 ","End":"00:11.325","Text":"we\u0027ve spoken about a non-ideal battery,"},{"Start":"00:11.325 ","End":"00:16.470","Text":"where we\u0027ve said that the non-ideal battery is made up of the EMF,"},{"Start":"00:16.470 ","End":"00:19.125","Text":"which is the ideal voltage source,"},{"Start":"00:19.125 ","End":"00:24.975","Text":"plus some internal resistance R,"},{"Start":"00:24.975 ","End":"00:30.285","Text":"and that this is our battery."},{"Start":"00:30.285 ","End":"00:35.527","Text":"Now we know that if at the terminal of this battery,"},{"Start":"00:35.527 ","End":"00:37.805","Text":"so from this point to this point,"},{"Start":"00:37.805 ","End":"00:42.328","Text":"we would take a voltmeter,"},{"Start":"00:42.328 ","End":"00:48.630","Text":"and we would measure out the voltage over or across the battery."},{"Start":"00:48.700 ","End":"00:58.340","Text":"The voltage which would appear on the volt meter will be equal to the terminal voltage,"},{"Start":"00:58.340 ","End":"01:02.180","Text":"which is just the voltage from this point to this point,"},{"Start":"01:02.180 ","End":"01:03.605","Text":"which as we know,"},{"Start":"01:03.605 ","End":"01:08.975","Text":"is equal to the EMF minus the current through"},{"Start":"01:08.975 ","End":"01:16.640","Text":"the battery multiplied by the internal resistance."},{"Start":"01:16.640 ","End":"01:19.730","Text":"If we take an ideal volt meter,"},{"Start":"01:19.730 ","End":"01:26.585","Text":"so what is an ideal voltmeter the resistance of the voltmeter is infinitely large."},{"Start":"01:26.585 ","End":"01:31.025","Text":"The resistance of the voltmeter is approaching infinity."},{"Start":"01:31.025 ","End":"01:34.145","Text":"If the resistance is approaching infinity,"},{"Start":"01:34.145 ","End":"01:39.080","Text":"then the current flowing through this circuit,"},{"Start":"01:39.080 ","End":"01:46.480","Text":"through the battery and through the volt meter is therefore going to be approaching 0."},{"Start":"01:46.480 ","End":"01:50.555","Text":"In this case, if our current I is approaching 0,"},{"Start":"01:50.555 ","End":"01:55.310","Text":"we can see that our terminal voltage is going to be equal"},{"Start":"01:55.310 ","End":"02:00.305","Text":"to Epsilon minus 0 multiplied by the internal resistance."},{"Start":"02:00.305 ","End":"02:03.860","Text":"Which means that our V terminal is simply going to be"},{"Start":"02:03.860 ","End":"02:08.345","Text":"equal to Epsilon or equal to our EMF."},{"Start":"02:08.345 ","End":"02:13.000","Text":"That is a very easy method of measuring the EMF."},{"Start":"02:13.000 ","End":"02:15.305","Text":"We just take our non-ideal battery,"},{"Start":"02:15.305 ","End":"02:18.140","Text":"we connect it to an ideal volt meter."},{"Start":"02:18.140 ","End":"02:21.425","Text":"Therefore, there is no current flowing through and therefore,"},{"Start":"02:21.425 ","End":"02:25.654","Text":"what our volt meter will show is our terminal voltage,"},{"Start":"02:25.654 ","End":"02:31.567","Text":"which in this specific case will be equal to the EMF."},{"Start":"02:31.567 ","End":"02:33.990","Text":"That\u0027s the end of this lesson."}],"ID":22306},{"Watched":false,"Name":"Exercise 9","Duration":"17m 28s","ChapterTopicVideoID":21490,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.205","Text":"Hello. In this lesson,"},{"Start":"00:02.205 ","End":"00:05.160","Text":"we\u0027re going to be answering the following questions."},{"Start":"00:05.160 ","End":"00:09.045","Text":"The following circuit is made up of 4 resistors,"},{"Start":"00:09.045 ","End":"00:14.760","Text":"an ideal volt meter and ammeter a non-ideal battery."},{"Start":"00:14.760 ","End":"00:20.220","Text":"Over here, we\u0027ll have a resistor to represent"},{"Start":"00:20.220 ","End":"00:25.935","Text":"the internal resistance of the battery and a switch."},{"Start":"00:25.935 ","End":"00:29.790","Text":"The value of the ammeter is recorded twice."},{"Start":"00:29.790 ","End":"00:32.490","Text":"Once when the switch is closed,"},{"Start":"00:32.490 ","End":"00:34.515","Text":"and again when it is open."},{"Start":"00:34.515 ","End":"00:41.925","Text":"The 2 readings that we got from the ammeter was 1.5 amps and 1.8 amps."},{"Start":"00:41.925 ","End":"00:45.260","Text":"Question number 1 is at what position was"},{"Start":"00:45.260 ","End":"00:48.799","Text":"the switch when the greater value for current was recorded?"},{"Start":"00:48.799 ","End":"00:51.360","Text":"Explain your answer."},{"Start":"00:52.640 ","End":"00:58.445","Text":"The first thing that we know from Ohm\u0027s law is that the voltage in"},{"Start":"00:58.445 ","End":"01:05.220","Text":"the general case is equal to the current multiplied by the resistance."},{"Start":"01:05.220 ","End":"01:11.000","Text":"In this case, our voltage is given from our battery,"},{"Start":"01:11.000 ","End":"01:14.435","Text":"which is this over here,"},{"Start":"01:14.435 ","End":"01:17.215","Text":"and it\u0027s a non-ideal battery."},{"Start":"01:17.215 ","End":"01:22.045","Text":"What we\u0027re getting is an EMF."},{"Start":"01:22.045 ","End":"01:27.125","Text":"Our EMF is simply equal to"},{"Start":"01:27.125 ","End":"01:29.750","Text":"the current flowing through the circuit"},{"Start":"01:29.750 ","End":"01:33.500","Text":"multiplied by the total resistance of the circuit."},{"Start":"01:33.500 ","End":"01:37.880","Text":"The total resistance of the circuit takes into account all of"},{"Start":"01:37.880 ","End":"01:45.630","Text":"the resistors and the internal resistance of the non-ideal battery."},{"Start":"01:46.190 ","End":"01:51.470","Text":"What we\u0027re trying to see is when the current was greater."},{"Start":"01:51.470 ","End":"01:55.940","Text":"What are we going to do is we\u0027re going to isolate out our current,"},{"Start":"01:55.940 ","End":"02:02.835","Text":"and this is simply equal to the EMF divided by our total resistance."},{"Start":"02:02.835 ","End":"02:10.560","Text":"What we can see is that as the resistance increases, our current decreases."},{"Start":"02:11.780 ","End":"02:15.240","Text":"The resistance is in the denominator,"},{"Start":"02:15.240 ","End":"02:17.345","Text":"so as it increases,"},{"Start":"02:17.345 ","End":"02:19.070","Text":"the fraction as a whole,"},{"Start":"02:19.070 ","End":"02:23.150","Text":"and so therefore the current will decrease."},{"Start":"02:23.150 ","End":"02:27.560","Text":"Therefore we can see that the more resistance we have,"},{"Start":"02:27.560 ","End":"02:29.315","Text":"the lower the current."},{"Start":"02:29.315 ","End":"02:33.485","Text":"The more resistors we have is when our switch is open."},{"Start":"02:33.485 ","End":"02:35.030","Text":"When our switch is open,"},{"Start":"02:35.030 ","End":"02:40.295","Text":"the current has no choice but to also bypass this resistor,"},{"Start":"02:40.295 ","End":"02:42.845","Text":"it will bypass this, and then this,"},{"Start":"02:42.845 ","End":"02:46.085","Text":"and then this resistor, and then this resistor and then back."},{"Start":"02:46.085 ","End":"02:49.115","Text":"However, when the switch is closed,"},{"Start":"02:49.115 ","End":"02:56.110","Text":"the current can therefore bypass this resistor and go via the switch."},{"Start":"02:56.110 ","End":"02:58.535","Text":"If it\u0027s bypassing this resistor,"},{"Start":"02:58.535 ","End":"03:01.670","Text":"that means that the total resistance will be"},{"Start":"03:01.670 ","End":"03:06.630","Text":"less and therefore the current will be greater."},{"Start":"03:07.880 ","End":"03:16.595","Text":"The lower the total resistance will result in a higher value for our current."},{"Start":"03:16.595 ","End":"03:21.095","Text":"That happens when our current can bypass 1 of the resistors,"},{"Start":"03:21.095 ","End":"03:24.840","Text":"which happens when the switch is closed."},{"Start":"03:25.430 ","End":"03:31.190","Text":"The answer to Question 1 is that the higher or"},{"Start":"03:31.190 ","End":"03:36.460","Text":"the greater value for current was recorded when the switch was closed."},{"Start":"03:36.460 ","End":"03:39.745","Text":"If you wanted to do the calculation,"},{"Start":"03:39.745 ","End":"03:44.105","Text":"EMF divided by the total resistance of all 4 resistors,"},{"Start":"03:44.105 ","End":"03:48.245","Text":"or the total resistance of the 3 resistors when the switch is closed,"},{"Start":"03:48.245 ","End":"03:53.230","Text":"you\u0027ll see that the current will obviously be higher when the switch is closed."},{"Start":"03:53.230 ","End":"03:57.445","Text":"Now let\u0027s take a look at Question number 2."},{"Start":"03:57.445 ","End":"04:04.590","Text":"What will be the value on the voltmeter when the switch is open and when it is closed?"},{"Start":"04:05.780 ","End":"04:09.395","Text":"Let\u0027s look at the first case,"},{"Start":"04:09.395 ","End":"04:13.490","Text":"which is when the switch is open."},{"Start":"04:13.490 ","End":"04:16.900","Text":"Let\u0027s go from here."},{"Start":"04:16.900 ","End":"04:19.200","Text":"This is our Point A."},{"Start":"04:19.200 ","End":"04:22.745","Text":"Then we\u0027re going all the way to here,"},{"Start":"04:22.745 ","End":"04:25.025","Text":"which is our Point B."},{"Start":"04:25.025 ","End":"04:29.599","Text":"The voltmeter is measuring the voltage across the battery,"},{"Start":"04:29.599 ","End":"04:36.700","Text":"which is the same as measuring the voltage across all of these components."},{"Start":"04:38.990 ","End":"04:46.845","Text":"When it\u0027s open, we\u0027re just going to use Kirchhoff\u0027s law. We start."},{"Start":"04:46.845 ","End":"04:49.790","Text":"As we remember, we start with V_A,"},{"Start":"04:49.790 ","End":"04:53.715","Text":"and then that\u0027s the total voltage."},{"Start":"04:53.715 ","End":"04:57.860","Text":"Then our voltage goes down as we go down each resistor."},{"Start":"04:57.860 ","End":"05:04.730","Text":"We have V_A minus our 3 ohm resistor."},{"Start":"05:04.730 ","End":"05:07.955","Text":"The voltage drop across our 3 ohm resistor."},{"Start":"05:07.955 ","End":"05:12.800","Text":"Right now, if our switch is open,"},{"Start":"05:12.800 ","End":"05:16.820","Text":"we said that that corresponds to the lower value for"},{"Start":"05:16.820 ","End":"05:23.280","Text":"current because we said that the higher value for current is when the switch is closed."},{"Start":"05:23.280 ","End":"05:28.560","Text":"We have V_A minus 1.5 amps"},{"Start":"05:28.560 ","End":"05:34.770","Text":"multiplied by 3 ohms and then it\u0027s open so we have to go via this 2 ohm resistor,"},{"Start":"05:34.770 ","End":"05:40.710","Text":"so minus 1.5 amps multiplied by"},{"Start":"05:40.710 ","End":"05:48.290","Text":"2 ohms minus the voltage drop across this other 2 ohm resistor."},{"Start":"05:48.290 ","End":"05:52.730","Text":"We can just multiply this by 2 and"},{"Start":"05:52.730 ","End":"05:59.010","Text":"then the voltage drop across the second 3 ohm resistor."},{"Start":"05:59.010 ","End":"06:03.020","Text":"We can multiply this by 2 as well."},{"Start":"06:03.020 ","End":"06:09.630","Text":"This is equal to the voltage at Point B, V_B."},{"Start":"06:10.340 ","End":"06:18.465","Text":"Now we can subtract V_B from both sides and add this expression onto both sides."},{"Start":"06:18.465 ","End":"06:21.920","Text":"What we\u0027ll get is that V_A minus"},{"Start":"06:21.920 ","End":"06:29.515","Text":"V_B=2 times 1.5 times 3,"},{"Start":"06:29.515 ","End":"06:36.135","Text":"plus 2 times 1.5 times 2."},{"Start":"06:36.135 ","End":"06:45.420","Text":"This is equal to 15 volts."},{"Start":"06:45.420 ","End":"06:52.655","Text":"This is what the voltmeter we\u0027ll show when the switch is open."},{"Start":"06:52.655 ","End":"06:57.830","Text":"Now, let\u0027s see what we will see when the switch is closed."},{"Start":"06:57.830 ","End":"06:59.480","Text":"Again, we\u0027re doing the same thing,"},{"Start":"06:59.480 ","End":"07:02.045","Text":"but this time here is closed."},{"Start":"07:02.045 ","End":"07:03.860","Text":"We have V_A, again,"},{"Start":"07:03.860 ","End":"07:06.180","Text":"we\u0027re starting from here,"},{"Start":"07:06.320 ","End":"07:08.910","Text":"and then we\u0027re subtracting,"},{"Start":"07:08.910 ","End":"07:10.550","Text":"so now the switch is closed,"},{"Start":"07:10.550 ","End":"07:13.730","Text":"which we saw A corresponded to the higher current,"},{"Start":"07:13.730 ","End":"07:17.885","Text":"which means that our current is equal to 1.8 amps."},{"Start":"07:17.885 ","End":"07:24.065","Text":"We have minus 1.8 multiplied by our 3 ohm resistor."},{"Start":"07:24.065 ","End":"07:28.475","Text":"Then we bypass this resistor because the switch is closed."},{"Start":"07:28.475 ","End":"07:31.420","Text":"Then we have minus."},{"Start":"07:31.420 ","End":"07:37.615","Text":"Again, 1.8 amps multiplied by our 2 ohm resistance."},{"Start":"07:37.615 ","End":"07:39.505","Text":"That\u0027s the voltage drop over that."},{"Start":"07:39.505 ","End":"07:43.000","Text":"Then we have a voltage drop over the 3 ohm resistor,"},{"Start":"07:43.000 ","End":"07:46.515","Text":"which is again 1.8 times 3."},{"Start":"07:46.515 ","End":"07:50.860","Text":"We can multiply this by 2."},{"Start":"07:50.960 ","End":"08:00.745","Text":"Then this is going to be equal to the voltage at this point over here, V_B."},{"Start":"08:00.745 ","End":"08:04.450","Text":"Again, I\u0027m going to subtract V_B from both sides"},{"Start":"08:04.450 ","End":"08:07.975","Text":"and add on this over here to both sides."},{"Start":"08:07.975 ","End":"08:11.515","Text":"What we\u0027ll get is that V_A minus V_B,"},{"Start":"08:11.515 ","End":"08:13.845","Text":"which is what our volt meter will show,"},{"Start":"08:13.845 ","End":"08:15.895","Text":"will be equal to,"},{"Start":"08:15.895 ","End":"08:24.910","Text":"so we just have 2 multiply it by 1.8 times 3 minus 1.8 times 2,"},{"Start":"08:24.910 ","End":"08:27.395","Text":"which when we plug it into the calculator,"},{"Start":"08:27.395 ","End":"08:32.700","Text":"is equal to 14.4 volts."},{"Start":"08:32.980 ","End":"08:36.470","Text":"This is the answer to Question number 2,"},{"Start":"08:36.470 ","End":"08:42.320","Text":"the value on the voltmeter when the switch is open and when the switch is closed."},{"Start":"08:42.320 ","End":"08:45.440","Text":"Now, let\u0027s move on to Question number 3."},{"Start":"08:45.440 ","End":"08:51.210","Text":"Calculate the EMF and the internal resistance of the battery."},{"Start":"08:52.180 ","End":"08:58.165","Text":"As we know, our EMF and our internal resistance is unknown."},{"Start":"08:58.165 ","End":"09:02.135","Text":"We\u0027re going to try and find them from what we do know."},{"Start":"09:02.135 ","End":"09:06.545","Text":"First of all, when our switch is open,"},{"Start":"09:06.545 ","End":"09:08.240","Text":"so in this case like,"},{"Start":"09:08.240 ","End":"09:11.105","Text":"so we know that this,"},{"Start":"09:11.105 ","End":"09:13.325","Text":"which is our terminal voltage,"},{"Start":"09:13.325 ","End":"09:15.590","Text":"is equal to 15 volts."},{"Start":"09:15.590 ","End":"09:17.495","Text":"Remember the equation."},{"Start":"09:17.495 ","End":"09:25.065","Text":"Our terminal voltage is equal to our EMF minus I times the internal resistance."},{"Start":"09:25.065 ","End":"09:27.455","Text":"When our switch is open,"},{"Start":"09:27.455 ","End":"09:29.800","Text":"our terminal voltage was 15 volts,"},{"Start":"09:29.800 ","End":"09:34.340","Text":"and this was equal to our EMF minus the current,"},{"Start":"09:34.340 ","End":"09:38.885","Text":"in this case, which was equal to 1.5"},{"Start":"09:38.885 ","End":"09:44.485","Text":"amps multiplied by our internal resistance, which we don\u0027t know."},{"Start":"09:44.485 ","End":"09:47.569","Text":"Then when our switch was closed,"},{"Start":"09:47.569 ","End":"09:51.920","Text":"our terminal voltage was 14.4 volts,"},{"Start":"09:51.920 ","End":"09:57.230","Text":"which was equal to our EMF minus our current in this case,"},{"Start":"09:57.230 ","End":"10:03.775","Text":"which was 1.8 amps multiplied by our internal resistance."},{"Start":"10:03.775 ","End":"10:07.440","Text":"Now let\u0027s scroll down."},{"Start":"10:07.450 ","End":"10:11.630","Text":"What we can do is we can isolate out our EMF."},{"Start":"10:11.630 ","End":"10:18.460","Text":"In this case, our EMF is equal to 15 plus 1.5r,"},{"Start":"10:18.460 ","End":"10:20.210","Text":"and in this case,"},{"Start":"10:20.210 ","End":"10:28.724","Text":"our EMF is equal to 14.4 plus 1.8r."},{"Start":"10:28.724 ","End":"10:34.480","Text":"Now what we can do is we can just play around with the algebra."},{"Start":"10:34.480 ","End":"10:38.845","Text":"We\u0027re going to subtract 14.4 from both sides,"},{"Start":"10:38.845 ","End":"10:42.088","Text":"and 1.5 off from both sides."},{"Start":"10:42.088 ","End":"10:46.420","Text":"What we\u0027ll have is 0.6 over here,"},{"Start":"10:46.420 ","End":"10:51.110","Text":"which is equal to 0.3r."},{"Start":"10:52.410 ","End":"10:56.920","Text":"Now what we\u0027re going to do is divide both sides by 0.3."},{"Start":"10:56.920 ","End":"11:04.285","Text":"We have 0.6 divided by 0.3,"},{"Start":"11:04.285 ","End":"11:07.765","Text":"which is going to be equal to our internal resistance,"},{"Start":"11:07.765 ","End":"11:12.820","Text":"which is equal to 2 volts."},{"Start":"11:12.820 ","End":"11:15.656","Text":"Now we have our internal resistance."},{"Start":"11:15.656 ","End":"11:21.145","Text":"We can just plug it into one of these equations in order to get our EMF."},{"Start":"11:21.145 ","End":"11:23.275","Text":"Let\u0027s go here."},{"Start":"11:23.275 ","End":"11:31.600","Text":"We have our EMF is equal to 15 plus 1.5 multiplied by our internal resistance,"},{"Start":"11:31.600 ","End":"11:36.895","Text":"which is 2, so that is just equal to 15 plus 3,"},{"Start":"11:36.895 ","End":"11:41.150","Text":"which is equal to 18 volts."},{"Start":"11:41.250 ","End":"11:47.123","Text":"Of course, our internal resistance is given in ohms."},{"Start":"11:47.123 ","End":"11:51.640","Text":"This is the answer to Question number 3,"},{"Start":"11:51.640 ","End":"11:56.905","Text":"and now let\u0027s go on to answer Question number 4."},{"Start":"11:56.905 ","End":"12:00.160","Text":"Question number 4 is if we scroll up,"},{"Start":"12:00.160 ","End":"12:07.580","Text":"what values with the ammeter and voltmeter show if they traded positions?"},{"Start":"12:07.980 ","End":"12:12.310","Text":"In this question, we\u0027re pretending like this,"},{"Start":"12:12.310 ","End":"12:15.520","Text":"instead of being the voltmeter is the ammeter,"},{"Start":"12:15.520 ","End":"12:19.645","Text":"and that this instead of being the ammeter is the voltmeter."},{"Start":"12:19.645 ","End":"12:27.487","Text":"We know that the resistance of the voltmeter is approaching infinity."},{"Start":"12:27.487 ","End":"12:29.965","Text":"It is an infinite resistance."},{"Start":"12:29.965 ","End":"12:32.425","Text":"If we have an infinite resistance,"},{"Start":"12:32.425 ","End":"12:36.145","Text":"that means that this will act as an open circuit."},{"Start":"12:36.145 ","End":"12:38.425","Text":"If we have an open circuit,"},{"Start":"12:38.425 ","End":"12:42.490","Text":"that means that the current isn\u0027t flowing over here."},{"Start":"12:42.490 ","End":"12:46.465","Text":"This is of course because we\u0027re being told in the question above,"},{"Start":"12:46.465 ","End":"12:53.395","Text":"that we have an ideal ammeter and voltmeter."},{"Start":"12:53.395 ","End":"12:56.095","Text":"Now on the other hand,"},{"Start":"12:56.095 ","End":"13:02.040","Text":"here we have the ammeter which is an ideal ammeter,"},{"Start":"13:02.040 ","End":"13:04.365","Text":"and we know that over here,"},{"Start":"13:04.365 ","End":"13:11.015","Text":"the resistance of the ammeter is equal to 0,"},{"Start":"13:11.015 ","End":"13:15.655","Text":"and as we know, that will act like a short circuit."},{"Start":"13:15.655 ","End":"13:19.968","Text":"Then it\u0027s as if the ammeter doesn\u0027t exist here,"},{"Start":"13:19.968 ","End":"13:25.760","Text":"and we just have some wire passing through over here."},{"Start":"13:26.280 ","End":"13:29.830","Text":"What we have is here an open circuit,"},{"Start":"13:29.830 ","End":"13:31.540","Text":"so no current is flowing,"},{"Start":"13:31.540 ","End":"13:34.375","Text":"and here we have a short circuit."},{"Start":"13:34.375 ","End":"13:39.355","Text":"Therefore, our circuit is going to look like so."},{"Start":"13:39.355 ","End":"13:46.270","Text":"We\u0027re going to have a battery over here where we have our EMF,"},{"Start":"13:46.270 ","End":"13:51.412","Text":"then over here we have our resistor small r,"},{"Start":"13:51.412 ","End":"13:58.160","Text":"and then it\u0027s just connected by a wire like so."},{"Start":"13:59.280 ","End":"14:06.985","Text":"Now let\u0027s scroll down so that we have more space and we can draw the circuit again."},{"Start":"14:06.985 ","End":"14:14.440","Text":"We have our battery connected to the internal resistor,"},{"Start":"14:14.440 ","End":"14:19.767","Text":"and then we just have this connected to a wire."},{"Start":"14:19.767 ","End":"14:26.658","Text":"What will the ammeter show and what will the voltmeter show in this case?"},{"Start":"14:26.658 ","End":"14:30.370","Text":"The ammeter, let\u0027s show this with a A."},{"Start":"14:30.370 ","End":"14:34.075","Text":"The ammeter will show the current flowing through."},{"Start":"14:34.075 ","End":"14:39.580","Text":"The current from Ohm\u0027s law is going to be the voltage,"},{"Start":"14:39.580 ","End":"14:40.870","Text":"which in this case,"},{"Start":"14:40.870 ","End":"14:45.130","Text":"our voltage is our EMF divided by the resistance."},{"Start":"14:45.130 ","End":"14:51.685","Text":"In this case, divided by the total resistance whenever we\u0027re dealing with EMF."},{"Start":"14:51.685 ","End":"14:56.555","Text":"The total resistance is just lowercase r, our internal resistance."},{"Start":"14:56.555 ","End":"14:59.890","Text":"Our EMF is equal to 18 volts,"},{"Start":"14:59.890 ","End":"15:04.435","Text":"and our internal resistance is equal to 2 ohms."},{"Start":"15:04.435 ","End":"15:11.125","Text":"Therefore, our ammeter will show us a reading of 9 amps."},{"Start":"15:11.125 ","End":"15:14.830","Text":"Now, what about the voltmeter?"},{"Start":"15:14.830 ","End":"15:18.385","Text":"Let\u0027s scroll up to our original diagram."},{"Start":"15:18.385 ","End":"15:20.770","Text":"Here we have our voltmeter."},{"Start":"15:20.770 ","End":"15:27.690","Text":"Now because we know that the resistance in the ideal voltmeter is infinity,"},{"Start":"15:27.690 ","End":"15:31.995","Text":"so we have no current flowing through here,"},{"Start":"15:31.995 ","End":"15:38.205","Text":"which means that the voltage over each one of these resistors is equal to 0."},{"Start":"15:38.205 ","End":"15:39.975","Text":"So V is equal to 0 here,"},{"Start":"15:39.975 ","End":"15:46.055","Text":"V=0 over here, regardless of if the switch is open or closed."},{"Start":"15:46.055 ","End":"15:48.415","Text":"Here, V=0,"},{"Start":"15:48.415 ","End":"15:53.365","Text":"and here the voltage is equal to 0."},{"Start":"15:53.365 ","End":"15:58.795","Text":"Therefore, if the voltage is equal to 0 over all of these components,"},{"Start":"15:58.795 ","End":"16:06.220","Text":"the voltmeter is measuring the voltage between these 2 points, and again,"},{"Start":"16:06.220 ","End":"16:09.445","Text":"because the voltage over all of these components is equal to 0,"},{"Start":"16:09.445 ","End":"16:14.919","Text":"the voltage between these 2 points is the same as the voltage between these 2 points,"},{"Start":"16:14.919 ","End":"16:17.470","Text":"between points A and B."},{"Start":"16:17.470 ","End":"16:20.965","Text":"Now in order to calculate the voltage between A and B,"},{"Start":"16:20.965 ","End":"16:26.275","Text":"we can either go via this route where we can use Kirchhoff\u0027s law,"},{"Start":"16:26.275 ","End":"16:29.935","Text":"or we can just go via this route,"},{"Start":"16:29.935 ","End":"16:33.760","Text":"because the ammeter is as we saw just a short circuit."},{"Start":"16:33.760 ","End":"16:40.100","Text":"We can consider the ammeter is just a wire connecting these 2 points."},{"Start":"16:40.200 ","End":"16:45.430","Text":"Then if these 2 points are connected just by a wire,"},{"Start":"16:45.430 ","End":"16:52.090","Text":"then therefore we can say that the voltage between these 2 points is equal to 0."},{"Start":"16:52.090 ","End":"16:56.200","Text":"Another way we could do it is by using that V is"},{"Start":"16:56.200 ","End":"17:00.865","Text":"equal to Epsilon minus Ir where in that case,"},{"Start":"17:00.865 ","End":"17:06.520","Text":"we would have that V is equal to Epsilon or the EMF,"},{"Start":"17:06.520 ","End":"17:10.195","Text":"which is equal to 18 volts minus Ir,"},{"Start":"17:10.195 ","End":"17:16.390","Text":"where I is 9 amps multiplied by 2 ohms,"},{"Start":"17:16.390 ","End":"17:19.825","Text":"and then again we get 0."},{"Start":"17:19.825 ","End":"17:22.150","Text":"This is just via Kirchhoff\u0027s."},{"Start":"17:22.150 ","End":"17:23.965","Text":"This is the answer,"},{"Start":"17:23.965 ","End":"17:25.870","Text":"therefore for Question number 4,"},{"Start":"17:25.870 ","End":"17:29.060","Text":"and that is the end of this lesson."}],"ID":22307},{"Watched":false,"Name":"Exercise 10","Duration":"10m 24s","ChapterTopicVideoID":21491,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:04.500","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.500 ","End":"00:09.900","Text":"In the diagram, there is an infinite ladder circuit containing resistors."},{"Start":"00:09.900 ","End":"00:12.493","Text":"Each resistor has a resistance R,"},{"Start":"00:12.493 ","End":"00:17.650","Text":"question Number 1 is to calculate the total resistance of the circuit."},{"Start":"00:17.650 ","End":"00:20.460","Text":"That means that between these 2 points,"},{"Start":"00:20.460 ","End":"00:25.330","Text":"we want to calculate the total resistance."},{"Start":"00:25.400 ","End":"00:30.870","Text":"The trick to answering questions dealing with infinite ladders is to"},{"Start":"00:30.870 ","End":"00:36.525","Text":"notice that the ladders will usually have some repetitive quality."},{"Start":"00:36.525 ","End":"00:43.025","Text":"For instance, we can look over here and we can see this is one unit in"},{"Start":"00:43.025 ","End":"00:46.340","Text":"our infinite ladder and we can see that we"},{"Start":"00:46.340 ","End":"00:51.245","Text":"have 2 resistors connecting like so and this will repeat."},{"Start":"00:51.245 ","End":"00:57.155","Text":"Then the next unit has 2 resistors connected in the same way,"},{"Start":"00:57.155 ","End":"00:59.830","Text":"and so on and so forth."},{"Start":"00:59.830 ","End":"01:06.200","Text":"What we\u0027re trying to find is we\u0027re trying to find some resistor that"},{"Start":"01:06.200 ","End":"01:12.005","Text":"we can connect over here with the resistance R total,"},{"Start":"01:12.005 ","End":"01:20.575","Text":"where R total is equal to all of these resistors connected like so in this ladder,"},{"Start":"01:20.575 ","End":"01:26.950","Text":"question number 1 is to find R total."},{"Start":"01:27.290 ","End":"01:34.479","Text":"Let\u0027s rub this out for 1 second,"},{"Start":"01:34.479 ","End":"01:37.890","Text":"here we said this is 1 unit."},{"Start":"01:37.890 ","End":"01:45.170","Text":"What do we want to do is we want to separate this first unit from all of the other units."},{"Start":"01:45.170 ","End":"01:49.385","Text":"I\u0027m making a cut over here,"},{"Start":"01:49.385 ","End":"01:51.860","Text":"this is now separated."},{"Start":"01:51.860 ","End":"02:00.630","Text":"And now what I want to do is I want to calculate the resistance of all of this over here."},{"Start":"02:02.030 ","End":"02:06.125","Text":"Because we were told that this is an infinite ladder,"},{"Start":"02:06.125 ","End":"02:11.299","Text":"so the difference of having infinite units"},{"Start":"02:11.299 ","End":"02:16.640","Text":"or infinity minus 1 units doesn\u0027t really make a difference."},{"Start":"02:16.640 ","End":"02:18.380","Text":"What we can see,"},{"Start":"02:18.380 ","End":"02:21.320","Text":"that\u0027s a trick that you use when dealing with anything with"},{"Start":"02:21.320 ","End":"02:25.340","Text":"infinity is that the resistance of everything over here"},{"Start":"02:25.340 ","End":"02:30.800","Text":"in blue so of all of the units and all of the rungs of"},{"Start":"02:30.800 ","End":"02:33.950","Text":"the ladder so the resistance in blue is"},{"Start":"02:33.950 ","End":"02:37.610","Text":"going to be the same as the resistance of everything,"},{"Start":"02:37.610 ","End":"02:42.125","Text":"all the rungs within this green section over here,"},{"Start":"02:42.125 ","End":"02:45.350","Text":"exactly because we have infinite rungs."},{"Start":"02:45.350 ","End":"02:50.970","Text":"If we have infinite rungs or infinity minus 1 rungs,"},{"Start":"02:50.970 ","End":"02:53.710","Text":"the resistance will be the same."},{"Start":"02:53.870 ","End":"02:57.045","Text":"Using that idea,"},{"Start":"02:57.045 ","End":"03:04.835","Text":"what we\u0027re going to do is we\u0027re going to take away all of these units over here in green,"},{"Start":"03:04.835 ","End":"03:11.510","Text":"imagine that they don\u0027t exist and then we\u0027re going to add in a resistor over here in red,"},{"Start":"03:11.510 ","End":"03:15.195","Text":"which has a resistance of our total."},{"Start":"03:15.195 ","End":"03:18.980","Text":"Because we said that the resistance of everything in blue is R total,"},{"Start":"03:18.980 ","End":"03:22.190","Text":"and the resistance of everything in green is R total,"},{"Start":"03:22.190 ","End":"03:29.800","Text":"I\u0027ve switched out everything in green with the resistance of R total."},{"Start":"03:30.500 ","End":"03:36.756","Text":"Now, we\u0027re going to use a very strange equation,"},{"Start":"03:36.756 ","End":"03:38.195","Text":"but it makes sense."},{"Start":"03:38.195 ","End":"03:43.555","Text":"As we said, everything in blue is equal to"},{"Start":"03:43.555 ","End":"03:49.070","Text":"R total and everything in green is also equal to R total."},{"Start":"03:49.070 ","End":"03:56.975","Text":"Because we\u0027re using the idea that infinity minus 1 is still equal to infinity,"},{"Start":"03:56.975 ","End":"04:03.660","Text":"we replaced everything in green with this resistor R total instead."},{"Start":"04:04.600 ","End":"04:09.695","Text":"Now, we know that if we add this resistance,"},{"Start":"04:09.695 ","End":"04:14.195","Text":"R total plus the resistance due to these 2 resistors,"},{"Start":"04:14.195 ","End":"04:17.375","Text":"this also has to be equal to R total."},{"Start":"04:17.375 ","End":"04:23.570","Text":"Because then we\u0027re taking into account all of the resistors within the blue circle,"},{"Start":"04:23.570 ","End":"04:26.585","Text":"which we know is equal to R total."},{"Start":"04:26.585 ","End":"04:31.895","Text":"We\u0027re going to add the red resistor to these 2 resistors"},{"Start":"04:31.895 ","End":"04:37.680","Text":"in the black circle and the total resistance is equal to R total."},{"Start":"04:40.060 ","End":"04:46.370","Text":"The first thing that we\u0027re going to do is we can see that the red resistor,"},{"Start":"04:46.370 ","End":"04:50.930","Text":"and this black resistor I\u0027m reminding you that both have"},{"Start":"04:50.930 ","End":"04:57.620","Text":"resistance of R. These 2 resistors over here,"},{"Start":"04:57.620 ","End":"05:03.120","Text":"the red and this black 1 are connected in parallel."},{"Start":"05:03.590 ","End":"05:10.425","Text":"Therefore we can say that the resistance R1,2 is equal to,"},{"Start":"05:10.425 ","End":"05:13.580","Text":"how do we connect 2 resistors in parallel?"},{"Start":"05:13.580 ","End":"05:21.020","Text":"It\u0027s equal to the resistance of this one multiplied by the resistance of this one so"},{"Start":"05:21.020 ","End":"05:24.770","Text":"multiplied by R total divided by the resistance of"},{"Start":"05:24.770 ","End":"05:30.180","Text":"this one plus the resistance of this one."},{"Start":"05:30.350 ","End":"05:35.120","Text":"All these 2 resistors are now equal to"},{"Start":"05:35.120 ","End":"05:41.225","Text":"R1,2 and then we\u0027re going to add that in series to this resistor R over here,"},{"Start":"05:41.225 ","End":"05:44.760","Text":"because it\u0027s now connected in series to this."},{"Start":"05:45.890 ","End":"05:49.685","Text":"When we add all of this resistance together,"},{"Start":"05:49.685 ","End":"05:52.940","Text":"we get R total because then we have"},{"Start":"05:52.940 ","End":"05:58.250","Text":"the resistance of this rung in the ladder plus all of these rungs,"},{"Start":"05:58.250 ","End":"06:01.840","Text":"which we said is equal to R total."},{"Start":"06:01.840 ","End":"06:07.940","Text":"We get that R total is equal to this resistor connected in"},{"Start":"06:07.940 ","End":"06:14.590","Text":"series to this resistor. R plus R1,2,"},{"Start":"06:14.590 ","End":"06:25.275","Text":"which is equal to R plus R multiplied by R total divided by R plus total."},{"Start":"06:25.275 ","End":"06:28.190","Text":"Now, what we\u0027re going to do is we\u0027re going to"},{"Start":"06:28.190 ","End":"06:31.405","Text":"multiply both sides by this denominator over here,"},{"Start":"06:31.405 ","End":"06:37.430","Text":"because we\u0027re multiplying both sides by R plus R total."},{"Start":"06:37.430 ","End":"06:45.245","Text":"Then what we\u0027ll have is R,R total plus R total squared,"},{"Start":"06:45.245 ","End":"06:51.620","Text":"which is equal to R multiplied by R,"},{"Start":"06:51.620 ","End":"06:58.550","Text":"or just R^2 plus R,"},{"Start":"06:58.550 ","End":"07:07.665","Text":"R total and then plus over here R, R total."},{"Start":"07:07.665 ","End":"07:10.875","Text":"Now, we can subtract from both sides,"},{"Start":"07:10.875 ","End":"07:12.615","Text":"R, R total."},{"Start":"07:12.615 ","End":"07:17.420","Text":"Then what we have is that R total squared,"},{"Start":"07:17.420 ","End":"07:21.215","Text":"and then we can do this so minus R,"},{"Start":"07:21.215 ","End":"07:28.745","Text":"R total minus R^2 is equal to 0."},{"Start":"07:28.745 ","End":"07:32.375","Text":"We have a quadratic equation,"},{"Start":"07:32.375 ","End":"07:40.345","Text":"which now we can solve via our quadratic equation solver,"},{"Start":"07:40.345 ","End":"07:46.513","Text":"we\u0027ll get that R total 1,2 is equal to,"},{"Start":"07:46.513 ","End":"07:50.940","Text":"we\u0027ll have R plus,"},{"Start":"07:50.940 ","End":"07:56.220","Text":"minus the square root of R^2 minus"},{"Start":"07:56.220 ","End":"08:05.980","Text":"4^2 and then we have this divided by 2."},{"Start":"08:06.590 ","End":"08:12.045","Text":"Now, what we can do is we can simplify this,"},{"Start":"08:12.045 ","End":"08:15.093","Text":"we take the R^2 out,"},{"Start":"08:15.093 ","End":"08:25.960","Text":"what we\u0027ll have is 1 plus minus the square root of 5 divided by 2R."},{"Start":"08:27.950 ","End":"08:31.850","Text":"Now, we can see we have 2 options of answers,"},{"Start":"08:31.850 ","End":"08:38.480","Text":"but the only answer that we can use is the answer with the plus,"},{"Start":"08:38.480 ","End":"08:44.330","Text":"because 1 minus root 5 is a negative number and then we get a negative resistance,"},{"Start":"08:44.330 ","End":"08:46.775","Text":"which doesn\u0027t make sense."},{"Start":"08:46.775 ","End":"08:52.700","Text":"We can only use the option for 1 plus root 5 so therefore the resistance is"},{"Start":"08:52.700 ","End":"08:58.980","Text":"1 plus root 5 divided by 2R."},{"Start":"09:00.890 ","End":"09:05.630","Text":"This is the answer to question number 1 and the trick is to"},{"Start":"09:05.630 ","End":"09:10.765","Text":"realize this that infinity minus 1 is still equal to infinity."},{"Start":"09:10.765 ","End":"09:14.705","Text":"To take out all the other rungs in the infinite ladder,"},{"Start":"09:14.705 ","End":"09:20.315","Text":"replace it by R total and then you say that this resistance,"},{"Start":"09:20.315 ","End":"09:28.013","Text":"R total is equal plus when we add it onto the other resistors and the circuit"},{"Start":"09:28.013 ","End":"09:36.550","Text":"is going to be equal to R total as well and then from there you just solve the equation."},{"Start":"09:36.550 ","End":"09:39.713","Text":"Now, let\u0027s take a look at question Number 2,"},{"Start":"09:39.713 ","End":"09:44.525","Text":"question Number 2 is to calculate the current in the battery."},{"Start":"09:44.525 ","End":"09:50.465","Text":"The current I is simply going to be equal to the voltage"},{"Start":"09:50.465 ","End":"09:56.960","Text":"divided by the total resistance of the circuit, R total."},{"Start":"09:56.960 ","End":"10:01.850","Text":"We haven\u0027t been given this value so we can just leave it as"},{"Start":"10:01.850 ","End":"10:07.100","Text":"v. We have V divided by R value for R total,"},{"Start":"10:07.100 ","End":"10:12.960","Text":"which is simply equal to 1 plus the square root of 5,"},{"Start":"10:12.960 ","End":"10:19.640","Text":"all of this multiplied by R and then R2 in the denominator comes up here."},{"Start":"10:19.640 ","End":"10:22.190","Text":"That\u0027s the answer to question Number 2,"},{"Start":"10:22.190 ","End":"10:24.990","Text":"and that is the end of this lesson."}],"ID":22308},{"Watched":false,"Name":"Exercise 11","Duration":"14m 39s","ChapterTopicVideoID":21492,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lesson,"},{"Start":"00:01.890 ","End":"00:04.295","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.295 ","End":"00:08.640","Text":"In the diagram, there is an infinite ladder circuit containing resistors."},{"Start":"00:08.640 ","End":"00:14.295","Text":"Each resistor has a resistance of R. Find an equation for the voltage on"},{"Start":"00:14.295 ","End":"00:21.760","Text":"each resistor and calculate the current on the 23rd perpendicular resistor."},{"Start":"00:22.070 ","End":"00:24.540","Text":"In the previous lesson,"},{"Start":"00:24.540 ","End":"00:30.540","Text":"we saw that the total resistance of this infinite ladder circuit"},{"Start":"00:30.540 ","End":"00:38.040","Text":"was equal to 1 plus the square root of 5 divided by 2R,"},{"Start":"00:38.040 ","End":"00:44.405","Text":"and this we calculated via using the idea that"},{"Start":"00:44.405 ","End":"00:51.960","Text":"the resistance of the whole ladder minus this first run is equal to R total,"},{"Start":"00:51.960 ","End":"00:55.460","Text":"and the resistance of everything including the first rung is"},{"Start":"00:55.460 ","End":"00:59.495","Text":"also equal to R total, which is this."},{"Start":"00:59.495 ","End":"01:04.260","Text":"Then we just solve that equation to get this."},{"Start":"01:05.150 ","End":"01:10.680","Text":"What we want to do is we want to know the voltage on each resistor."},{"Start":"01:10.750 ","End":"01:15.725","Text":"If we put in a volt meter over here,"},{"Start":"01:15.725 ","End":"01:19.080","Text":"we would measure V_1,"},{"Start":"01:19.080 ","End":"01:22.140","Text":"the voltage on the first rung."},{"Start":"01:22.140 ","End":"01:27.485","Text":"Then if we would put a voltmeter over here,"},{"Start":"01:27.485 ","End":"01:30.685","Text":"this is the voltage on the 2nd run,"},{"Start":"01:30.685 ","End":"01:37.470","Text":"V_2, and it\u0027s actually measuring the voltage on this resistor over here."},{"Start":"01:37.660 ","End":"01:42.710","Text":"Then over here, we\u0027ll have our V_3,"},{"Start":"01:42.710 ","End":"01:45.335","Text":"which is the voltage on the 3rd rung,"},{"Start":"01:45.335 ","End":"01:48.150","Text":"and then so on and so forth."},{"Start":"01:48.530 ","End":"01:50.580","Text":"As we can see,"},{"Start":"01:50.580 ","End":"01:53.960","Text":"V_1 is connected to a voltage source."},{"Start":"01:53.960 ","End":"01:58.595","Text":"We can see that V_1 is going to be equal to V over here."},{"Start":"01:58.595 ","End":"02:02.575","Text":"Now we want to find an equation for V_2 and V_3,"},{"Start":"02:02.575 ","End":"02:07.410","Text":"and then eventually we\u0027ll be able to find an equation for V_n."},{"Start":"02:08.030 ","End":"02:16.115","Text":"The first thing that we\u0027re going to calculate is the voltage across this resistor."},{"Start":"02:16.115 ","End":"02:18.425","Text":"Let\u0027s say over here,"},{"Start":"02:18.425 ","End":"02:20.180","Text":"we have a current flowing,"},{"Start":"02:20.180 ","End":"02:22.750","Text":"and let\u0027s call this I_1."},{"Start":"02:22.750 ","End":"02:25.775","Text":"What is I_1 equal to?"},{"Start":"02:25.775 ","End":"02:29.510","Text":"I_1 is equal to the voltage at this point,"},{"Start":"02:29.510 ","End":"02:30.725","Text":"which is V_1,"},{"Start":"02:30.725 ","End":"02:36.465","Text":"which is in fact V divided by the total resistance of the circuit,"},{"Start":"02:36.465 ","End":"02:39.760","Text":"so divided by R total."},{"Start":"02:40.190 ","End":"02:43.610","Text":"The fact that I put V_2 over here,"},{"Start":"02:43.610 ","End":"02:47.145","Text":"I could have also connected it over here."},{"Start":"02:47.145 ","End":"02:49.290","Text":"This is still V_2."},{"Start":"02:49.290 ","End":"02:55.940","Text":"We would get the same voltage measurement before this resistor or after this resistor."},{"Start":"02:55.940 ","End":"02:59.630","Text":"If I wanted an equation for V_2,"},{"Start":"02:59.630 ","End":"03:04.310","Text":"it would be equal to the voltage at this point over here,"},{"Start":"03:04.310 ","End":"03:06.965","Text":"which is V_1 or V,"},{"Start":"03:06.965 ","End":"03:11.575","Text":"minus the voltage drop over this resistor."},{"Start":"03:11.575 ","End":"03:19.145","Text":"The voltage drop over this resistor is just equal to I_R, and of course,"},{"Start":"03:19.145 ","End":"03:24.610","Text":"each resistor has a resistance of R. It\u0027s just going to be equal to I_1,"},{"Start":"03:24.610 ","End":"03:26.540","Text":"the current across over here,"},{"Start":"03:26.540 ","End":"03:29.936","Text":"multiplied by the resistance of this resistor."},{"Start":"03:29.936 ","End":"03:33.080","Text":"That\u0027s the voltage drop between"},{"Start":"03:33.080 ","End":"03:39.090","Text":"this point to this point or this point and it makes no difference."},{"Start":"03:39.740 ","End":"03:42.510","Text":"Now I can rewrite this,"},{"Start":"03:42.510 ","End":"03:47.100","Text":"as being equal to V_1 minus my I_1 is"},{"Start":"03:47.100 ","End":"03:52.260","Text":"equal to V_1 divided by R total and then multiplied by"},{"Start":"03:52.260 ","End":"03:56.855","Text":"R. Or I can just write this as being equal to V_1"},{"Start":"03:56.855 ","End":"04:03.390","Text":"and then 1 minus R divided by R total."},{"Start":"04:04.550 ","End":"04:11.315","Text":"The equation for V_3 over here is going to be exactly the same thing."},{"Start":"04:11.315 ","End":"04:21.095","Text":"V_3 is going to be equal to the voltage V_2 minus the voltage drop over this resistor,"},{"Start":"04:21.095 ","End":"04:29.690","Text":"where this resistor is a resistance R. It\u0027s going to be minus I_2 multiplied by R,"},{"Start":"04:29.690 ","End":"04:35.280","Text":"where this over here is my current I_2."},{"Start":"04:36.110 ","End":"04:40.265","Text":"Now I have to know what I_2 is equal to."},{"Start":"04:40.265 ","End":"04:49.380","Text":"I_2 is going to be equal to V_2 divided by R total."},{"Start":"04:50.150 ","End":"04:53.420","Text":"Why is that? In the previous lesson,"},{"Start":"04:53.420 ","End":"04:56.600","Text":"we saw that when dealing with an infinite ladder circuit,"},{"Start":"04:56.600 ","End":"05:02.030","Text":"the resistance from the 2nd rung and onwards is the same as"},{"Start":"05:02.030 ","End":"05:05.625","Text":"the resistance from the first rung and"},{"Start":"05:05.625 ","End":"05:09.735","Text":"onwards in this ladder circuit, because it\u0027s infinite."},{"Start":"05:09.735 ","End":"05:13.790","Text":"Therefore, the total resistance of this section"},{"Start":"05:13.790 ","End":"05:17.795","Text":"over here from V_2 and all the way down is R total,"},{"Start":"05:17.795 ","End":"05:21.800","Text":"just like it was from the first rung and all the way down."},{"Start":"05:21.800 ","End":"05:28.325","Text":"Then, of course, the current is equal to the voltage divided by the total resistance,"},{"Start":"05:28.325 ","End":"05:31.290","Text":"so that\u0027s why we get this equation."},{"Start":"05:31.760 ","End":"05:36.360","Text":"Then I can plug this over here."},{"Start":"05:36.360 ","End":"05:41.645","Text":"I\u0027ll get that V_3 is equal to V_2 minus I_2,"},{"Start":"05:41.645 ","End":"05:49.505","Text":"which is V_2 divided by R total and then multiplied by R. Or in other words,"},{"Start":"05:49.505 ","End":"05:57.520","Text":"I have V_2 and then I have 1 minus R divided by R total."},{"Start":"05:58.160 ","End":"06:00.515","Text":"Therefore, what I can say,"},{"Start":"06:00.515 ","End":"06:01.775","Text":"I\u0027ll write it over here."},{"Start":"06:01.775 ","End":"06:05.255","Text":"We can see that the same pattern is going on."},{"Start":"06:05.255 ","End":"06:10.950","Text":"What I\u0027ll have is that at V_n plus 1,"},{"Start":"06:10.950 ","End":"06:14.850","Text":"the next rung up will be equal to"},{"Start":"06:14.850 ","End":"06:22.515","Text":"V_n multiplied by 1 minus R divided by R total."},{"Start":"06:22.515 ","End":"06:25.695","Text":"If I had V_4,"},{"Start":"06:25.695 ","End":"06:28.485","Text":"so that would be n plus 1."},{"Start":"06:28.485 ","End":"06:36.520","Text":"V_4 would be equal to V_3 multiplied by what\u0027s in the brackets."},{"Start":"06:36.520 ","End":"06:43.550","Text":"Now, of course, we can simplify this more because here we have I divided by R total,"},{"Start":"06:43.550 ","End":"06:45.260","Text":"where I noticed in R total,"},{"Start":"06:45.260 ","End":"06:49.655","Text":"we also have this factor R over here that we can cancel out."},{"Start":"06:49.655 ","End":"06:55.280","Text":"What we\u0027ll actually have is V_n multiplied by 1 minus,"},{"Start":"06:55.280 ","End":"06:59.495","Text":"and then this fraction gets flipped around because R total is in the denominator,"},{"Start":"06:59.495 ","End":"07:08.296","Text":"so it\u0027s just going to be minus 2 divided by 1 plus the square root of 5."},{"Start":"07:08.296 ","End":"07:14.515","Text":"Now we can even simplify this more further."},{"Start":"07:14.515 ","End":"07:16.540","Text":"We can make a common denominator."},{"Start":"07:16.540 ","End":"07:18.775","Text":"What we\u0027ll have is V_n,"},{"Start":"07:18.775 ","End":"07:22.120","Text":"and then we have the common denominator,"},{"Start":"07:22.120 ","End":"07:26.740","Text":"so we\u0027ll have 1 plus the square root of 5 and then minus"},{"Start":"07:26.740 ","End":"07:31.855","Text":"2 divided by 1 plus the square root of 5."},{"Start":"07:31.855 ","End":"07:40.705","Text":"Then therefore I will get that the equation for V_n plus 1 is equal to V_n multiplied by,"},{"Start":"07:40.705 ","End":"07:49.335","Text":"so we have root 5 minus 1 divided by root 5 plus 1."},{"Start":"07:49.335 ","End":"07:53.800","Text":"This is our equation for the voltage on each resistor."},{"Start":"07:53.900 ","End":"08:00.075","Text":"What I\u0027m going to do now is I\u0027m going to call what\u0027s going on over here in the brackets,"},{"Start":"08:00.075 ","End":"08:04.300","Text":"I\u0027m going to call this q. I\u0027m calling this q"},{"Start":"08:04.300 ","End":"08:09.655","Text":"because this equation is an example for a geometric progression."},{"Start":"08:09.655 ","End":"08:17.285","Text":"The general equation for a geometric progression is a_n plus 1."},{"Start":"08:17.285 ","End":"08:22.575","Text":"It\u0027s always an equation for the next term in the series and the progression,"},{"Start":"08:22.575 ","End":"08:26.250","Text":"which is equal to the previous term,"},{"Start":"08:26.250 ","End":"08:31.630","Text":"so a_n multiplied by some constant q."},{"Start":"08:32.220 ","End":"08:34.645","Text":"Or in other words,"},{"Start":"08:34.645 ","End":"08:39.151","Text":"I can say that it is equal to the first term,"},{"Start":"08:39.151 ","End":"08:43.670","Text":"a_1, multiplied by a constant q^n."},{"Start":"08:45.990 ","End":"08:51.790","Text":"What we\u0027re doing is we\u0027re taking our first term and multiplying it by our constant."},{"Start":"08:51.790 ","End":"08:53.605","Text":"Then to get to the second term,"},{"Start":"08:53.605 ","End":"08:56.215","Text":"we just multiply by the constant again."},{"Start":"08:56.215 ","End":"09:01.990","Text":"Let\u0027s say, we can see that V_1 is just equal to"},{"Start":"09:01.990 ","End":"09:09.770","Text":"V. Then we have that V_2 is equal to V_1,"},{"Start":"09:10.050 ","End":"09:17.335","Text":"or let\u0027s, just write V_1 multiplied by our q."},{"Start":"09:17.335 ","End":"09:19.405","Text":"Let\u0027s just leave it at that."},{"Start":"09:19.405 ","End":"09:22.405","Text":"Then I want to get to my V_3."},{"Start":"09:22.405 ","End":"09:24.880","Text":"Actually, let\u0027s just leave that as V_2."},{"Start":"09:24.880 ","End":"09:27.115","Text":"Now I want to get to my V_3,"},{"Start":"09:27.115 ","End":"09:35.420","Text":"so my V_3 is equal to my V_2 multiplied by q."},{"Start":"09:35.420 ","End":"09:37.935","Text":"All of this we saw, we got to q,"},{"Start":"09:37.935 ","End":"09:39.885","Text":"but what is my V_2?"},{"Start":"09:39.885 ","End":"09:43.005","Text":"My V_2 is equal to this,"},{"Start":"09:43.005 ","End":"09:46.065","Text":"which is V_1 multiplied by q,"},{"Start":"09:46.065 ","End":"09:48.910","Text":"and then again multiplied by q."},{"Start":"09:49.200 ","End":"09:55.270","Text":"That is how we get to this power of n. What would be V_4?"},{"Start":"09:55.270 ","End":"09:58.990","Text":"V_4 would be my V_3 q,"},{"Start":"09:58.990 ","End":"10:02.515","Text":"and V_3 would be V_2 q,"},{"Start":"10:02.515 ","End":"10:08.590","Text":"which is just V_1 q^2 then again"},{"Start":"10:08.590 ","End":"10:15.070","Text":"multiplied by this q over here, so q cubed."},{"Start":"10:15.070 ","End":"10:18.295","Text":"We can see that my n is my 3,"},{"Start":"10:18.295 ","End":"10:20.245","Text":"that\u0027s why it\u0027s to the power of n,"},{"Start":"10:20.245 ","End":"10:23.330","Text":"and my n plus 1 is my 4."},{"Start":"10:24.450 ","End":"10:27.220","Text":"In this case, this of course,"},{"Start":"10:27.220 ","End":"10:29.305","Text":"is a correct equation."},{"Start":"10:29.305 ","End":"10:35.695","Text":"However, if we want to make it into this version like a geometric progression."},{"Start":"10:35.695 ","End":"10:38.365","Text":"We can rub this out to make a bit more space."},{"Start":"10:38.365 ","End":"10:45.205","Text":"We can say that V_n plus 1 is simply equal to our first term,"},{"Start":"10:45.205 ","End":"10:46.960","Text":"which we saw was V_1,"},{"Start":"10:46.960 ","End":"10:52.150","Text":"where we said that V_1 is equal to V multiplied by our q,"},{"Start":"10:52.150 ","End":"10:53.529","Text":"which is our constant,"},{"Start":"10:53.529 ","End":"11:01.720","Text":"which is this, which is root 5 minus 1 divided by root 5 plus 1."},{"Start":"11:01.720 ","End":"11:09.925","Text":"All of this is to the power of n. Both versions are correct,"},{"Start":"11:09.925 ","End":"11:12.175","Text":"but we\u0027re going to go with this one because it looks"},{"Start":"11:12.175 ","End":"11:15.880","Text":"a bit neater and is easier to calculate."},{"Start":"11:15.880 ","End":"11:19.900","Text":"The next part of the question is to calculate the"},{"Start":"11:19.900 ","End":"11:23.965","Text":"current on the 23rd perpendicular resistor."},{"Start":"11:23.965 ","End":"11:27.325","Text":"This is the first perpendicular resistor,"},{"Start":"11:27.325 ","End":"11:30.145","Text":"this is the second perpendicular resistor,"},{"Start":"11:30.145 ","End":"11:32.890","Text":"this is the third perpendicular resistor,"},{"Start":"11:32.890 ","End":"11:39.310","Text":"and we\u0027re trying to find the current on the 23rd one."},{"Start":"11:39.310 ","End":"11:46.195","Text":"First of all what we can see is that the voltage on the resistors,"},{"Start":"11:46.195 ","End":"11:53.410","Text":"let\u0027s say if this is the second perpendicular resistor,"},{"Start":"11:53.410 ","End":"11:57.890","Text":"we can see that the voltage across it is V_3."},{"Start":"11:57.890 ","End":"12:02.410","Text":"It\u0027s the voltage for the third rung,"},{"Start":"12:02.410 ","End":"12:06.115","Text":"even though this is the second perpendicular resistor."},{"Start":"12:06.115 ","End":"12:09.580","Text":"This is the first perpendicular resistor and the voltage across"},{"Start":"12:09.580 ","End":"12:14.180","Text":"it is the voltage for the next rung, V_2."},{"Start":"12:15.270 ","End":"12:23.935","Text":"Let\u0027s imagine I wanted to see the current going over here."},{"Start":"12:23.935 ","End":"12:26.980","Text":"We see that I_1 is the current traveling here,"},{"Start":"12:26.980 ","End":"12:31.450","Text":"and then at this point it splits into I_2."},{"Start":"12:31.450 ","End":"12:36.610","Text":"Let\u0027s call the current going over here flowing across this resistor,"},{"Start":"12:36.610 ","End":"12:40.340","Text":"let\u0027s call it I_1 tilde."},{"Start":"12:40.680 ","End":"12:44.395","Text":"What is I_1 tilde equal to?"},{"Start":"12:44.395 ","End":"12:48.205","Text":"I_1 tilde is equal to the voltage"},{"Start":"12:48.205 ","End":"12:53.680","Text":"across this resistor divided by the resistance of the resistor."},{"Start":"12:53.680 ","End":"12:55.930","Text":"The voltage going across, we said,"},{"Start":"12:55.930 ","End":"12:58.630","Text":"is this voltage over here,"},{"Start":"12:58.630 ","End":"13:03.235","Text":"V_2 divided by the resistance of the resistor,"},{"Start":"13:03.235 ","End":"13:08.410","Text":"which is R. Now if I want the"},{"Start":"13:08.410 ","End":"13:14.665","Text":"current therefore on the 23rd perpendicular resistor,"},{"Start":"13:14.665 ","End":"13:22.765","Text":"that will equal to the voltage on the 23rd perpendicular resistor."},{"Start":"13:22.765 ","End":"13:28.075","Text":"But here it would be 24 because the voltage across"},{"Start":"13:28.075 ","End":"13:35.110","Text":"the 23rd perpendicular resistor is going to be equal to V_24."},{"Start":"13:35.110 ","End":"13:38.110","Text":"Let\u0027s make that a bit clearer, V_24,"},{"Start":"13:38.110 ","End":"13:41.020","Text":"just like we saw I_1 will be V_2,"},{"Start":"13:41.020 ","End":"13:47.365","Text":"so I_23 will be V_24 divided by R. What is V_24?"},{"Start":"13:47.365 ","End":"13:49.645","Text":"It\u0027s what we just calculated over here."},{"Start":"13:49.645 ","End":"13:56.305","Text":"So it\u0027s going to be equal to V divided by the resistance,"},{"Start":"13:56.305 ","End":"13:59.305","Text":"multiplied by this over here."},{"Start":"13:59.305 ","End":"14:06.100","Text":"The square root of 5 minus 1 divided by the square root of 5 plus 1."},{"Start":"14:06.100 ","End":"14:08.770","Text":"Then it\u0027s going to be to the power of what?"},{"Start":"14:08.770 ","End":"14:12.085","Text":"Notice the V is n plus 1,"},{"Start":"14:12.085 ","End":"14:17.365","Text":"then this over here is to the power of n. V over here is 24."},{"Start":"14:17.365 ","End":"14:19.855","Text":"24 is equal to n plus 1,"},{"Start":"14:19.855 ","End":"14:23.230","Text":"which means that n is equal to 3,"},{"Start":"14:23.230 ","End":"14:28.720","Text":"because 24 minus 1 is what n is equal to, which is 23."},{"Start":"14:28.720 ","End":"14:33.280","Text":"Here it\u0027s 1 less than what we have here."},{"Start":"14:33.280 ","End":"14:36.665","Text":"That is the answer to this question,"},{"Start":"14:36.665 ","End":"14:39.630","Text":"and this is the end of the lesson."}],"ID":22309},{"Watched":false,"Name":"Exercise 12","Duration":"27m 38s","ChapterTopicVideoID":21493,"CourseChapterTopicPlaylistID":99474,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.085","Text":"Hello. In this lesson,"},{"Start":"00:02.085 ","End":"00:04.830","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.830 ","End":"00:09.585","Text":"We\u0027re being told to calculate the resistance in the following circuit."},{"Start":"00:09.585 ","End":"00:15.600","Text":"We can see that this is an infinite ladder circuit or in other words,"},{"Start":"00:15.600 ","End":"00:19.710","Text":"calculate the total resistance between points A and point B."},{"Start":"00:19.710 ","End":"00:25.260","Text":"Then we\u0027re being told that R_1 and R_2 are equal and to therefore,"},{"Start":"00:25.260 ","End":"00:28.870","Text":"calculate the current in each resistor."},{"Start":"00:29.420 ","End":"00:34.365","Text":"Again, we have this infinite ladder circuit."},{"Start":"00:34.365 ","End":"00:40.485","Text":"Our first rung is this over here, this X shape."},{"Start":"00:40.485 ","End":"00:45.150","Text":"Our second rung is this X shape,"},{"Start":"00:45.150 ","End":"00:48.170","Text":"our third rung is this over here,"},{"Start":"00:48.170 ","End":"00:50.770","Text":"and so on and so forth."},{"Start":"00:50.770 ","End":"00:56.915","Text":"The trick that we learned in the previous lesson or 2 lessons ago,"},{"Start":"00:56.915 ","End":"01:01.340","Text":"is when we\u0027re dealing with an infinite ladder circuit."},{"Start":"01:01.340 ","End":"01:09.070","Text":"The total resistance of all of the rungs in the ladder up until infinity,"},{"Start":"01:09.200 ","End":"01:13.140","Text":"is equal to R total."},{"Start":"01:13.140 ","End":"01:17.465","Text":"Our trick was because we\u0027re dealing with infinity,"},{"Start":"01:17.465 ","End":"01:24.555","Text":"that means that infinity minus 1 is also infinity."},{"Start":"01:24.555 ","End":"01:30.300","Text":"Therefore, we can take all of the rungs minus the first rung."},{"Start":"01:30.300 ","End":"01:33.245","Text":"All of this over here,"},{"Start":"01:33.245 ","End":"01:38.930","Text":"up until infinity and the resistance of all of these rungs minus the"},{"Start":"01:38.930 ","End":"01:45.540","Text":"first is also going to be equal to R total because we are dealing with infinity."},{"Start":"01:46.700 ","End":"01:50.150","Text":"Let\u0027s just redraw this diagram,"},{"Start":"01:50.150 ","End":"01:52.750","Text":"taking this into account."},{"Start":"01:52.750 ","End":"01:57.575","Text":"What I\u0027ve done is I\u0027ve taken my first rung over here."},{"Start":"01:57.575 ","End":"02:02.735","Text":"Then instead of all of the rungs enclosed in the red,"},{"Start":"02:02.735 ","End":"02:05.450","Text":"whose resistance I know is R total,"},{"Start":"02:05.450 ","End":"02:16.290","Text":"I\u0027m just going to replace this with 1 resistor over here with the resistance of R total."},{"Start":"02:16.700 ","End":"02:23.690","Text":"Now of course, the total resistance of all of this is also equal to R total,"},{"Start":"02:23.690 ","End":"02:26.540","Text":"as we saw 2 lessons ago but first,"},{"Start":"02:26.540 ","End":"02:32.360","Text":"let\u0027s calculate the resistance of everything in here."},{"Start":"02:32.360 ","End":"02:36.210","Text":"Then we\u0027ll equate that back again to R total."},{"Start":"02:37.570 ","End":"02:42.700","Text":"It\u0027s a little bit hard to understand what\u0027s going on from this diagram."},{"Start":"02:42.700 ","End":"02:45.325","Text":"Let\u0027s draw it in an easier way."},{"Start":"02:45.325 ","End":"02:48.100","Text":"We\u0027re starting at point A."},{"Start":"02:48.100 ","End":"02:50.365","Text":"Then we can see that from A,"},{"Start":"02:50.365 ","End":"02:53.175","Text":"we go into R_1."},{"Start":"02:53.175 ","End":"02:56.195","Text":"What we\u0027re trying to do is we\u0027re trying to find a route"},{"Start":"02:56.195 ","End":"02:59.900","Text":"from A to B and then we\u0027re going to build on it."},{"Start":"02:59.900 ","End":"03:04.595","Text":"I can go up R_1 and then down to R_2,"},{"Start":"03:04.595 ","End":"03:06.595","Text":"and I get to B."},{"Start":"03:06.595 ","End":"03:08.507","Text":"Let\u0027s do this."},{"Start":"03:08.507 ","End":"03:14.405","Text":"Here I have my R_1 and then here,"},{"Start":"03:14.405 ","End":"03:18.160","Text":"I have my R_2."},{"Start":"03:18.160 ","End":"03:22.870","Text":"Then I get to this point over here, B."},{"Start":"03:22.910 ","End":"03:29.150","Text":"Another route that I could have gotten from A to B is to go up R_1,"},{"Start":"03:29.150 ","End":"03:35.070","Text":"up R total and then get to another R_1 and"},{"Start":"03:35.070 ","End":"03:41.175","Text":"get to B. I could have gone up R_1, then R total."},{"Start":"03:41.175 ","End":"03:47.610","Text":"I can see that my R total splits or so between my R_1 and R_2."},{"Start":"03:47.610 ","End":"03:49.835","Text":"I can go down here."},{"Start":"03:49.835 ","End":"03:53.460","Text":"Let\u0027s say this is my R total."},{"Start":"03:54.070 ","End":"03:59.430","Text":"Then it goes down R_1,"},{"Start":"03:59.680 ","End":"04:06.570","Text":"like so and then it gets to B."},{"Start":"04:06.570 ","End":"04:10.100","Text":"Then of course I could also go from A to B via"},{"Start":"04:10.100 ","End":"04:14.630","Text":"going from A straight through this R_2 over here,"},{"Start":"04:14.630 ","End":"04:16.585","Text":"and then down R_1."},{"Start":"04:16.585 ","End":"04:24.735","Text":"That means I could go like so from A and then straight down to resistor over here,"},{"Start":"04:24.735 ","End":"04:31.630","Text":"R_2 and then I get to R_1 and back to B."},{"Start":"04:33.320 ","End":"04:39.125","Text":"Now what I can do just to give us a little bit more space is I can just"},{"Start":"04:39.125 ","End":"04:46.370","Text":"extend this over here and call this A and B. I haven\u0027t changed anything."},{"Start":"04:46.370 ","End":"04:49.115","Text":"Now let\u0027s look at the currents."},{"Start":"04:49.115 ","End":"04:54.990","Text":"Let\u0027s say here I have a current I flowing through."},{"Start":"04:54.990 ","End":"04:58.220","Text":"Then my current splits."},{"Start":"04:58.220 ","End":"05:03.585","Text":"It splits when it goes up R_1 or up R_2."},{"Start":"05:03.585 ","End":"05:06.830","Text":"Let\u0027s say that the current through here,"},{"Start":"05:06.830 ","End":"05:15.695","Text":"let\u0027s call this I_1 and that the current through here is I_2."},{"Start":"05:15.695 ","End":"05:23.095","Text":"Then I can say that the current going through here is"},{"Start":"05:23.095 ","End":"05:31.145","Text":"I_3 and I can say that the current over here is I_4 and over here is I_5."},{"Start":"05:31.145 ","End":"05:34.490","Text":"Then we get back to here where of course,"},{"Start":"05:34.490 ","End":"05:38.100","Text":"I\u0027m going to have the same current I flowing through."},{"Start":"05:39.020 ","End":"05:43.850","Text":"Now it will help if we imagine that our points A and"},{"Start":"05:43.850 ","End":"05:49.300","Text":"B are connected to some voltage source."},{"Start":"05:49.300 ","End":"05:57.460","Text":"Something that looks like so with a voltage V_AB,"},{"Start":"05:57.460 ","End":"06:02.232","Text":"the voltage between points AB is V_AB."},{"Start":"06:02.232 ","End":"06:08.490","Text":"Of course, here we have our current I flowing through."},{"Start":"06:09.320 ","End":"06:11.830","Text":"What I can do is I can use"},{"Start":"06:11.830 ","End":"06:16.450","Text":"Kirchhoff\u0027s law in order to try and solve the currents over here."},{"Start":"06:16.450 ","End":"06:21.490","Text":"However, what we can see is that the current over this resistor,"},{"Start":"06:21.490 ","End":"06:27.095","Text":"R_1 over here is going to be the current in this resistor, R_1 over here."},{"Start":"06:27.095 ","End":"06:29.890","Text":"Similarly over here, we have symmetry."},{"Start":"06:29.890 ","End":"06:33.520","Text":"The current in the resistor R_2 over here is going"},{"Start":"06:33.520 ","End":"06:37.675","Text":"to be the same as the current in this resistor, R_2 over here."},{"Start":"06:37.675 ","End":"06:44.530","Text":"Therefore, we can see that I_1 is equal to I_4,"},{"Start":"06:44.530 ","End":"06:53.980","Text":"and similarly, I_2 is equal to I_5."},{"Start":"06:53.980 ","End":"07:02.495","Text":"Regarding I_3, we can see that we have I_1 going into this node over here."},{"Start":"07:02.495 ","End":"07:09.930","Text":"Then we have I_2 coming out from over here."},{"Start":"07:09.930 ","End":"07:15.500","Text":"Therefore, if whatever is leftover from this split over here,"},{"Start":"07:15.500 ","End":"07:19.435","Text":"this 3-way split, is going to be our I_3."},{"Start":"07:19.435 ","End":"07:26.835","Text":"In other words, I_3 is equal to I_1 minus I_2,"},{"Start":"07:26.835 ","End":"07:30.795","Text":"because I_1 turns into I_3 and I_2."},{"Start":"07:30.795 ","End":"07:36.600","Text":"Therefore, I_1 minus I_2 will give us what\u0027s leftover, which is I_3."},{"Start":"07:37.250 ","End":"07:44.015","Text":"What I did now is I just switched out so that I get these equations for current,"},{"Start":"07:44.015 ","End":"07:46.980","Text":"with only I_1, I_2."},{"Start":"07:46.980 ","End":"07:50.495","Text":"Then what I can notice that from symmetry,"},{"Start":"07:50.495 ","End":"07:56.410","Text":"if I flip this around so that my current was going in the opposite direction,"},{"Start":"07:56.410 ","End":"08:00.315","Text":"instead of clockwise it would start going counterclockwise."},{"Start":"08:00.315 ","End":"08:07.980","Text":"The voltage that I would have would be the same voltage because of the symmetry."},{"Start":"08:08.270 ","End":"08:13.215","Text":"The only difference is that in the diagram,"},{"Start":"08:13.215 ","End":"08:17.385","Text":"here my I_2 is at the bottom and here I_1 is at the top,"},{"Start":"08:17.385 ","End":"08:21.130","Text":"and here I_1 is at the bottom and I_2 is at the top."},{"Start":"08:21.590 ","End":"08:27.260","Text":"We\u0027ll, speak later about how we can prove that will have"},{"Start":"08:27.260 ","End":"08:32.075","Text":"the same voltage here as over here but in the meantime,"},{"Start":"08:32.075 ","End":"08:35.840","Text":"let\u0027s carry on answering the question."},{"Start":"08:35.840 ","End":"08:42.230","Text":"First of all, we can say that the voltage V_AB is equal to our"},{"Start":"08:42.230 ","End":"08:49.175","Text":"current I multiplied by the total resistance of this circuit."},{"Start":"08:49.175 ","End":"08:53.795","Text":"Now I can also say that V_AB,"},{"Start":"08:53.795 ","End":"09:00.210","Text":"remember I can just find a route from A to B. I can also say that"},{"Start":"09:00.210 ","End":"09:07.280","Text":"V_AB is equal to the current on this resistor multiplied by its resistance,"},{"Start":"09:07.280 ","End":"09:14.960","Text":"I_2 multiplied by R_2 plus the current on this resistor multiplied by its resistance."},{"Start":"09:14.960 ","End":"09:18.950","Text":"Plus I_1, R_1."},{"Start":"09:18.950 ","End":"09:21.485","Text":"V_AB is something that I just made off."},{"Start":"09:21.485 ","End":"09:22.895","Text":"I don\u0027t know what it is,"},{"Start":"09:22.895 ","End":"09:25.055","Text":"isn\u0027t given to me in the question."},{"Start":"09:25.055 ","End":"09:28.145","Text":"Now I could just get rid of that."},{"Start":"09:28.145 ","End":"09:30.930","Text":"So I\u0027m left with this equation."},{"Start":"09:31.130 ","End":"09:37.445","Text":"The next equation that I\u0027m going to do is I\u0027m going to choose some loop in here."},{"Start":"09:37.445 ","End":"09:44.220","Text":"Let\u0027s imagine that I\u0027m choosing this loop, like so."},{"Start":"09:46.570 ","End":"09:51.170","Text":"I have some voltage over here."},{"Start":"09:51.170 ","End":"09:56.255","Text":"Then I\u0027m going in the direction of the currents that I\u0027ve defined as the direction."},{"Start":"09:56.255 ","End":"10:00.035","Text":"That means I\u0027m going to have a voltage drop over this resistor."},{"Start":"10:00.035 ","End":"10:04.940","Text":"What I have is negative I_1, R_1."},{"Start":"10:04.940 ","End":"10:07.790","Text":"Then I go down this resistor and again,"},{"Start":"10:07.790 ","End":"10:09.470","Text":"I\u0027m in the direction of the current."},{"Start":"10:09.470 ","End":"10:11.720","Text":"I\u0027m going to have a voltage drop again,"},{"Start":"10:11.720 ","End":"10:21.635","Text":"so negative and then the current is I_1 minus I_2 multiplied by the resistance R total."},{"Start":"10:21.635 ","End":"10:24.935","Text":"Then I go across this resistor,"},{"Start":"10:24.935 ","End":"10:29.975","Text":"where now, I\u0027m going in the opposite direction to the current,"},{"Start":"10:29.975 ","End":"10:34.665","Text":"which means that my voltage drop is going to be positive,"},{"Start":"10:34.665 ","End":"10:40.274","Text":"plus I_2, R_2."},{"Start":"10:40.274 ","End":"10:42.670","Text":"All of this, of course,"},{"Start":"10:42.670 ","End":"10:44.800","Text":"because I\u0027ve come back to the same point,"},{"Start":"10:44.800 ","End":"10:47.240","Text":"is equal to 0."},{"Start":"10:47.940 ","End":"10:51.445","Text":"Now let\u0027s add in 1 more equation,"},{"Start":"10:51.445 ","End":"10:55.165","Text":"and that\u0027s the equation over here at this intersection."},{"Start":"10:55.165 ","End":"10:57.115","Text":"I have a current going in,"},{"Start":"10:57.115 ","End":"10:58.405","Text":"which is I,"},{"Start":"10:58.405 ","End":"11:03.130","Text":"and my I is equal to the currents coming out of this intersection,"},{"Start":"11:03.130 ","End":"11:08.410","Text":"so it\u0027s equal to the current I_1 plus I_2."},{"Start":"11:08.410 ","End":"11:11.695","Text":"Now let\u0027s scroll over here to give us a bit more space."},{"Start":"11:11.695 ","End":"11:15.380","Text":"I can plug that in over here."},{"Start":"11:15.570 ","End":"11:18.850","Text":"I have, therefore,"},{"Start":"11:18.850 ","End":"11:24.415","Text":"I multiplied by R_T is in fact equal to this over here."},{"Start":"11:24.415 ","End":"11:32.090","Text":"It\u0027s equal to I_1 plus I_2 multiplied by my R_T."},{"Start":"11:33.360 ","End":"11:37.690","Text":"As we can see, I have 3 unknowns, I_1,"},{"Start":"11:37.690 ","End":"11:42.625","Text":"I_2, and R total, and 2 equations."},{"Start":"11:42.625 ","End":"11:47.665","Text":"All you have to do is you have to isolate either I_1 or I_2."},{"Start":"11:47.665 ","End":"11:52.120","Text":"Then when you substitute that into the equation,"},{"Start":"11:52.120 ","End":"11:58.000","Text":"you\u0027re going to be left with an equation that just has R total."},{"Start":"11:58.000 ","End":"12:01.780","Text":"If you isolate out I_1 and then you plug that in,"},{"Start":"12:01.780 ","End":"12:05.080","Text":"you\u0027ll have an equation just with I_2 and R total,"},{"Start":"12:05.080 ","End":"12:09.115","Text":"the I_2\u0027s will cancel out and you\u0027ll just have something left with R total."},{"Start":"12:09.115 ","End":"12:11.830","Text":"This happens always."},{"Start":"12:11.830 ","End":"12:15.505","Text":"Let\u0027s isolate out our I_1."},{"Start":"12:15.505 ","End":"12:18.370","Text":"Let\u0027s do it from this equation over here."},{"Start":"12:18.370 ","End":"12:22.540","Text":"What we\u0027ll have is that if we move stuff around,"},{"Start":"12:22.540 ","End":"12:32.710","Text":"so we\u0027ll see that I_2 multiplies our R_2 over here."},{"Start":"12:32.710 ","End":"12:40.375","Text":"Also, R_T over here with a minus and a minus becomes a plus."},{"Start":"12:40.375 ","End":"12:43.780","Text":"Plus our R total."},{"Start":"12:43.780 ","End":"12:47.140","Text":"This is equal to our I_1,"},{"Start":"12:47.140 ","End":"12:49.345","Text":"which I\u0027ve moved to the other side."},{"Start":"12:49.345 ","End":"12:52.240","Text":"It\u0027s a positive and then this negative when moved to"},{"Start":"12:52.240 ","End":"12:57.055","Text":"the other side also becomes positive."},{"Start":"12:57.055 ","End":"13:04.495","Text":"Then we have that I_1 multiplies R_1 plus R total."},{"Start":"13:04.495 ","End":"13:07.435","Text":"I\u0027ve just rearranged this equation."},{"Start":"13:07.435 ","End":"13:10.885","Text":"Now if I want to isolate out I_1,"},{"Start":"13:10.885 ","End":"13:15.970","Text":"it\u0027s simply equal to I_2 multiplied by R_2"},{"Start":"13:15.970 ","End":"13:24.350","Text":"plus R total divided by R_1 plus R total."},{"Start":"13:24.930 ","End":"13:31.990","Text":"Now let\u0027s plug this into this equation over here."},{"Start":"13:31.990 ","End":"13:36.670","Text":"What we\u0027ll be left with is I_2,"},{"Start":"13:36.670 ","End":"13:40.990","Text":"R_2 plus I_1R,"},{"Start":"13:40.990 ","End":"13:44.830","Text":"so we have plus I_2,"},{"Start":"13:44.830 ","End":"13:55.580","Text":"R_2 plus R total divided by R_1 plus R total."},{"Start":"13:55.680 ","End":"14:01.330","Text":"This is equal to I_1 plus I_2, R total."},{"Start":"14:01.330 ","End":"14:05.050","Text":"What I can do is I can write this as I_2,"},{"Start":"14:05.050 ","End":"14:10.945","Text":"R total multiplied by 1 plus R_2"},{"Start":"14:10.945 ","End":"14:18.230","Text":"plus R total divided by R_1 plus R total."},{"Start":"14:19.230 ","End":"14:22.930","Text":"Now we can divide both sides by I_2,"},{"Start":"14:22.930 ","End":"14:25.180","Text":"so I_2 here cancels,"},{"Start":"14:25.180 ","End":"14:27.775","Text":"here cancels, and here cancels."},{"Start":"14:27.775 ","End":"14:34.030","Text":"Now we\u0027re left with an equation that only has R total as an unknown."},{"Start":"14:34.030 ","End":"14:36.730","Text":"We have 1 equation with 1 unknown."},{"Start":"14:36.730 ","End":"14:41.990","Text":"First of all, I forgot to multiply it over here by R_1."},{"Start":"14:42.030 ","End":"14:48.535","Text":"Now what we can do is we can multiply both sides by the denominator to get rid of this."},{"Start":"14:48.535 ","End":"14:52.810","Text":"We\u0027re multiplying both sides by R_1 plus R total."},{"Start":"14:52.810 ","End":"14:58.240","Text":"What we\u0027ll have is R_2 multiplied by R_1"},{"Start":"14:58.240 ","End":"15:04.000","Text":"plus R_2 multiplied by R total plus,"},{"Start":"15:04.000 ","End":"15:09.250","Text":"so here we have R_2 multiplied by R_1."},{"Start":"15:09.250 ","End":"15:11.620","Text":"That\u0027s over here."},{"Start":"15:11.620 ","End":"15:14.080","Text":"We can multiply this by 2,"},{"Start":"15:14.080 ","End":"15:19.250","Text":"and R total multiplied by R_1."},{"Start":"15:20.940 ","End":"15:23.920","Text":"This is equal to,"},{"Start":"15:23.920 ","End":"15:28.480","Text":"so here we have R total multiplied by R_1,"},{"Start":"15:28.480 ","End":"15:31.555","Text":"so R_1, R total."},{"Start":"15:31.555 ","End":"15:35.890","Text":"Then R total multiplied by R total."},{"Start":"15:35.890 ","End":"15:40.340","Text":"So R total squared."},{"Start":"15:41.520 ","End":"15:47.300","Text":"Then we also have plus"},{"Start":"15:49.980 ","End":"15:57.355","Text":"R_2 multiplied by R total"},{"Start":"15:57.355 ","End":"16:01.345","Text":"plus another R total multiplied by R total."},{"Start":"16:01.345 ","End":"16:04.730","Text":"I can put a 2 over here as well."},{"Start":"16:05.010 ","End":"16:09.070","Text":"Now we can see that we can cancel out some terms."},{"Start":"16:09.070 ","End":"16:12.564","Text":"First, we have R_2, R total."},{"Start":"16:12.564 ","End":"16:15.175","Text":"We can subtract from both sides R_2,"},{"Start":"16:15.175 ","End":"16:19.540","Text":"R total, and we can subtract from both sides R_1,"},{"Start":"16:19.540 ","End":"16:25.625","Text":"R total, and then we\u0027re left with 1 term on this side and 1 term on this side,"},{"Start":"16:25.625 ","End":"16:28.510","Text":"so we can divide both sides by 2."},{"Start":"16:28.510 ","End":"16:37.930","Text":"Then what we\u0027re left with is that R total squared is equal to R_1, R_2."},{"Start":"16:37.930 ","End":"16:40.749","Text":"Or in other words,"},{"Start":"16:40.749 ","End":"16:46.180","Text":"R total is equal to the square root"},{"Start":"16:46.180 ","End":"16:55.660","Text":"of R_1, R_2."},{"Start":"16:55.660 ","End":"16:59.515","Text":"This is the answer to question Number 1."},{"Start":"16:59.515 ","End":"17:04.059","Text":"If we go all the way back to the beginning,"},{"Start":"17:04.059 ","End":"17:06.640","Text":"we can see that our question Number 1 was to calculate"},{"Start":"17:06.640 ","End":"17:10.910","Text":"the total resistance of the following circuit."},{"Start":"17:11.070 ","End":"17:17.950","Text":"We answered that and now we\u0027re being told that R_1 is equal to R_2."},{"Start":"17:17.950 ","End":"17:22.339","Text":"We want to calculate the current in each resistor."},{"Start":"17:22.590 ","End":"17:25.180","Text":"I cleared some space."},{"Start":"17:25.180 ","End":"17:29.605","Text":"We\u0027re being told that R_1 is equal to R_2."},{"Start":"17:29.605 ","End":"17:32.125","Text":"We\u0027re answering question Number 2."},{"Start":"17:32.125 ","End":"17:35.725","Text":"R_1 is equal to R_2."},{"Start":"17:35.725 ","End":"17:40.160","Text":"Therefore, our total resistance,"},{"Start":"17:40.770 ","End":"17:43.570","Text":"and let\u0027s just call this R,"},{"Start":"17:43.570 ","End":"17:46.855","Text":"so is equal to the square root of R_1,"},{"Start":"17:46.855 ","End":"17:50.770","Text":"R_2, which is in this case just the square root of R^2,"},{"Start":"17:50.770 ","End":"17:59.230","Text":"which means it\u0027s just equal to R. If we have an infinite ladder circuit like so,"},{"Start":"17:59.230 ","End":"18:02.425","Text":"where all of these resistors are equal to R,"},{"Start":"18:02.425 ","End":"18:07.240","Text":"so the total resistance of the entire circuit will just be equal to"},{"Start":"18:07.240 ","End":"18:11.635","Text":"R. When we were solving question Number 1,"},{"Start":"18:11.635 ","End":"18:13.360","Text":"we got this equation,"},{"Start":"18:13.360 ","End":"18:16.210","Text":"for this loop over here in blue."},{"Start":"18:16.210 ","End":"18:21.650","Text":"Then we just sorted out the algebra where we got this."},{"Start":"18:22.770 ","End":"18:28.540","Text":"In the case where R_1 is equal to R_2 and we just called it R,"},{"Start":"18:28.540 ","End":"18:38.455","Text":"we saw that R total is also equal to R. What we therefore get when R_1 is equal to R_2,"},{"Start":"18:38.455 ","End":"18:41.335","Text":"which is equal to R total,"},{"Start":"18:41.335 ","End":"18:42.775","Text":"which is equal to R,"},{"Start":"18:42.775 ","End":"18:52.810","Text":"we get that I_2 is multiplied by R plus R,"},{"Start":"18:52.810 ","End":"18:54.895","Text":"so it\u0027s multiplied by 2R,"},{"Start":"18:54.895 ","End":"19:00.880","Text":"which is equal to I_1 multiplied by R plus R, just 2R."},{"Start":"19:00.880 ","End":"19:04.420","Text":"Then we can divide both sides by 2R,"},{"Start":"19:04.420 ","End":"19:09.980","Text":"and therefore we get that I_1 is equal to I_2."},{"Start":"19:10.350 ","End":"19:15.760","Text":"Now, what\u0027s interesting is that if I_1 is equal to I_2,"},{"Start":"19:15.760 ","End":"19:22.660","Text":"then the current going through R total over here is I_1 minus I_2,"},{"Start":"19:22.660 ","End":"19:25.660","Text":"which is equal to 0 if they\u0027re the same."},{"Start":"19:25.660 ","End":"19:31.285","Text":"That means that the current going through R total is equal to 0,"},{"Start":"19:31.285 ","End":"19:38.480","Text":"which means that the current going through all of this section of the circuit in red."},{"Start":"19:40.620 ","End":"19:45.295","Text":"If I_1 is equal to I_2,"},{"Start":"19:45.295 ","End":"19:52.610","Text":"therefore, the current going through the red is equal to 0."},{"Start":"19:52.610 ","End":"20:02.960","Text":"That means that there\u0027s current flowing through only these 4 resistors."},{"Start":"20:04.500 ","End":"20:11.840","Text":"These 4, and the current is going to be identical through each resistor."},{"Start":"20:12.120 ","End":"20:19.465","Text":"The answer to question number 2 is that to calculate the current in each resistor,"},{"Start":"20:19.465 ","End":"20:22.030","Text":"the current is only flowing through"},{"Start":"20:22.030 ","End":"20:27.205","Text":"the first 4 resistors where our current is just going to be equal to I."},{"Start":"20:27.205 ","End":"20:30.264","Text":"There\u0027s an identical current in each resistor,"},{"Start":"20:30.264 ","End":"20:34.525","Text":"and there is 0 current in all of the other resistors,"},{"Start":"20:34.525 ","End":"20:39.025","Text":"all of the other infinite resistors in this circuit."},{"Start":"20:39.025 ","End":"20:45.350","Text":"Now, earlier we said that I is equal to I_1 plus I_2."},{"Start":"20:45.450 ","End":"20:53.480","Text":"I is equal to I_1 plus I_2."},{"Start":"20:53.550 ","End":"20:56.965","Text":"If the currents are identical,"},{"Start":"20:56.965 ","End":"21:02.065","Text":"so this is equal to, let\u0027s say 2I_1."},{"Start":"21:02.065 ","End":"21:08.080","Text":"Therefore, we can say 2I_1 is equal"},{"Start":"21:08.080 ","End":"21:16.190","Text":"to the voltage V divided by the resistance."},{"Start":"21:16.260 ","End":"21:26.510","Text":"We can see that the current I_1or I_2 is simply equal to V divided by 2_R."},{"Start":"21:27.150 ","End":"21:31.690","Text":"That\u0027s the answer to our question."},{"Start":"21:31.690 ","End":"21:36.055","Text":"This will be the current in each one of the first 4 resistors."},{"Start":"21:36.055 ","End":"21:40.360","Text":"Earlier I said that I will go into prove"},{"Start":"21:40.360 ","End":"21:44.455","Text":"this symmetry of why the voltage in each half is the same."},{"Start":"21:44.455 ","End":"21:47.690","Text":"That\u0027s what I\u0027m going to do now."},{"Start":"21:48.690 ","End":"21:55.510","Text":"Here we have the diagram that we saw before."},{"Start":"21:55.510 ","End":"22:05.800","Text":"We\u0027re assuming that over here we have current I_1 and that over here we have current I_2."},{"Start":"22:05.800 ","End":"22:10.690","Text":"What we want to prove is that also over here we have"},{"Start":"22:10.690 ","End":"22:16.910","Text":"current I_1 and that also over here we have current I_2."},{"Start":"22:17.000 ","End":"22:23.065","Text":"We\u0027re going to prove this via proof by contradiction."},{"Start":"22:23.065 ","End":"22:28.075","Text":"Our contradiction is we\u0027ll say that I_1 is equal to, let\u0027s say,"},{"Start":"22:28.075 ","End":"22:32.170","Text":"2 amps and that over here this isn\u0027t I_1,"},{"Start":"22:32.170 ","End":"22:34.195","Text":"but this is some I_3."},{"Start":"22:34.195 ","End":"22:38.735","Text":"Let\u0027s say that this is equal to 3 amps."},{"Start":"22:38.735 ","End":"22:41.680","Text":"We have our voltage source over here."},{"Start":"22:41.680 ","End":"22:44.335","Text":"We did some calculations and we found"},{"Start":"22:44.335 ","End":"22:48.235","Text":"that the current flowing through this resistor is 2 amps,"},{"Start":"22:48.235 ","End":"22:52.850","Text":"and the current flowing through this resistor is 3 amps."},{"Start":"22:53.280 ","End":"22:56.725","Text":"Now I have the same voltage over here,"},{"Start":"22:56.725 ","End":"22:59.740","Text":"but I switched the sides."},{"Start":"22:59.740 ","End":"23:02.965","Text":"Now the short side is over here,"},{"Start":"23:02.965 ","End":"23:06.170","Text":"and the long side is over here."},{"Start":"23:07.170 ","End":"23:10.225","Text":"Now, what I\u0027m allowed to do,"},{"Start":"23:10.225 ","End":"23:16.210","Text":"because we\u0027re just looking at this drawing as it is"},{"Start":"23:16.210 ","End":"23:23.470","Text":"now but someone who\u0027s standing behind this diagram will see it in the opposite way."},{"Start":"23:23.470 ","End":"23:29.200","Text":"I can just flip this around along this axis."},{"Start":"23:29.200 ","End":"23:35.776","Text":"I\u0027m going to rotate it and then instead of having my I_1 here and my R_2 here,"},{"Start":"23:35.776 ","End":"23:40.700","Text":"they\u0027re just going to flip and the same with this side."},{"Start":"23:40.700 ","End":"23:47.600","Text":"This is the exact same thing that we had a second ago."},{"Start":"23:47.940 ","End":"23:54.280","Text":"This is the diagram that we were looking at just before we flipped the direction."},{"Start":"23:54.280 ","End":"23:58.315","Text":"We can see that both diagrams are the same."},{"Start":"23:58.315 ","End":"24:03.175","Text":"In the first diagram, we were looking width from here."},{"Start":"24:03.175 ","End":"24:07.360","Text":"The current flow is in the clockwise direction then it reaches"},{"Start":"24:07.360 ","End":"24:12.415","Text":"this junction where it goes to R_1 and to R_2 at the bottom."},{"Start":"24:12.415 ","End":"24:15.865","Text":"Here, it\u0027s the exact mirror image."},{"Start":"24:15.865 ","End":"24:19.075","Text":"You can really see that it\u0027s a mirror image."},{"Start":"24:19.075 ","End":"24:22.240","Text":"Here we\u0027re going anticlockwise."},{"Start":"24:22.240 ","End":"24:27.280","Text":"We reach this junction and then just like here where we had R_1"},{"Start":"24:27.280 ","End":"24:32.216","Text":"going in this direction and R_2 going in this direction,"},{"Start":"24:32.216 ","End":"24:33.970","Text":"we have the exact same thing."},{"Start":"24:33.970 ","End":"24:37.794","Text":"What we have is the exact same circuit,"},{"Start":"24:37.794 ","End":"24:40.225","Text":"just mirror images of it."},{"Start":"24:40.225 ","End":"24:45.880","Text":"In that case, I get that the current over here fits 2 amps."},{"Start":"24:45.880 ","End":"24:52.675","Text":"The current over here also has to be 2 amps."},{"Start":"24:52.675 ","End":"24:55.285","Text":"If the current over here is 3 amps,"},{"Start":"24:55.285 ","End":"25:05.095","Text":"so that means that the current over here also has to be 3 amps."},{"Start":"25:05.095 ","End":"25:11.275","Text":"To get to this, I just flipped it around this axis over here."},{"Start":"25:11.275 ","End":"25:14.200","Text":"We said it\u0027s the exact same problem."},{"Start":"25:14.200 ","End":"25:18.490","Text":"Now what I\u0027m going to do is I\u0027m going to flip it back."},{"Start":"25:18.490 ","End":"25:21.790","Text":"Here I just flipped it back."},{"Start":"25:21.790 ","End":"25:27.550","Text":"I have my R_2 back on top on this side and my R_1 back on top on this side."},{"Start":"25:27.550 ","End":"25:30.280","Text":"Then if I flipped it like so,"},{"Start":"25:30.280 ","End":"25:33.055","Text":"so here where I had my 3 amps,"},{"Start":"25:33.055 ","End":"25:34.630","Text":"so it was on the R_1."},{"Start":"25:34.630 ","End":"25:39.115","Text":"Then over here, I\u0027m going to have again my current of 3 amps."},{"Start":"25:39.115 ","End":"25:42.075","Text":"Over here where I have my 2 amps,"},{"Start":"25:42.075 ","End":"25:44.785","Text":"so my 2 amps is going to be over here."},{"Start":"25:44.785 ","End":"25:46.714","Text":"I just flipped it around,"},{"Start":"25:46.714 ","End":"25:49.555","Text":"that\u0027s completely a legal move."},{"Start":"25:49.555 ","End":"25:53.935","Text":"I have the exact same circuit."},{"Start":"25:53.935 ","End":"25:56.365","Text":"Now what do we see?"},{"Start":"25:56.365 ","End":"26:02.290","Text":"I can see that my voltage source is still opposite to the original voltage source."},{"Start":"26:02.290 ","End":"26:05.650","Text":"What I\u0027m saying is that if I took the original voltage source,"},{"Start":"26:05.650 ","End":"26:06.955","Text":"so I have here,"},{"Start":"26:06.955 ","End":"26:09.875","Text":"2 amps and here 3 amps,"},{"Start":"26:09.875 ","End":"26:13.105","Text":"if I switch the direction of the voltage source,"},{"Start":"26:13.105 ","End":"26:15.715","Text":"then I\u0027ll get a different circuit."},{"Start":"26:15.715 ","End":"26:18.370","Text":"Over here, I\u0027ll have 3 amps instead of down"},{"Start":"26:18.370 ","End":"26:22.765","Text":"here and here I\u0027ll have 2 amps instead of up here."},{"Start":"26:22.765 ","End":"26:28.105","Text":"But that doesn\u0027t make sense because when we switch the direction of the voltage source,"},{"Start":"26:28.105 ","End":"26:29.590","Text":"all that that does,"},{"Start":"26:29.590 ","End":"26:34.465","Text":"is it just changes the sign in front of the current."},{"Start":"26:34.465 ","End":"26:38.350","Text":"If in one case we had plus 2 amps,"},{"Start":"26:38.350 ","End":"26:42.355","Text":"if we switch the sign or if we switched the voltage source,"},{"Start":"26:42.355 ","End":"26:46.855","Text":"the direction of it, we would just have a current of negative 2 amps."},{"Start":"26:46.855 ","End":"26:50.470","Text":"That\u0027s all it is. It doesn\u0027t make sense that if"},{"Start":"26:50.470 ","End":"26:53.650","Text":"our voltage source is like so we have 2 amperes here,"},{"Start":"26:53.650 ","End":"26:56.830","Text":"but if it\u0027s just the opposite way around,"},{"Start":"26:56.830 ","End":"27:00.220","Text":"we have a current of 3 amperes."},{"Start":"27:00.220 ","End":"27:04.570","Text":"That is our contradiction and so therefore,"},{"Start":"27:04.570 ","End":"27:08.920","Text":"if the current over here in this case is 2 amps,"},{"Start":"27:08.920 ","End":"27:14.380","Text":"then over here, the current also has to be equal to 2 amps,"},{"Start":"27:14.380 ","End":"27:18.235","Text":"because otherwise our system contradicts itself."},{"Start":"27:18.235 ","End":"27:23.965","Text":"All right, so now we\u0027ve proved that we have the same voltage on both sides."},{"Start":"27:23.965 ","End":"27:28.150","Text":"That is why we can use symmetry and that\u0027s why straight away,"},{"Start":"27:28.150 ","End":"27:30.760","Text":"we could have said that the current over here is the"},{"Start":"27:30.760 ","End":"27:33.490","Text":"same as the current over here and the same for this."},{"Start":"27:33.490 ","End":"27:36.550","Text":"The current over here is the same as the current over here."},{"Start":"27:36.550 ","End":"27:39.560","Text":"That is the end of this lesson."}],"ID":22310}],"Thumbnail":null,"ID":99474}]

[{"ID":99474,"Videos":[22296,22297,22298,21390,22299,22300,22301,22293,21389,22294,22295,22302,22303,22304,21391,22305,22306,22307,22308,22309,22310]}];

[22296,22297,22298,21390,22299,22300,22301,22293,21389,22294,22295,22302,22303,22304,21391,22305,22306,22307,22308,22309,22310];

1

3