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[{"Name":"Introduction to Inductance","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro to Inductance","Duration":"19m 36s","ChapterTopicVideoID":21539,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21539.jpeg","UploadDate":"2020-04-21T16:23:14.4530000","DurationForVideoObject":"PT19M36S","Description":null,"MetaTitle":"Intro to Inductance: Video + Workbook | Proprep","MetaDescription":"Inductance - Introduction to Inductance. 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/inductance/introduction-to-inductance/vid22369","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.455","Text":"Hello. In this lesson,"},{"Start":"00:01.455 ","End":"00:03.750","Text":"we\u0027re going to be learning about inductance."},{"Start":"00:03.750 ","End":"00:05.723","Text":"We\u0027re going to learn what it is,"},{"Start":"00:05.723 ","End":"00:14.280","Text":"we\u0027ll do a short little example to see how to use inductance and the equations involved."},{"Start":"00:14.280 ","End":"00:22.049","Text":"Then we\u0027re also going to see how inductance can help us to calculate all sorts of things."},{"Start":"00:22.400 ","End":"00:31.740","Text":"First of all, inductance is denoted by the letter L. So this is inductance."},{"Start":"00:32.120 ","End":"00:41.340","Text":"It is equal to the magnetic flux divided by the current."},{"Start":"00:41.740 ","End":"00:47.690","Text":"Let\u0027s imagine we have this coil and we want to calculate its inductance."},{"Start":"00:47.690 ","End":"00:51.800","Text":"The first thing through a conducting coil,"},{"Start":"00:51.800 ","End":"00:59.405","Text":"we have or we can put a current I or induce a current I to flow through it."},{"Start":"00:59.405 ","End":"01:03.110","Text":"Then as we know, if we have a current I,"},{"Start":"01:03.110 ","End":"01:06.905","Text":"then that means we\u0027re going to have a magnetic field."},{"Start":"01:06.905 ","End":"01:09.185","Text":"From the right-hand rule,"},{"Start":"01:09.185 ","End":"01:12.995","Text":"if the current is flowing like so,"},{"Start":"01:12.995 ","End":"01:16.955","Text":"then we\u0027re going to have a magnetic field going"},{"Start":"01:16.955 ","End":"01:24.850","Text":"through the coil like so, magnetic field B."},{"Start":"01:25.760 ","End":"01:30.575","Text":"Then if we know the current and we know the magnetic field,"},{"Start":"01:30.575 ","End":"01:33.050","Text":"we can calculate the magnetic flux."},{"Start":"01:33.050 ","End":"01:37.370","Text":"The magnetic flux is just how much magnetic field if you will,"},{"Start":"01:37.370 ","End":"01:41.140","Text":"passes through a closed loop."},{"Start":"01:41.140 ","End":"01:45.320","Text":"Here we basically have a closed circle,"},{"Start":"01:45.320 ","End":"01:47.520","Text":"so we can just calculate the surface area of that,"},{"Start":"01:47.520 ","End":"01:51.380","Text":"but we\u0027ll look at that in a few moments."},{"Start":"01:51.380 ","End":"01:57.035","Text":"Then we can calculate the magnetic flux and we can calculate then the inductance."},{"Start":"01:57.035 ","End":"02:01.430","Text":"Now, what is interesting and what we\u0027ll see is that the inductance,"},{"Start":"02:01.430 ","End":"02:05.330","Text":"even though this is its definition and this is its equation,"},{"Start":"02:05.330 ","End":"02:10.039","Text":"it isn\u0027t dependent on the magnetic field or on the current itself,"},{"Start":"02:10.039 ","End":"02:15.065","Text":"it\u0027s solely dependent on the geometric build of"},{"Start":"02:15.065 ","End":"02:23.150","Text":"the electrical component that we\u0027re trying to calculate the conductance for."},{"Start":"02:23.480 ","End":"02:25.987","Text":"Inductance, if you will,"},{"Start":"02:25.987 ","End":"02:29.120","Text":"in that sentence is very similar to the capacitance."},{"Start":"02:29.120 ","End":"02:38.385","Text":"If we remember, we said that capacitance is equal to the charge divided by the voltage,"},{"Start":"02:38.385 ","End":"02:42.245","Text":"and what we saw is that even though capacitance is defined as this,"},{"Start":"02:42.245 ","End":"02:46.835","Text":"we saw that the capacitance of a capacitor is very much dependent"},{"Start":"02:46.835 ","End":"02:53.120","Text":"on the shape of the capacitor itself, its geometric build."},{"Start":"02:53.120 ","End":"03:00.900","Text":"In that sense, the inductance is very similar to what we saw in the capacitance."},{"Start":"03:01.760 ","End":"03:07.495","Text":"In other words, we can look at the inductance as like capacitance,"},{"Start":"03:07.495 ","End":"03:09.280","Text":"but with the magnetic field,"},{"Start":"03:09.280 ","End":"03:12.295","Text":"whereas capacitance is using electricity."},{"Start":"03:12.295 ","End":"03:16.220","Text":"Here inductance is working with magnetism."},{"Start":"03:17.000 ","End":"03:21.580","Text":"Now let\u0027s see how we actually calculate the inductance."},{"Start":"03:21.580 ","End":"03:22.900","Text":"The first method is,"},{"Start":"03:22.900 ","End":"03:25.510","Text":"of course, using the definition."},{"Start":"03:25.510 ","End":"03:30.715","Text":"You calculate the magnetic flux divided by the current, and then you have it."},{"Start":"03:30.715 ","End":"03:33.310","Text":"But just like what we saw in capacitance,"},{"Start":"03:33.310 ","End":"03:38.380","Text":"where there were a few ways to calculate a component\u0027s capacitance."},{"Start":"03:38.380 ","End":"03:42.400","Text":"In the same way, there\u0027s a few ways to calculate the inductance."},{"Start":"03:42.400 ","End":"03:48.020","Text":"The first way is by using just exactly the definition."},{"Start":"03:48.020 ","End":"03:51.580","Text":"So we\u0027re going to go over this."},{"Start":"03:51.580 ","End":"03:57.060","Text":"Let\u0027s see. This is by the definition."},{"Start":"03:57.060 ","End":"03:59.785","Text":"What we do, step number 1,"},{"Start":"03:59.785 ","End":"04:04.230","Text":"is assume that there\u0027s a current I flowing through."},{"Start":"04:05.390 ","End":"04:08.150","Text":"What we can do is we can assume that there\u0027s"},{"Start":"04:08.150 ","End":"04:11.869","Text":"some general current I flowing through the components,"},{"Start":"04:11.869 ","End":"04:14.420","Text":"let\u0027s say if it\u0027s a coil and of course,"},{"Start":"04:14.420 ","End":"04:16.250","Text":"there could be a current flowing through,"},{"Start":"04:16.250 ","End":"04:20.360","Text":"and either way this is going to be correct for any current."},{"Start":"04:20.360 ","End":"04:25.420","Text":"But also even if you have a coil and there\u0027s no current flowing through it,"},{"Start":"04:25.420 ","End":"04:30.109","Text":"because we\u0027ve already said that the inductance is independent on the current,"},{"Start":"04:30.109 ","End":"04:37.325","Text":"it\u0027s solely dependent on the geometric shape of the electrical component."},{"Start":"04:37.325 ","End":"04:41.640","Text":"It doesn\u0027t matter if there is or isn\u0027t a current flowing through."},{"Start":"04:41.640 ","End":"04:43.240","Text":"That\u0027s why I say assume."},{"Start":"04:43.240 ","End":"04:47.580","Text":"You don\u0027t have to actually have a current flowing through."},{"Start":"04:47.800 ","End":"04:55.175","Text":"Step number 2 is to calculate the magnetic field induced by this current I."},{"Start":"04:55.175 ","End":"04:59.510","Text":"Let\u0027s say we\u0027re looking at this example over here where we have a coil."},{"Start":"04:59.510 ","End":"05:04.310","Text":"We\u0027ve already seen that the equation for the magnetic fields going through"},{"Start":"05:04.310 ","End":"05:10.490","Text":"a coil is equal to Mu naught multiplied by I. Mu naught is a constant I is,"},{"Start":"05:10.490 ","End":"05:15.440","Text":"of course, this current multiplied by n,"},{"Start":"05:15.440 ","End":"05:23.160","Text":"where n is of course equal to the density of these wraps in the coil."},{"Start":"05:23.160 ","End":"05:27.410","Text":"The density of the wraps is just equal to the number of"},{"Start":"05:27.410 ","End":"05:33.095","Text":"turns of the coil divided by the length of the coil."},{"Start":"05:33.095 ","End":"05:35.405","Text":"Here, this is 1 turn, 2,"},{"Start":"05:35.405 ","End":"05:36.890","Text":"3, 4, 5,"},{"Start":"05:36.890 ","End":"05:38.495","Text":"6, 7, 8."},{"Start":"05:38.495 ","End":"05:42.350","Text":"Here let\u0027s say in this section we have 8 turns,"},{"Start":"05:42.350 ","End":"05:46.025","Text":"so N would be 8 divided by l,"},{"Start":"05:46.025 ","End":"05:47.810","Text":"whatever length this is,"},{"Start":"05:47.810 ","End":"05:56.400","Text":"let\u0027s say half a meter and that would give us lowercase n. This is the length,"},{"Start":"05:56.400 ","End":"06:04.520","Text":"l. Now let\u0027s say that the radius of each circle made by this loop is of radius,"},{"Start":"06:04.520 ","End":"06:06.410","Text":"let\u0027s say it\u0027s a."},{"Start":"06:06.410 ","End":"06:10.969","Text":"Step number 3 is to calculate the magnetic flux."},{"Start":"06:10.969 ","End":"06:16.280","Text":"As we know, the magnetic flux is simply equal"},{"Start":"06:16.280 ","End":"06:21.680","Text":"to the magnetic field multiplied by the surface area,"},{"Start":"06:21.680 ","End":"06:32.000","Text":"or more generally, we have the integral of the magnetic field dot ds."},{"Start":"06:32.000 ","End":"06:33.545","Text":"This is the magnetic flux."},{"Start":"06:33.545 ","End":"06:40.580","Text":"However, if we have that b is equal to a constant or b is uniform,"},{"Start":"06:40.580 ","End":"06:43.520","Text":"then we can just write it as B.S."},{"Start":"06:43.520 ","End":"06:49.940","Text":"Then here our B is equal to Mu naught I."},{"Start":"06:49.940 ","End":"06:53.755","Text":"Then we\u0027ll just write out the whole equation for n,"},{"Start":"06:53.755 ","End":"06:56.775","Text":"so N divided by l,"},{"Start":"06:56.775 ","End":"07:00.991","Text":"the number of turns divided by the length,"},{"Start":"07:00.991 ","End":"07:04.040","Text":"and then we multiply it by the surface area."},{"Start":"07:04.040 ","End":"07:08.375","Text":"We\u0027re dealing with the surface area of this closed loop,"},{"Start":"07:08.375 ","End":"07:09.860","Text":"which is a circle."},{"Start":"07:09.860 ","End":"07:14.570","Text":"The surface area is going to be Pi multiplied by the radius squared,"},{"Start":"07:14.570 ","End":"07:19.620","Text":"where the radius over here is a, so Pi a^2."},{"Start":"07:20.780 ","End":"07:28.760","Text":"This is of course the magnetic flux of just 1 of these closed loops."},{"Start":"07:28.760 ","End":"07:32.060","Text":"However, you can look at this coil as if you have"},{"Start":"07:32.060 ","End":"07:34.850","Text":"a ring with a magnetic field flowing"},{"Start":"07:34.850 ","End":"07:37.850","Text":"through it and another ring that\u0027s also current-carrying."},{"Start":"07:37.850 ","End":"07:39.335","Text":"We have a current-carrying ring,"},{"Start":"07:39.335 ","End":"07:40.760","Text":"and another one, and another one,"},{"Start":"07:40.760 ","End":"07:45.290","Text":"and another one, and so each one has its own magnetic flux."},{"Start":"07:45.290 ","End":"07:49.290","Text":"This is the magnetic flux of 1 ring."},{"Start":"07:49.870 ","End":"07:53.165","Text":"I have 1 ring or 1 turn in the coil."},{"Start":"07:53.165 ","End":"07:57.305","Text":"In order to find the total magnetic flux,"},{"Start":"07:57.305 ","End":"08:02.240","Text":"what we want to do is we want to find the magnetic flux of 1 ring,"},{"Start":"08:02.240 ","End":"08:03.620","Text":"which we\u0027ve done,"},{"Start":"08:03.620 ","End":"08:08.405","Text":"and multiply it by the number of rings that we have, which as we said,"},{"Start":"08:08.405 ","End":"08:14.395","Text":"is N. What we\u0027re left with is Mu naught I"},{"Start":"08:14.395 ","End":"08:21.510","Text":"N^2 divided by l multiplied by Pi a^2."},{"Start":"08:23.160 ","End":"08:31.460","Text":"Now step number 4 is to calculate the inductance itself."},{"Start":"08:31.890 ","End":"08:37.495","Text":"Before we do that, I just added over here calculate total magnetic flux."},{"Start":"08:37.495 ","End":"08:38.560","Text":"If you\u0027re writing notes,"},{"Start":"08:38.560 ","End":"08:40.525","Text":"please include this total,"},{"Start":"08:40.525 ","End":"08:44.200","Text":"so you remember that specifically in a coil,"},{"Start":"08:44.200 ","End":"08:48.985","Text":"the magnetic flux you are calculating for 1 ring, and then you have,"},{"Start":"08:48.985 ","End":"08:53.470","Text":"of course, to get the total magnetic flux calculated for all of the rings."},{"Start":"08:53.470 ","End":"08:56.740","Text":"You have to multiply it by the number of turns."},{"Start":"08:56.740 ","End":"08:59.320","Text":"Please remember it\u0027s total magnetic flux."},{"Start":"08:59.320 ","End":"09:01.610","Text":"This is very important."},{"Start":"09:02.280 ","End":"09:06.925","Text":"Of course, step number 4 is calculate L,"},{"Start":"09:06.925 ","End":"09:09.235","Text":"which is of course the inductance."},{"Start":"09:09.235 ","End":"09:14.800","Text":"L, the inductance is equal to the magnetic flux,"},{"Start":"09:14.800 ","End":"09:17.080","Text":"remember this is total magnetic flux."},{"Start":"09:17.080 ","End":"09:19.015","Text":"Just remember this."},{"Start":"09:19.015 ","End":"09:23.320","Text":"I added a T for total divided by the current,"},{"Start":"09:23.320 ","End":"09:32.995","Text":"so we have Mu naught I N^2 divided by L Pi a squared."},{"Start":"09:32.995 ","End":"09:36.850","Text":"This is the total magnetic flux divided by the current."},{"Start":"09:36.850 ","End":"09:38.860","Text":"We can see that the current cancels out,"},{"Start":"09:38.860 ","End":"09:45.775","Text":"and so that is exactly why if we just look 1 second back at step number 1 when we said,"},{"Start":"09:45.775 ","End":"09:48.550","Text":"assume there is a current I flowing through,"},{"Start":"09:48.550 ","End":"09:53.485","Text":"why it didn\u0027t matter and it was independent of the current because we can see that"},{"Start":"09:53.485 ","End":"09:59.740","Text":"whether there is current or no current at all or any value for current,"},{"Start":"09:59.740 ","End":"10:01.555","Text":"the current anyway cancels out."},{"Start":"10:01.555 ","End":"10:04.690","Text":"We can see that at least in this example,"},{"Start":"10:04.690 ","End":"10:07.300","Text":"it\u0027s very clear to see that current really"},{"Start":"10:07.300 ","End":"10:10.510","Text":"has nothing to do with this because it cancels out."},{"Start":"10:10.510 ","End":"10:20.230","Text":"What we get is the inductance is equal to Mu naught N^2 Pi a^2 divided by L,"},{"Start":"10:20.230 ","End":"10:26.260","Text":"and so this is first of all how you work out the inductance."},{"Start":"10:26.260 ","End":"10:29.800","Text":"On top of that, this is also an equation for"},{"Start":"10:29.800 ","End":"10:35.000","Text":"your equation sheets for the inductance of a coil."},{"Start":"10:35.250 ","End":"10:39.880","Text":"Write this in your equation sheets and also this equation for"},{"Start":"10:39.880 ","End":"10:43.795","Text":"the magnetic field in a coil,"},{"Start":"10:43.795 ","End":"10:48.160","Text":"we\u0027re in a ring, it\u0027s also very important in a coil."},{"Start":"10:48.160 ","End":"10:52.015","Text":"What we can see here from the result that we got is that"},{"Start":"10:52.015 ","End":"10:56.050","Text":"the inductance for this coil is dependent on Mu naught, which as we know,"},{"Start":"10:56.050 ","End":"10:58.105","Text":"is just a constant,"},{"Start":"10:58.105 ","End":"11:00.730","Text":"and then it\u0027s dependent on N,"},{"Start":"11:00.730 ","End":"11:03.775","Text":"which is the number of turns in the coil."},{"Start":"11:03.775 ","End":"11:07.420","Text":"Pi a^2, which is the surface area of the coil,"},{"Start":"11:07.420 ","End":"11:10.390","Text":"and L, which is the length of the coil."},{"Start":"11:10.390 ","End":"11:16.390","Text":"As we said before, even though this is the definition for inductance,"},{"Start":"11:16.390 ","End":"11:20.260","Text":"we would have thought it\u0027s dependent on the magnetic flux and the current."},{"Start":"11:20.260 ","End":"11:26.500","Text":"As we can see, there\u0027s no memory here to the magnetic flux,"},{"Start":"11:26.500 ","End":"11:28.930","Text":"and to the current,"},{"Start":"11:28.930 ","End":"11:30.250","Text":"these aren\u0027t uneven factors,"},{"Start":"11:30.250 ","End":"11:33.445","Text":"and here we have a constant multiplied by"},{"Start":"11:33.445 ","End":"11:39.340","Text":"different elements which tell us about the geometric build of the coil."},{"Start":"11:39.340 ","End":"11:42.025","Text":"Its length, the number of turns that we have,"},{"Start":"11:42.025 ","End":"11:45.685","Text":"and the surface area enclosed over here."},{"Start":"11:45.685 ","End":"11:52.640","Text":"The inductance is dependent on the geometric build as we can see here."},{"Start":"11:53.670 ","End":"11:59.455","Text":"We\u0027ve seen how to calculate it according to the definition."},{"Start":"11:59.455 ","End":"12:03.955","Text":"We\u0027ve seen that it\u0027s very similar to the capacitance in the sense that it\u0027s"},{"Start":"12:03.955 ","End":"12:10.015","Text":"independent of these elements which are in its definition,"},{"Start":"12:10.015 ","End":"12:15.295","Text":"and it\u0027s just dependent on its geometric shape so that we saw."},{"Start":"12:15.295 ","End":"12:17.275","Text":"Now what we\u0027re going to do is we\u0027re going to speak"},{"Start":"12:17.275 ","End":"12:19.900","Text":"about inductance and what it\u0027s good for,"},{"Start":"12:19.900 ","End":"12:22.790","Text":"what do we use it for."},{"Start":"12:22.860 ","End":"12:26.485","Text":"Right now, we can call it inductance."},{"Start":"12:26.485 ","End":"12:30.970","Text":"But it is actually what we\u0027re speaking about now is self-inductance and"},{"Start":"12:30.970 ","End":"12:32.860","Text":"later we\u0027ll speak about a different type of"},{"Start":"12:32.860 ","End":"12:35.620","Text":"inductance but we\u0027re going to leave it at this in the meantime,"},{"Start":"12:35.620 ","End":"12:39.700","Text":"so inductance or self-inductance at the meantime,"},{"Start":"12:39.700 ","End":"12:41.810","Text":"at the present moment."},{"Start":"12:42.780 ","End":"12:53.150","Text":"Inductance or self-inductance is used in order to work as a shortcut to Faraday\u0027s law."},{"Start":"12:53.670 ","End":"13:02.455","Text":"If you haven\u0027t gone over the chapter dealing with Faraday\u0027s law, please do."},{"Start":"13:02.455 ","End":"13:05.875","Text":"Just as a reminder for Faraday."},{"Start":"13:05.875 ","End":"13:08.710","Text":"Faraday\u0027s law says that the EMF,"},{"Start":"13:08.710 ","End":"13:11.815","Text":"this is the EMF it\u0027s voltage."},{"Start":"13:11.815 ","End":"13:19.640","Text":"The EMF is equal to the negative time derivative of the magnetic flux."},{"Start":"13:21.120 ","End":"13:30.310","Text":"I can say as we\u0027ve already seen that the magnetic flux from this equation for inductance,"},{"Start":"13:30.310 ","End":"13:32.605","Text":"if I isolate out the magnetic flux,"},{"Start":"13:32.605 ","End":"13:34.900","Text":"I can get that it is equal to L,"},{"Start":"13:34.900 ","End":"13:38.890","Text":"the inductance multiplied by I, the current."},{"Start":"13:38.890 ","End":"13:42.830","Text":"I could say that this is equal to LI."},{"Start":"13:43.610 ","End":"13:47.745","Text":"Now if I want to calculate the EMF,"},{"Start":"13:47.745 ","End":"13:53.850","Text":"I know that it\u0027s equal to the negative time derivative of the magnetic flux."},{"Start":"13:53.850 ","End":"13:57.350","Text":"That means it\u0027s equal to negative the time derivative of this."},{"Start":"13:57.350 ","End":"14:00.490","Text":"This L is the inductance,"},{"Start":"14:00.490 ","End":"14:03.865","Text":"and as we saw, the inductance is constant."},{"Start":"14:03.865 ","End":"14:08.485","Text":"First of all, we can see it from this equation and also because we said that it\u0027s"},{"Start":"14:08.485 ","End":"14:13.450","Text":"only dependent on the geometric builds of our components,"},{"Start":"14:13.450 ","End":"14:16.809","Text":"so of course the build is constant,"},{"Start":"14:16.809 ","End":"14:20.215","Text":"it\u0027s uniform, it\u0027s not changing with respect to time."},{"Start":"14:20.215 ","End":"14:22.195","Text":"L is a constant,"},{"Start":"14:22.195 ","End":"14:28.030","Text":"so we just have L and our current as we know does have a time derivative,"},{"Start":"14:28.030 ","End":"14:34.670","Text":"so multiplied by I dot the time derivative of the current."},{"Start":"14:35.040 ","End":"14:37.825","Text":"Why is this equation useful?"},{"Start":"14:37.825 ","End":"14:42.550","Text":"If you look back in the chapter dealing with Faraday\u0027s law,"},{"Start":"14:42.550 ","End":"14:46.270","Text":"we have to do quite a few calculations depending on what\u0027s going on"},{"Start":"14:46.270 ","End":"14:50.515","Text":"in order to calculate the EMF, the electromotive force."},{"Start":"14:50.515 ","End":"14:53.215","Text":"However, as we can see,"},{"Start":"14:53.215 ","End":"14:56.230","Text":"it\u0027s relatively easy to calculate the inductance,"},{"Start":"14:56.230 ","End":"14:59.245","Text":"so it\u0027s just dependent on the geometric shape."},{"Start":"14:59.245 ","End":"15:05.845","Text":"In order to calculate the EMF and makes it much easier because I could just take"},{"Start":"15:05.845 ","End":"15:13.550","Text":"the inductance and multiply it by the negative time derivative of the current."},{"Start":"15:14.070 ","End":"15:21.295","Text":"Another definition quickly before we end the lesson for the inductance,"},{"Start":"15:21.295 ","End":"15:24.880","Text":"which is a parallel definition to what we saw at the beginning of the lesson,"},{"Start":"15:24.880 ","End":"15:28.570","Text":"is instead of having the magnetic flux divided by the current,"},{"Start":"15:28.570 ","End":"15:31.960","Text":"we can also write that it\u0027s the change in"},{"Start":"15:31.960 ","End":"15:37.060","Text":"the magnetic flux with respect to the change in the current."},{"Start":"15:37.060 ","End":"15:43.750","Text":"This is because the magnetic flux is linearly independent on the current."},{"Start":"15:43.750 ","End":"15:46.340","Text":"Let\u0027s exactly see why."},{"Start":"15:47.640 ","End":"15:51.340","Text":"Just so you know, generally it\u0027s easier to work"},{"Start":"15:51.340 ","End":"15:55.615","Text":"with this form of the equation for the inductance."},{"Start":"15:55.615 ","End":"16:02.120","Text":"However, you could also use this equation so just so that you know it."},{"Start":"16:02.340 ","End":"16:07.375","Text":"The reason that these two are linear to"},{"Start":"16:07.375 ","End":"16:11.995","Text":"one another depends or comes from Biot–Savart law."},{"Start":"16:11.995 ","End":"16:17.140","Text":"Just as a reminder, Biot–Savart\u0027s law is dB is equal"},{"Start":"16:17.140 ","End":"16:23.680","Text":"to Mu naught multiplied by the current divided by 4Pi,"},{"Start":"16:23.680 ","End":"16:30.565","Text":"and then multiplied by dl vector cross-product with"},{"Start":"16:30.565 ","End":"16:37.970","Text":"the r vector divided by the magnitude of the r vector cubed."},{"Start":"16:38.850 ","End":"16:43.270","Text":"Let\u0027s take some electrical components."},{"Start":"16:43.270 ","End":"16:46.885","Text":"It doesn\u0027t have to be too complex and it can be any shape,"},{"Start":"16:46.885 ","End":"16:50.755","Text":"and so this is our electrical components,"},{"Start":"16:50.755 ","End":"16:56.070","Text":"and let\u0027s say I want to calculate the magnetic field at this point over here."},{"Start":"16:56.730 ","End":"16:59.090","Text":"Through this component,"},{"Start":"16:59.090 ","End":"17:02.600","Text":"I have some current I flowing through it,"},{"Start":"17:02.600 ","End":"17:09.780","Text":"and what I do is I cut up this component into lots of different pieces."},{"Start":"17:10.170 ","End":"17:15.740","Text":"What I do is I look at this tiny little piece with this current I flowing through it,"},{"Start":"17:15.740 ","End":"17:23.820","Text":"and I look at the magnetic field that it is inducing at this point."},{"Start":"17:24.750 ","End":"17:27.330","Text":"From Biot–Savart\u0027s law,"},{"Start":"17:27.330 ","End":"17:32.760","Text":"in order to see what the magnetic field is at this point due to this piece,"},{"Start":"17:32.760 ","End":"17:37.070","Text":"I have to multiply over here by the length of"},{"Start":"17:37.070 ","End":"17:41.810","Text":"this piece and its distance away from this arbitrary point that I chose."},{"Start":"17:41.810 ","End":"17:46.510","Text":"However, either way, I\u0027m going to get some constant,"},{"Start":"17:46.510 ","End":"17:50.330","Text":"that\u0027s dependent on the length of this piece and its distance"},{"Start":"17:50.330 ","End":"17:55.265","Text":"away multiplied by another constant multiplied by I."},{"Start":"17:55.265 ","End":"18:00.740","Text":"What I get is this linear dependence with I."},{"Start":"18:01.980 ","End":"18:05.930","Text":"Then when I do this integral in order to calculate"},{"Start":"18:05.930 ","End":"18:11.390","Text":"the total magnetic field at this point from all of the pieces in this component,"},{"Start":"18:11.390 ","End":"18:16.280","Text":"what I\u0027ll get is the magnetic field"},{"Start":"18:16.280 ","End":"18:20.870","Text":"will be equal to the current multiplied by some kind of constant."},{"Start":"18:20.870 ","End":"18:27.084","Text":"We can see that the magnetic field is linearly dependent on the current."},{"Start":"18:27.084 ","End":"18:34.710","Text":"Then from this, we can see that the flux is also linearly dependent on the current,"},{"Start":"18:34.710 ","End":"18:41.450","Text":"and so that is why you can use this version of the equation to calculate the inductance,"},{"Start":"18:41.450 ","End":"18:44.915","Text":"and you can use this version of the equation as well."},{"Start":"18:44.915 ","End":"18:46.655","Text":"This version, as I said before,"},{"Start":"18:46.655 ","End":"18:48.605","Text":"is easier to use."},{"Start":"18:48.605 ","End":"18:52.040","Text":"However, you never know what type of question you\u0027ll get and you"},{"Start":"18:52.040 ","End":"18:55.850","Text":"might have to use this or this might be slightly easier,"},{"Start":"18:55.850 ","End":"18:59.235","Text":"so also write down this equation."},{"Start":"18:59.235 ","End":"19:07.780","Text":"Please write this equation for inductance as well as this equation for inductance,"},{"Start":"19:07.780 ","End":"19:11.405","Text":"and this is the equation for the inductance of the coil,"},{"Start":"19:11.405 ","End":"19:14.930","Text":"and if you want also write down over here"},{"Start":"19:14.930 ","End":"19:19.495","Text":"the steps to calculate the induction via using this definition."},{"Start":"19:19.495 ","End":"19:24.020","Text":"Remember that step number 3 is to calculate the total magnetic flux,"},{"Start":"19:24.020 ","End":"19:27.770","Text":"so not just the magnetic flux of 1 ring or"},{"Start":"19:27.770 ","End":"19:32.615","Text":"depending on what other shapes of electrical components you\u0027ll get."},{"Start":"19:32.615 ","End":"19:37.050","Text":"That is it, that is the end of this lesson."}],"ID":22369},{"Watched":false,"Name":"Exercise 1","Duration":"12m 26s","ChapterTopicVideoID":21540,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.810","Text":"Hello. In this lesson we\u0027re going to be answering the following question,"},{"Start":"00:03.810 ","End":"00:10.260","Text":"2 infinitely long wires are distance d away from each other\u0027s center."},{"Start":"00:10.260 ","End":"00:14.760","Text":"The radius of each wire is a,"},{"Start":"00:14.760 ","End":"00:19.155","Text":"and at infinity the 2 wires are joined together."},{"Start":"00:19.155 ","End":"00:25.815","Text":"Somehow the edges of these 2 infinitely long wires are joined,"},{"Start":"00:25.815 ","End":"00:29.940","Text":"and the current I flows clockwise through the wires."},{"Start":"00:29.940 ","End":"00:34.025","Text":"Over here we\u0027ll have a current going in this direction,"},{"Start":"00:34.025 ","End":"00:39.645","Text":"and through this wire we\u0027ll have a current going in this downwards direction."},{"Start":"00:39.645 ","End":"00:42.390","Text":"Calculate the inductance,"},{"Start":"00:42.390 ","End":"00:47.165","Text":"and we\u0027re being told to ignore the magnetic flux through the wires themselves."},{"Start":"00:47.165 ","End":"00:51.875","Text":"We can imagine that there\u0027s this current flowing on the surface of the wire,"},{"Start":"00:51.875 ","End":"00:55.630","Text":"and so we don\u0027t have to look at the flux inside the wire."},{"Start":"00:55.630 ","End":"00:58.070","Text":"In later lessons, we will deal with that,"},{"Start":"00:58.070 ","End":"01:00.455","Text":"but for the meantime, we don\u0027t have to look at that."},{"Start":"01:00.455 ","End":"01:02.915","Text":"We\u0027re asked to calculate the inductance,"},{"Start":"01:02.915 ","End":"01:08.350","Text":"and we\u0027re asked what is the inductance per unit length?"},{"Start":"01:09.350 ","End":"01:12.365","Text":"Let\u0027s begin. As we saw,"},{"Start":"01:12.365 ","End":"01:14.615","Text":"the inductance, L,"},{"Start":"01:14.615 ","End":"01:21.085","Text":"is equal to the magnetic flux divided by the current."},{"Start":"01:21.085 ","End":"01:23.565","Text":"We already know the current."},{"Start":"01:23.565 ","End":"01:27.680","Text":"What we want to do is we want to calculate the magnetic flux,"},{"Start":"01:27.680 ","End":"01:30.920","Text":"and as we\u0027ve seen in previous lessons in previous chapters,"},{"Start":"01:30.920 ","End":"01:34.685","Text":"the magnetic flux is given by the integral of B,"},{"Start":"01:34.685 ","End":"01:37.730","Text":"the magnetic field.ds,"},{"Start":"01:37.730 ","End":"01:41.530","Text":"the surface area of the closed loop."},{"Start":"01:41.530 ","End":"01:45.720","Text":"Over here, that\u0027s the surface enclosed in all of this."},{"Start":"01:45.720 ","End":"01:52.560","Text":"We\u0027re looking at all of this surface area here like so."},{"Start":"01:53.260 ","End":"01:55.420","Text":"The first thing we want to do,"},{"Start":"01:55.420 ","End":"01:58.640","Text":"is we want to calculate the magnetic field."},{"Start":"01:58.640 ","End":"02:02.690","Text":"Let\u0027s choose this arbitrary point over here,"},{"Start":"02:02.690 ","End":"02:06.200","Text":"and we want to know the magnetic field at this point."},{"Start":"02:06.200 ","End":"02:16.577","Text":"Let\u0027s say that the distance this point is away from the center of the left wire,"},{"Start":"02:16.577 ","End":"02:19.980","Text":"let\u0027s call that distance x."},{"Start":"02:20.710 ","End":"02:24.665","Text":"From the chapter dealing with magnetism,"},{"Start":"02:24.665 ","End":"02:28.625","Text":"we saw that the magnetic field caused by"},{"Start":"02:28.625 ","End":"02:36.230","Text":"an infinitely long wire is equal to Mu_naught multiplied by the current flowing through,"},{"Start":"02:36.230 ","End":"02:38.345","Text":"divided by 2Pir,"},{"Start":"02:38.345 ","End":"02:43.710","Text":"where r is the distance the point is away from the wire."},{"Start":"02:44.270 ","End":"02:49.075","Text":"This is the magnetic field at this point due to this left wire."},{"Start":"02:49.075 ","End":"02:50.380","Text":"Let\u0027s call this BL,"},{"Start":"02:50.380 ","End":"02:53.004","Text":"and of course over here,"},{"Start":"02:53.004 ","End":"02:55.495","Text":"r is x because that\u0027s how we defined it."},{"Start":"02:55.495 ","End":"03:01.885","Text":"We can call this Mu_naught I, divided by 2Pix."},{"Start":"03:01.885 ","End":"03:04.645","Text":"From the right-hand rule,"},{"Start":"03:04.645 ","End":"03:08.875","Text":"we can see that at this point over here,"},{"Start":"03:08.875 ","End":"03:14.450","Text":"the magnetic field is going into the page."},{"Start":"03:15.530 ","End":"03:18.405","Text":"This is from the left wire,"},{"Start":"03:18.405 ","End":"03:23.110","Text":"and now we want to calculate the magnetic field due to this wire over here,"},{"Start":"03:23.110 ","End":"03:25.930","Text":"the right wire, so B_R."},{"Start":"03:25.930 ","End":"03:31.025","Text":"Again, we have the same equation here for the magnetic field due to an infinite wire."},{"Start":"03:31.025 ","End":"03:35.570","Text":"However, this time instead of r being equal to x,"},{"Start":"03:35.570 ","End":"03:42.240","Text":"where this time measuring this distance over here."},{"Start":"03:42.240 ","End":"03:47.240","Text":"This distance is this whole distance between the 2 wires,"},{"Start":"03:47.240 ","End":"03:54.470","Text":"which is d, minus this distance over here, minus x."},{"Start":"03:54.470 ","End":"04:01.280","Text":"This time, the magnetic field is Mu_naught I divided by 2Pi,"},{"Start":"04:01.280 ","End":"04:03.080","Text":"and then its distance,"},{"Start":"04:03.080 ","End":"04:06.330","Text":"which is d minus x,"},{"Start":"04:07.070 ","End":"04:11.690","Text":"and of course for the direction here again through the right hand rule,"},{"Start":"04:11.690 ","End":"04:14.450","Text":"our thumb points in the direction of the current,"},{"Start":"04:14.450 ","End":"04:17.270","Text":"and our fingers curl in the direction of the magnetic field,"},{"Start":"04:17.270 ","End":"04:20.120","Text":"and again, we can see that in this region over here,"},{"Start":"04:20.120 ","End":"04:23.575","Text":"the magnetic field is going into the page."},{"Start":"04:23.575 ","End":"04:30.690","Text":"The magnetic field due to both the left and the right hand wires is into the page,"},{"Start":"04:30.690 ","End":"04:32.220","Text":"they\u0027re in the same direction."},{"Start":"04:32.220 ","End":"04:36.920","Text":"The total magnetic field is just going to be adding these 2 up."},{"Start":"04:36.920 ","End":"04:41.700","Text":"We have Mu_naught i divided by 2Pi,"},{"Start":"04:41.700 ","End":"04:45.060","Text":"and then we have 1 divided by x,"},{"Start":"04:45.060 ","End":"04:50.260","Text":"plus 1 divided by d minus x."},{"Start":"04:51.140 ","End":"04:54.290","Text":"Now that we have the total magnetic field,"},{"Start":"04:54.290 ","End":"04:56.540","Text":"we want to calculate the flux,"},{"Start":"04:56.540 ","End":"04:58.475","Text":"which as we\u0027ve seen,"},{"Start":"04:58.475 ","End":"04:59.914","Text":"we\u0027re going to have to integrate."},{"Start":"04:59.914 ","End":"05:02.240","Text":"Why are we integrating?"},{"Start":"05:02.240 ","End":"05:07.460","Text":"Because we can see that our B is dependent on distance."},{"Start":"05:07.460 ","End":"05:10.235","Text":"It\u0027s dependent on x. It has a variable."},{"Start":"05:10.235 ","End":"05:12.710","Text":"If our B was uniform or constant,"},{"Start":"05:12.710 ","End":"05:14.720","Text":"then we could just write B,"},{"Start":"05:14.720 ","End":"05:18.935","Text":"the uniform a number or the constant multiplied by the surface area."},{"Start":"05:18.935 ","End":"05:20.630","Text":"But because B is changing,"},{"Start":"05:20.630 ","End":"05:22.805","Text":"we have to integrate."},{"Start":"05:22.805 ","End":"05:25.040","Text":"Because we\u0027re integrating along ds,"},{"Start":"05:25.040 ","End":"05:26.465","Text":"which is the area,"},{"Start":"05:26.465 ","End":"05:29.225","Text":"we have to do a double integral."},{"Start":"05:29.225 ","End":"05:32.015","Text":"Now, although these wires are infinitely long,"},{"Start":"05:32.015 ","End":"05:35.075","Text":"let\u0027s just call them for the sake of the integral,"},{"Start":"05:35.075 ","End":"05:42.645","Text":"let\u0027s say that each wire is of length l. We have the integral of B,"},{"Start":"05:42.645 ","End":"05:47.250","Text":"we have Mu_naught I divided by 2Pi,"},{"Start":"05:47.250 ","End":"05:54.645","Text":"and then we have 1 divided by x plus 1 divided by d minus x,"},{"Start":"05:54.645 ","End":"06:00.605","Text":"and all of this is multiplied by the surface area."},{"Start":"06:00.605 ","End":"06:05.330","Text":"Let\u0027s say that this direction is the y-direction,"},{"Start":"06:05.330 ","End":"06:10.170","Text":"and that over here is the x-direction,"},{"Start":"06:10.170 ","End":"06:11.625","Text":"which is what we\u0027ve been doing now."},{"Start":"06:11.625 ","End":"06:17.930","Text":"The surface area is the change in x multiplied by the change in y dx,"},{"Start":"06:17.930 ","End":"06:20.407","Text":"dy, and now let\u0027s do our bounds,"},{"Start":"06:20.407 ","End":"06:25.610","Text":"so on y we\u0027re going from 0 until the length of the wire,"},{"Start":"06:25.610 ","End":"06:29.045","Text":"which is l. On x,"},{"Start":"06:29.045 ","End":"06:35.120","Text":"what we\u0027re doing is we want to calculate the magnetic flux,"},{"Start":"06:35.120 ","End":"06:38.960","Text":"where we ignore the magnetic flux through the wires themselves."},{"Start":"06:38.960 ","End":"06:43.835","Text":"We\u0027re not starting the integral from here in the center of the wire,"},{"Start":"06:43.835 ","End":"06:50.990","Text":"but we\u0027re starting the integral from the edge of this wire until this edge of this wire."},{"Start":"06:50.990 ","End":"06:53.615","Text":"We\u0027re integrating over here."},{"Start":"06:53.615 ","End":"06:56.810","Text":"That means that this is a distance a,"},{"Start":"06:56.810 ","End":"07:00.560","Text":"from what we saw here, because the wire has a radius a."},{"Start":"07:00.560 ","End":"07:03.620","Text":"We\u0027re integrating from a, and again,"},{"Start":"07:03.620 ","End":"07:08.135","Text":"we\u0027re not going up until d over here,"},{"Start":"07:08.135 ","End":"07:11.885","Text":"but we\u0027re going from d is right in the center,"},{"Start":"07:11.885 ","End":"07:13.895","Text":"and then we have this again,"},{"Start":"07:13.895 ","End":"07:16.055","Text":"distance, this radius a."},{"Start":"07:16.055 ","End":"07:19.605","Text":"We\u0027re going up until d minus a."},{"Start":"07:19.605 ","End":"07:21.250","Text":"We\u0027re integrating up until d,"},{"Start":"07:21.250 ","End":"07:23.600","Text":"and then back to this edge over here,"},{"Start":"07:23.600 ","End":"07:26.570","Text":"so d minus a."},{"Start":"07:26.570 ","End":"07:32.760","Text":"This way we\u0027re integrating along all of this."},{"Start":"07:36.080 ","End":"07:40.050","Text":"Now let\u0027s do the integral."},{"Start":"07:40.050 ","End":"07:44.700","Text":"First of all, Mu_naught I 2Pi are constants, so we can take them out."},{"Start":"07:44.700 ","End":"07:48.800","Text":"Mu_naught I divided by 2Pi,"},{"Start":"07:48.800 ","End":"07:53.050","Text":"and then we can see that our variable y isn\u0027t over here."},{"Start":"07:53.050 ","End":"07:55.545","Text":"When we integrate along dy,"},{"Start":"07:55.545 ","End":"07:57.900","Text":"we\u0027re just going to get l minus 0,"},{"Start":"07:57.900 ","End":"07:59.505","Text":"which is l,"},{"Start":"07:59.505 ","End":"08:03.805","Text":"and now what we want to do is we want to integrate 1 divided by x,"},{"Start":"08:03.805 ","End":"08:06.715","Text":"plus 1 divided by d minus x."},{"Start":"08:06.715 ","End":"08:12.850","Text":"The integral of 1 divided by x is, as we know,"},{"Start":"08:12.890 ","End":"08:18.945","Text":"ln(x) in the bounds of a to d minus a,"},{"Start":"08:18.945 ","End":"08:26.040","Text":"and then we\u0027re adding the integral of 1 divided by d minus x."},{"Start":"08:26.260 ","End":"08:34.550","Text":"Again, we\u0027ll have ln(d) minus x this time."},{"Start":"08:34.550 ","End":"08:38.390","Text":"However, because of this minus before the x,"},{"Start":"08:38.390 ","End":"08:41.210","Text":"we\u0027ll have a negative sign over here."},{"Start":"08:41.210 ","End":"08:44.720","Text":"We can subtract this and we can just put in the negative,"},{"Start":"08:44.720 ","End":"08:51.365","Text":"and this is of course in the bounds also of a to d minus a."},{"Start":"08:51.365 ","End":"08:53.435","Text":"Then we can carry this on."},{"Start":"08:53.435 ","End":"08:58.190","Text":"We have Mu_naught IL divided by 2Pi,"},{"Start":"08:58.190 ","End":"09:06.125","Text":"and then over here we have ln(d) minus a minus ln(a),"},{"Start":"09:06.125 ","End":"09:09.440","Text":"which is just going to give us from the law of ln,"},{"Start":"09:09.440 ","End":"09:15.655","Text":"ln of d minus a divided by a,"},{"Start":"09:15.655 ","End":"09:20.105","Text":"and then what we can do here is we can,"},{"Start":"09:20.105 ","End":"09:23.960","Text":"instead of having this as a minus and our bounds like so,"},{"Start":"09:23.960 ","End":"09:26.150","Text":"we can make this a plus,"},{"Start":"09:26.150 ","End":"09:29.390","Text":"and then we can just flip the bounds over."},{"Start":"09:29.390 ","End":"09:33.125","Text":"Here we\u0027ll have a and here we\u0027ll have d minus a."},{"Start":"09:33.125 ","End":"09:40.550","Text":"We\u0027ll have plus ln(d) minus x,"},{"Start":"09:40.550 ","End":"09:42.560","Text":"which here we substitute an a,"},{"Start":"09:42.560 ","End":"09:44.465","Text":"and then minus ln,"},{"Start":"09:44.465 ","End":"09:51.665","Text":"so then we can just do by law of ln divided by d minus x,"},{"Start":"09:51.665 ","End":"09:53.345","Text":"where x is d minus a."},{"Start":"09:53.345 ","End":"10:03.545","Text":"D minus a, which of course this will give us d minus d, which is 0,"},{"Start":"10:03.545 ","End":"10:06.440","Text":"minus minus a is positive a,"},{"Start":"10:06.440 ","End":"10:10.845","Text":"we can just rub this all out as like so,"},{"Start":"10:10.845 ","End":"10:12.577","Text":"and we can just put brackets,"},{"Start":"10:12.577 ","End":"10:17.200","Text":"we don\u0027t have to write the absolute value because d is obviously greater than a."},{"Start":"10:17.200 ","End":"10:19.880","Text":"It\u0027s always going to be a positive number."},{"Start":"10:19.880 ","End":"10:24.410","Text":"Then we can just add these 2 together."},{"Start":"10:24.410 ","End":"10:31.360","Text":"This answer also makes sense because these wires are symmetrical."},{"Start":"10:31.360 ","End":"10:34.220","Text":"We have the same current flowing through them,"},{"Start":"10:34.220 ","End":"10:36.365","Text":"they\u0027re the same dimensions,"},{"Start":"10:36.365 ","End":"10:42.575","Text":"and so it makes sense that each wire will have"},{"Start":"10:42.575 ","End":"10:46.310","Text":"the same amount of effect on the magnetic field at"},{"Start":"10:46.310 ","End":"10:50.765","Text":"a specific point and therefore the magnetic flux throughout."},{"Start":"10:50.765 ","End":"10:57.190","Text":"What do we get? Here we\u0027ll just have 2 times this over here,"},{"Start":"10:57.190 ","End":"10:59.985","Text":"that will cancel out with this 2 in the denominator."},{"Start":"10:59.985 ","End":"11:04.005","Text":"We have Mu_naught IL,"},{"Start":"11:04.005 ","End":"11:11.770","Text":"divided by Pi multiplied by ln(d) minus a divided by a."},{"Start":"11:13.430 ","End":"11:15.805","Text":"Then, of course,"},{"Start":"11:15.805 ","End":"11:17.920","Text":"in order to get what we want,"},{"Start":"11:17.920 ","End":"11:21.475","Text":"which is for the first part at least we want the inductance."},{"Start":"11:21.475 ","End":"11:25.480","Text":"We take the magnetic flux and we divide it by the current."},{"Start":"11:25.480 ","End":"11:27.685","Text":"The current over here on the numerator,"},{"Start":"11:27.685 ","End":"11:31.295","Text":"we\u0027ll divide by this current in the denominator."},{"Start":"11:31.295 ","End":"11:34.695","Text":"We\u0027ll be left with Mu_naught,"},{"Start":"11:34.695 ","End":"11:36.150","Text":"the I will cancel out,"},{"Start":"11:36.150 ","End":"11:45.100","Text":"l divided by Pi multiplied by ln(d) minus a divided by a."},{"Start":"11:45.290 ","End":"11:48.715","Text":"This is the inductance,"},{"Start":"11:48.715 ","End":"11:52.405","Text":"and now we want the inductance per unit length."},{"Start":"11:52.405 ","End":"11:57.225","Text":"This is just going to be the inductance per unit length,"},{"Start":"11:57.225 ","End":"11:59.070","Text":"divided by the length,"},{"Start":"11:59.070 ","End":"12:03.975","Text":"the l. It\u0027s just going to be this with this l taken away,"},{"Start":"12:03.975 ","End":"12:10.070","Text":"because we\u0027ve divided both sides by l. We have Mu_naught divided by Pi,"},{"Start":"12:10.070 ","End":"12:15.960","Text":"and then multiplied by ln d minus a divided by a."},{"Start":"12:17.270 ","End":"12:20.180","Text":"This is the answer to the question."},{"Start":"12:20.180 ","End":"12:21.935","Text":"This is the inductance,"},{"Start":"12:21.935 ","End":"12:26.130","Text":"and this is the inductance per unit length."}],"ID":22370},{"Watched":false,"Name":"Enery Stored in a Coil","Duration":"9m 42s","ChapterTopicVideoID":21541,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.920","Text":"Hello. In this lesson we\u0027re going to be speaking about the energy stored in a coil."},{"Start":"00:04.920 ","End":"00:09.000","Text":"Here we have a coil and we\u0027re told that it\u0027s inductance is equal"},{"Start":"00:09.000 ","End":"00:13.200","Text":"to L and it\u0027s connected to this electrical circuit,"},{"Start":"00:13.200 ","End":"00:17.280","Text":"and there is a current I flowing through,"},{"Start":"00:17.280 ","End":"00:24.915","Text":"so we\u0027re told that the energy stored due to inductance is"},{"Start":"00:24.915 ","End":"00:34.120","Text":"equal to half of the inductance multiplied by the current flowing through squared."},{"Start":"00:34.310 ","End":"00:37.880","Text":"If we have an electrical components,"},{"Start":"00:37.880 ","End":"00:41.690","Text":"so in this case we have a coil so that means that it\u0027s going"},{"Start":"00:41.690 ","End":"00:45.560","Text":"to have an inductance and there\u0027s a current flowing through it,"},{"Start":"00:45.560 ","End":"00:47.285","Text":"so if there\u0027s a current flowing through,"},{"Start":"00:47.285 ","End":"00:50.120","Text":"we\u0027re going to have energy stored."},{"Start":"00:50.120 ","End":"00:53.060","Text":"Energy is going to be stored in this coil,"},{"Start":"00:53.060 ","End":"00:58.500","Text":"which means that later we can use this energy."},{"Start":"00:58.520 ","End":"01:04.400","Text":"How can we access this energy that\u0027s stored if we were to cut"},{"Start":"01:04.400 ","End":"01:08.150","Text":"this electrical circuit so that would"},{"Start":"01:08.150 ","End":"01:12.185","Text":"mean that there won\u0027t be any more current flowing through."},{"Start":"01:12.185 ","End":"01:14.930","Text":"All of the energy is stored in the coil."},{"Start":"01:14.930 ","End":"01:18.905","Text":"Well, start to flow through what is left,"},{"Start":"01:18.905 ","End":"01:22.835","Text":"so it\u0027s going to oppose the change in the current."},{"Start":"01:22.835 ","End":"01:24.290","Text":"If the current stops,"},{"Start":"01:24.290 ","End":"01:27.530","Text":"we\u0027re going to have this energy that is flowing in place"},{"Start":"01:27.530 ","End":"01:32.020","Text":"of the current to oppose this change in flow."},{"Start":"01:32.020 ","End":"01:35.869","Text":"If we cut this electrical circuit,"},{"Start":"01:35.869 ","End":"01:44.135","Text":"the current isn\u0027t going to stop in one go so go from a value of sum I until 0,"},{"Start":"01:44.135 ","End":"01:48.890","Text":"and that is because of this coil and the energy stored in it,"},{"Start":"01:48.890 ","End":"01:51.140","Text":"so the current will slowly,"},{"Start":"01:51.140 ","End":"01:56.220","Text":"slowly start to reduce until it reaches a 0 value."},{"Start":"01:56.420 ","End":"01:59.075","Text":"Now, this equation, obviously,"},{"Start":"01:59.075 ","End":"02:01.490","Text":"right in your equation sheets,"},{"Start":"02:01.490 ","End":"02:07.020","Text":"and it is very similar to the equation for the energy stored in the capacitor,"},{"Start":"02:07.020 ","End":"02:10.879","Text":"and the whole idea is also very similar to that of a capacitor,"},{"Start":"02:10.879 ","End":"02:15.365","Text":"where the energy stored on a capacitor is equal to half like here."},{"Start":"02:15.365 ","End":"02:22.190","Text":"Then q squared divided by c. Here instead of q squared,"},{"Start":"02:22.190 ","End":"02:24.215","Text":"we have I squared, and here,"},{"Start":"02:24.215 ","End":"02:26.330","Text":"instead of dividing by c,"},{"Start":"02:26.330 ","End":"02:29.915","Text":"we multiply by L, the inductance."},{"Start":"02:29.915 ","End":"02:34.380","Text":"We can see that these 2 equations are also very, very similar."},{"Start":"02:35.300 ","End":"02:39.515","Text":"Now let\u0027s expand on this equation over here."},{"Start":"02:39.515 ","End":"02:41.600","Text":"It\u0027s going to make it easier for us"},{"Start":"02:41.600 ","End":"02:44.935","Text":"to work with it and also to understand what\u0027s going on."},{"Start":"02:44.935 ","End":"02:46.550","Text":"As we\u0027ve already seen,"},{"Start":"02:46.550 ","End":"02:50.705","Text":"the EMF or the voltage is equal to"},{"Start":"02:50.705 ","End":"02:56.030","Text":"negative l multiplied by I dot where I dot is of course,"},{"Start":"02:56.030 ","End":"02:59.170","Text":"the time derivative of the current."},{"Start":"02:59.170 ","End":"03:04.810","Text":"This equation, remember a few lessons ago we saw how we got to it."},{"Start":"03:04.810 ","End":"03:07.925","Text":"We used the equation for"},{"Start":"03:07.925 ","End":"03:14.405","Text":"the inductance and we just rearranged it by isolating out the magnetic flux."},{"Start":"03:14.405 ","End":"03:20.405","Text":"We saw that the magnetic flux is equal to the inductance multiplied by the current."},{"Start":"03:20.405 ","End":"03:23.090","Text":"Then we used Faraday\u0027s law for EMF,"},{"Start":"03:23.090 ","End":"03:31.650","Text":"which is equal to the negative time derivative of the magnetic flux."},{"Start":"03:31.650 ","End":"03:35.000","Text":"Then once we substitute this into this equation,"},{"Start":"03:35.000 ","End":"03:38.070","Text":"we get this equation over here."},{"Start":"03:39.230 ","End":"03:46.490","Text":"Let\u0027s begin, so a current starts flowing through our circuit over here,"},{"Start":"03:46.490 ","End":"03:51.215","Text":"and the current doesn\u0027t go from 0 till i in one stop,"},{"Start":"03:51.215 ","End":"03:53.800","Text":"it\u0027s a gradual increase in current."},{"Start":"03:53.800 ","End":"04:02.280","Text":"We start a current of 0 and gradually it becomes a current of value."},{"Start":"04:03.410 ","End":"04:08.265","Text":"What we can see is that our i is increasing,"},{"Start":"04:08.265 ","End":"04:11.260","Text":"so if our current is increasing,"},{"Start":"04:11.260 ","End":"04:13.750","Text":"this means for now increasing."},{"Start":"04:13.750 ","End":"04:23.500","Text":"Then that means that the EMF inside of the coil is going to oppose this change."},{"Start":"04:23.500 ","End":"04:26.920","Text":"It\u0027s going to impose this increase in current,"},{"Start":"04:26.920 ","End":"04:29.440","Text":"and this is what we saw from Faraday\u0027s Law."},{"Start":"04:29.440 ","End":"04:34.727","Text":"Faraday\u0027s law, we\u0027re trying to oppose the change."},{"Start":"04:34.727 ","End":"04:40.090","Text":"As we saw, current and the magnetic fields are linearly dependent,"},{"Start":"04:40.090 ","End":"04:46.010","Text":"and so also, the same goes with the magnetic flux and current."},{"Start":"04:47.390 ","End":"04:53.035","Text":"What we want to know, is we want to know the work done by this coil,"},{"Start":"04:53.035 ","End":"05:02.240","Text":"and the coil is called L. The work done or the rate of work done, dw by dt."},{"Start":"05:02.240 ","End":"05:05.210","Text":"The rate of work done is, as we know,"},{"Start":"05:05.210 ","End":"05:07.933","Text":"equal to the power of the coil."},{"Start":"05:07.933 ","End":"05:11.165","Text":"Another equation for the power of the coil,"},{"Start":"05:11.165 ","End":"05:12.545","Text":"or power in general,"},{"Start":"05:12.545 ","End":"05:18.000","Text":"is the current multiplied by the voltage."},{"Start":"05:18.470 ","End":"05:20.580","Text":"Here is voltage,"},{"Start":"05:20.580 ","End":"05:24.005","Text":"and we said that when we\u0027re dealing with coils,"},{"Start":"05:24.005 ","End":"05:27.687","Text":"the voltage that we\u0027re dealing with is called EMF."},{"Start":"05:27.687 ","End":"05:29.910","Text":"We can substitute this in."},{"Start":"05:29.910 ","End":"05:32.720","Text":"We have I multiplied by the EMF,"},{"Start":"05:32.720 ","End":"05:34.460","Text":"which is voltage,"},{"Start":"05:34.460 ","End":"05:39.005","Text":"and then we can substitute in the value for the EMF."},{"Start":"05:39.005 ","End":"05:42.665","Text":"We have I multiplied by,"},{"Start":"05:42.665 ","End":"05:48.090","Text":"over here, negative L and then I dot."},{"Start":"05:48.090 ","End":"05:49.700","Text":"Instead of writing I dot,"},{"Start":"05:49.700 ","End":"05:52.550","Text":"let\u0027s write di by dt."},{"Start":"05:52.550 ","End":"05:56.940","Text":"Means the same thing, the time derivative of the current."},{"Start":"05:57.500 ","End":"06:01.965","Text":"This is the power of the coil,"},{"Start":"06:01.965 ","End":"06:05.960","Text":"and now what we want to know is we wanted to know what is"},{"Start":"06:05.960 ","End":"06:10.655","Text":"the value of the external power which is opposing this."},{"Start":"06:10.655 ","End":"06:17.220","Text":"The external power which is opposing this increase in current."},{"Start":"06:17.960 ","End":"06:23.570","Text":"In order to negate this power from the coil,"},{"Start":"06:23.570 ","End":"06:29.525","Text":"our external power is going to have to be equal and opposite to the power of the coil."},{"Start":"06:29.525 ","End":"06:32.990","Text":"It\u0027s equal to negative pl."},{"Start":"06:32.990 ","End":"06:35.850","Text":"We just get rid of this negative."},{"Start":"06:35.850 ","End":"06:43.830","Text":"The external power will be equal to IL multiplied by di by dt."},{"Start":"06:44.270 ","End":"06:48.140","Text":"Now if I want to calculate the work,"},{"Start":"06:48.140 ","End":"06:54.095","Text":"I have to integrate on this power of the external force."},{"Start":"06:54.095 ","End":"06:58.940","Text":"The work done is going to be equal to the integral"},{"Start":"06:58.940 ","End":"07:05.250","Text":"of my external power with respect to time, so dt."},{"Start":"07:05.900 ","End":"07:09.045","Text":"That\u0027s how I get my work."},{"Start":"07:09.045 ","End":"07:13.910","Text":"Now I can substitute this inside of the integral of IL,"},{"Start":"07:13.910 ","End":"07:23.430","Text":"di by dt, dt."},{"Start":"07:23.430 ","End":"07:26.500","Text":"Mathematicians won\u0027t like what I\u0027m about to do."},{"Start":"07:26.500 ","End":"07:28.240","Text":"But for us physicists,"},{"Start":"07:28.240 ","End":"07:29.560","Text":"we can do this."},{"Start":"07:29.560 ","End":"07:32.880","Text":"We can just cancel out the dt\u0027s."},{"Start":"07:32.880 ","End":"07:38.295","Text":"Here we have a dt in the numerator and here a dt in the denominator, so this is fine."},{"Start":"07:38.295 ","End":"07:42.910","Text":"Now what we\u0027re integrating as IL DI, or in other words,"},{"Start":"07:42.910 ","End":"07:49.810","Text":"the L is a constant so we have L and then we\u0027re integrating along I DI."},{"Start":"07:49.810 ","End":"07:52.880","Text":"Of course, we can also put in the bounds,"},{"Start":"07:52.880 ","End":"07:57.220","Text":"so we\u0027re integrating from the initial current,"},{"Start":"07:57.220 ","End":"08:02.290","Text":"which is 0 until our final value for current, which is i."},{"Start":"08:02.290 ","End":"08:05.060","Text":"Here we can do tags if we wanted to just"},{"Start":"08:05.060 ","End":"08:09.605","Text":"differentiates between our bound I and our variable i."},{"Start":"08:09.605 ","End":"08:17.440","Text":"Then what we get is that this is equal to L multiplied by I^2 divided by 2."},{"Start":"08:17.440 ","End":"08:20.360","Text":"This is the work done."},{"Start":"08:20.360 ","End":"08:27.685","Text":"Notice it\u0027s the same value as what we have over here, half LI^2,"},{"Start":"08:27.685 ","End":"08:35.735","Text":"so this is the work done by the coil in order to oppose this increase in the current,"},{"Start":"08:35.735 ","End":"08:42.470","Text":"so once we switch off the current or we cut this wire over here,"},{"Start":"08:42.470 ","End":"08:48.200","Text":"what we\u0027re going to see is that this work done by the coil is going to be released back"},{"Start":"08:48.200 ","End":"08:54.675","Text":"into the circuit to oppose the change now in the decrease in current,"},{"Start":"08:54.675 ","End":"08:57.950","Text":"and that is why we can say that as"},{"Start":"08:57.950 ","End":"09:01.070","Text":"soon as there\u0027s some kind of current flowing through the circuit,"},{"Start":"09:01.070 ","End":"09:03.570","Text":"there\u0027s going to be energy stored,"},{"Start":"09:03.570 ","End":"09:06.665","Text":"potential energy stored in the coil,"},{"Start":"09:06.665 ","End":"09:10.520","Text":"such that once we switch off the current source or cut"},{"Start":"09:10.520 ","End":"09:15.440","Text":"off or cut the wire in the electric circuits,"},{"Start":"09:15.440 ","End":"09:24.716","Text":"the potential energy stored in the coil is going to be released back into the circuits."},{"Start":"09:24.716 ","End":"09:29.225","Text":"That\u0027s why the second we have current flowing through the circuit,"},{"Start":"09:29.225 ","End":"09:34.105","Text":"we have this potential energy which is stored in the coil,"},{"Start":"09:34.105 ","End":"09:38.825","Text":"and we just saw how to derive this equation."},{"Start":"09:38.825 ","End":"09:42.389","Text":"That is the end of this lesson."}],"ID":22371},{"Watched":false,"Name":"Energy Stored in a Magnetic Field","Duration":"6m 2s","ChapterTopicVideoID":21542,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.825","Text":"Hello. In this lesson,"},{"Start":"00:01.825 ","End":"00:07.430","Text":"we\u0027re going to speak very shortly about the energy stored in a magnetic field."},{"Start":"00:07.430 ","End":"00:13.750","Text":"The equation for the energy stored in a magnetic field is equal"},{"Start":"00:13.750 ","End":"00:20.870","Text":"to the integral of the whole area or the whole region."},{"Start":"00:21.320 ","End":"00:25.050","Text":"The integral is of B^2,"},{"Start":"00:25.050 ","End":"00:26.940","Text":"where B is the magnetic field."},{"Start":"00:26.940 ","End":"00:29.200","Text":"In other words, we\u0027re just taking the magnitude of"},{"Start":"00:29.200 ","End":"00:31.600","Text":"the magnetic field because it\u0027s squared."},{"Start":"00:31.600 ","End":"00:33.160","Text":"If there is a minus,"},{"Start":"00:33.160 ","End":"00:37.045","Text":"it doesn\u0027t make a difference divided by 2Mu naught,"},{"Start":"00:37.045 ","End":"00:40.965","Text":"Mu naught is of course a constant and dv."},{"Start":"00:40.965 ","End":"00:47.500","Text":"Here, of course, this is the volume of the whole region."},{"Start":"00:47.690 ","End":"00:52.580","Text":"When we looked at the energy stored in an electric field,"},{"Start":"00:52.580 ","End":"00:54.365","Text":"we saw that u was equal to,"},{"Start":"00:54.365 ","End":"01:00.360","Text":"again, the integral of the whole region of Epsilon naught E^2."},{"Start":"01:00.360 ","End":"01:04.460","Text":"The magnetic field squared divided by 2dv."},{"Start":"01:04.460 ","End":"01:07.250","Text":"It\u0027s the exact same integral."},{"Start":"01:07.250 ","End":"01:10.775","Text":"Just instead of having Mu naught in the denominator,"},{"Start":"01:10.775 ","End":"01:13.985","Text":"we had Epsilon naught in the numerator,"},{"Start":"01:13.985 ","End":"01:15.530","Text":"but we can see that it\u0027s the same thing."},{"Start":"01:15.530 ","End":"01:18.455","Text":"We\u0027re integrating in the whole region divided by 2."},{"Start":"01:18.455 ","End":"01:20.195","Text":"We\u0027re taking our field,"},{"Start":"01:20.195 ","End":"01:24.470","Text":"the magnitude of the field, and squaring it."},{"Start":"01:24.800 ","End":"01:27.160","Text":"First of all, this equation,"},{"Start":"01:27.160 ","End":"01:29.440","Text":"write in your equation books or sheets,"},{"Start":"01:29.440 ","End":"01:31.360","Text":"and of course this as well if you don\u0027t have it"},{"Start":"01:31.360 ","End":"01:35.660","Text":"written from the section where we dealt with electric fields."},{"Start":"01:35.660 ","End":"01:38.320","Text":"When we were dealing just with electric fields,"},{"Start":"01:38.320 ","End":"01:40.960","Text":"we still haven\u0027t learned of the magnetic field."},{"Start":"01:40.960 ","End":"01:43.855","Text":"As far as we\u0027re concerned,"},{"Start":"01:43.855 ","End":"01:46.090","Text":"this was the energy of the system."},{"Start":"01:46.090 ","End":"01:52.150","Text":"However, now that we\u0027ve learned that electricity and"},{"Start":"01:52.150 ","End":"01:55.585","Text":"magnetism are so closely related that the topic is called"},{"Start":"01:55.585 ","End":"01:59.110","Text":"electromagnetism we can see that in actual fact,"},{"Start":"01:59.110 ","End":"02:02.380","Text":"the total energy of a system."},{"Start":"02:02.380 ","End":"02:06.980","Text":"Let\u0027s write u total is going to be equal to the addition of"},{"Start":"02:06.980 ","End":"02:13.205","Text":"the energy from the electric field and the energy stored in the magnetic field."},{"Start":"02:13.205 ","End":"02:15.545","Text":"We can just write this out as one equation,"},{"Start":"02:15.545 ","End":"02:18.860","Text":"Epsilon naught E^2 divided by"},{"Start":"02:18.860 ","End":"02:25.695","Text":"2 plus B^2 divided by 2Mu naught,"},{"Start":"02:25.695 ","End":"02:30.350","Text":"and all of these dv,"},{"Start":"02:30.350 ","End":"02:36.630","Text":"where of course, this is an integral on the whole region."},{"Start":"02:38.180 ","End":"02:43.415","Text":"This is the equation to find the energy stored."},{"Start":"02:43.415 ","End":"02:48.445","Text":"Of course, if we do actually only have a magnetic fields present."},{"Start":"02:48.445 ","End":"02:51.290","Text":"This, the electric field will be equal to 0,"},{"Start":"02:51.290 ","End":"02:56.295","Text":"so we can just ignore this expression."},{"Start":"02:56.295 ","End":"03:02.090","Text":"Similarly, if we only have an electric field and no magnetic field,"},{"Start":"03:02.090 ","End":"03:04.180","Text":"then the magnetic field will be equal to 0."},{"Start":"03:04.180 ","End":"03:06.875","Text":"Then this expression, we can just ignore,"},{"Start":"03:06.875 ","End":"03:10.189","Text":"or we can just use each one of these equations separately."},{"Start":"03:10.189 ","End":"03:11.704","Text":"It doesn\u0027t make a difference."},{"Start":"03:11.704 ","End":"03:15.150","Text":"This is really the one to save."},{"Start":"03:16.550 ","End":"03:19.245","Text":"Now, just like we saw,"},{"Start":"03:19.245 ","End":"03:20.910","Text":"if we have Q,"},{"Start":"03:20.910 ","End":"03:22.595","Text":"which is the charge,"},{"Start":"03:22.595 ","End":"03:30.790","Text":"the equation for it is equal to the integral of if we\u0027re dealing with the volume."},{"Start":"03:30.790 ","End":"03:38.090","Text":"It\u0027s the charge density per unit volume multiplied by the volume."},{"Start":"03:38.090 ","End":"03:39.755","Text":"This, as we could see,"},{"Start":"03:39.755 ","End":"03:42.500","Text":"was the charge density."},{"Start":"03:42.500 ","End":"03:46.490","Text":"Rho is the charge density here specifically per unit volume."},{"Start":"03:46.490 ","End":"03:52.730","Text":"But it could also be if we had a different charge per unit area."},{"Start":"03:52.730 ","End":"03:56.880","Text":"That would have been Sigma ds."},{"Start":"03:56.880 ","End":"04:03.785","Text":"Again, this is the charge density just this time per unit area."},{"Start":"04:03.785 ","End":"04:08.680","Text":"If the same way, we can see that with these equations as well."},{"Start":"04:08.680 ","End":"04:12.060","Text":"Here we\u0027re integrating dv."},{"Start":"04:12.060 ","End":"04:20.910","Text":"We can consider all of this over here as the energy density."},{"Start":"04:21.790 ","End":"04:27.020","Text":"The exact same thing over here for the electric field."},{"Start":"04:27.020 ","End":"04:31.645","Text":"This is also the energy density of the electric field."},{"Start":"04:31.645 ","End":"04:33.425","Text":"Here, of course,"},{"Start":"04:33.425 ","End":"04:38.645","Text":"this whole section is the total energy density."},{"Start":"04:38.645 ","End":"04:45.730","Text":"Taking account the energy density from the electric field and from the magnetic field."},{"Start":"04:45.730 ","End":"04:52.205","Text":"Sometimes we will call this energy density by the letter Mu."},{"Start":"04:52.205 ","End":"04:56.435","Text":"Mu_E is the energy density of the electric field."},{"Start":"04:56.435 ","End":"04:58.115","Text":"That\u0027s just equal to this."},{"Start":"04:58.115 ","End":"05:03.060","Text":"As we saw, Epsilon naught E^2 divided by 2."},{"Start":"05:03.060 ","End":"05:05.690","Text":"If you just remember that this is the energy density,"},{"Start":"05:05.690 ","End":"05:09.635","Text":"you don\u0027t have to write that this is also sometimes called Mu_E."},{"Start":"05:09.635 ","End":"05:12.440","Text":"But if you want, you can write that down."},{"Start":"05:12.440 ","End":"05:16.340","Text":"This is Mu_E. Of course,"},{"Start":"05:16.340 ","End":"05:19.470","Text":"we have the magnetic energy density Mu_B,"},{"Start":"05:19.470 ","End":"05:22.815","Text":"which is equal to this over here."},{"Start":"05:22.815 ","End":"05:24.955","Text":"We can call this Mu_B,"},{"Start":"05:24.955 ","End":"05:31.070","Text":"which is equal to B^2 divided by 2Mu naught,"},{"Start":"05:31.070 ","End":"05:35.100","Text":"which is of course this section over here, Mu_B."},{"Start":"05:35.270 ","End":"05:39.950","Text":"I\u0027m not going to derive these equations because they\u0027re a bit complicated."},{"Start":"05:39.950 ","End":"05:41.780","Text":"But what\u0027s important to notice is"},{"Start":"05:41.780 ","End":"05:46.265","Text":"the similarity between the magnetic field and the electric field,"},{"Start":"05:46.265 ","End":"05:49.790","Text":"which we can see in many different places,"},{"Start":"05:49.790 ","End":"05:53.250","Text":"but here specifically when dealing with energy."},{"Start":"05:53.250 ","End":"05:58.910","Text":"Just remember the magnetic field and the electric fields are very closely related."},{"Start":"05:58.910 ","End":"06:02.880","Text":"That is the end of this lesson."}],"ID":22372},{"Watched":false,"Name":"Calculating Inductance from Energy Stored in a Magnetic Field","Duration":"6m 23s","ChapterTopicVideoID":21543,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.755","Text":"Hello. In this lesson,"},{"Start":"00:01.755 ","End":"00:06.165","Text":"we\u0027re going to be looking at another method for calculating the inductance."},{"Start":"00:06.165 ","End":"00:09.980","Text":"That is very useful and in some cases can be a lot,"},{"Start":"00:09.980 ","End":"00:11.890","Text":"lot easier to use."},{"Start":"00:11.890 ","End":"00:18.450","Text":"We\u0027ve already seen in the previous lesson that the energy stored in an inductor is equal"},{"Start":"00:18.450 ","End":"00:27.550","Text":"to half of the inductance multiplied by the current squared."},{"Start":"00:27.740 ","End":"00:37.680","Text":"We\u0027ve also seen that the energy stored in a magnetic field is equal to the integral over"},{"Start":"00:37.680 ","End":"00:42.218","Text":"the whole region of B^2"},{"Start":"00:42.218 ","End":"00:48.590","Text":"divided by 2 Mu naught dv,"},{"Start":"00:48.590 ","End":"00:52.200","Text":"where of course this v is for volume."},{"Start":"00:52.520 ","End":"00:56.240","Text":"This energy over here stored in the magnetic field"},{"Start":"00:56.240 ","End":"00:59.430","Text":"is derived from this equation over here."},{"Start":"00:59.430 ","End":"01:05.180","Text":"What we can see is that these equations are actually in fact the same,"},{"Start":"01:05.180 ","End":"01:13.474","Text":"so these are 2 different ways to calculate the energy or the same energy exactly."},{"Start":"01:13.474 ","End":"01:17.600","Text":"In which case we can equate these 2 equations."},{"Start":"01:17.620 ","End":"01:22.010","Text":"What we can do is we can calculate the energy from"},{"Start":"01:22.010 ","End":"01:27.680","Text":"the magnetic field and then we can divide the answer that we get to this"},{"Start":"01:27.680 ","End":"01:32.555","Text":"by I^2 and usually we\u0027ll have the I is the current"},{"Start":"01:32.555 ","End":"01:37.430","Text":"will cancel out because of the magnetic field squared over here and multiply it by 2,"},{"Start":"01:37.430 ","End":"01:39.260","Text":"which of course will cancel out with this 2 over"},{"Start":"01:39.260 ","End":"01:42.395","Text":"here and that will give us the inductance."},{"Start":"01:42.395 ","End":"01:45.730","Text":"Let\u0027s now take a look at an example."},{"Start":"01:45.730 ","End":"01:55.130","Text":"Let\u0027s look at the basic example of a coil so here we have our coil,"},{"Start":"01:55.130 ","End":"01:59.735","Text":"and we\u0027re given in this question,"},{"Start":"01:59.735 ","End":"02:01.895","Text":"the length of the coil,"},{"Start":"02:01.895 ","End":"02:07.760","Text":"the total number of turns in the coil, and the radius."},{"Start":"02:07.760 ","End":"02:14.840","Text":"This distance over here as r. We saw"},{"Start":"02:14.840 ","End":"02:18.500","Text":"that the magnetic field of a coil is equal"},{"Start":"02:18.500 ","End":"02:22.475","Text":"to Mu naught multiplied by the current flowing through it,"},{"Start":"02:22.475 ","End":"02:24.590","Text":"multiplied by lowercase n,"},{"Start":"02:24.590 ","End":"02:29.870","Text":"where lowercase n is the total number of turns,"},{"Start":"02:29.870 ","End":"02:32.915","Text":"capital N divided by the total length."},{"Start":"02:32.915 ","End":"02:40.670","Text":"In case of the density of times the density of the coil."},{"Start":"02:40.670 ","End":"02:46.265","Text":"A magnetic field going through this coil is going to be"},{"Start":"02:46.265 ","End":"02:53.895","Text":"a constant magnetic field that goes right through the coil like so."},{"Start":"02:53.895 ","End":"02:59.420","Text":"Now we can do the integral for u so u is equal to the integral in"},{"Start":"02:59.420 ","End":"03:05.690","Text":"the whole region of B^2 so we have Mu naught ^2 I^2,"},{"Start":"03:05.690 ","End":"03:12.095","Text":"lowercase n ^2 divided by 2 Mu naught,"},{"Start":"03:12.095 ","End":"03:16.730","Text":"so this and this can cancel out dv."},{"Start":"03:16.730 ","End":"03:21.515","Text":"Now we know that the magnetic field due to"},{"Start":"03:21.515 ","End":"03:27.395","Text":"a current carrying coil is going to be constant and only inside the coil."},{"Start":"03:27.395 ","End":"03:33.290","Text":"Outside the coil, there\u0027s no magnetic field lines in which case our integral,"},{"Start":"03:33.290 ","End":"03:36.487","Text":"all of these are constants,"},{"Start":"03:36.487 ","End":"03:41.495","Text":"and our integral dv is just on the volume inside"},{"Start":"03:41.495 ","End":"03:48.110","Text":"the coil so we have Mu naught I^2 n^2."},{"Start":"03:48.110 ","End":"03:52.880","Text":"We can put already capital N^2 divided by L^2,"},{"Start":"03:52.880 ","End":"03:58.842","Text":"so this was lower-n squared and then we can put here also the 2,"},{"Start":"03:58.842 ","End":"04:03.050","Text":"and then our integral dv is just the volume inside"},{"Start":"04:03.050 ","End":"04:07.925","Text":"this tube so that\u0027s of course equal to the surface area of"},{"Start":"04:07.925 ","End":"04:13.910","Text":"each term so that\u0027s going to be Pi r^2 so"},{"Start":"04:13.910 ","End":"04:21.845","Text":"the surface area of the circle enclosed multiplied by the whole length of the coil,"},{"Start":"04:21.845 ","End":"04:29.645","Text":"which is l. Now we have the energy stored in the magnetic field."},{"Start":"04:29.645 ","End":"04:34.600","Text":"Now what we can do is we can,"},{"Start":"04:34.600 ","End":"04:37.820","Text":"first of all, it cancel out with one of these L\u0027s."},{"Start":"04:37.820 ","End":"04:39.260","Text":"We\u0027re left with"},{"Start":"04:39.260 ","End":"04:46.130","Text":"Mu naught I^2 capital N^2 Pi r^2"},{"Start":"04:46.130 ","End":"04:50.660","Text":"divided by 2L."},{"Start":"04:52.820 ","End":"04:56.685","Text":"Now as we said, these 2 equations are equal."},{"Start":"04:56.685 ","End":"05:03.980","Text":"Let\u0027s equate them so this is equal to half LI^2."},{"Start":"05:03.980 ","End":"05:07.400","Text":"Then the I^2 over here cancels out with"},{"Start":"05:07.400 ","End":"05:11.540","Text":"the I^2 over here and the half over here cancels out with the half over"},{"Start":"05:11.540 ","End":"05:16.355","Text":"here and therefore we get that the inductance is equal to"},{"Start":"05:16.355 ","End":"05:25.715","Text":"Mu naught capital N^2 Pi r^2 divided by L,"},{"Start":"05:25.715 ","End":"05:30.800","Text":"which is the exact same equation for the inductance that we"},{"Start":"05:30.800 ","End":"05:37.650","Text":"calculated in a previous lesson via using the definition for inductance."},{"Start":"05:37.960 ","End":"05:42.680","Text":"In this type of example that we looked at,"},{"Start":"05:42.680 ","End":"05:47.900","Text":"it is actually easier to work out the inductance via the definition."},{"Start":"05:47.900 ","End":"05:50.270","Text":"However, we\u0027ll see later on in"},{"Start":"05:50.270 ","End":"05:54.290","Text":"this chapter that there are some questions where it\u0027s not exactly"},{"Start":"05:54.290 ","End":"06:02.105","Text":"clear how to work out the magnetic flux and in those cases,"},{"Start":"06:02.105 ","End":"06:05.165","Text":"when we can\u0027t work out the magnetic flux,"},{"Start":"06:05.165 ","End":"06:08.090","Text":"this is the equation to use."},{"Start":"06:08.090 ","End":"06:12.185","Text":"You work out the energy is stored"},{"Start":"06:12.185 ","End":"06:16.460","Text":"in the magnetic field and you equate it to this equation over here,"},{"Start":"06:16.460 ","End":"06:19.910","Text":"and then you just isolate out the inductance."},{"Start":"06:19.910 ","End":"06:23.190","Text":"That\u0027s the end of this lesson."}],"ID":22373},{"Watched":false,"Name":"Mutual Inductance","Duration":"16m 8s","ChapterTopicVideoID":21544,"CourseChapterTopicPlaylistID":99483,"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":"00:05.400","Text":"we\u0027re going to be speaking about mutual inductance."},{"Start":"00:05.400 ","End":"00:12.130","Text":"Mutual inductance is denoted by the letter capital M_1,2."},{"Start":"00:12.680 ","End":"00:23.440","Text":"It is equal to Phi 1 divided by I_2."},{"Start":"00:23.440 ","End":"00:32.280","Text":"Where we see that Phi over here is the flux and I is the current."},{"Start":"00:32.280 ","End":"00:34.395","Text":"Just a reminder, up until now,"},{"Start":"00:34.395 ","End":"00:37.230","Text":"we\u0027ve been speaking about inductance L,"},{"Start":"00:37.230 ","End":"00:46.490","Text":"which was given by the magnetic flux divided by the current in that 1 inductor."},{"Start":"00:46.490 ","End":"00:49.010","Text":"Okay, so we, a lot of the time,"},{"Start":"00:49.010 ","End":"00:51.950","Text":"spoke about the coil being an inductor."},{"Start":"00:51.950 ","End":"00:55.340","Text":"Over here, we\u0027re looking at the mutual inductance."},{"Start":"00:55.340 ","End":"01:00.320","Text":"We\u0027re looking at the relationship between the magnetic flux in"},{"Start":"01:00.320 ","End":"01:07.325","Text":"1 inductor compared to the current flowing through the other inductor."},{"Start":"01:07.325 ","End":"01:11.896","Text":"What we have is 2 inductors and we\u0027re looking at the relationship,"},{"Start":"01:11.896 ","End":"01:15.815","Text":"if 1 inductor has a current I_2 flowing through it,"},{"Start":"01:15.815 ","End":"01:21.750","Text":"then what will be the magnetic flux through the other inductor?"},{"Start":"01:22.610 ","End":"01:26.170","Text":"Let\u0027s take a look at what we\u0027re talking about,"},{"Start":"01:26.170 ","End":"01:30.825","Text":"so let\u0027s imagine that we have a coil."},{"Start":"01:30.825 ","End":"01:35.010","Text":"It\u0027s going something like so,"},{"Start":"01:35.010 ","End":"01:36.920","Text":"and at the top,"},{"Start":"01:36.920 ","End":"01:46.625","Text":"we have that the current over here is going into the page like so,"},{"Start":"01:46.625 ","End":"01:51.355","Text":"and that means that over here at the bottom,"},{"Start":"01:51.355 ","End":"01:58.260","Text":"the current is coming rather out of the page."},{"Start":"01:58.260 ","End":"02:01.740","Text":"Now, this is what the current looks like,"},{"Start":"02:01.740 ","End":"02:07.300","Text":"so I\u0027m just going to erase the coil just to make this diagram a bit more clear."},{"Start":"02:07.300 ","End":"02:13.010","Text":"Now let\u0027s imagine that around this green coil,"},{"Start":"02:13.010 ","End":"02:16.220","Text":"I have another larger coil wrapped around,"},{"Start":"02:16.220 ","End":"02:18.334","Text":"so let\u0027s draw it in black."},{"Start":"02:18.334 ","End":"02:22.385","Text":"Here, the direction of the current is the same,"},{"Start":"02:22.385 ","End":"02:27.020","Text":"so the current at the top is also going into the page,"},{"Start":"02:27.020 ","End":"02:35.120","Text":"and the current at the bottom is going out of the page."},{"Start":"02:35.120 ","End":"02:39.710","Text":"Then I\u0027m just going to add in some extra over here"},{"Start":"02:39.710 ","End":"02:44.750","Text":"just so that we have a different value for capital N."},{"Start":"02:44.750 ","End":"02:49.730","Text":"This is what it looks like so let\u0027s say that the length of"},{"Start":"02:49.730 ","End":"02:56.265","Text":"both of the inductors is L,"},{"Start":"02:56.265 ","End":"02:59.460","Text":"and then let\u0027s say that the radius of the outer coil,"},{"Start":"02:59.460 ","End":"03:02.955","Text":"the black coil, is a_1,"},{"Start":"03:02.955 ","End":"03:08.135","Text":"and let\u0027s say that the radius of the inner coil, the green coil,"},{"Start":"03:08.135 ","End":"03:15.208","Text":"is a_2 and then let\u0027s imagine that the outer coil,"},{"Start":"03:15.208 ","End":"03:20.180","Text":"the black coil has N_1 wraps of the coil,"},{"Start":"03:20.180 ","End":"03:27.630","Text":"and that the inner coil has N_2 wraps of the coil."},{"Start":"03:28.760 ","End":"03:36.740","Text":"Now what we want to do is we want to calculate the mutual inductance of these 2 coils."},{"Start":"03:36.740 ","End":"03:40.250","Text":"So let\u0027s imagine that we have a current I_2 flowing"},{"Start":"03:40.250 ","End":"03:45.227","Text":"through this green inductor, where everything is 2."},{"Start":"03:45.227 ","End":"03:50.661","Text":"So we have a current flowing through the inner inductor,"},{"Start":"03:50.661 ","End":"03:54.500","Text":"the green inductor, and what we want to do for mutual inductance,"},{"Start":"03:54.500 ","End":"04:01.830","Text":"we want to see what magnetic flux we have through the outer black inductor,"},{"Start":"04:01.830 ","End":"04:04.955","Text":"the outer black coil,"},{"Start":"04:04.955 ","End":"04:08.600","Text":"due to this current flowing through the inner coil."},{"Start":"04:08.600 ","End":"04:12.980","Text":"What we have over here is"},{"Start":"04:12.980 ","End":"04:17.585","Text":"we have our current I_2 flowing through the coil, which as we know,"},{"Start":"04:17.585 ","End":"04:23.690","Text":"is going to form a magnetic field in the leftwards direction in this green coil,"},{"Start":"04:23.690 ","End":"04:28.415","Text":"and then what we want to do is we want to see how this magnetic field,"},{"Start":"04:28.415 ","End":"04:31.745","Text":"due to the current flowing through the inner coil,"},{"Start":"04:31.745 ","End":"04:36.510","Text":"is going to affect the magnetic flux in the outer coil."},{"Start":"04:36.510 ","End":"04:38.600","Text":"So whenever we\u0027re looking at mutual inductance,"},{"Start":"04:38.600 ","End":"04:42.545","Text":"we\u0027re looking at the current through 1 of the components"},{"Start":"04:42.545 ","End":"04:47.820","Text":"and how it affects the magnetic flux in the other component."},{"Start":"04:48.020 ","End":"04:51.425","Text":"What we can see therefore,"},{"Start":"04:51.425 ","End":"04:57.470","Text":"is that the magnetic flux in the second component and"},{"Start":"04:57.470 ","End":"05:05.230","Text":"the second coil is going to be as a function of the current in the inner coil,"},{"Start":"05:05.230 ","End":"05:11.675","Text":"and what we\u0027ll see is that it\u0027s always going to be linearly related."},{"Start":"05:11.675 ","End":"05:15.530","Text":"So the magnetic flux in component number 1 is as"},{"Start":"05:15.530 ","End":"05:20.675","Text":"a function of the current and component number 2 and it\u0027s going to be equal to A,"},{"Start":"05:20.675 ","End":"05:22.310","Text":"which is a constant,"},{"Start":"05:22.310 ","End":"05:28.277","Text":"multiplied by the current in the second component."},{"Start":"05:28.277 ","End":"05:31.860","Text":"So that is going to be the relationship."},{"Start":"05:32.570 ","End":"05:38.585","Text":"This is always going to be a constant and what we\u0027re going to see is that"},{"Start":"05:38.585 ","End":"05:47.550","Text":"the mutual inductance is also always dependent on the geometric shape of the inductor."},{"Start":"05:48.710 ","End":"05:53.750","Text":"Now let\u0027s work out the mutual inductance."},{"Start":"05:53.750 ","End":"05:57.470","Text":"The first step that we want to do when working out"},{"Start":"05:57.470 ","End":"06:05.890","Text":"the mutual inductance is to assume that we have this current I_2 flowing."},{"Start":"06:05.890 ","End":"06:11.380","Text":"So we\u0027re assuming that a current I_2 is flowing through inductor number 2,"},{"Start":"06:11.380 ","End":"06:14.569","Text":"which over here is the green inductor."},{"Start":"06:14.569 ","End":"06:20.050","Text":"Next, the second step is what we\u0027re going to do is we\u0027re going to"},{"Start":"06:20.050 ","End":"06:26.235","Text":"calculate the magnetic field inside of body number 1,"},{"Start":"06:26.235 ","End":"06:30.580","Text":"so inside of the outer inductor."},{"Start":"06:31.130 ","End":"06:34.935","Text":"Here let\u0027s do this calculation,"},{"Start":"06:34.935 ","End":"06:39.745","Text":"so the magnetic field is going to be B is"},{"Start":"06:39.745 ","End":"06:44.830","Text":"equal to Mu naught multiplied by the current flowing"},{"Start":"06:44.830 ","End":"06:48.670","Text":"through which we know is just I_2 and"},{"Start":"06:48.670 ","End":"06:55.940","Text":"then multiplied by the number"},{"Start":"06:55.940 ","End":"06:59.300","Text":"of turns divided by the length,"},{"Start":"06:59.300 ","End":"07:01.325","Text":"so the number of turns."},{"Start":"07:01.325 ","End":"07:06.360","Text":"Here we are speaking about the number of turns in the green inductor and why is that?"},{"Start":"07:06.360 ","End":"07:10.560","Text":"Because I_2 is flowing through the green inductor,"},{"Start":"07:10.560 ","End":"07:13.400","Text":"so we want to see how many turns it goes through."},{"Start":"07:13.400 ","End":"07:16.940","Text":"How many turns does this current flow through?"},{"Start":"07:16.940 ","End":"07:20.990","Text":"That is relating to the inductor that the current is flowing through,"},{"Start":"07:20.990 ","End":"07:23.845","Text":"which over here is the green inductor."},{"Start":"07:23.845 ","End":"07:27.645","Text":"We\u0027re multiplying this by lowercase n_2,"},{"Start":"07:27.645 ","End":"07:32.630","Text":"where lowercase n_2 is just equal to the density of turns,"},{"Start":"07:32.630 ","End":"07:35.690","Text":"so that\u0027s the total number of turns and inductor number"},{"Start":"07:35.690 ","End":"07:39.930","Text":"2 divided by the total length of the inductor."},{"Start":"07:41.700 ","End":"07:45.805","Text":"Now we\u0027re going to do step number 3,"},{"Start":"07:45.805 ","End":"07:52.820","Text":"which is to calculate the magnetic flux in inductor number 1."},{"Start":"07:53.310 ","End":"07:56.920","Text":"We\u0027re calculating the magnetic flux and inductor 1,"},{"Start":"07:56.920 ","End":"08:00.305","Text":"or in other words Phi_1 over here."},{"Start":"08:00.305 ","End":"08:06.790","Text":"The magnetic flux, Phi_B,"},{"Start":"08:06.790 ","End":"08:13.195","Text":"of course, we\u0027re calculating through 1 ring of the coil."},{"Start":"08:13.195 ","End":"08:15.220","Text":"Remember that. Whenever we do this first,"},{"Start":"08:15.220 ","End":"08:18.460","Text":"we calculate for 1 ring in the coil and then"},{"Start":"08:18.460 ","End":"08:22.315","Text":"we multiply by the total amount of rings that we have."},{"Start":"08:22.315 ","End":"08:26.515","Text":"This is, of course, equal to the magnetic field"},{"Start":"08:26.515 ","End":"08:33.140","Text":"multiplied by the surface area of 1 ring."},{"Start":"08:33.630 ","End":"08:37.540","Text":"What I said now is almost right,"},{"Start":"08:37.540 ","End":"08:39.310","Text":"the surface area of 1 ring."},{"Start":"08:39.310 ","End":"08:45.670","Text":"However, we have to notice that the magnetic field that we just calculated is,"},{"Start":"08:45.670 ","End":"08:52.885","Text":"of course, the magnetic field due to the current flowing through this inner coil."},{"Start":"08:52.885 ","End":"08:57.865","Text":"Which means that the magnetic field is only going to be located,"},{"Start":"08:57.865 ","End":"08:59.365","Text":"as we\u0027ve already seen,"},{"Start":"08:59.365 ","End":"09:04.390","Text":"within the confines of the inner coil."},{"Start":"09:04.390 ","End":"09:10.070","Text":"Only inside the inner coil are we going to have this magnetic field."},{"Start":"09:10.170 ","End":"09:15.040","Text":"When we\u0027re calculating the surface area for the flux,"},{"Start":"09:15.040 ","End":"09:16.480","Text":"we have to bear in mind that it\u0027s"},{"Start":"09:16.480 ","End":"09:21.310","Text":"the surface area that the magnetic field is passing through."},{"Start":"09:21.310 ","End":"09:29.875","Text":"Over here, our magnetic field is equal to Mu_0 I_2 n_2."},{"Start":"09:29.875 ","End":"09:32.995","Text":"The surface area, as we can see,"},{"Start":"09:32.995 ","End":"09:37.270","Text":"the magnetic field is only flowing through the inner coil,"},{"Start":"09:37.270 ","End":"09:42.219","Text":"so the surface area is equal to Pi multiplied by the radius"},{"Start":"09:42.219 ","End":"09:47.515","Text":"of the inner coil squared, so Pi a_2^2."},{"Start":"09:47.515 ","End":"09:52.615","Text":"If somehow the magnetic field was flowing through the whole area,"},{"Start":"09:52.615 ","End":"09:55.990","Text":"including inside the outer coil,"},{"Start":"09:55.990 ","End":"09:58.400","Text":"then we would write A_1^2."},{"Start":"09:59.300 ","End":"10:02.250","Text":"That\u0027s very important to remember."},{"Start":"10:02.250 ","End":"10:07.440","Text":"This surface area, although we\u0027re working it out for the outer coil,"},{"Start":"10:07.440 ","End":"10:11.085","Text":"we have to realize where the magnetic field is"},{"Start":"10:11.085 ","End":"10:15.615","Text":"located and we can see it\u0027s only inside the inner coil."},{"Start":"10:15.615 ","End":"10:20.580","Text":"Now what we want to do is we want to find the magnetic flux,"},{"Start":"10:20.580 ","End":"10:25.920","Text":"the total magnetic flux through not just 1 loop,"},{"Start":"10:25.920 ","End":"10:28.990","Text":"but through all of the loops."},{"Start":"10:28.990 ","End":"10:31.120","Text":"Through the entire coil."},{"Start":"10:31.120 ","End":"10:34.315","Text":"Let\u0027s write that as Phi_1."},{"Start":"10:34.315 ","End":"10:39.820","Text":"This is the magnetic flux through component number 1, the black coil."},{"Start":"10:39.820 ","End":"10:45.579","Text":"That means we\u0027re going to have to multiply all of this by the number of turns,"},{"Start":"10:45.579 ","End":"10:49.105","Text":"but the number of turns in the outer coil."},{"Start":"10:49.105 ","End":"10:53.335","Text":"We multiply this by N_1,"},{"Start":"10:53.335 ","End":"10:56.710","Text":"so this is multiplied by N_1."},{"Start":"10:56.710 ","End":"11:00.790","Text":"Why is that?"},{"Start":"11:00.790 ","End":"11:06.265","Text":"We\u0027re trying to calculate the magnetic flux through the outer coil."},{"Start":"11:06.265 ","End":"11:11.815","Text":"We\u0027ve calculated the magnetic field,"},{"Start":"11:11.815 ","End":"11:16.090","Text":"which is confined to a certain area and we\u0027ve worked out"},{"Start":"11:16.090 ","End":"11:20.485","Text":"the flux for just 1 of these loops in the coil."},{"Start":"11:20.485 ","End":"11:23.410","Text":"But now we want to calculate for the entire coil,"},{"Start":"11:23.410 ","End":"11:30.980","Text":"so we\u0027re going to multiply by the total number of turns in the outer coil."},{"Start":"11:32.130 ","End":"11:37.480","Text":"Now if we want to work out the mutual inductance, so that is,"},{"Start":"11:37.480 ","End":"11:40.645","Text":"of course, step number 4,"},{"Start":"11:40.645 ","End":"11:44.000","Text":"calculate the mutual inductance."},{"Start":"11:44.520 ","End":"11:52.600","Text":"Then we can write that M_1,2 is equal to the magnetic flux."},{"Start":"11:52.600 ","End":"12:01.780","Text":"So we have Mu_0 I_2 n_2"},{"Start":"12:01.780 ","End":"12:11.990","Text":"multiplied by Pi a_2^2 multiplied by N_1 divided by I_2."},{"Start":"12:12.650 ","End":"12:17.340","Text":"These 2 cancel out and then we can also"},{"Start":"12:17.340 ","End":"12:21.660","Text":"substitute n if we want what our lowercase n was equal to."},{"Start":"12:21.660 ","End":"12:28.080","Text":"We have Mu_0, n_2,"},{"Start":"12:28.080 ","End":"12:32.640","Text":"which is N_2 divided by"},{"Start":"12:32.640 ","End":"12:39.670","Text":"L and then multiplied by Pi a_2^2 N_1."},{"Start":"12:41.520 ","End":"12:44.170","Text":"This is the mutual inductance."},{"Start":"12:44.170 ","End":"12:48.520","Text":"And now let\u0027s see how can I calculate it, how can I use it?"},{"Start":"12:48.520 ","End":"12:51.910","Text":"First of all, we can see that indeed,"},{"Start":"12:51.910 ","End":"12:56.035","Text":"the mutual inductance is dependent on the geometric shape."},{"Start":"12:56.035 ","End":"12:58.450","Text":"So we have Mu_0, which is a constant."},{"Start":"12:58.450 ","End":"13:03.940","Text":"N_2 and N_1 are simply the amount of turns in each coil and"},{"Start":"13:03.940 ","End":"13:10.825","Text":"Pi a^2 is just the surface area through which the magnetic field passes through,"},{"Start":"13:10.825 ","End":"13:14.739","Text":"and L is just the length of the coils."},{"Start":"13:14.739 ","End":"13:20.680","Text":"It\u0027s solely dependent on the geometric shape and that is it."},{"Start":"13:20.680 ","End":"13:23.530","Text":"How can I use that?"},{"Start":"13:23.530 ","End":"13:27.475","Text":"If I multiply both sides by I_2,"},{"Start":"13:27.475 ","End":"13:31.330","Text":"so what I can see is that I can say that"},{"Start":"13:31.330 ","End":"13:39.370","Text":"my magnetic flux through the outer component is simply equal to the mutual inductance,"},{"Start":"13:39.370 ","End":"13:41.260","Text":"which is a constant,"},{"Start":"13:41.260 ","End":"13:46.825","Text":"multiplied by the current flowing through the inner component."},{"Start":"13:46.825 ","End":"13:50.635","Text":"I can just use this equation and as we can see,"},{"Start":"13:50.635 ","End":"13:53.590","Text":"it really is a constant multiplied by I_2,"},{"Start":"13:53.590 ","End":"13:58.315","Text":"just like we said before, the magnetic flux."},{"Start":"13:58.315 ","End":"14:01.990","Text":"Another thing that I can use this for is then, therefore,"},{"Start":"14:01.990 ","End":"14:07.915","Text":"I can say that the EMF through component number 1 is equal to negative"},{"Start":"14:07.915 ","End":"14:15.440","Text":"the mutual inductance multiplied by I_2 dot."},{"Start":"14:16.560 ","End":"14:21.010","Text":"Just to remind you that EMF is equal to"},{"Start":"14:21.010 ","End":"14:26.155","Text":"the negative time derivative of the magnetic flux."},{"Start":"14:26.155 ","End":"14:29.620","Text":"We can see that if I have this current,"},{"Start":"14:29.620 ","End":"14:35.740","Text":"I can get that the EMF in the second component or in the outer component,"},{"Start":"14:35.740 ","End":"14:41.930","Text":"is dependent on the current of the inner component."},{"Start":"14:42.420 ","End":"14:46.930","Text":"We can use this to help us solve a lot of questions."},{"Start":"14:46.930 ","End":"14:50.725","Text":"Now, the most important thing to remember when dealing with"},{"Start":"14:50.725 ","End":"14:57.700","Text":"mutual inductance is that M_1,2 is equal to M_2,1."},{"Start":"14:57.700 ","End":"15:02.995","Text":"In other words, if I had a current flowing through the outer component,"},{"Start":"15:02.995 ","End":"15:08.710","Text":"I could calculate the magnetic flux through the inner components."},{"Start":"15:08.710 ","End":"15:15.745","Text":"I could inverse this question where I have current in outer and flux in inner."},{"Start":"15:15.745 ","End":"15:19.135","Text":"These 2 are equal."},{"Start":"15:19.135 ","End":"15:27.065","Text":"Of course, here that would mean that we\u0027d have Phi_2 divided by I_1."},{"Start":"15:27.065 ","End":"15:28.865","Text":"These are equal."},{"Start":"15:28.865 ","End":"15:30.860","Text":"This is important to note,"},{"Start":"15:30.860 ","End":"15:33.379","Text":"especially because in some questions,"},{"Start":"15:33.379 ","End":"15:37.490","Text":"we won\u0027t know the current in"},{"Start":"15:37.490 ","End":"15:39.470","Text":"the other component or the flux in"},{"Start":"15:39.470 ","End":"15:42.799","Text":"the other component because it will be too difficult to calculate."},{"Start":"15:42.799 ","End":"15:45.020","Text":"We can work out the inverse."},{"Start":"15:45.020 ","End":"15:52.590","Text":"We can work out M_2,1 and then just know that it\u0027s exactly equal to M_1,2."},{"Start":"15:52.590 ","End":"15:55.250","Text":"That\u0027s why a lot of the time mutual inductance is"},{"Start":"15:55.250 ","End":"16:00.240","Text":"just denoted as M without these numbers."},{"Start":"16:01.230 ","End":"16:05.800","Text":"All of this is mutual inductance."},{"Start":"16:05.800 ","End":"16:09.170","Text":"That is the end of this lesson."}],"ID":22374},{"Watched":false,"Name":"Exercise 2","Duration":"17m 42s","ChapterTopicVideoID":21545,"CourseChapterTopicPlaylistID":99483,"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":"00:04.320","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.320 ","End":"00:09.990","Text":"A ring of radius b is placed on the x-y plane at the origin."},{"Start":"00:09.990 ","End":"00:15.520","Text":"A ring of radius a is placed with a center at z."},{"Start":"00:15.620 ","End":"00:22.115","Text":"The height of its center is z,"},{"Start":"00:22.115 ","End":"00:29.255","Text":"and at an angle of Phi to the z-axis so this is the z-axis."},{"Start":"00:29.255 ","End":"00:36.565","Text":"Number 1 is to calculate the mutual inductance."},{"Start":"00:36.565 ","End":"00:44.750","Text":"What we could do is we can assume that a current I flows through 1 of the bodies."},{"Start":"00:44.750 ","End":"00:49.295","Text":"Now if we assume that the current I flows through the smaller ring,"},{"Start":"00:49.295 ","End":"00:52.370","Text":"then we would have to calculate the magnetic field in"},{"Start":"00:52.370 ","End":"00:56.830","Text":"the bigger ring and then calculate the flux through the bigger ring,"},{"Start":"00:56.830 ","End":"01:02.420","Text":"then we could substitute it into the equation to calculate the mutual inductance."},{"Start":"01:02.420 ","End":"01:07.810","Text":"However, if we assume that the current is flowing through"},{"Start":"01:07.810 ","End":"01:13.180","Text":"the smaller ring because the smaller ring is at this angle of Theta,"},{"Start":"01:13.180 ","End":"01:19.565","Text":"the magnetic field that we would find would be in this diagonal direction."},{"Start":"01:19.565 ","End":"01:25.795","Text":"That is going to complicate our calculations by quite a lot."},{"Start":"01:25.795 ","End":"01:31.510","Text":"Then it will be very difficult to calculate the flux through the larger ring."},{"Start":"01:31.510 ","End":"01:35.845","Text":"We could maybe try and calculate using the idea of dipoles."},{"Start":"01:35.845 ","End":"01:38.215","Text":"However, this is very complicated."},{"Start":"01:38.215 ","End":"01:42.520","Text":"What we\u0027re going to do instead is we\u0027re going to assume that there\u0027s"},{"Start":"01:42.520 ","End":"01:47.640","Text":"this current I_b flowing through the larger ring instead."},{"Start":"01:47.640 ","End":"01:54.140","Text":"Then we can assume that there\u0027s a magnetic field in the z direction."},{"Start":"01:54.140 ","End":"01:59.315","Text":"This is B_z flowing through the center of"},{"Start":"01:59.315 ","End":"02:06.960","Text":"the small ring due to the current flowing through the large ring."},{"Start":"02:07.660 ","End":"02:11.300","Text":"Because there was a current flowing through the larger ring,"},{"Start":"02:11.300 ","End":"02:16.650","Text":"there\u0027s going to be this magnetic field through the center of the smaller ring."},{"Start":"02:16.650 ","End":"02:22.685","Text":"Then using this, we\u0027re going to be able to calculate the mutual inductance."},{"Start":"02:22.685 ","End":"02:28.610","Text":"Now, another important point is that we can assume that z,"},{"Start":"02:28.610 ","End":"02:30.935","Text":"so this height over here,"},{"Start":"02:30.935 ","End":"02:36.830","Text":"and b the radius of this outer ring is much smaller than a,"},{"Start":"02:36.830 ","End":"02:39.680","Text":"the radius of the smaller ring."},{"Start":"02:39.680 ","End":"02:43.520","Text":"That means that we can assume that the magnetic field throughout"},{"Start":"02:43.520 ","End":"02:48.130","Text":"the smaller ring is constant or is uniform because"},{"Start":"02:48.130 ","End":"02:50.765","Text":"the equation for the magnetic field is"},{"Start":"02:50.765 ","End":"02:57.540","Text":"a constant equation and each time we move a little bit to the left,"},{"Start":"02:57.540 ","End":"03:03.300","Text":"we can assume that the magnetic field is uniform inside the smaller ring."},{"Start":"03:03.800 ","End":"03:08.915","Text":"Let\u0027s calculate the magnetic field at the center of this ring."},{"Start":"03:08.915 ","End":"03:12.140","Text":"We\u0027re going to do it using the Biot-Savart\u0027s law."},{"Start":"03:12.140 ","End":"03:13.340","Text":"I\u0027m not going to calculate it."},{"Start":"03:13.340 ","End":"03:16.114","Text":"I\u0027m just going to write out the onset."},{"Start":"03:16.114 ","End":"03:19.850","Text":"If you don\u0027t remember how to do it or you want to do it yourself,"},{"Start":"03:19.850 ","End":"03:22.610","Text":"please pause the video over here and check your answer."},{"Start":"03:22.610 ","End":"03:24.020","Text":"If you don\u0027t know how to do it,"},{"Start":"03:24.020 ","End":"03:27.080","Text":"please go back to the chapter where we look at"},{"Start":"03:27.080 ","End":"03:31.375","Text":"calculating magnetic fields using Biot-Savart\u0027s law."},{"Start":"03:31.375 ","End":"03:36.200","Text":"The magnetic field B_z over here is going to be equal to"},{"Start":"03:36.200 ","End":"03:40.850","Text":"Mu naught multiplied by the current in the larger ring."},{"Start":"03:40.850 ","End":"03:46.340","Text":"That we said was I_b multiplied by the radius of the larger ring squared,"},{"Start":"03:46.340 ","End":"03:50.910","Text":"that\u0027s b^2 divided by 2."},{"Start":"03:54.020 ","End":"03:59.670","Text":"All of this is multiplied by the radius squared,"},{"Start":"03:59.670 ","End":"04:03.240","Text":"so b^2 plus this height squared,"},{"Start":"04:03.240 ","End":"04:10.050","Text":"so plus z^2 to the power of negative 3 divided by 2."},{"Start":"04:10.050 ","End":"04:14.070","Text":"This is what the magnetic field over here is equal to."},{"Start":"04:15.380 ","End":"04:18.285","Text":"We\u0027ve done steps 1 and 2."},{"Start":"04:18.285 ","End":"04:20.630","Text":"The next thing that we want to do is step number 3,"},{"Start":"04:20.630 ","End":"04:27.020","Text":"calculate the magnetic flux in the inductor over here."},{"Start":"04:27.020 ","End":"04:30.470","Text":"What we know is that the magnetic flux,"},{"Start":"04:30.470 ","End":"04:33.170","Text":"so through this inductor,"},{"Start":"04:33.170 ","End":"04:41.430","Text":"so through the ring of radius a is equal to the integral of the magnetic field B_z.ds,"},{"Start":"04:42.140 ","End":"04:49.020","Text":"where ds is the unit of area."},{"Start":"04:49.020 ","End":"04:54.300","Text":"As we\u0027ve already said due to z and b being much greater than a,"},{"Start":"04:54.300 ","End":"04:58.960","Text":"and due to the magnetic field being a continuous equation."},{"Start":"04:58.960 ","End":"05:06.180","Text":"We saw that our magnetic field is constant or is uniform throughout this ring."},{"Start":"05:07.700 ","End":"05:10.350","Text":"This also is a vector."},{"Start":"05:10.350 ","End":"05:13.470","Text":"The magnetic field is uniform throughout the ring,"},{"Start":"05:13.470 ","End":"05:21.620","Text":"but our ds now is just going to be the surface area of the ring,"},{"Start":"05:21.620 ","End":"05:26.815","Text":"our ds vector is perpendicular to the ring,"},{"Start":"05:26.815 ","End":"05:32.205","Text":"so our ds vector we can assume is like so."},{"Start":"05:32.205 ","End":"05:34.935","Text":"This is the direction of the vector."},{"Start":"05:34.935 ","End":"05:38.915","Text":"What we have to notice over here is our dot-product."},{"Start":"05:38.915 ","End":"05:43.740","Text":"What this is just going to be equal to is B_zds,"},{"Start":"05:44.750 ","End":"05:48.330","Text":"so B_z multiplied by ds,"},{"Start":"05:48.330 ","End":"05:55.005","Text":"and then the dot product will give us cosine of the angle between the 2."},{"Start":"05:55.005 ","End":"06:02.560","Text":"The angle between the 2 is this Phi."},{"Start":"06:05.630 ","End":"06:10.875","Text":"I\u0027ve written it a bit differently so that we don\u0027t get confused with the flux."},{"Start":"06:10.875 ","End":"06:14.970","Text":"It\u0027s the angle between these 2."},{"Start":"06:14.970 ","End":"06:21.200","Text":"This is the ds vector"},{"Start":"06:21.200 ","End":"06:27.825","Text":"perpendicular to the plane of the ring."},{"Start":"06:27.825 ","End":"06:31.390","Text":"The angle between the 2 is cosine of Phi."},{"Start":"06:31.390 ","End":"06:34.835","Text":"Now notice that if we were integrating,"},{"Start":"06:34.835 ","End":"06:38.000","Text":"the angle that we would be integrating over here would be"},{"Start":"06:38.000 ","End":"06:41.150","Text":"Theta whereas this angle over here,"},{"Start":"06:41.150 ","End":"06:43.585","Text":"Phi is a constant."},{"Start":"06:43.585 ","End":"06:48.590","Text":"What we can see is that our whole integral over here is just the constant."},{"Start":"06:48.590 ","End":"06:51.470","Text":"We\u0027re assuming that our B_z is constant,"},{"Start":"06:51.470 ","End":"06:52.879","Text":"that\u0027s our whole assumption."},{"Start":"06:52.879 ","End":"06:55.745","Text":"Ds is just the surface area of the ring,"},{"Start":"06:55.745 ","End":"07:00.995","Text":"and Phi is constant so the cosine of Phi is also going to be a constant."},{"Start":"07:00.995 ","End":"07:05.015","Text":"What we get is that the magnetic flux through"},{"Start":"07:05.015 ","End":"07:09.380","Text":"the smaller ring is equal to B_z, this over here,"},{"Start":"07:09.380 ","End":"07:13.900","Text":"multiplied by cosine of this angle Phi,"},{"Start":"07:13.900 ","End":"07:16.715","Text":"and then multiplied by the surface area of the ring,"},{"Start":"07:16.715 ","End":"07:21.065","Text":"which is Pi a^2."},{"Start":"07:21.065 ","End":"07:24.700","Text":"Now the final step we have to do is step number"},{"Start":"07:24.700 ","End":"07:27.970","Text":"4 to just calculate the mutual inductance."},{"Start":"07:27.970 ","End":"07:31.465","Text":"As we know, the mutual inductance"},{"Start":"07:31.465 ","End":"07:37.000","Text":"between 1 and 2 is equal to the mutual inductance of 2 on 1,"},{"Start":"07:37.000 ","End":"07:41.020","Text":"so we can just call this M. This is just equal"},{"Start":"07:41.020 ","End":"07:45.580","Text":"to the magnetic flux in the body that we calculated,"},{"Start":"07:45.580 ","End":"07:51.070","Text":"so a divided by the current in the other body."},{"Start":"07:51.070 ","End":"07:52.702","Text":"Over here it\u0027s b."},{"Start":"07:52.702 ","End":"07:57.475","Text":"Now what we can do is we can just plug everything in,"},{"Start":"07:57.475 ","End":"08:00.490","Text":"so the flux is going to be equal to this."},{"Start":"08:00.490 ","End":"08:07.450","Text":"We have B_z, so we have Mu_naught I_b b^2 divided by"},{"Start":"08:07.450 ","End":"08:16.015","Text":"2 multiplied by b^2 plus z^2 to the power of negative 3 over 2."},{"Start":"08:16.015 ","End":"08:26.390","Text":"That\u0027s our B_z multiplied by cosine of the angle multiplied by Pi a^2."},{"Start":"08:26.940 ","End":"08:31.120","Text":"This is Phi a, the flux."},{"Start":"08:31.120 ","End":"08:33.700","Text":"That\u0027s that, and then divide it by I_b."},{"Start":"08:33.700 ","End":"08:38.395","Text":"All of this is divided by I_b."},{"Start":"08:38.395 ","End":"08:43.015","Text":"As we can see, the current cancels out as we would have expected,"},{"Start":"08:43.015 ","End":"08:50.320","Text":"and this will just give us Mu_naught b^2 multiplied by cos of"},{"Start":"08:50.320 ","End":"08:55.105","Text":"Phi Pi a^2 b^2"},{"Start":"08:55.105 ","End":"09:00.475","Text":"multiplied by b^2 plus z^2,"},{"Start":"09:00.475 ","End":"09:06.025","Text":"and all of this is divided by 2."},{"Start":"09:06.025 ","End":"09:09.715","Text":"This is the answer to question number 1."},{"Start":"09:09.715 ","End":"09:11.440","Text":"This is the mutual inductance."},{"Start":"09:11.440 ","End":"09:16.645","Text":"Now let\u0027s go on to answer question number 2."},{"Start":"09:16.645 ","End":"09:23.973","Text":"Now let\u0027s answer question number 2 and question number 2 is asking us to calculate I_a,"},{"Start":"09:23.973 ","End":"09:27.955","Text":"so the current through the smaller ring, given I_b,"},{"Start":"09:27.955 ","End":"09:30.625","Text":"the current through the larger ring,"},{"Start":"09:30.625 ","End":"09:35.140","Text":"and R_a, the resistance of the smoldering."},{"Start":"09:35.140 ","End":"09:38.320","Text":"From the usual equation relating voltage,"},{"Start":"09:38.320 ","End":"09:39.940","Text":"current, and resistance,"},{"Start":"09:39.940 ","End":"09:44.215","Text":"we know that I_a is going to be equal to the voltage through a,"},{"Start":"09:44.215 ","End":"09:45.588","Text":"so the EMF,"},{"Start":"09:45.588 ","End":"09:48.520","Text":"divided by the resistance of a,"},{"Start":"09:48.520 ","End":"09:50.710","Text":"which we\u0027re given in the question."},{"Start":"09:50.710 ","End":"09:54.039","Text":"What we want to do is we want to calculate Epsilon_a,"},{"Start":"09:54.039 ","End":"09:56.170","Text":"so the EMF or the voltage."},{"Start":"09:56.170 ","End":"10:05.360","Text":"We know that that\u0027s equal to the negative time derivative of the magnetic flux through a."},{"Start":"10:06.570 ","End":"10:09.940","Text":"The magnetic flux through a,"},{"Start":"10:09.940 ","End":"10:11.980","Text":"let\u0027s just write it over here,"},{"Start":"10:11.980 ","End":"10:13.630","Text":"is equal to,"},{"Start":"10:13.630 ","End":"10:16.480","Text":"so from this equation over here,"},{"Start":"10:16.480 ","End":"10:18.805","Text":"the magnetic flux, we can just rearrange it."},{"Start":"10:18.805 ","End":"10:21.085","Text":"If we isolate out Phi of a,"},{"Start":"10:21.085 ","End":"10:23.695","Text":"we get that it\u0027s equal to M,"},{"Start":"10:23.695 ","End":"10:27.550","Text":"the mutual inductance, multiplied by I_b."},{"Start":"10:27.550 ","End":"10:32.320","Text":"We can say that this is equal to the negative time derivative of this."},{"Start":"10:32.320 ","End":"10:34.345","Text":"M is a constant,"},{"Start":"10:34.345 ","End":"10:41.245","Text":"and I_b is of course something that we can take the derivative of."},{"Start":"10:41.245 ","End":"10:45.400","Text":"Then we can say that this is equal to negative M and then"},{"Start":"10:45.400 ","End":"10:50.320","Text":"the time derivative of I_b is simply from this equation going to be"},{"Start":"10:50.320 ","End":"10:54.700","Text":"equal to I_naught multiplied by"},{"Start":"10:54.700 ","End":"10:59.500","Text":"negative Omega sine of"},{"Start":"10:59.500 ","End":"11:05.605","Text":"Omega t. The negatives and the negatives equal a positive."},{"Start":"11:05.605 ","End":"11:10.645","Text":"We get that Epsilon_a is equal to"},{"Start":"11:10.645 ","End":"11:16.120","Text":"MI_naught Omega multiplied by"},{"Start":"11:16.120 ","End":"11:21.580","Text":"sine of Omega t. Now we have Epsilon_a,"},{"Start":"11:21.580 ","End":"11:24.505","Text":"so all we have to do is plug it into this equation."},{"Start":"11:24.505 ","End":"11:28.420","Text":"We get that I_a is equal to Epsilon_a,"},{"Start":"11:28.420 ","End":"11:35.050","Text":"so that\u0027s MI_naught Omega sine of Omega t,"},{"Start":"11:35.050 ","End":"11:39.745","Text":"and all of this is divided by R_a."},{"Start":"11:39.745 ","End":"11:42.295","Text":"This is the answer to question number 2,"},{"Start":"11:42.295 ","End":"11:46.390","Text":"to calculate the current through the smaller loop and as we can see,"},{"Start":"11:46.390 ","End":"11:48.595","Text":"it\u0027s also in the positive direction."},{"Start":"11:48.595 ","End":"11:51.220","Text":"If this is the direction of the z-axis,"},{"Start":"11:51.220 ","End":"11:55.765","Text":"we can see that our I_a is in the positive Theta direction,"},{"Start":"11:55.765 ","End":"11:59.710","Text":"so it\u0027s going in the same direction as our I_b."},{"Start":"11:59.710 ","End":"12:03.340","Text":"Before we continue on to the next question,"},{"Start":"12:03.340 ","End":"12:05.695","Text":"I just want to go over something."},{"Start":"12:05.695 ","End":"12:13.270","Text":"What would happen if my current over here, I_b was constant?"},{"Start":"12:13.270 ","End":"12:17.050","Text":"Imagine that cosine Omega t was crossed out,"},{"Start":"12:17.050 ","End":"12:19.000","Text":"so I_b was just equal to I_naught;"},{"Start":"12:19.000 ","End":"12:20.634","Text":"it was a constant."},{"Start":"12:20.634 ","End":"12:25.885","Text":"However, my angle over here Phi was dependent on time."},{"Start":"12:25.885 ","End":"12:34.210","Text":"We can imagine that the ring would be rotating around with this changing angle of Phi."},{"Start":"12:34.210 ","End":"12:37.540","Text":"That would mean that over here, my M,"},{"Start":"12:37.540 ","End":"12:43.540","Text":"my mutual inductance would be time-dependent because Phi over here would be changing."},{"Start":"12:43.540 ","End":"12:46.248","Text":"In that case, what I would do,"},{"Start":"12:46.248 ","End":"12:48.417","Text":"so I\u0027m just going to draw this in red,"},{"Start":"12:48.417 ","End":"12:53.500","Text":"what I would do is instead of taking the time derivative of I_b,"},{"Start":"12:53.500 ","End":"12:58.930","Text":"I would take the time derivative of M. That would be"},{"Start":"12:58.930 ","End":"13:02.080","Text":"the only difference and of course that would be very easy"},{"Start":"13:02.080 ","End":"13:06.235","Text":"because all I would have to do is take the derivative of cosine of Phi."},{"Start":"13:06.235 ","End":"13:12.490","Text":"That\u0027s if my current was constant but my angle Phi was changing and if"},{"Start":"13:12.490 ","End":"13:19.240","Text":"both my current was dependent on time and my angle Phi was dependent on time."},{"Start":"13:19.240 ","End":"13:23.650","Text":"At this stage, I would do the time derivative of both."},{"Start":"13:23.650 ","End":"13:29.395","Text":"We take the derivative of both of them with respect to time and we would just use"},{"Start":"13:29.395 ","End":"13:35.890","Text":"the chain rule in order to take the derivative of these 2 functions."},{"Start":"13:35.890 ","End":"13:41.785","Text":"Another option that we could have is that either the current is still"},{"Start":"13:41.785 ","End":"13:47.290","Text":"dependent on time or the angle Phi is still dependent on time."},{"Start":"13:47.290 ","End":"13:52.390","Text":"It doesn\u0027t make a difference, but another option that we could add to everything is that"},{"Start":"13:52.390 ","End":"14:00.505","Text":"the ring is moving in some sort of z-direction so that the ring is moving upwards,"},{"Start":"14:00.505 ","End":"14:05.620","Text":"in which case our z would be dependent on time and it would be equal to"},{"Start":"14:05.620 ","End":"14:11.740","Text":"V_naught t and then might be plus some z_naught."},{"Start":"14:11.740 ","End":"14:14.950","Text":"In that case, in this equation over here,"},{"Start":"14:14.950 ","End":"14:17.110","Text":"our z is also dependent on t,"},{"Start":"14:17.110 ","End":"14:21.850","Text":"so when we would need to take the time derivative of M,"},{"Start":"14:21.850 ","End":"14:25.240","Text":"we would either just take the derivative with respect to z or"},{"Start":"14:25.240 ","End":"14:28.795","Text":"with respect to z and Phi using the chain rule"},{"Start":"14:28.795 ","End":"14:31.720","Text":"depending on what\u0027s going on in the question"},{"Start":"14:31.720 ","End":"14:37.190","Text":"and also the time derivative of current if need be."},{"Start":"14:38.130 ","End":"14:41.720","Text":"All of these are just possible options that is"},{"Start":"14:41.720 ","End":"14:46.550","Text":"useful to know how to solve them in an exam."},{"Start":"14:46.550 ","End":"14:51.360","Text":"Now we\u0027re going to move on to question number 3."},{"Start":"14:52.110 ","End":"14:54.375","Text":"Question number 3,"},{"Start":"14:54.375 ","End":"14:56.540","Text":"let\u0027s solve it over here,"},{"Start":"14:56.540 ","End":"15:02.015","Text":"is to work out tau_b or the torque of b."},{"Start":"15:02.015 ","End":"15:06.590","Text":"We\u0027re looking at this ring over here."},{"Start":"15:06.990 ","End":"15:09.700","Text":"This is what I\u0027m trying to find."},{"Start":"15:09.700 ","End":"15:13.945","Text":"Now, I can remember that my magnetic torque,"},{"Start":"15:13.945 ","End":"15:16.000","Text":"or the torque, the moment of force,"},{"Start":"15:16.000 ","End":"15:21.280","Text":"on this is equal to Mu,"},{"Start":"15:21.280 ","End":"15:24.910","Text":"which is a vector cross B,"},{"Start":"15:24.910 ","End":"15:31.555","Text":"where Mu is the magnetic moment and B is the magnetic field."},{"Start":"15:31.555 ","End":"15:35.980","Text":"Working with this is going to be very difficult because I not"},{"Start":"15:35.980 ","End":"15:40.240","Text":"only will have to find the magnetic moment on this ring b,"},{"Start":"15:40.240 ","End":"15:45.400","Text":"but also I have to calculate the magnetic field acting here, which I don\u0027t know."},{"Start":"15:45.400 ","End":"15:51.834","Text":"The first thing that I\u0027m going to do is I\u0027m going to work out tau_a,"},{"Start":"15:51.834 ","End":"15:58.730","Text":"the torque on ring a over here because that is going to be equal to Mu_a,"},{"Start":"15:58.730 ","End":"16:00.823","Text":"which I don\u0027t know and I have to calculate it,"},{"Start":"16:00.823 ","End":"16:04.565","Text":"cross product with the magnetic field on a,"},{"Start":"16:04.565 ","End":"16:08.640","Text":"which I know I calculated it over here."},{"Start":"16:08.790 ","End":"16:17.430","Text":"Then I\u0027m going to say that tau_a is equal to negative tau_b."},{"Start":"16:17.430 ","End":"16:20.180","Text":"This comes from Newton\u0027s third law,"},{"Start":"16:20.180 ","End":"16:26.550","Text":"which of course works also on torques just like it does on forces."},{"Start":"16:26.550 ","End":"16:30.170","Text":"In the question, I\u0027m not asked for the directions,"},{"Start":"16:30.170 ","End":"16:33.440","Text":"I\u0027m just asked for the magnitude."},{"Start":"16:33.440 ","End":"16:37.520","Text":"I\u0027m going to calculate the magnitude of all of this."},{"Start":"16:37.520 ","End":"16:44.780","Text":"This is just going to be equal to Mu_a multiplied by the magnetic field,"},{"Start":"16:44.780 ","End":"16:47.360","Text":"which over here we denoted as B_z."},{"Start":"16:47.360 ","End":"16:52.760","Text":"It\u0027s the same thing, it\u0027s the magnetic field through ring a and then multiplied by"},{"Start":"16:52.760 ","End":"16:58.630","Text":"sine of the angle between the 2 vectors."},{"Start":"16:58.630 ","End":"17:00.605","Text":"That\u0027s what we have over here,"},{"Start":"17:00.605 ","End":"17:02.270","Text":"and then Mu_a,"},{"Start":"17:02.270 ","End":"17:05.060","Text":"so let\u0027s see what this is."},{"Start":"17:05.060 ","End":"17:09.380","Text":"Mu_a is simply equal to,"},{"Start":"17:09.380 ","End":"17:11.240","Text":"we saw the equation before,"},{"Start":"17:11.240 ","End":"17:15.950","Text":"it\u0027s equal to the current going through the rings,"},{"Start":"17:15.950 ","End":"17:18.320","Text":"so I_a, which we know what this is,"},{"Start":"17:18.320 ","End":"17:20.870","Text":"multiplied by the surface area of the ring."},{"Start":"17:20.870 ","End":"17:25.790","Text":"It\u0027s a circle, so it\u0027s Pi_a squared because the radius is a."},{"Start":"17:25.790 ","End":"17:30.840","Text":"We just substitute in Mu_a over here,"},{"Start":"17:30.840 ","End":"17:35.800","Text":"B_z over here, and then sine of the angle, and that\u0027s it."},{"Start":"17:35.800 ","End":"17:38.930","Text":"That is our answer to question number 3."},{"Start":"17:38.930 ","End":"17:42.450","Text":"That is the end of this lesson."}],"ID":22375},{"Watched":false,"Name":"Exercise 3","Duration":"7m 15s","ChapterTopicVideoID":21546,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.095","Text":"Hello. In this lesson,"},{"Start":"00:02.095 ","End":"00:06.420","Text":"we\u0027re going to be answering the following question about a transformer."},{"Start":"00:06.420 ","End":"00:10.140","Text":"A transformer is constructed of 2 coils,"},{"Start":"00:10.140 ","End":"00:12.629","Text":"Coil 1 and Coil 2,"},{"Start":"00:12.629 ","End":"00:15.750","Text":"each of a different number of turns."},{"Start":"00:15.750 ","End":"00:17.760","Text":"Different amount of wraps,"},{"Start":"00:17.760 ","End":"00:23.955","Text":"so here we have N_1 wraps and here we have N_2 wraps."},{"Start":"00:23.955 ","End":"00:29.085","Text":"The coil is wrap around opposite sides of a rectangular core."},{"Start":"00:29.085 ","End":"00:31.115","Text":"We\u0027re meant to assume that"},{"Start":"00:31.115 ","End":"00:35.360","Text":"all the magnetic field lines are through the core or in other words,"},{"Start":"00:35.360 ","End":"00:39.440","Text":"the magnetic flux is uniform through the core."},{"Start":"00:39.440 ","End":"00:44.258","Text":"All the magnetic field lines are just going like so,"},{"Start":"00:44.258 ","End":"00:48.020","Text":"they remain within the core,"},{"Start":"00:48.020 ","End":"00:52.560","Text":"and they just go around like so."},{"Start":"00:54.140 ","End":"01:01.095","Text":"We just have all of these lines like so."},{"Start":"01:01.095 ","End":"01:06.620","Text":"If I would cut the core over here or over here,"},{"Start":"01:07.880 ","End":"01:11.800","Text":"or over here or over here, I would get the exact same value for"},{"Start":"01:11.800 ","End":"01:15.295","Text":"the magnetic flux because it\u0027s a uniform throughout."},{"Start":"01:15.295 ","End":"01:18.190","Text":"That\u0027s what this is saying."},{"Start":"01:18.190 ","End":"01:23.125","Text":"We\u0027re being told that the voltage on the left coil is"},{"Start":"01:23.125 ","End":"01:28.270","Text":"an alternating voltage that\u0027s equal to V naught multiplied by sine"},{"Start":"01:28.270 ","End":"01:32.920","Text":"of Omega t. We\u0027re being asked to calculate the voltage on the coil on"},{"Start":"01:32.920 ","End":"01:37.748","Text":"the right as a function of the voltage on the left,"},{"Start":"01:37.748 ","End":"01:40.470","Text":"and we\u0027re being told that N_1 and N_2,"},{"Start":"01:40.470 ","End":"01:44.830","Text":"the number of turns on each coil are given."},{"Start":"01:45.710 ","End":"01:49.040","Text":"In other words, we have this core over here."},{"Start":"01:49.040 ","End":"01:52.550","Text":"It\u0027s a rectangle and around the left side,"},{"Start":"01:52.550 ","End":"01:59.735","Text":"we have this coil with N_1 number of turns that\u0027s wrapped around this left side."},{"Start":"01:59.735 ","End":"02:00.890","Text":"Wrapped around the right side,"},{"Start":"02:00.890 ","End":"02:04.565","Text":"we have this coil of N_2 number of turns."},{"Start":"02:04.565 ","End":"02:06.480","Text":"We\u0027re being told that on the left side,"},{"Start":"02:06.480 ","End":"02:07.790","Text":"we have this voltage,"},{"Start":"02:07.790 ","End":"02:11.983","Text":"V naught sine Omega t. It\u0027s an alternating voltage,"},{"Start":"02:11.983 ","End":"02:14.480","Text":"and we\u0027re being asked to calculate the voltage on"},{"Start":"02:14.480 ","End":"02:19.800","Text":"the right coil as a function of this voltage."},{"Start":"02:20.740 ","End":"02:26.930","Text":"Let\u0027s look at the flux through the coil number 1."},{"Start":"02:26.930 ","End":"02:28.700","Text":"As we\u0027ve seen in the previous lessons,"},{"Start":"02:28.700 ","End":"02:31.985","Text":"that\u0027s equal to the number of turns in the coil"},{"Start":"02:31.985 ","End":"02:37.230","Text":"multiplied by the flux in each ring, so the flux through each ring."},{"Start":"02:38.590 ","End":"02:44.100","Text":"If we want to calculate the flux through Coil 2,"},{"Start":"02:44.100 ","End":"02:45.470","Text":"it\u0027s going to be the same thing."},{"Start":"02:45.470 ","End":"02:51.560","Text":"It\u0027s the number of turns in Coil 2 multiplied by the flux through each ring,"},{"Start":"02:51.560 ","End":"02:54.580","Text":"which as we know, is uniform."},{"Start":"02:54.580 ","End":"02:57.300","Text":"We were told that in the question."},{"Start":"02:57.300 ","End":"03:00.530","Text":"In other words, what I can do is I can isolate out"},{"Start":"03:00.530 ","End":"03:04.850","Text":"the flux and each ring and I will get therefore,"},{"Start":"03:04.850 ","End":"03:09.640","Text":"if I isolate the flux in each ring and of course Phi ring is equal to Phi ring,"},{"Start":"03:09.640 ","End":"03:13.650","Text":"then what I\u0027ll get is that Phi_1 divided by"},{"Start":"03:13.650 ","End":"03:20.235","Text":"Phi_2=N_1 divided by N_2."},{"Start":"03:20.235 ","End":"03:23.045","Text":"You can just do the algebra. It\u0027s very easy."},{"Start":"03:23.045 ","End":"03:28.260","Text":"Isolate out the Phi ring and equate the 2 equations to one another."},{"Start":"03:28.940 ","End":"03:34.025","Text":"Let\u0027s isolate the various Phi\u0027s, the various fluxes."},{"Start":"03:34.025 ","End":"03:42.585","Text":"The flux through Coil 1 is going to be equal to N_1 divided by N_2 multiplied by Phi_2."},{"Start":"03:42.585 ","End":"03:51.250","Text":"The flux through the coil number 2 is going to be equal to N_2 divided by N_1 Phi_1."},{"Start":"03:51.740 ","End":"03:54.890","Text":"What I want to do is I want to calculate"},{"Start":"03:54.890 ","End":"03:58.505","Text":"the voltage on the coil in the right on this coil."},{"Start":"03:58.505 ","End":"04:03.770","Text":"In order to calculate the voltage, so EMF_2,"},{"Start":"04:03.770 ","End":"04:09.149","Text":"I have to take the time derivative of the flux,"},{"Start":"04:09.149 ","End":"04:13.140","Text":"which is going to be equal to the time derivative of this."},{"Start":"04:13.140 ","End":"04:17.018","Text":"N_2 divided by N_1 is obviously a constant,"},{"Start":"04:17.018 ","End":"04:20.210","Text":"and then I have the time derivative of this."},{"Start":"04:20.210 ","End":"04:24.890","Text":"This is going to be equal to N_2 divided by N_1 and"},{"Start":"04:24.890 ","End":"04:30.670","Text":"the time derivative of the flux in this coil is equal to Epsilon_1."},{"Start":"04:30.670 ","End":"04:33.090","Text":"Epsilon_1 is this over here."},{"Start":"04:33.090 ","End":"04:36.905","Text":"Epsilon_1 is the voltage through the left coil,"},{"Start":"04:36.905 ","End":"04:38.955","Text":"which we were given is equal to this."},{"Start":"04:38.955 ","End":"04:45.500","Text":"What we get is that it\u0027s equal to N_2 divided by N_1 multiplied by"},{"Start":"04:45.500 ","End":"04:54.845","Text":"V naught sine of Omega t. We can see that in the right coil,"},{"Start":"04:54.845 ","End":"04:58.460","Text":"we\u0027re also going to have an alternating voltage."},{"Start":"04:58.460 ","End":"04:59.840","Text":"The only difference is that"},{"Start":"04:59.840 ","End":"05:05.036","Text":"its maximal value is going to be different to the maximal value on N_1"},{"Start":"05:05.036 ","End":"05:08.330","Text":"because here the maximum value on"},{"Start":"05:08.330 ","End":"05:12.810","Text":"the right one is going to be N_2 divided by N_1 V naught,"},{"Start":"05:12.810 ","End":"05:15.270","Text":"whereas the maximum value on the left coil,"},{"Start":"05:15.270 ","End":"05:20.920","Text":"on the N_1 coil is just going to be V naught."},{"Start":"05:21.650 ","End":"05:26.390","Text":"This is how to calculate a question dealing with a transformer."},{"Start":"05:26.390 ","End":"05:29.960","Text":"But what exactly is a transformer good for?"},{"Start":"05:29.960 ","End":"05:40.730","Text":"Here we have our electricity factory that produces electricity to the city."},{"Start":"05:40.730 ","End":"05:48.275","Text":"Here I am with my phone and I want to plug it in to charge it."},{"Start":"05:48.275 ","End":"05:50.705","Text":"Now, depending on what country you\u0027re in,"},{"Start":"05:50.705 ","End":"05:57.610","Text":"the factory is going to supply either 220 volts or 110 volts."},{"Start":"05:57.610 ","End":"06:03.795","Text":"These are the most general voltage supplies."},{"Start":"06:03.795 ","End":"06:08.870","Text":"Let\u0027s imagine that my phone only needs 5 volts."},{"Start":"06:08.870 ","End":"06:15.380","Text":"What we do is we connect in transformers and then the transformers will"},{"Start":"06:15.380 ","End":"06:22.955","Text":"transform this 220 or 110 volts into the 5 volts which are suitable for my phone."},{"Start":"06:22.955 ","End":"06:30.475","Text":"What I need is I need to find out the ratio of the turns in each coil."},{"Start":"06:30.475 ","End":"06:34.470","Text":"This is the ratio and that is going to"},{"Start":"06:34.470 ","End":"06:39.310","Text":"be of course equal to Epsilon_2 divided by Epsilon_1."},{"Start":"06:41.960 ","End":"06:48.420","Text":"Here, what my phone needs is 5 volts as opposed to,"},{"Start":"06:48.420 ","End":"06:51.720","Text":"let\u0027s say 220 volts."},{"Start":"06:51.720 ","End":"06:56.960","Text":"This is what is going to give us this ratio over here of the number of"},{"Start":"06:56.960 ","End":"07:02.275","Text":"turns that I needed in order to charge a phone safely."},{"Start":"07:02.275 ","End":"07:06.305","Text":"Each electrical device has its transformer,"},{"Start":"07:06.305 ","End":"07:13.100","Text":"which allows the device to be charged safely without having excess voltage."},{"Start":"07:13.100 ","End":"07:16.200","Text":"That is the end of this lesson."}],"ID":22376},{"Watched":false,"Name":"Exercise 4","Duration":"9m 25s","ChapterTopicVideoID":21353,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"Hello. In this lesson,"},{"Start":"00:01.620 ","End":"00:05.040","Text":"we\u0027re going to be answering the following question."},{"Start":"00:05.040 ","End":"00:13.200","Text":"A coaxial cable consists of an inner wire of radius a and an outer envelope of radius b."},{"Start":"00:13.200 ","End":"00:15.015","Text":"The length of the wire is l,"},{"Start":"00:15.015 ","End":"00:21.255","Text":"such that l is much greater than the value for either a or b."},{"Start":"00:21.255 ","End":"00:26.310","Text":"A current I flows through the inner wire so we can write here"},{"Start":"00:26.310 ","End":"00:31.980","Text":"I and an equal and opposite current flows through the outer envelope."},{"Start":"00:31.980 ","End":"00:36.000","Text":"A current of the same magnitude but in the opposite direction,"},{"Start":"00:36.000 ","End":"00:40.170","Text":"so we can call it negative I flows through the outer envelope."},{"Start":"00:40.170 ","End":"00:43.340","Text":"Calculate the self-inductance per unit length."},{"Start":"00:43.340 ","End":"00:47.255","Text":"Ignore the magnetic field through the inner wire."},{"Start":"00:47.255 ","End":"00:53.440","Text":"Basically, we\u0027re trying to calculate the inductance of this coaxial cable."},{"Start":"00:53.480 ","End":"00:55.970","Text":"In this example over here,"},{"Start":"00:55.970 ","End":"00:58.520","Text":"it\u0027s a little bit difficult for us to calculate"},{"Start":"00:58.520 ","End":"01:01.655","Text":"the magnetic flux because we\u0027re not exactly sure"},{"Start":"01:01.655 ","End":"01:10.130","Text":"which region is contained within this as like a closed circuit."},{"Start":"01:10.130 ","End":"01:13.925","Text":"Calculating the magnetic flux here is difficult because we don\u0027t know"},{"Start":"01:13.925 ","End":"01:19.900","Text":"exactly the borders that were meant to be integrating upon for the magnetic flux."},{"Start":"01:19.900 ","End":"01:22.850","Text":"What we\u0027re going to do is we\u0027re going to use"},{"Start":"01:22.850 ","End":"01:26.915","Text":"this equation to calculate the energy stored in the magnetic field."},{"Start":"01:26.915 ","End":"01:34.215","Text":"That\u0027s the integral b^2 divided by 2Mu naught dv."},{"Start":"01:34.215 ","End":"01:37.100","Text":"Then we\u0027re going to equate it to the energy"},{"Start":"01:37.100 ","End":"01:41.880","Text":"stored in an inductor which is equal to 1/2LI^2."},{"Start":"01:43.270 ","End":"01:49.620","Text":"We\u0027re expecting the 1/2 and the I^2 to just cancel out."},{"Start":"01:50.090 ","End":"01:53.180","Text":"In order to calculate the magnetic field,"},{"Start":"01:53.180 ","End":"01:55.775","Text":"we\u0027re going to use Ampere\u0027s Law."},{"Start":"01:55.775 ","End":"01:59.490","Text":"We\u0027re going to make this loop over here,"},{"Start":"01:59.490 ","End":"02:02.960","Text":"Ampere\u0027s loop where it\u0027s of a radius,"},{"Start":"02:02.960 ","End":"02:07.940","Text":"lowercase r. What we can say is that we can just"},{"Start":"02:07.940 ","End":"02:13.320","Text":"assume that the current over here is going into the page."},{"Start":"02:13.320 ","End":"02:16.895","Text":"That means that the current over here is coming out of the page."},{"Start":"02:16.895 ","End":"02:21.110","Text":"If it is going into the page over here,"},{"Start":"02:21.110 ","End":"02:24.455","Text":"then using the right-hand rule,"},{"Start":"02:24.455 ","End":"02:29.430","Text":"our Ampere\u0027s loop is going to be in this direction."},{"Start":"02:29.740 ","End":"02:35.825","Text":"Or in other words, this is the direction that the magnetic field will be flowing through."},{"Start":"02:35.825 ","End":"02:38.800","Text":"Now, what we\u0027re going to do is Ampere\u0027s law."},{"Start":"02:38.800 ","End":"02:45.380","Text":"We have the closed loop integral of B.dl which is equal"},{"Start":"02:45.380 ","End":"02:52.000","Text":"to Mu naught multiplied by I_in."},{"Start":"02:52.000 ","End":"02:57.980","Text":"Our magnetic field is constant because our I is constant and our dl is,"},{"Start":"02:57.980 ","End":"02:59.615","Text":"of course, a fixed length."},{"Start":"02:59.615 ","End":"03:06.155","Text":"What we have is b multiplied by the circumference of this circle."},{"Start":"03:06.155 ","End":"03:07.655","Text":"This is meant to be a circle,"},{"Start":"03:07.655 ","End":"03:10.620","Text":"so multiplied by 2Pir."},{"Start":"03:11.120 ","End":"03:15.940","Text":"This is equal to Mu naught multiplied by I_in."},{"Start":"03:15.940 ","End":"03:18.950","Text":"The only current that we have inside here is"},{"Start":"03:18.950 ","End":"03:22.060","Text":"this current that we\u0027re given in the question I."},{"Start":"03:22.060 ","End":"03:26.940","Text":"Then what we can do is we can isolate out our b and we get that"},{"Start":"03:26.940 ","End":"03:31.725","Text":"it\u0027s equal to Mu naught I divided by 2Pir."},{"Start":"03:31.725 ","End":"03:33.890","Text":"Its direction is, of course,"},{"Start":"03:33.890 ","End":"03:38.160","Text":"in the positive Theta direction."},{"Start":"03:38.900 ","End":"03:46.440","Text":"This is in the region where r is greater than a but smaller than b,"},{"Start":"03:46.440 ","End":"03:50.655","Text":"what about in the region where r is greater than b?"},{"Start":"03:50.655 ","End":"03:55.260","Text":"In this region, we have equal and opposite current."},{"Start":"03:55.260 ","End":"03:58.235","Text":"We have a current going into the page and coming"},{"Start":"03:58.235 ","End":"04:01.550","Text":"out of the page which means that the total I_in is equal to 0."},{"Start":"04:01.550 ","End":"04:09.840","Text":"This is equal to 0 so let\u0027s do r is greater than b so I_in,"},{"Start":"04:09.840 ","End":"04:14.225","Text":"in this case, is equal to 0 because we have plus I minus I and"},{"Start":"04:14.225 ","End":"04:19.350","Text":"therefore our magnetic field will also be equal to 0."},{"Start":"04:19.350 ","End":"04:22.400","Text":"Of course, we were told to ignore the magnetic field"},{"Start":"04:22.400 ","End":"04:26.905","Text":"through the inner wire so we\u0027re not even looking at that region."},{"Start":"04:26.905 ","End":"04:32.300","Text":"What we know is just inside the coaxial cable itself,"},{"Start":"04:32.300 ","End":"04:34.535","Text":"so between the radius of a and b,"},{"Start":"04:34.535 ","End":"04:38.040","Text":"we have a magnetic field and that is it."},{"Start":"04:39.370 ","End":"04:44.315","Text":"What we\u0027ve gotten from this step is the magnetic field"},{"Start":"04:44.315 ","End":"04:49.760","Text":"itself and also the region that we\u0027re going to be integrating."},{"Start":"04:49.760 ","End":"04:53.180","Text":"Of course, we have to take into account also the length."},{"Start":"04:53.180 ","End":"04:55.430","Text":"Let\u0027s do this equation."},{"Start":"04:55.430 ","End":"04:59.620","Text":"We have that the energy is equal to the integral dvs."},{"Start":"04:59.620 ","End":"05:05.315","Text":"We can already see it\u0027s going to be a triple integral of b^2, so b^2,"},{"Start":"05:05.315 ","End":"05:14.040","Text":"we have Mu naught squared I^2 divided by 2^2."},{"Start":"05:14.040 ","End":"05:22.190","Text":"We\u0027ll just leave it at that, Pi^2r^2 divided by 2Mu naught,"},{"Start":"05:22.190 ","End":"05:24.700","Text":"so divided by another 2."},{"Start":"05:24.700 ","End":"05:30.135","Text":"Then here, Mu naught and dv."},{"Start":"05:30.135 ","End":"05:33.540","Text":"We\u0027re working in cylindrical coordinates."},{"Start":"05:33.540 ","End":"05:37.020","Text":"We have r dr,"},{"Start":"05:37.020 ","End":"05:41.110","Text":"d Theta, and dz."},{"Start":"05:41.420 ","End":"05:46.390","Text":"We can see that the magnetic field is dependent on the radius."},{"Start":"05:46.390 ","End":"05:47.800","Text":"As our radius increases,"},{"Start":"05:47.800 ","End":"05:51.820","Text":"the magnetic field decreases so that\u0027s why"},{"Start":"05:51.820 ","End":"05:57.480","Text":"we have to integrate along this direction as well."},{"Start":"05:57.480 ","End":"06:01.780","Text":"First of all, this Mu squared can cancel with this Mu in"},{"Start":"06:01.780 ","End":"06:08.680","Text":"the denominator in the meantime and this r over here can cancel with this r over here."},{"Start":"06:08.680 ","End":"06:12.890","Text":"Now we can also plug in the bounds."},{"Start":"06:12.890 ","End":"06:16.455","Text":"For dz, that\u0027s just the length of the wire."},{"Start":"06:16.455 ","End":"06:21.175","Text":"We\u0027re going from 0 until the length L. For d Theta,"},{"Start":"06:21.175 ","End":"06:26.510","Text":"we\u0027re doing a full circle so from 0 until 2Pi."},{"Start":"06:26.510 ","End":"06:32.725","Text":"For r, we\u0027re going from the inner radius which is a,"},{"Start":"06:32.725 ","End":"06:38.000","Text":"until the outer radius which is b."},{"Start":"06:38.060 ","End":"06:45.170","Text":"We can see that also now we can take out some of the constants and also we"},{"Start":"06:45.170 ","End":"06:51.635","Text":"don\u0027t have the variable Theta or z over here so we can just multiply out by the balance."},{"Start":"06:51.635 ","End":"06:54.575","Text":"What we have is Mu naught I^2"},{"Start":"06:54.575 ","End":"07:03.130","Text":"divided by 2^3 Pi^2."},{"Start":"07:03.170 ","End":"07:11.560","Text":"Then we\u0027re also multiplying by L and by 2Pi."},{"Start":"07:14.240 ","End":"07:21.690","Text":"First of all, here we have 2^3 so let\u0027s cancel it out with this 2,"},{"Start":"07:21.690 ","End":"07:24.030","Text":"and then what we\u0027re left with is 2^2."},{"Start":"07:24.030 ","End":"07:28.935","Text":"I\u0027m just going to rub out this 2 so that it doesn\u0027t get confusing."},{"Start":"07:28.935 ","End":"07:34.520","Text":"Then we have this Pi over here which could cancel out with this."},{"Start":"07:34.520 ","End":"07:39.470","Text":"Then we have the integral of what is left over here which"},{"Start":"07:39.470 ","End":"07:45.405","Text":"is 1 divided by r dr between a and b."},{"Start":"07:45.405 ","End":"07:49.895","Text":"That as we know, is just going to be equal to ln of b divided by a."},{"Start":"07:49.895 ","End":"07:55.530","Text":"What we\u0027re left with is Mu naught I^2 L divided by 4Pi of ln b divided by a."},{"Start":"08:07.550 ","End":"08:11.040","Text":"Now let\u0027s look at this."},{"Start":"08:11.040 ","End":"08:13.825","Text":"We know that this equation over here,"},{"Start":"08:13.825 ","End":"08:18.295","Text":"what we\u0027ve just calculated is equal to this equation over here."},{"Start":"08:18.295 ","End":"08:26.817","Text":"We have that Mu naught I^2 L divided by 4Pi ln"},{"Start":"08:26.817 ","End":"08:35.305","Text":"of b over a is equal to 1/2 of LI^2."},{"Start":"08:35.305 ","End":"08:38.590","Text":"First of all, the I^2 can cancel out and this"},{"Start":"08:38.590 ","End":"08:42.160","Text":"1/2 can cancel out with this quarter so we\u0027re left with this."},{"Start":"08:42.160 ","End":"08:48.517","Text":"Therefore what we get is that the self-inductance is equal to Mu naught l,"},{"Start":"08:48.517 ","End":"08:50.360","Text":"lower case l, of course,"},{"Start":"08:50.360 ","End":"08:58.810","Text":"the length divided by 2Pi multiplied by ln of b divided by a."},{"Start":"08:58.810 ","End":"09:01.650","Text":"This is the self-inductance."},{"Start":"09:01.650 ","End":"09:05.135","Text":"If we want the self-inductance per unit length,"},{"Start":"09:05.135 ","End":"09:10.360","Text":"we just of course divide both sides by this lowercase l and we\u0027re left with"},{"Start":"09:10.360 ","End":"09:17.240","Text":"Mu naught divided by 2Pi of ln b divided by a."},{"Start":"09:18.560 ","End":"09:25.860","Text":"This is the answer to this question and that is the end of this lesson."}],"ID":21433},{"Watched":false,"Name":"Exercise 5","Duration":"9m 25s","ChapterTopicVideoID":21547,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"Hello, in this lesson,"},{"Start":"00:01.650 ","End":"00:04.095","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.095 ","End":"00:09.165","Text":"A toroid of inner radius a and outer radius"},{"Start":"00:09.165 ","End":"00:15.345","Text":"b and width h has N turns in its coil."},{"Start":"00:15.345 ","End":"00:19.815","Text":"Question number 1 is to calculate the inductance."},{"Start":"00:19.815 ","End":"00:23.955","Text":"In order to calculate inductance,"},{"Start":"00:23.955 ","End":"00:28.245","Text":"we have to use the equation which is that inductance is"},{"Start":"00:28.245 ","End":"00:33.575","Text":"equal to the magnetic flux divided by the current."},{"Start":"00:33.575 ","End":"00:36.649","Text":"In order to calculate the magnetic flux,"},{"Start":"00:36.649 ","End":"00:42.515","Text":"we have to use this equation which is the integral of B.ds,"},{"Start":"00:42.515 ","End":"00:47.525","Text":"how much magnetic fields passes through the given area."},{"Start":"00:47.525 ","End":"00:49.775","Text":"Then order to calculate that,"},{"Start":"00:49.775 ","End":"00:52.300","Text":"we need to know the magnetic field."},{"Start":"00:52.300 ","End":"00:54.140","Text":"We\u0027ve done this in a previous lesson,"},{"Start":"00:54.140 ","End":"01:00.575","Text":"and the magnetic field of a toroid is equal to Mu naught multiplied by the current,"},{"Start":"01:00.575 ","End":"01:08.105","Text":"multiplied by the number of turns in the coil divided by 2Pir,"},{"Start":"01:08.105 ","End":"01:14.340","Text":"and then from the right-hand rule we know that it\u0027s in the Theta direction."},{"Start":"01:14.450 ","End":"01:20.125","Text":"This equation we actually derived when we were doing Ampere\u0027s law."},{"Start":"01:20.125 ","End":"01:24.730","Text":"In that chapter, if you can\u0027t remember how we got to this equation,"},{"Start":"01:24.730 ","End":"01:28.460","Text":"please go back to that lesson."},{"Start":"01:28.910 ","End":"01:33.370","Text":"Now we want to calculate the magnetic flux."},{"Start":"01:33.370 ","End":"01:36.160","Text":"As we said, it\u0027s the integral on b,"},{"Start":"01:36.160 ","End":"01:44.590","Text":"so Mu naught IN divided by 2Pir.ds."},{"Start":"01:44.590 ","End":"01:51.590","Text":"So ds has to be perpendicular to the direction of travel of the magnetic field."},{"Start":"01:52.160 ","End":"01:56.080","Text":"If we saw that this is in the Theta direction,"},{"Start":"01:56.080 ","End":"02:02.425","Text":"so let\u0027s say the magnetic field is in this direction like so."},{"Start":"02:02.425 ","End":"02:05.925","Text":"Our ds has to be perpendicular to this."},{"Start":"02:05.925 ","End":"02:08.330","Text":"What we\u0027re going to do is we\u0027re going to take"},{"Start":"02:08.330 ","End":"02:17.430","Text":"this little square over here through this loop."},{"Start":"02:17.430 ","End":"02:22.065","Text":"We\u0027re taking just this chunk over here,"},{"Start":"02:22.065 ","End":"02:25.205","Text":"like so, and then we go down here,"},{"Start":"02:25.205 ","End":"02:29.295","Text":"like so, like this."},{"Start":"02:29.295 ","End":"02:35.765","Text":"We\u0027re integrating therefore along this edge."},{"Start":"02:35.765 ","End":"02:37.655","Text":"We want to integrate along the height,"},{"Start":"02:37.655 ","End":"02:39.780","Text":"so that\u0027s dz,"},{"Start":"02:40.630 ","End":"02:45.870","Text":"and we\u0027re integrating of course,"},{"Start":"02:45.870 ","End":"02:48.440","Text":"this width over here,"},{"Start":"02:48.440 ","End":"02:51.105","Text":"so this width like so."},{"Start":"02:51.105 ","End":"02:55.890","Text":"That is going to be also di so we want these 2,"},{"Start":"02:55.890 ","End":"02:58.595","Text":"and then our bounds."},{"Start":"02:58.595 ","End":"03:05.885","Text":"For r we\u0027re going from the inner radius of a until the outer radius of b, and for dz,"},{"Start":"03:05.885 ","End":"03:11.195","Text":"we\u0027re going from a height of 0 and to a height or width of"},{"Start":"03:11.195 ","End":"03:20.740","Text":"h. Now let\u0027s calculate this integral.. We can see that we don\u0027t have a variable z,"},{"Start":"03:20.740 ","End":"03:29.490","Text":"so we can just multiply everything by h. We have Mu naught INh divided by 2Pi,"},{"Start":"03:29.490 ","End":"03:35.995","Text":"and then we have our integral from a to b of 1 divided by dr,"},{"Start":"03:35.995 ","End":"03:39.325","Text":"which of course we know is going to come up as a ln."},{"Start":"03:39.325 ","End":"03:48.060","Text":"We have Mu naught INh divided by 2Pi multiplied by ln and from the ln\u0027s,"},{"Start":"03:48.060 ","End":"03:51.435","Text":"it\u0027s ln of b divided by a."},{"Start":"03:51.435 ","End":"03:56.130","Text":"In actual fact it\u0027s ln of b minus ln of a,"},{"Start":"03:56.130 ","End":"04:00.405","Text":"which according to law of ln just works out like this."},{"Start":"04:00.405 ","End":"04:02.700","Text":"This is our magnetic flux,"},{"Start":"04:02.700 ","End":"04:05.030","Text":"so therefore we can say that"},{"Start":"04:05.030 ","End":"04:09.080","Text":"our inductance is equal to a magnetic flux divided by the current."},{"Start":"04:09.080 ","End":"04:16.900","Text":"Our magnetic flux is Mu naught INh divided by 2Pi,"},{"Start":"04:16.900 ","End":"04:19.700","Text":"ln of b divided by a,"},{"Start":"04:19.700 ","End":"04:22.090","Text":"and all of this is divided by I."},{"Start":"04:22.090 ","End":"04:23.770","Text":"These cancel out,"},{"Start":"04:23.770 ","End":"04:30.530","Text":"and so what we get is that this is just equal to"},{"Start":"04:30.530 ","End":"04:38.160","Text":"Mu naught Nh divided"},{"Start":"04:38.160 ","End":"04:43.925","Text":"by 2Pi multiplied by ln of b divided by a."},{"Start":"04:43.925 ","End":"04:52.460","Text":"Now, notice that this is just the inductance for 1 of these terms,"},{"Start":"04:52.460 ","End":"04:55.200","Text":"just for this section."},{"Start":"04:55.550 ","End":"05:01.340","Text":"This is what we saw also in a previous lesson where we were dealing with a coil."},{"Start":"05:01.340 ","End":"05:03.785","Text":"This is just for 1 turn in the coil,"},{"Start":"05:03.785 ","End":"05:05.240","Text":"but we have N turns."},{"Start":"05:05.240 ","End":"05:07.325","Text":"Each one can be thought of as a ring."},{"Start":"05:07.325 ","End":"05:09.815","Text":"We have to sum up on all the rings,"},{"Start":"05:09.815 ","End":"05:11.600","Text":"and so we have N rings."},{"Start":"05:11.600 ","End":"05:13.805","Text":"This is the inductance for 1 ring,"},{"Start":"05:13.805 ","End":"05:17.420","Text":"and therefore the inductance for N rings,"},{"Start":"05:17.420 ","End":"05:21.470","Text":"so for the whole toroid is just going to be"},{"Start":"05:21.470 ","End":"05:27.275","Text":"this L_1 multiplied by the total number of turns."},{"Start":"05:27.275 ","End":"05:30.810","Text":"What we\u0027ll have is Mu naught N multiplied by N,"},{"Start":"05:30.810 ","End":"05:38.590","Text":"so N^2 h divided by 2Pi ln of b divided by a."},{"Start":"05:39.530 ","End":"05:46.055","Text":"What\u0027s important to remember is that when you find the inductance,"},{"Start":"05:46.055 ","End":"05:49.400","Text":"you\u0027re just finding the inductance for just one"},{"Start":"05:49.400 ","End":"05:53.285","Text":"of these turns and just to remember to multiply it out."},{"Start":"05:53.285 ","End":"05:55.835","Text":"Now let\u0027s take a look at question number 2,"},{"Start":"05:55.835 ","End":"06:01.380","Text":"calculate the energy stored in the toroid of current I flows through it."},{"Start":"06:02.120 ","End":"06:06.720","Text":"The energy for a coil is,"},{"Start":"06:06.720 ","End":"06:15.440","Text":"so U_L for coil is equal to 1/2 multiplied by the inductance multiplied by I^2."},{"Start":"06:15.440 ","End":"06:17.795","Text":"This is the simple way."},{"Start":"06:17.795 ","End":"06:22.370","Text":"Just plug in what your L is over here and I squared"},{"Start":"06:22.370 ","End":"06:27.125","Text":"we\u0027re given in the question and then you get the energy stored."},{"Start":"06:27.125 ","End":"06:32.310","Text":"Another way that you should know but it\u0027s much more complicated is this equation."},{"Start":"06:32.310 ","End":"06:35.285","Text":"This is the energy stored in a magnetic field,"},{"Start":"06:35.285 ","End":"06:42.400","Text":"which is also equal to the energy stored in the toroid or the energy stored in a coil,"},{"Start":"06:42.400 ","End":"06:52.410","Text":"and it is equal to the integral of b squared divided by 2Mu naught dv."},{"Start":"06:52.660 ","End":"06:55.790","Text":"What we\u0027re doing is we\u0027re integrating"},{"Start":"06:55.790 ","End":"07:00.785","Text":"the whole region or the whole volume where there is a magnetic field."},{"Start":"07:00.785 ","End":"07:04.190","Text":"Here we obviously are dealing with 3-dimensions,"},{"Start":"07:04.190 ","End":"07:07.325","Text":"so we\u0027re integrating with respect to 3 dimensions,"},{"Start":"07:07.325 ","End":"07:09.890","Text":"and then we have B squared."},{"Start":"07:09.890 ","End":"07:13.665","Text":"That is over here, this is our B."},{"Start":"07:13.665 ","End":"07:18.680","Text":"We have Mu naught squared I^2 N^2"},{"Start":"07:18.680 ","End":"07:24.220","Text":"divided by 4Pi^2 I^2,"},{"Start":"07:24.220 ","End":"07:28.575","Text":"so 4Pi^2 I^2,"},{"Start":"07:28.575 ","End":"07:36.810","Text":"and then this is multiplied by 1 divided by 2Mu naught."},{"Start":"07:36.810 ","End":"07:38.580","Text":"We can of course cross this out,"},{"Start":"07:38.580 ","End":"07:39.970","Text":"and then we\u0027re doing dv,"},{"Start":"07:39.970 ","End":"07:43.460","Text":"so we\u0027re using cylindrical coordinates over here."},{"Start":"07:43.460 ","End":"07:49.735","Text":"It\u0027s easier, so we have r dr d Theta dz."},{"Start":"07:49.735 ","End":"07:52.430","Text":"Of course it could cancel stuff out."},{"Start":"07:52.430 ","End":"07:54.440","Text":"Now the bounds,"},{"Start":"07:54.440 ","End":"07:58.230","Text":"for r we\u0027re going from the inner radius to the outer radius."},{"Start":"07:58.230 ","End":"08:00.754","Text":"The whole volume of the toroids,"},{"Start":"08:00.754 ","End":"08:02.540","Text":"we\u0027re going from A to B,"},{"Start":"08:02.540 ","End":"08:07.380","Text":"and then in Theta we\u0027re going from 0 until 2Pi,"},{"Start":"08:07.380 ","End":"08:08.670","Text":"so full circle,"},{"Start":"08:08.670 ","End":"08:15.260","Text":"and for z we\u0027re going from a height of 0 until our maximum height of the toroid,"},{"Start":"08:15.260 ","End":"08:20.480","Text":"which is h. Then all you have to do is you"},{"Start":"08:20.480 ","End":"08:25.985","Text":"have to integrate this and you will get the exact same answer for these 2 equations."},{"Start":"08:25.985 ","End":"08:32.340","Text":"They\u0027re exactly the same and there\u0027s even a lesson where we show that."},{"Start":"08:33.020 ","End":"08:36.650","Text":"This is a very easy integral to do,"},{"Start":"08:36.650 ","End":"08:41.990","Text":"so if you want to feel free to do it and also plug in this"},{"Start":"08:41.990 ","End":"08:47.885","Text":"into this equation and see that really these 2 are equal to the same thing,"},{"Start":"08:47.885 ","End":"08:49.400","Text":"it will be very helpful."},{"Start":"08:49.400 ","End":"08:51.305","Text":"But anyway in the meantime,"},{"Start":"08:51.305 ","End":"08:56.735","Text":"we can just write that the energy stored is equal to half multiplied by"},{"Start":"08:56.735 ","End":"09:01.665","Text":"Mu naught N^2 h divided by"},{"Start":"09:01.665 ","End":"09:10.110","Text":"2 pi multiplied by ln of b divided by a multiplied by I^2."},{"Start":"09:10.630 ","End":"09:13.955","Text":"This is the answer to question number 2,"},{"Start":"09:13.955 ","End":"09:15.710","Text":"this is the energy stored,"},{"Start":"09:15.710 ","End":"09:18.680","Text":"and please do work out this"},{"Start":"09:18.680 ","End":"09:22.415","Text":"integral and you\u0027ll see that you\u0027ll get the exact same answer."},{"Start":"09:22.415 ","End":"09:25.470","Text":"That\u0027s the end of this lesson."}],"ID":22377},{"Watched":false,"Name":"Exercise 6","Duration":"19m 48s","ChapterTopicVideoID":21354,"CourseChapterTopicPlaylistID":99483,"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":"00:04.500","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.500 ","End":"00:08.085","Text":"The EMF, resistances,"},{"Start":"00:08.085 ","End":"00:10.485","Text":"and inductance are given."},{"Start":"00:10.485 ","End":"00:19.630","Text":"Calculate the current through the coil as a function of time if the initial current is 0."},{"Start":"00:20.780 ","End":"00:23.850","Text":"In other words, our initial current,"},{"Start":"00:23.850 ","End":"00:27.900","Text":"so our current at is equal to 0,"},{"Start":"00:27.900 ","End":"00:33.360","Text":"and we want to find the equation for the current as a function of time."},{"Start":"00:33.360 ","End":"00:37.720","Text":"We have a slightly more complicated circuit over here,"},{"Start":"00:37.720 ","End":"00:40.190","Text":"so what we\u0027re going to do is we\u0027re going to write out"},{"Start":"00:40.190 ","End":"00:44.850","Text":"these circuit equations using Kirchhoff\u0027s laws."},{"Start":"00:45.140 ","End":"00:50.498","Text":"Let\u0027s say that the current going up here is called I_2,"},{"Start":"00:50.498 ","End":"00:55.650","Text":"just because this is the current that\u0027s going to go through this I_2 resistor,"},{"Start":"00:55.650 ","End":"00:59.120","Text":"and let\u0027s say this is the positive and this is the negative end."},{"Start":"00:59.120 ","End":"01:00.470","Text":"Then over here, of course,"},{"Start":"01:00.470 ","End":"01:01.670","Text":"our current splits,"},{"Start":"01:01.670 ","End":"01:06.790","Text":"so let\u0027s say that the current going through resistor R_1 is called I_1."},{"Start":"01:06.790 ","End":"01:10.040","Text":"Again, this is the positive and this is the negative end."},{"Start":"01:10.040 ","End":"01:14.930","Text":"Then we have our current going through the inductor or the coil,"},{"Start":"01:14.930 ","End":"01:17.410","Text":"and let\u0027s call it I_3."},{"Start":"01:17.410 ","End":"01:19.745","Text":"Now using Kirchhoff\u0027s law,"},{"Start":"01:19.745 ","End":"01:23.090","Text":"let\u0027s choose an arbitrary direction like so."},{"Start":"01:23.090 ","End":"01:24.950","Text":"Then we can start over here,"},{"Start":"01:24.950 ","End":"01:31.340","Text":"so we go from the battery that has a voltage or EMF of Epsilon,"},{"Start":"01:31.340 ","End":"01:35.915","Text":"and then we go across the resistor from the positive to the negative end,"},{"Start":"01:35.915 ","End":"01:41.490","Text":"so we have minus I_2, R_2."},{"Start":"01:41.810 ","End":"01:46.530","Text":"Then we go across the R_1 resistor from positive to negative,"},{"Start":"01:46.530 ","End":"01:50.510","Text":"so we have minus I_1, R_1."},{"Start":"01:50.600 ","End":"01:55.480","Text":"Then we carry on and we come back to our starting point."},{"Start":"01:55.480 ","End":"01:58.610","Text":"That means that this is equal to 0."},{"Start":"01:58.970 ","End":"02:04.890","Text":"Then, similarly, for this side of the circuit over here."},{"Start":"02:04.890 ","End":"02:10.180","Text":"Let\u0027s say that this is the positive and this is the negative end of the coil,"},{"Start":"02:10.180 ","End":"02:12.519","Text":"so we go across the coil,"},{"Start":"02:12.519 ","End":"02:15.170","Text":"we\u0027ll start from this point over here."},{"Start":"02:15.170 ","End":"02:18.010","Text":"We go down the coils."},{"Start":"02:18.010 ","End":"02:21.970","Text":"We have negative LI dot,"},{"Start":"02:21.970 ","End":"02:23.970","Text":"so that\u0027s the voltage across the coil."},{"Start":"02:23.970 ","End":"02:28.150","Text":"We carry on then we go up over here,"},{"Start":"02:28.150 ","End":"02:31.130","Text":"so from negative to positive on the resistor,"},{"Start":"02:31.130 ","End":"02:35.935","Text":"so we have plus I_1, R_1."},{"Start":"02:35.935 ","End":"02:39.610","Text":"Then we continue around in the circle and we get back to our starting points,"},{"Start":"02:39.610 ","End":"02:43.860","Text":"so all of this is also equal to 0."},{"Start":"02:44.960 ","End":"02:48.620","Text":"I have, in the meantime, 2 equations,"},{"Start":"02:48.620 ","End":"02:52.535","Text":"but I have technically 3 unknowns,"},{"Start":"02:52.535 ","End":"02:54.320","Text":"I_1, I_2, and I_3."},{"Start":"02:54.320 ","End":"03:00.410","Text":"I can say that I_2 is equal to the splitting over here,"},{"Start":"03:00.410 ","End":"03:04.580","Text":"so it\u0027s equal to I_1 plus I_3."},{"Start":"03:04.580 ","End":"03:07.380","Text":"Now I have 1, 2,"},{"Start":"03:07.380 ","End":"03:12.220","Text":"3 equation and 3 unknowns."},{"Start":"03:13.010 ","End":"03:18.355","Text":"Just as a reminder, this I dot is actually I_3 dot,"},{"Start":"03:18.355 ","End":"03:21.230","Text":"because that\u0027s the current passing through."},{"Start":"03:21.230 ","End":"03:27.065","Text":"What I want to do is I want to now isolate all of these equations and solve for I_3,"},{"Start":"03:27.065 ","End":"03:30.740","Text":"because the question is calculate the current through the coil."},{"Start":"03:30.740 ","End":"03:32.405","Text":"What is the current through the coil?"},{"Start":"03:32.405 ","End":"03:34.360","Text":"It\u0027s I_3."},{"Start":"03:34.360 ","End":"03:38.520","Text":"I\u0027m going to want to get everything in terms of I_3."},{"Start":"03:39.670 ","End":"03:43.790","Text":"The first thing that I\u0027m going to do is I\u0027m going to get"},{"Start":"03:43.790 ","End":"03:47.210","Text":"rid of this I_2 over here in the first equation."},{"Start":"03:47.210 ","End":"03:51.975","Text":"I\u0027m putting 3 into equation number 1."},{"Start":"03:51.975 ","End":"03:56.105","Text":"What I\u0027ll get is Epsilon minus and then instead of I_2,"},{"Start":"03:56.105 ","End":"04:00.240","Text":"I have I_1 plus I_3,"},{"Start":"04:00.240 ","End":"04:05.400","Text":"so I_1 plus I_3 multiplied"},{"Start":"04:05.400 ","End":"04:10.890","Text":"by R_2 minus I_1,"},{"Start":"04:10.890 ","End":"04:13.750","Text":"R_1, which is equal to 0."},{"Start":"04:14.480 ","End":"04:19.775","Text":"My next step is to try and get rid of this R_1 over here,"},{"Start":"04:19.775 ","End":"04:21.170","Text":"or rather the I_1."},{"Start":"04:21.170 ","End":"04:23.345","Text":"I want to get rid of the I_1."},{"Start":"04:23.345 ","End":"04:28.325","Text":"I\u0027m just going to move out my I_1\u0027s to the other side."},{"Start":"04:28.325 ","End":"04:32.420","Text":"I\u0027ll have epsilon minus I_3,"},{"Start":"04:32.420 ","End":"04:37.335","Text":"R_2 is equal to,"},{"Start":"04:37.335 ","End":"04:44.530","Text":"and then I have I_1 multiplied by R_1 plus R_2."},{"Start":"04:44.750 ","End":"04:50.610","Text":"Now, what I want to do is I want to isolate out my I_1 as I said."},{"Start":"04:50.610 ","End":"04:55.160","Text":"I\u0027ll get that my I_1 is equal to epsilon minus"},{"Start":"04:55.160 ","End":"05:02.280","Text":"I_3 R_2 divided by R_1 plus R_2."},{"Start":"05:02.280 ","End":"05:08.015","Text":"Now, I\u0027m going to plug this into this equation number 2."},{"Start":"05:08.015 ","End":"05:12.325","Text":"Basically, what I have here is 3 into 1,"},{"Start":"05:12.325 ","End":"05:16.385","Text":"so this is now pretty much equation number 1,"},{"Start":"05:16.385 ","End":"05:21.305","Text":"and now I\u0027m taking 1 and putting it into 2."},{"Start":"05:21.305 ","End":"05:23.705","Text":"Where this is 1."},{"Start":"05:23.705 ","End":"05:28.240","Text":"I have negative L,"},{"Start":"05:28.240 ","End":"05:31.850","Text":"I3 dot plus I_1, R_1."},{"Start":"05:31.850 ","End":"05:37.605","Text":"I could say that this is equal to negative I_1, R_1,"},{"Start":"05:37.605 ","End":"05:42.495","Text":"so I_1 is Epsilon minus I_3,"},{"Start":"05:42.495 ","End":"05:48.550","Text":"R_2 divided by R_1 plus R_2,"},{"Start":"05:48.550 ","End":"05:49.950","Text":"so that was I_1,"},{"Start":"05:49.950 ","End":"05:52.380","Text":"R_1 so this is all multiplied by"},{"Start":"05:52.380 ","End":"05:56.315","Text":"R_1 and now I can get rid of the negatives from both sides."},{"Start":"05:56.315 ","End":"05:58.055","Text":"Now I have LI_3."},{"Start":"05:58.055 ","End":"06:00.480","Text":"is equal to this."},{"Start":"06:00.860 ","End":"06:05.435","Text":"Now you just solve this as a differential equation,"},{"Start":"06:05.435 ","End":"06:08.285","Text":"like we saw in the previous lesson."},{"Start":"06:08.285 ","End":"06:13.025","Text":"What we get is that I_3 as a function of time,"},{"Start":"06:13.025 ","End":"06:19.475","Text":"is equal to Epsilon or the EMF divided by R_2 multiplied by"},{"Start":"06:19.475 ","End":"06:27.360","Text":"1 minus e to the power of negative R total divided by L t,"},{"Start":"06:27.470 ","End":"06:35.355","Text":"where R total is equal to R_1,"},{"Start":"06:35.355 ","End":"06:41.700","Text":"R_2 divided by R_1 plus R_2."},{"Start":"06:41.700 ","End":"06:48.900","Text":"Here we can see that we\u0027ve just joined on the resistors in parallel."},{"Start":"06:50.900 ","End":"06:54.410","Text":"This is our onset and what we want to do now is we want"},{"Start":"06:54.410 ","End":"06:57.855","Text":"to see if this answer makes sense."},{"Start":"06:57.855 ","End":"07:01.670","Text":"We were told that the initial current is 0."},{"Start":"07:01.670 ","End":"07:05.450","Text":"That means that I at t is equal to 0,"},{"Start":"07:05.450 ","End":"07:07.859","Text":"should work out to be equal to 0."},{"Start":"07:07.859 ","End":"07:10.715","Text":"Let\u0027s plug in t=0."},{"Start":"07:10.715 ","End":"07:15.980","Text":"Then we have 0 multiplied by R t divided by L, which is 0."},{"Start":"07:15.980 ","End":"07:19.925","Text":"Then we have e to the power of negative 0,"},{"Start":"07:19.925 ","End":"07:23.870","Text":"e^0, and e^0 is of course 1."},{"Start":"07:23.870 ","End":"07:26.315","Text":"Then here we have 1 minus 1,"},{"Start":"07:26.315 ","End":"07:29.570","Text":"which is 0 and then 0 times this constant is,"},{"Start":"07:29.570 ","End":"07:34.080","Text":"of course, 0 and we get 0. That\u0027s great."},{"Start":"07:34.080 ","End":"07:37.722","Text":"Now, what about t at infinity,"},{"Start":"07:37.722 ","End":"07:40.835","Text":"after a long period of time."},{"Start":"07:40.835 ","End":"07:43.760","Text":"If we plug in over here infinity,"},{"Start":"07:43.760 ","End":"07:46.949","Text":"then we\u0027ll have negative infinity."},{"Start":"07:46.949 ","End":"07:54.150","Text":"So e to the power of negative infinity is approaching 0."},{"Start":"07:57.230 ","End":"08:03.020","Text":"That means that this whole expression over here is approaching 0."},{"Start":"08:03.020 ","End":"08:04.910","Text":"If t is infinity,"},{"Start":"08:04.910 ","End":"08:09.010","Text":"e to the power of negative infinity is approaching 0."},{"Start":"08:09.010 ","End":"08:11.715","Text":"Then we have 1 minus 0, which is 1,"},{"Start":"08:11.715 ","End":"08:15.190","Text":"and then 1 multiplied by this constant is just equal to this constant."},{"Start":"08:15.190 ","End":"08:19.790","Text":"We get that the current at t is equal to infinity,"},{"Start":"08:19.790 ","End":"08:25.775","Text":"or after a very long period of time is approximately Epsilon divided by R_2."},{"Start":"08:25.775 ","End":"08:28.940","Text":"We get the maximum value for current,"},{"Start":"08:28.940 ","End":"08:32.135","Text":"which is exactly what we would\u0027ve expected as seen in"},{"Start":"08:32.135 ","End":"08:37.230","Text":"the previous lesson explaining the RL circuit."},{"Start":"08:37.610 ","End":"08:44.345","Text":"In this case, we get that the circuit is actually like a short circuit."},{"Start":"08:44.345 ","End":"08:47.840","Text":"What we\u0027ll have is that all the current will go through"},{"Start":"08:47.840 ","End":"08:51.485","Text":"the coil so we can imagine that there\u0027s no resistance in the coil,"},{"Start":"08:51.485 ","End":"08:55.005","Text":"all the current is flowing through the coil."},{"Start":"08:55.005 ","End":"08:56.960","Text":"In this case, we can see that we\u0027re dealing with"},{"Start":"08:56.960 ","End":"09:00.184","Text":"a circuit that consists just of the battery,"},{"Start":"09:00.184 ","End":"09:04.055","Text":"the I_2 resistor, and the coil or the inductor."},{"Start":"09:04.055 ","End":"09:07.925","Text":"This R_1 resistor doesn\u0027t"},{"Start":"09:07.925 ","End":"09:13.820","Text":"actually play any role in the circuit after a very long period of time."},{"Start":"09:16.330 ","End":"09:19.315","Text":"This is the answer to this question."},{"Start":"09:19.315 ","End":"09:23.230","Text":"If they would ask you a question such as calculate the current through"},{"Start":"09:23.230 ","End":"09:27.880","Text":"the coil when it\u0027s at a stable state or after a very long period of time,"},{"Start":"09:27.880 ","End":"09:34.360","Text":"then you could just give the answer of a regular circuit that just has the battery,"},{"Start":"09:34.360 ","End":"09:41.780","Text":"a resistor, and this over here and you wouldn\u0027t even have to look at the R_1 resistor."},{"Start":"09:43.310 ","End":"09:46.405","Text":"Now we\u0027ve finished solving the question."},{"Start":"09:46.405 ","End":"09:48.610","Text":"What I\u0027m now going to do is I\u0027m going to solve"},{"Start":"09:48.610 ","End":"09:53.110","Text":"this differential equation in order to see how we got to this solution."},{"Start":"09:53.110 ","End":"09:56.085","Text":"If you already know how to solve differential equations,"},{"Start":"09:56.085 ","End":"10:00.060","Text":"then you can finish the lesson over here."},{"Start":"10:01.250 ","End":"10:04.550","Text":"Just like before, we saw that it\u0027s best to write"},{"Start":"10:04.550 ","End":"10:07.790","Text":"out this equation instead of with this I dot,"},{"Start":"10:07.790 ","End":"10:11.130","Text":"with dI_3 by dt."},{"Start":"10:11.130 ","End":"10:17.205","Text":"We have L multiplied by dI_3 by dt,"},{"Start":"10:17.205 ","End":"10:19.335","Text":"which is equal 2,"},{"Start":"10:19.335 ","End":"10:23.430","Text":"and then we have Epsilon."},{"Start":"10:23.430 ","End":"10:28.935","Text":"Let\u0027s divide both sides by L first of all."},{"Start":"10:28.935 ","End":"10:33.070","Text":"We have R_1 divided by L,"},{"Start":"10:33.320 ","End":"10:44.560","Text":"and then this is multiplied by Epsilon divided by R_1 plus R_2 minus,"},{"Start":"10:44.560 ","End":"10:50.000","Text":"and then we have I_3,"},{"Start":"10:50.000 ","End":"10:54.200","Text":"R_2 divided by R_1 plus R_2,"},{"Start":"10:54.200 ","End":"10:58.770","Text":"so I_3, R_2 divided by"},{"Start":"10:58.770 ","End":"11:05.850","Text":"R_1 plus R_2 and of course here we have brackets."},{"Start":"11:07.770 ","End":"11:12.130","Text":"In fact, I\u0027m just going to write this out separately."},{"Start":"11:12.130 ","End":"11:14.605","Text":"Here we have I_3, don\u0027t forget."},{"Start":"11:14.605 ","End":"11:17.755","Text":"Then here we\u0027ll also again multiply by R_1 divided by L,"},{"Start":"11:17.755 ","End":"11:20.365","Text":"just to make it a little bit easier."},{"Start":"11:20.365 ","End":"11:24.517","Text":"Now, what I\u0027m going to do is I\u0027m going to multiply both sides by this dt."},{"Start":"11:24.517 ","End":"11:28.840","Text":"Because remember, I want to get all my I_3s on 1 side and"},{"Start":"11:28.840 ","End":"11:34.430","Text":"all my t variables on the other side of the equal sign."},{"Start":"11:34.430 ","End":"11:37.000","Text":"Just to simplify it,"},{"Start":"11:37.000 ","End":"11:38.770","Text":"I\u0027m going to call this expression over"},{"Start":"11:38.770 ","End":"11:40.915","Text":"here so that I don\u0027t have to keep writing this out."},{"Start":"11:40.915 ","End":"11:44.414","Text":"I\u0027m going to call this C for constant."},{"Start":"11:44.414 ","End":"11:47.245","Text":"Don\u0027t get confused with capacitance,"},{"Start":"11:47.245 ","End":"11:49.225","Text":"this is just a constant."},{"Start":"11:49.225 ","End":"11:56.644","Text":"This over here, I don\u0027t want to get my I_3 in,"},{"Start":"11:56.644 ","End":"12:02.020","Text":"so this expression over here I\u0027m going to call as R total divided by"},{"Start":"12:02.020 ","End":"12:08.485","Text":"L. I have R_1 R_2 divided by R_1 plus R_2."},{"Start":"12:08.485 ","End":"12:10.495","Text":"That\u0027s the R total,"},{"Start":"12:10.495 ","End":"12:12.085","Text":"just like we have over here."},{"Start":"12:12.085 ","End":"12:13.720","Text":"Then we have L in the denominator,"},{"Start":"12:13.720 ","End":"12:19.135","Text":"so it\u0027s divided by L. Now,"},{"Start":"12:19.135 ","End":"12:22.360","Text":"as I just said, I\u0027m multiplying both sides by dt,"},{"Start":"12:22.360 ","End":"12:27.490","Text":"so I have dI_3 is equal to C"},{"Start":"12:27.490 ","End":"12:34.660","Text":"minus R total divided by L I_3,"},{"Start":"12:34.660 ","End":"12:38.900","Text":"and all of this is multiplied by dt."},{"Start":"12:39.810 ","End":"12:43.465","Text":"I have my side for I_3 and I have my side for"},{"Start":"12:43.465 ","End":"12:50.665","Text":"t. I can see that I have this I_3 variable on the wrong side of the equation."},{"Start":"12:50.665 ","End":"12:55.165","Text":"What I\u0027m going to do is I\u0027m going to divide both sides by these brackets."},{"Start":"12:55.165 ","End":"13:01.825","Text":"What I\u0027m going to have is 1 divided by C minus"},{"Start":"13:01.825 ","End":"13:09.010","Text":"R_T divided by L I_3,"},{"Start":"13:09.010 ","End":"13:12.740","Text":"dI_3, and this is going to be equal to dt."},{"Start":"13:13.260 ","End":"13:17.020","Text":"Now, all my variables are on the correct side of the equation,"},{"Start":"13:17.020 ","End":"13:19.989","Text":"so what I\u0027m going to do is I\u0027m going to integrate."},{"Start":"13:19.989 ","End":"13:24.250","Text":"I\u0027m going to plug in my integration signs and I\u0027m going to integrate from t is equal to"},{"Start":"13:24.250 ","End":"13:29.920","Text":"0 until t. Then from the current t is equal to 0,"},{"Start":"13:29.920 ","End":"13:35.260","Text":"so I know that my current at t is equal to 0 is just equal to 0."},{"Start":"13:35.260 ","End":"13:37.780","Text":"That I was told in the question,"},{"Start":"13:37.780 ","End":"13:39.280","Text":"the initial current is 0."},{"Start":"13:39.280 ","End":"13:43.930","Text":"I\u0027m going to integrate until I get the current at some time t,"},{"Start":"13:43.930 ","End":"13:47.825","Text":"where of course, these 2ts correspond."},{"Start":"13:47.825 ","End":"13:52.020","Text":"What I\u0027m going to have over here on this side,"},{"Start":"13:52.020 ","End":"13:55.965","Text":"I can see that by I_3 is to the power of 1,"},{"Start":"13:55.965 ","End":"13:58.035","Text":"and it\u0027s in the denominator."},{"Start":"13:58.035 ","End":"14:00.015","Text":"I\u0027m going to have ln."},{"Start":"14:00.015 ","End":"14:03.690","Text":"Now, what I do is I add a minus over here because"},{"Start":"14:03.690 ","End":"14:07.230","Text":"the coefficient of I_3 is negative R_T divided by L,"},{"Start":"14:07.230 ","End":"14:11.830","Text":"so I have negative L divided by R_T,"},{"Start":"14:12.450 ","End":"14:17.350","Text":"and that is all I have over here."},{"Start":"14:17.350 ","End":"14:20.950","Text":"Then I have this is ln of,"},{"Start":"14:20.950 ","End":"14:25.285","Text":"and so then I have this where I plug in I_T."},{"Start":"14:25.285 ","End":"14:34.480","Text":"I have C minus R_T divided by L I as a function of t divided by,"},{"Start":"14:34.480 ","End":"14:36.115","Text":"so this is from law of ln\u0027s,"},{"Start":"14:36.115 ","End":"14:41.890","Text":"C minus R_T divided by L multiplied by 0, which is 0,"},{"Start":"14:41.890 ","End":"14:46.555","Text":"so just divided by C. This is simply equal to"},{"Start":"14:46.555 ","End":"14:55.060","Text":"t. Here I\u0027ve just done law of ln\u0027s and integrated."},{"Start":"14:55.060 ","End":"14:59.830","Text":"Now, what I\u0027m going to do is I\u0027m going to divide both sides by"},{"Start":"14:59.830 ","End":"15:04.165","Text":"this negative L divided by R_T."},{"Start":"15:04.165 ","End":"15:09.025","Text":"What I\u0027m going to be left with is ln of C minus"},{"Start":"15:09.025 ","End":"15:14.845","Text":"R_T divided by L I as a function of t divided by C,"},{"Start":"15:14.845 ","End":"15:24.790","Text":"which is going to be equal to negative tR_T divided by L. Now,"},{"Start":"15:24.790 ","End":"15:29.755","Text":"what I\u0027m going to do is I\u0027m going to raise each side by e. Let\u0027s put e over here."},{"Start":"15:29.755 ","End":"15:32.140","Text":"Then of course when I have e and a ln,"},{"Start":"15:32.140 ","End":"15:33.850","Text":"they just cancel each other out."},{"Start":"15:33.850 ","End":"15:42.748","Text":"What I have is C minus R_T divided by L multiplied by I as a function of t,"},{"Start":"15:42.748 ","End":"15:44.620","Text":"all of this divided by C,"},{"Start":"15:44.620 ","End":"15:53.980","Text":"which is equal to e^negative R total divided by L multiplied by t. I just rearranged it."},{"Start":"15:53.980 ","End":"15:58.735","Text":"e^ln just cancels each of them out."},{"Start":"15:58.735 ","End":"16:01.900","Text":"Now, of course, I want to isolate out my I,"},{"Start":"16:01.900 ","End":"16:04.540","Text":"and of course this is I_3."},{"Start":"16:04.540 ","End":"16:09.520","Text":"Let\u0027s just plug in I_3 everywhere just so that it\u0027s very clear"},{"Start":"16:09.520 ","End":"16:15.355","Text":"that it is in fact the current going through our inductor."},{"Start":"16:15.355 ","End":"16:17.935","Text":"That\u0027s I_3. Anyway,"},{"Start":"16:17.935 ","End":"16:22.120","Text":"now what we want to do is we want to isolate out the I_3."},{"Start":"16:22.120 ","End":"16:31.855","Text":"Let\u0027s multiply both sides by C. What we have is C minus R_T divided by L multiplied by"},{"Start":"16:31.855 ","End":"16:35.320","Text":"I_3 as a function of t is equal to"},{"Start":"16:35.320 ","End":"16:44.380","Text":"Ce^negative R_T divided by L multiplied by t. Now,"},{"Start":"16:44.380 ","End":"16:47.095","Text":"let\u0027s also just isolate out this I_3."},{"Start":"16:47.095 ","End":"16:50.305","Text":"What we have is I_3 as a function of t,"},{"Start":"16:50.305 ","End":"16:51.550","Text":"which is equal to,"},{"Start":"16:51.550 ","End":"16:56.095","Text":"so we have C over here."},{"Start":"16:56.095 ","End":"17:05.785","Text":"Then minus Ce^negative R_T divided by L t. Then we\u0027re going to"},{"Start":"17:05.785 ","End":"17:10.180","Text":"divide both sides by R_T divided by L."},{"Start":"17:10.180 ","End":"17:16.640","Text":"All of this is going to be multiplied by L divided by R_T."},{"Start":"17:18.090 ","End":"17:28.255","Text":"Then we can just write that I_3 as a function of t is equal to L divided by R_T"},{"Start":"17:28.255 ","End":"17:34.750","Text":"multiplied by C. Then all this is multiplied by 1 minus e^negative"},{"Start":"17:34.750 ","End":"17:41.770","Text":"R_T divided by L multiplied by t. Then,"},{"Start":"17:41.770 ","End":"17:46.360","Text":"remember that our C was this over here,"},{"Start":"17:46.360 ","End":"17:51.415","Text":"R_1 Epsilon L multiplied by R_1 plus R_2."},{"Start":"17:51.415 ","End":"17:56.470","Text":"If you plug that into this equation over here,"},{"Start":"17:56.470 ","End":"17:58.634","Text":"let\u0027s just do this,"},{"Start":"17:58.634 ","End":"18:04.105","Text":"we have L divided by R_T multiplied by C,"},{"Start":"18:04.105 ","End":"18:05.770","Text":"where C, as we said,"},{"Start":"18:05.770 ","End":"18:15.205","Text":"was R_1 multiplied by Epsilon divided by L multiplied by R_1 plus R_2,"},{"Start":"18:15.205 ","End":"18:21.807","Text":"then R total is of course equal to,"},{"Start":"18:21.807 ","End":"18:24.670","Text":"we\u0027ll just leave this as L over here,"},{"Start":"18:24.670 ","End":"18:27.325","Text":"and then we have L divided by R total."},{"Start":"18:27.325 ","End":"18:38.320","Text":"The reciprocal of R total is R_1 plus R_2 divided by R_1 R_2."},{"Start":"18:38.320 ","End":"18:44.560","Text":"This over here is just 1 divided by R total, which was this over here."},{"Start":"18:44.560 ","End":"18:46.840","Text":"Then we have R_1 plus R_2,"},{"Start":"18:46.840 ","End":"18:49.135","Text":"which cancels out with this R_1 plus R_2."},{"Start":"18:49.135 ","End":"18:52.810","Text":"This R_1 cancels out with this R_1."},{"Start":"18:52.810 ","End":"19:02.290","Text":"This L cancels out with this L. What we\u0027re just left with Epsilon over here."},{"Start":"19:02.290 ","End":"19:06.760","Text":"Of course this R_2 in then denominator,"},{"Start":"19:06.760 ","End":"19:09.920","Text":"so Epsilon divided by R_2,"},{"Start":"19:09.920 ","End":"19:12.325","Text":"Epsilon divided by R_2."},{"Start":"19:12.325 ","End":"19:20.170","Text":"Therefore, we can say that I_3 as a function of time is equal to all of this over here,"},{"Start":"19:20.170 ","End":"19:26.664","Text":"we\u0027re left with Epsilon divided by R_2 multiplied by 1 minus e to"},{"Start":"19:26.664 ","End":"19:30.910","Text":"the negative R total divided by L multiplied by"},{"Start":"19:30.910 ","End":"19:37.705","Text":"t. You can write this down and now let\u0027s just look at the solution that we had earlier."},{"Start":"19:37.705 ","End":"19:42.450","Text":"It\u0027s the exact same solution where of course this is R total,"},{"Start":"19:42.450 ","End":"19:45.525","Text":"R_1 R_2 divided by R_1 plus R_2."},{"Start":"19:45.525 ","End":"19:49.300","Text":"That is the end of this lesson."}],"ID":21434},{"Watched":false,"Name":"Exercise 7","Duration":"16m 21s","ChapterTopicVideoID":21548,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.695","Text":"Hello, in this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.695 ","End":"00:09.050","Text":"A conducting ring of radius a and resistance R,"},{"Start":"00:09.050 ","End":"00:12.630","Text":"so imagine that there\u0027s some kind of resistor here of R,"},{"Start":"00:12.630 ","End":"00:15.495","Text":"is located inside a magnetic field,"},{"Start":"00:15.495 ","End":"00:20.160","Text":"which is dependent only on time and it\u0027s equal to At,"},{"Start":"00:20.160 ","End":"00:23.035","Text":"where A is a positive constant."},{"Start":"00:23.035 ","End":"00:28.920","Text":"The field is perpendicular to the plane of the ring,"},{"Start":"00:28.920 ","End":"00:33.960","Text":"so in other words we have maximum flux."},{"Start":"00:33.960 ","End":"00:41.090","Text":"Here we have a side view of the ring and here we have a bird\u0027s eye view of the ring."},{"Start":"00:41.090 ","End":"00:46.970","Text":"We\u0027re looking at it from different angles and of course don\u0027t get confused between a,"},{"Start":"00:46.970 ","End":"00:48.545","Text":"which is the radius of the ring,"},{"Start":"00:48.545 ","End":"00:51.895","Text":"and A, which is this constant."},{"Start":"00:51.895 ","End":"00:56.660","Text":"Question number 1 is to calculate the total EMF or"},{"Start":"00:56.660 ","End":"01:02.180","Text":"voltage acting on the ring as a function of current given that"},{"Start":"01:02.180 ","End":"01:07.330","Text":"the inductance of the ring is L. The"},{"Start":"01:07.330 ","End":"01:13.670","Text":"second that we have a magnetic field and we\u0027re being asked to calculate the EMF,"},{"Start":"01:13.670 ","End":"01:17.000","Text":"Faraday\u0027s law should jump up immediately."},{"Start":"01:17.000 ","End":"01:20.990","Text":"Faraday\u0027s law says that the EMF is equal to"},{"Start":"01:20.990 ","End":"01:28.065","Text":"the negative time derivative of the magnetic flux."},{"Start":"01:28.065 ","End":"01:31.200","Text":"What magnetic flux do we have?"},{"Start":"01:31.200 ","End":"01:35.510","Text":"Of course we have the magnetic flux due to this field over here,"},{"Start":"01:35.510 ","End":"01:38.135","Text":"this external magnetic field,"},{"Start":"01:38.135 ","End":"01:42.600","Text":"however, there\u0027s also another magnetic flux."},{"Start":"01:42.770 ","End":"01:47.090","Text":"Because our ring is a conducting ring,"},{"Start":"01:47.090 ","End":"01:51.110","Text":"the moment that we have this external magnetic field B,"},{"Start":"01:51.110 ","End":"01:56.885","Text":"it\u0027s going to cause a current to flow through this ring,"},{"Start":"01:56.885 ","End":"02:02.615","Text":"and then this current is in turn going to create its own magnetic field."},{"Start":"02:02.615 ","End":"02:04.835","Text":"We\u0027re going to have 2 magnetic fields."},{"Start":"02:04.835 ","End":"02:06.950","Text":"The magnetic field given in the question,"},{"Start":"02:06.950 ","End":"02:12.644","Text":"and the magnetic field induced in the ring due to this magnetic field,"},{"Start":"02:12.644 ","End":"02:15.715","Text":"this external magnetic field."},{"Start":"02:15.715 ","End":"02:20.885","Text":"What we can do is we can write out the equation for the magnetic flux."},{"Start":"02:20.885 ","End":"02:25.160","Text":"We have the magnetic flux of the external fields,"},{"Start":"02:25.160 ","End":"02:27.185","Text":"so that\u0027s this over here,"},{"Start":"02:27.185 ","End":"02:31.775","Text":"plus the magnetic flux of the internal field,"},{"Start":"02:31.775 ","End":"02:35.270","Text":"which is the magnetic field induced."},{"Start":"02:35.270 ","End":"02:43.685","Text":"First of all, let\u0027s calculate the magnetic flux of the external magnetic field."},{"Start":"02:43.685 ","End":"02:52.490","Text":"As we know, the equation for flux is the integral of the magnetic field.ds."},{"Start":"02:52.490 ","End":"02:58.040","Text":"What we\u0027re doing is we\u0027re integrating with respect to surface area."},{"Start":"02:58.040 ","End":"03:02.645","Text":"Now as we can see, the magnetic field is dependent on time,"},{"Start":"03:02.645 ","End":"03:05.265","Text":"but it\u0027s uniform in space."},{"Start":"03:05.265 ","End":"03:10.570","Text":"In other words, even though the magnetic field is changing in time,"},{"Start":"03:10.570 ","End":"03:16.880","Text":"for this integral it\u0027s irrelevant because we\u0027re just integrating with respect to ds,"},{"Start":"03:16.880 ","End":"03:18.050","Text":"with respect to space,"},{"Start":"03:18.050 ","End":"03:22.370","Text":"so here time and this integral is a constant."},{"Start":"03:22.370 ","End":"03:23.734","Text":"If we have a constant,"},{"Start":"03:23.734 ","End":"03:25.460","Text":"it just becomes B,"},{"Start":"03:25.460 ","End":"03:28.385","Text":"the magnetic field, multiplied by S,"},{"Start":"03:28.385 ","End":"03:30.125","Text":"the total surface area."},{"Start":"03:30.125 ","End":"03:32.540","Text":"The magnetic field is At,"},{"Start":"03:32.540 ","End":"03:37.700","Text":"and the surface area is the surface area of a circle of radius a,"},{"Start":"03:37.700 ","End":"03:40.740","Text":"so that\u0027s just Pia^2."},{"Start":"03:42.040 ","End":"03:49.965","Text":"Now we want to calculate the magnetic flux of the internal or induced field."},{"Start":"03:49.965 ","End":"03:56.480","Text":"What we\u0027re going to do is we\u0027re going to remember the equation for inductance,"},{"Start":"03:56.480 ","End":"04:03.905","Text":"which is inductance L is equal to the flux divided by the current."},{"Start":"04:03.905 ","End":"04:09.470","Text":"Remember that inductance is something that isn\u0027t actually dependent on the current,"},{"Start":"04:09.470 ","End":"04:15.490","Text":"it\u0027s just dependent on the geometric shape of the component."},{"Start":"04:15.490 ","End":"04:18.470","Text":"We\u0027re already given inductance in the question,"},{"Start":"04:18.470 ","End":"04:19.640","Text":"so what we can do,"},{"Start":"04:19.640 ","End":"04:22.100","Text":"is we can isolate the flux, of course,"},{"Start":"04:22.100 ","End":"04:24.590","Text":"this is magnetic flux,"},{"Start":"04:24.590 ","End":"04:31.590","Text":"and this is simply equal to L multiplied by I, the current."},{"Start":"04:31.590 ","End":"04:38.090","Text":"Now what we can do is we can add these 2 together and take the time derivative."},{"Start":"04:38.090 ","End":"04:44.375","Text":"We get that EMF is equal to the negative time derivative of the magnetic flux,"},{"Start":"04:44.375 ","End":"04:47.510","Text":"so the negative time derivative of this,"},{"Start":"04:47.510 ","End":"04:49.970","Text":"everything here is a constant aside from t,"},{"Start":"04:49.970 ","End":"04:51.275","Text":"so the t drops,"},{"Start":"04:51.275 ","End":"04:57.800","Text":"so you have APia^2 and then minus the time derivative of this,"},{"Start":"04:57.800 ","End":"05:01.595","Text":"or plus the negative time derivative of this,"},{"Start":"05:01.595 ","End":"05:05.640","Text":"which will give us LI dot."},{"Start":"05:05.770 ","End":"05:10.850","Text":"This is the answer to question number 1 and we\u0027ve already"},{"Start":"05:10.850 ","End":"05:17.175","Text":"seen this type of equation before for the EMF."},{"Start":"05:17.175 ","End":"05:19.425","Text":"That\u0027s the answer to question number 1."},{"Start":"05:19.425 ","End":"05:22.870","Text":"Let\u0027s answer question number 2."},{"Start":"05:23.990 ","End":"05:30.725","Text":"Question number 2 wants us to find an equation for the current as a function of time,"},{"Start":"05:30.725 ","End":"05:32.450","Text":"and then to solve it."},{"Start":"05:32.450 ","End":"05:39.290","Text":"It gives us the clue to use the solution of a coil in a charging circuit."},{"Start":"05:39.290 ","End":"05:43.520","Text":"We want to find the equation for the current as a function of time first of all."},{"Start":"05:43.520 ","End":"05:47.000","Text":"What we have over here from question number 1,"},{"Start":"05:47.000 ","End":"05:50.915","Text":"we have an equation linking the EMF,"},{"Start":"05:50.915 ","End":"05:54.935","Text":"which as we know, is voltage to current."},{"Start":"05:54.935 ","End":"05:58.910","Text":"What we can use is our usual equation that"},{"Start":"05:58.910 ","End":"06:03.586","Text":"voltage is equal to current multiplied by resistance."},{"Start":"06:03.586 ","End":"06:08.300","Text":"Where, of course, the resistance is given to us in the question."},{"Start":"06:08.300 ","End":"06:12.230","Text":"Now let\u0027s sub in everything that we have."},{"Start":"06:12.230 ","End":"06:15.645","Text":"Our EMF is equal to"},{"Start":"06:15.645 ","End":"06:24.180","Text":"negative APia^2 minus Ll dot and this is equal to IR."},{"Start":"06:24.180 ","End":"06:27.290","Text":"Now what we can do is we can rearrange it to have"},{"Start":"06:27.290 ","End":"06:32.180","Text":"the Is together, the currents together."},{"Start":"06:32.180 ","End":"06:34.830","Text":"What we have is"},{"Start":"06:37.690 ","End":"06:44.955","Text":"negative APia^2 is equal to IR plus LI dot."},{"Start":"06:44.955 ","End":"06:53.490","Text":"I just added LI dot to both sides and now let\u0027s call this V_0."},{"Start":"06:53.490 ","End":"06:57.550","Text":"Let\u0027s just give this a random name."},{"Start":"06:57.560 ","End":"07:02.210","Text":"What we\u0027ve seen is that the equation when we\u0027re"},{"Start":"07:02.210 ","End":"07:07.005","Text":"using the equation for coil in a charging circuit,"},{"Start":"07:07.005 ","End":"07:16.570","Text":"we\u0027ve seen that the equation is V_0 is equal to IR plus LI dot,"},{"Start":"07:16.570 ","End":"07:18.170","Text":"which is exactly what we have here,"},{"Start":"07:18.170 ","End":"07:19.805","Text":"just without the minus,"},{"Start":"07:19.805 ","End":"07:23.675","Text":"but we\u0027ll substitute in the minus a little bit later."},{"Start":"07:23.675 ","End":"07:31.220","Text":"We\u0027ve seen that the solution to this equation is just I as a function of t,"},{"Start":"07:31.220 ","End":"07:36.290","Text":"which is equal to V_0 divided by R multiplied by"},{"Start":"07:36.290 ","End":"07:42.350","Text":"1 minus e^negative t divided by tau,"},{"Start":"07:42.350 ","End":"07:47.280","Text":"where tau is equal to L divided by"},{"Start":"07:47.280 ","End":"07:54.305","Text":"R. Now if we substitute in the minus and what V_0 is,"},{"Start":"07:54.305 ","End":"08:00.835","Text":"what we get is that I as a function of time,"},{"Start":"08:00.835 ","End":"08:07.530","Text":"is equal to negative APia^2 divided"},{"Start":"08:07.530 ","End":"08:14.260","Text":"by R multiplied by 1 minus e^negative t,"},{"Start":"08:14.260 ","End":"08:16.710","Text":"and then I\u0027ll put in tau,"},{"Start":"08:16.710 ","End":"08:18.420","Text":"but because it\u0027s 1 divided by tau,"},{"Start":"08:18.420 ","End":"08:26.470","Text":"so we have R divided by L. This is the answer to question number 2."},{"Start":"08:26.470 ","End":"08:30.130","Text":"This is the equation for the current as a function of"},{"Start":"08:30.130 ","End":"08:35.030","Text":"time and this is the solved equation."},{"Start":"08:35.030 ","End":"08:36.955","Text":"With respect to the minus,"},{"Start":"08:36.955 ","End":"08:41.140","Text":"it just gives us the direction of the current."},{"Start":"08:41.140 ","End":"08:43.595","Text":"If we look over here at the bird\u0027s eye view,"},{"Start":"08:43.595 ","End":"08:48.020","Text":"we can see that the z-axis is coming out of the page,"},{"Start":"08:48.020 ","End":"08:50.285","Text":"so if we use the right-hand rule,"},{"Start":"08:50.285 ","End":"08:52.850","Text":"if the z-axis is coming out of the page"},{"Start":"08:52.850 ","End":"09:00.330","Text":"then the positive direction for current would be in this anticlockwise direction."},{"Start":"09:00.330 ","End":"09:03.460","Text":"However, we have this negative current,"},{"Start":"09:03.460 ","End":"09:06.055","Text":"so it\u0027s going in the opposite direction."},{"Start":"09:06.055 ","End":"09:13.075","Text":"In that case, our current is traveling in the clockwise direction."},{"Start":"09:13.075 ","End":"09:15.880","Text":"This we can see also from the right-hand rule,"},{"Start":"09:15.880 ","End":"09:18.556","Text":"but also from Lenz\u0027s law,"},{"Start":"09:18.556 ","End":"09:20.695","Text":"we could have seen this."},{"Start":"09:20.695 ","End":"09:25.150","Text":"Lenz\u0027s law states that the current is going to act in"},{"Start":"09:25.150 ","End":"09:30.475","Text":"the direction as to oppose the change made by this external field."},{"Start":"09:30.475 ","End":"09:34.690","Text":"If the external field is coming out of the page,"},{"Start":"09:34.690 ","End":"09:36.370","Text":"according to Lenz\u0027s law,"},{"Start":"09:36.370 ","End":"09:40.375","Text":"the induced magnetic field is going to be going into the page."},{"Start":"09:40.375 ","End":"09:44.995","Text":"Then, in that case, the direction of current that we would need in"},{"Start":"09:44.995 ","End":"09:49.719","Text":"order to induce this magnetic fields in the opposite direction."},{"Start":"09:49.719 ","End":"09:55.705","Text":"The direction inside the page would be this current going in the negative direction,"},{"Start":"09:55.705 ","End":"09:57.760","Text":"which would be, in this case,"},{"Start":"09:57.760 ","End":"10:01.855","Text":"in the clockwise direction."},{"Start":"10:01.855 ","End":"10:04.630","Text":"That\u0027s the answer to Question Number 2."},{"Start":"10:04.630 ","End":"10:07.955","Text":"Now let\u0027s go on to Question Number 3."},{"Start":"10:07.955 ","End":"10:11.730","Text":"Now let\u0027s answer Question Number 3."},{"Start":"10:11.730 ","End":"10:14.699","Text":"Let\u0027s scroll down."},{"Start":"10:14.699 ","End":"10:19.245","Text":"We\u0027re asked to calculate the current"},{"Start":"10:19.245 ","End":"10:23.880","Text":"and flux as a function of time when R is approaching 0."},{"Start":"10:23.880 ","End":"10:30.745","Text":"When the resistance is approaching 0 and we\u0027re being told to ignore the first moments."},{"Start":"10:30.745 ","End":"10:37.945","Text":"If we look over here and we substitute in 0 for R,"},{"Start":"10:37.945 ","End":"10:41.950","Text":"then what we get is e^-0,"},{"Start":"10:41.950 ","End":"10:43.750","Text":"which is just equal to 1."},{"Start":"10:43.750 ","End":"10:50.290","Text":"Then we get a current of this constant over here multiplied by 1 minus 1, which is 0."},{"Start":"10:50.290 ","End":"10:54.805","Text":"We get a current of 0, which isn\u0027t correct."},{"Start":"10:54.805 ","End":"11:01.015","Text":"What we do when we have a situation like this where our resistance is very small,"},{"Start":"11:01.015 ","End":"11:03.819","Text":"but we know that there is a current."},{"Start":"11:03.819 ","End":"11:07.825","Text":"What we do is we use the Taylor approximations."},{"Start":"11:07.825 ","End":"11:13.120","Text":"One of the Taylor approximations is the approximation for an exponent."},{"Start":"11:13.120 ","End":"11:16.345","Text":"For e to the power of some x,"},{"Start":"11:16.345 ","End":"11:20.210","Text":"when x is very small, it\u0027s approaching 0."},{"Start":"11:21.030 ","End":"11:25.270","Text":"The first term is 1."},{"Start":"11:25.270 ","End":"11:29.020","Text":"We saw that that doesn\u0027t work because then we get a current of 0."},{"Start":"11:29.020 ","End":"11:31.810","Text":"What we have to do is we have to go to the next term,"},{"Start":"11:31.810 ","End":"11:35.900","Text":"so the next order, which is x."},{"Start":"11:36.530 ","End":"11:38.810","Text":"Let\u0027s plug this in."},{"Start":"11:38.810 ","End":"11:46.720","Text":"What we\u0027ll get is that we have this e^-tR divided by L,"},{"Start":"11:46.720 ","End":"11:54.220","Text":"which will then be approximately equal to 1 plus x,"},{"Start":"11:54.220 ","End":"12:01.360","Text":"where x is negative tR divided by L. This is equal to"},{"Start":"12:01.360 ","End":"12:10.105","Text":"1 minus tr divided by l. Then we can plug this in over here in place of this."},{"Start":"12:10.105 ","End":"12:14.890","Text":"What we\u0027ll get is that our current as a function of t is equal to"},{"Start":"12:14.890 ","End":"12:23.019","Text":"negative A_Pi_a^2 divided by R multiplied by 1 minus the exponent."},{"Start":"12:23.019 ","End":"12:27.655","Text":"1 minus 1 minus tR divided by"},{"Start":"12:27.655 ","End":"12:34.735","Text":"L. The ones will cancel out and this minus with this minus becomes a positive."},{"Start":"12:34.735 ","End":"12:38.080","Text":"What we\u0027ll have is that this is equal to"},{"Start":"12:38.080 ","End":"12:45.340","Text":"negative A_Pi_a^2_tR"},{"Start":"12:45.340 ","End":"12:48.910","Text":"divided by RL."},{"Start":"12:48.910 ","End":"12:52.600","Text":"Of course, these R\u0027s will cancel out."},{"Start":"12:52.600 ","End":"12:55.060","Text":"This is our current."},{"Start":"12:55.060 ","End":"12:57.970","Text":"We can just rub this out."},{"Start":"12:57.970 ","End":"13:05.690","Text":"This is our current when R is approaching 0."},{"Start":"13:06.510 ","End":"13:08.890","Text":"This is the current."},{"Start":"13:08.890 ","End":"13:11.200","Text":"That was the first part of this question."},{"Start":"13:11.200 ","End":"13:14.470","Text":"Now we want to find the total flux."},{"Start":"13:14.470 ","End":"13:20.530","Text":"Let\u0027s just scroll up and remember that we wrote this equation for the total flux."},{"Start":"13:20.530 ","End":"13:24.865","Text":"We had that the magnetic flux was equal to"},{"Start":"13:24.865 ","End":"13:29.900","Text":"the external magnetic flux"},{"Start":"13:30.000 ","End":"13:36.860","Text":"plus the internal magnetic flux."},{"Start":"13:37.980 ","End":"13:43.195","Text":"We wrote that over here we have the things."},{"Start":"13:43.195 ","End":"13:50.215","Text":"Our external magnetic flux was equal to A_t_Pi_a^2,"},{"Start":"13:50.215 ","End":"13:52.300","Text":"as we can see from here."},{"Start":"13:52.300 ","End":"13:57.190","Text":"Our internal magnetic flux came from this over here."},{"Start":"13:57.190 ","End":"14:00.860","Text":"It was plus LI."},{"Start":"14:01.890 ","End":"14:05.575","Text":"This is our internal magnetic flux."},{"Start":"14:05.575 ","End":"14:12.310","Text":"Now what I can do is I can plug in my current into this over here."},{"Start":"14:12.310 ","End":"14:16.465","Text":"What I\u0027ll get is that my total magnetic flux is equal to"},{"Start":"14:16.465 ","End":"14:24.265","Text":"A_t_Pi_a^2 plus L multiplied by I,"},{"Start":"14:24.265 ","End":"14:32.860","Text":"which is negative A_Pi_a^2_t divided by L, from here."},{"Start":"14:32.860 ","End":"14:41.965","Text":"This L and this L cancel out and then I have A_t_Pi_a^2."},{"Start":"14:41.965 ","End":"14:44.328","Text":"Here are positive and here negative."},{"Start":"14:44.328 ","End":"14:48.145","Text":"In terms of magnetic flux is equal to 0."},{"Start":"14:48.145 ","End":"14:52.134","Text":"This is the answer to the total flux."},{"Start":"14:52.134 ","End":"14:54.790","Text":"These 2 are the answers to Question Number 3."},{"Start":"14:54.790 ","End":"14:58.840","Text":"Now, why do we get this value of 0?"},{"Start":"14:58.840 ","End":"15:01.705","Text":"Notice that if there\u0027s no resistance,"},{"Start":"15:01.705 ","End":"15:04.840","Text":"here the resistance is approaching 0,"},{"Start":"15:04.840 ","End":"15:10.330","Text":"then that means that the current that we\u0027re going to have is simply equal"},{"Start":"15:10.330 ","End":"15:15.685","Text":"to the magnetic flux of the external field."},{"Start":"15:15.685 ","End":"15:23.600","Text":"As we saw before, the magnetic flux of the external field divided by the inductance."},{"Start":"15:24.000 ","End":"15:29.769","Text":"In other words, the current is changing in relation"},{"Start":"15:29.769 ","End":"15:35.570","Text":"to the change of the external magnetic flux."},{"Start":"15:35.700 ","End":"15:38.079","Text":"When we have resistance,"},{"Start":"15:38.079 ","End":"15:45.639","Text":"there\u0027s some delay in the current and the change in the magnetic flux."},{"Start":"15:45.639 ","End":"15:48.640","Text":"However, when there\u0027s no resistance,"},{"Start":"15:48.640 ","End":"15:52.570","Text":"then there\u0027s no delay in the current catching up with"},{"Start":"15:52.570 ","End":"15:59.330","Text":"this change in magnetic flux of the external magnetic field."},{"Start":"15:59.760 ","End":"16:02.529","Text":"Then at any given moment,"},{"Start":"16:02.529 ","End":"16:05.665","Text":"we get that this equation is correct."},{"Start":"16:05.665 ","End":"16:10.105","Text":"The total magnetic flux is equal to 0."},{"Start":"16:10.105 ","End":"16:17.260","Text":"We get the same magnetic flux coming out of the page as we have going into the page."},{"Start":"16:17.260 ","End":"16:21.200","Text":"That is the end of this lesson."}],"ID":22378},{"Watched":false,"Name":"RL Circuit","Duration":"24m 39s","ChapterTopicVideoID":21355,"CourseChapterTopicPlaylistID":99483,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello. In this lesson,"},{"Start":"00:02.055 ","End":"00:07.980","Text":"we\u0027re going to be learning about this circuit that contains a resistor and an inductor."},{"Start":"00:07.980 ","End":"00:12.765","Text":"Over here, we have a circuit with a voltage source of V_naught,"},{"Start":"00:12.765 ","End":"00:16.274","Text":"a resistor with resistance R, and an inductor,"},{"Start":"00:16.274 ","End":"00:20.190","Text":"so a coil with inductance of L. Here,"},{"Start":"00:20.190 ","End":"00:27.045","Text":"we have the switch S. Let\u0027s imagine that we shut this switch,"},{"Start":"00:27.045 ","End":"00:33.560","Text":"and now what we want to know is what is going on over here inside the circuit."},{"Start":"00:33.560 ","End":"00:36.344","Text":"At t is equal to naught,"},{"Start":"00:36.344 ","End":"00:39.000","Text":"let\u0027s imagine that we shut the switch."},{"Start":"00:39.000 ","End":"00:40.745","Text":"At this exact moment,"},{"Start":"00:40.745 ","End":"00:45.745","Text":"we can assume that the current at t is equal to naught,"},{"Start":"00:45.745 ","End":"00:51.090","Text":"is also going to be equal to naught at that exact moment."},{"Start":"00:51.530 ","End":"00:55.250","Text":"Now we want to see what is going on in the circuit,"},{"Start":"00:55.250 ","End":"00:58.250","Text":"so we\u0027ll write out an equation for the voltage."},{"Start":"00:58.250 ","End":"01:00.170","Text":"We start at this point over here,"},{"Start":"01:00.170 ","End":"01:05.460","Text":"and then we go up over here in a positive directions."},{"Start":"01:05.460 ","End":"01:09.135","Text":"We\u0027re adding on this voltage of V_naught."},{"Start":"01:09.135 ","End":"01:13.025","Text":"Then we carry on over here and then we go past the resistor."},{"Start":"01:13.025 ","End":"01:18.515","Text":"Let\u0027s say that this is the at direction where we have a voltage drops."},{"Start":"01:18.515 ","End":"01:20.780","Text":"We have negative IR,"},{"Start":"01:20.780 ","End":"01:22.220","Text":"where IR is of course,"},{"Start":"01:22.220 ","End":"01:24.185","Text":"the voltage across the resistor,"},{"Start":"01:24.185 ","End":"01:28.265","Text":"and then we go down the inductor."},{"Start":"01:28.265 ","End":"01:31.663","Text":"Again, we have a voltage drop over the inductor,"},{"Start":"01:31.663 ","End":"01:36.950","Text":"and the voltage over an inductor is equal to L multiplied by I dot."},{"Start":"01:36.950 ","End":"01:40.595","Text":"Then we continue and we end up at the same point."},{"Start":"01:40.595 ","End":"01:44.135","Text":"We\u0027ve just completed a full circuit and so of course,"},{"Start":"01:44.135 ","End":"01:46.560","Text":"this is equal to 0."},{"Start":"01:47.900 ","End":"01:54.615","Text":"Here we have an equation for the voltage in this circuit."},{"Start":"01:54.615 ","End":"01:57.409","Text":"What we\u0027re left with is this differential equation,"},{"Start":"01:57.409 ","End":"02:01.580","Text":"which at the end of the lesson I\u0027ll show how to solve but in the meantime,"},{"Start":"02:01.580 ","End":"02:06.395","Text":"I\u0027m just going to write out the solution just so that we can talk about that."},{"Start":"02:06.395 ","End":"02:07.513","Text":"We have that,"},{"Start":"02:07.513 ","End":"02:09.755","Text":"the current as a function of time,"},{"Start":"02:09.755 ","End":"02:13.253","Text":"is equal to V_naught divided by R,"},{"Start":"02:13.253 ","End":"02:21.200","Text":"multiplied by 1 minus e^negative t divided by Tau,"},{"Start":"02:21.200 ","End":"02:28.130","Text":"where Tau is equal to L divided R. Now,"},{"Start":"02:28.130 ","End":"02:32.690","Text":"let\u0027s take a look at what the solution means graphically."},{"Start":"02:32.920 ","End":"02:35.780","Text":"If we look at this graph over here,"},{"Start":"02:35.780 ","End":"02:39.890","Text":"we have a graph of current versus time."},{"Start":"02:39.890 ","End":"02:45.755","Text":"What we can see is that the current starts off at t equal to 0,"},{"Start":"02:45.755 ","End":"02:47.990","Text":"which is also what we expected."},{"Start":"02:47.990 ","End":"02:53.720","Text":"Then slowly, it increases up until it"},{"Start":"02:53.720 ","End":"02:59.840","Text":"reaches or at least it\u0027s approaching this limit over here of V_naught divided by R,"},{"Start":"02:59.840 ","End":"03:01.840","Text":"which is of course, a constant."},{"Start":"03:01.840 ","End":"03:09.730","Text":"It\u0027s slowly approaching and we can see that from this equation over here."},{"Start":"03:09.730 ","End":"03:13.160","Text":"This is of course the graph with our coil."},{"Start":"03:13.160 ","End":"03:18.597","Text":"Now, what happened if we had this second without the coil, without an inductor?"},{"Start":"03:18.597 ","End":"03:22.375","Text":"This is what the graph would look like."},{"Start":"03:22.375 ","End":"03:25.590","Text":"I added that there\u0027s no inductor."},{"Start":"03:25.590 ","End":"03:28.085","Text":"If we took out the inductor,"},{"Start":"03:28.085 ","End":"03:31.130","Text":"before we flip the switch,"},{"Start":"03:31.130 ","End":"03:33.875","Text":"we would have this 0 current."},{"Start":"03:33.875 ","End":"03:36.440","Text":"Then at the exact moment we flip the current,"},{"Start":"03:36.440 ","End":"03:41.420","Text":"we would have this sudden jump in the current from 0 until"},{"Start":"03:41.420 ","End":"03:47.150","Text":"this constant value of V_naught divided by R. This of course,"},{"Start":"03:47.150 ","End":"03:51.890","Text":"is something that we don\u0027t want because we don\u0027t like having these jumps."},{"Start":"03:51.890 ","End":"03:54.950","Text":"What the coil does or what an inductor does,"},{"Start":"03:54.950 ","End":"04:03.365","Text":"is it slowly increases the current up until it gets to its final constant value."},{"Start":"04:03.365 ","End":"04:07.100","Text":"Oftentimes, people will say that an inductor or"},{"Start":"04:07.100 ","End":"04:12.605","Text":"the coil acts like an open and a short circuit."},{"Start":"04:12.605 ","End":"04:17.500","Text":"Right at the beginning is the second that the switch is flipped,"},{"Start":"04:17.500 ","End":"04:20.509","Text":"so we have this complete circuit."},{"Start":"04:20.509 ","End":"04:26.555","Text":"However, we can see that the current at that exact moment is still equal to 0."},{"Start":"04:26.555 ","End":"04:28.265","Text":"Right at the beginning,"},{"Start":"04:28.265 ","End":"04:33.780","Text":"you can say that an inductor acts like an open circuit."},{"Start":"04:34.010 ","End":"04:36.845","Text":"Later, or much later,"},{"Start":"04:36.845 ","End":"04:38.960","Text":"after a certain period of time,"},{"Start":"04:38.960 ","End":"04:43.355","Text":"the inductor acts like a short circuit."},{"Start":"04:43.355 ","End":"04:48.755","Text":"What we can see is that in a circuit without the inductor or without a coil,"},{"Start":"04:48.755 ","End":"04:51.490","Text":"we have this big change in the current."},{"Start":"04:51.490 ","End":"04:55.925","Text":"We go from a current being equal to 0 to a current of V_naught"},{"Start":"04:55.925 ","End":"05:01.265","Text":"divided by R. What we can see is,"},{"Start":"05:01.265 ","End":"05:03.980","Text":"as we know with inductors,"},{"Start":"05:03.980 ","End":"05:08.705","Text":"what they do is they act to oppose the change in current."},{"Start":"05:08.705 ","End":"05:12.260","Text":"When we flip the switch we have, just as we saw,"},{"Start":"05:12.260 ","End":"05:16.156","Text":"this big change in current."},{"Start":"05:16.156 ","End":"05:20.410","Text":"So the inductor is acting to resist this giant leap."},{"Start":"05:20.410 ","End":"05:23.225","Text":"It resists it, which means that we can see that we have"},{"Start":"05:23.225 ","End":"05:28.190","Text":"a much slower change in the current."},{"Start":"05:28.190 ","End":"05:31.055","Text":"Eventually, after a certain amount of time,"},{"Start":"05:31.055 ","End":"05:34.043","Text":"the circuit will act,"},{"Start":"05:34.043 ","End":"05:35.228","Text":"so the second with the coil,"},{"Start":"05:35.228 ","End":"05:40.010","Text":"with the inductor, will act like a circuit without an inductor."},{"Start":"05:40.010 ","End":"05:41.450","Text":"After a certain amount of time,"},{"Start":"05:41.450 ","End":"05:45.169","Text":"we can see that the current rarely does approach"},{"Start":"05:45.169 ","End":"05:49.130","Text":"this maximal value of V_naught divided by I and we get"},{"Start":"05:49.130 ","End":"05:53.720","Text":"the exact same current after a certain amount of time as"},{"Start":"05:53.720 ","End":"05:59.400","Text":"we would if we had the same circuit just without the inductor."},{"Start":"05:59.600 ","End":"06:02.375","Text":"Another way to explain this is,"},{"Start":"06:02.375 ","End":"06:04.775","Text":"as we can see at the beginning,"},{"Start":"06:04.775 ","End":"06:08.180","Text":"this circuit over here with the coil,"},{"Start":"06:08.180 ","End":"06:13.775","Text":"we can see that the change in current as a factor of time right at the beginning,"},{"Start":"06:13.775 ","End":"06:15.605","Text":"it\u0027s a very great change."},{"Start":"06:15.605 ","End":"06:19.400","Text":"We can say that di by dt is very big."},{"Start":"06:19.400 ","End":"06:22.918","Text":"We have this very steep gradient over here,"},{"Start":"06:22.918 ","End":"06:29.690","Text":"and it gradually evens out until we get to this value over here"},{"Start":"06:29.690 ","End":"06:33.170","Text":"where we\u0027re approaching V_naught divided by R. Here we can see"},{"Start":"06:33.170 ","End":"06:37.265","Text":"that the gradient of the curve in this region,"},{"Start":"06:37.265 ","End":"06:42.365","Text":"or in other words, di by dt is very small."},{"Start":"06:42.365 ","End":"06:46.280","Text":"We have a very, very small change in current with"},{"Start":"06:46.280 ","End":"06:50.825","Text":"respect to a relatively large change in time."},{"Start":"06:50.825 ","End":"06:53.435","Text":"In other words, in this region,"},{"Start":"06:53.435 ","End":"06:57.140","Text":"our di by dt is very small."},{"Start":"06:57.140 ","End":"06:59.690","Text":"What is di by dt?"},{"Start":"06:59.690 ","End":"07:03.245","Text":"It\u0027s the time derivative of the current."},{"Start":"07:03.245 ","End":"07:05.970","Text":"It\u0027s I dot. Over here,"},{"Start":"07:05.970 ","End":"07:07.385","Text":"we can say we have I dot."},{"Start":"07:07.385 ","End":"07:09.950","Text":"This is di by d_t."},{"Start":"07:09.950 ","End":"07:16.890","Text":"From this equation, which of course describes this circuit with the inductor,"},{"Start":"07:16.890 ","End":"07:18.755","Text":"we could see that at the beginning,"},{"Start":"07:18.755 ","End":"07:20.990","Text":"when t is relatively small,"},{"Start":"07:20.990 ","End":"07:27.965","Text":"we have this equation of irregular resistor circuit minus L,"},{"Start":"07:27.965 ","End":"07:32.465","Text":"the inductance, multiplied by this di by dt,"},{"Start":"07:32.465 ","End":"07:33.695","Text":"this gradient over here,"},{"Start":"07:33.695 ","End":"07:36.230","Text":"which here we can see is very large."},{"Start":"07:36.230 ","End":"07:39.850","Text":"It\u0027s very steep but as time goes by,"},{"Start":"07:39.850 ","End":"07:46.245","Text":"our I dot or our di by dt in this region is so small that it\u0027s approaching 0."},{"Start":"07:46.245 ","End":"07:48.575","Text":"The gradient is almost flat."},{"Start":"07:48.575 ","End":"07:50.265","Text":"It\u0027s almost a flat line."},{"Start":"07:50.265 ","End":"07:57.445","Text":"Then we have minus L multiplied by a very small value which is approaching 0."},{"Start":"07:57.445 ","End":"08:00.165","Text":"If we want to,"},{"Start":"08:00.165 ","End":"08:02.270","Text":"after a very large amount of time,"},{"Start":"08:02.270 ","End":"08:07.160","Text":"we can even discount this expression over here, this Ll dot."},{"Start":"08:07.160 ","End":"08:14.570","Text":"Then we get V_naught minus IR is equal to 0 after a long period of time."},{"Start":"08:14.570 ","End":"08:19.460","Text":"Then we can see really that this equation is very similar"},{"Start":"08:19.460 ","End":"08:25.254","Text":"to the equation of the circuit where we\u0027re to have no coil."},{"Start":"08:25.254 ","End":"08:30.475","Text":"This is of course a charging circuit."},{"Start":"08:30.475 ","End":"08:34.780","Text":"Here we can see that our coil is charging."},{"Start":"08:34.780 ","End":"08:39.940","Text":"Before I start doing this equation,"},{"Start":"08:39.940 ","End":"08:44.725","Text":"solving this differential equation in order to show how we get to the solution,"},{"Start":"08:44.725 ","End":"08:47.155","Text":"this I will do it later in this lesson."},{"Start":"08:47.155 ","End":"08:51.800","Text":"First, I want to speak about the discharge."},{"Start":"08:51.960 ","End":"08:55.870","Text":"Here we saw how we close the switch and then"},{"Start":"08:55.870 ","End":"09:00.745","Text":"this battery is charging up our coil or our inductor."},{"Start":"09:00.745 ","End":"09:04.060","Text":"Now what we want to do is we want to pretend that our inductor was now"},{"Start":"09:04.060 ","End":"09:07.690","Text":"fully charged and now we take away the battery."},{"Start":"09:07.690 ","End":"09:13.630","Text":"You could short circuit this by adding another wire over here so that it\u0027s"},{"Start":"09:13.630 ","End":"09:16.570","Text":"short-circuited and then here you have a switch to open"},{"Start":"09:16.570 ","End":"09:19.728","Text":"it so then the battery just isn\u0027t connected,"},{"Start":"09:19.728 ","End":"09:21.595","Text":"or just to make this a little bit simpler,"},{"Start":"09:21.595 ","End":"09:26.755","Text":"just imagine that there\u0027s no battery and this is just a wire going through."},{"Start":"09:26.755 ","End":"09:30.740","Text":"Of course, our inductor is fully charged."},{"Start":"09:31.140 ","End":"09:34.060","Text":"Now we\u0027re speaking about discharging."},{"Start":"09:34.060 ","End":"09:41.575","Text":"In the charging, we spoke about the initial current I at t is equal to 0,"},{"Start":"09:41.575 ","End":"09:47.185","Text":"exactly when we flip the switch and we said that the current was equal to 0."},{"Start":"09:47.185 ","End":"09:49.465","Text":"Now if we\u0027re dealing with discharging,"},{"Start":"09:49.465 ","End":"09:53.650","Text":"that means that we\u0027ve reached the maximum current,"},{"Start":"09:53.650 ","End":"09:56.320","Text":"so our coil is fully charged."},{"Start":"09:56.320 ","End":"09:59.890","Text":"That means that our I at t is equal to naught,"},{"Start":"09:59.890 ","End":"10:05.245","Text":"where now we\u0027re saying that t is equal to naught is at the moment we open the switch"},{"Start":"10:05.245 ","End":"10:11.140","Text":"up again or rather not open up the switch,"},{"Start":"10:11.140 ","End":"10:15.890","Text":"t is equal to naught is the second we remove the battery."},{"Start":"10:16.800 ","End":"10:19.090","Text":"When the battery is removed,"},{"Start":"10:19.090 ","End":"10:20.980","Text":"this is now t is equal to naught,"},{"Start":"10:20.980 ","End":"10:22.900","Text":"this situation over here."},{"Start":"10:22.900 ","End":"10:25.975","Text":"As we said, the current at this moment,"},{"Start":"10:25.975 ","End":"10:29.410","Text":"we\u0027re dealing with a fully charged inductor."},{"Start":"10:29.410 ","End":"10:32.080","Text":"That means that the current is at its maxima value,"},{"Start":"10:32.080 ","End":"10:37.015","Text":"which as we saw its maxima value is this over here,"},{"Start":"10:37.015 ","End":"10:39.160","Text":"which is over here,"},{"Start":"10:39.160 ","End":"10:49.040","Text":"V_naught divided by R. Then we can also call this just I naught, the initial current."},{"Start":"10:49.410 ","End":"10:55.195","Text":"Now what we want to do is we want to see what its equation is."},{"Start":"10:55.195 ","End":"10:59.290","Text":"What is here in this red box is the equation for"},{"Start":"10:59.290 ","End":"11:07.285","Text":"the current for a charging inductor but now we\u0027re dealing with a discharging inductor,"},{"Start":"11:07.285 ","End":"11:10.675","Text":"so we want to calculate the solution."},{"Start":"11:10.675 ","End":"11:16.450","Text":"What we have is the same circuit just without the battery."},{"Start":"11:16.450 ","End":"11:18.715","Text":"We can look at this equation over here,"},{"Start":"11:18.715 ","End":"11:22.345","Text":"it\u0027s the same equation just without the battery, without the V_naught."},{"Start":"11:22.345 ","End":"11:26.185","Text":"What we have is negative IR,"},{"Start":"11:26.185 ","End":"11:32.665","Text":"negative LI dot is equal to 0."},{"Start":"11:32.665 ","End":"11:35.710","Text":"Now what I can do is I can isolate out"},{"Start":"11:35.710 ","End":"11:41.395","Text":"my I and obviously solve this differential equation."},{"Start":"11:41.395 ","End":"11:43.990","Text":"Which once I solve this differential equation,"},{"Start":"11:43.990 ","End":"11:45.850","Text":"I\u0027ll get that my I, my current,"},{"Start":"11:45.850 ","End":"11:49.975","Text":"as a function of time is equal to V_naught divided by"},{"Start":"11:49.975 ","End":"11:55.510","Text":"R e to the power of negative t divided by Tau,"},{"Start":"11:55.510 ","End":"11:58.435","Text":"again, where Tau is this, it\u0027s L,"},{"Start":"11:58.435 ","End":"12:02.780","Text":"the inductance divided by R, the resistance."},{"Start":"12:03.420 ","End":"12:08.875","Text":"Again, this is the equation for the charging and this is the equation"},{"Start":"12:08.875 ","End":"12:14.605","Text":"for the discharging where the Tau is over here are the same thing."},{"Start":"12:14.605 ","End":"12:20.740","Text":"Now let\u0027s take a look at what this would look like on the graph."},{"Start":"12:20.740 ","End":"12:30.445","Text":"This dotted line represents the moment that we take the battery out."},{"Start":"12:30.445 ","End":"12:36.220","Text":"What we can see is that the graph is going to go down"},{"Start":"12:36.220 ","End":"12:42.445","Text":"like this exponentially with respect to time."},{"Start":"12:42.445 ","End":"12:50.035","Text":"This is the continuation of the time axis and this is what the graph will look like."},{"Start":"12:50.035 ","End":"12:53.800","Text":"Then if we look at the same graph over here,"},{"Start":"12:53.800 ","End":"12:57.835","Text":"if we were dealing with a circuit without the inductor,"},{"Start":"12:57.835 ","End":"13:01.390","Text":"then the current would go from its maximum value till"},{"Start":"13:01.390 ","End":"13:05.960","Text":"its minimum value immediately in 1 jump."},{"Start":"13:08.010 ","End":"13:12.175","Text":"Now we can see what happens with the discharge."},{"Start":"13:12.175 ","End":"13:18.565","Text":"What\u0027s important to note is that here when dealing with an inductor, the current,"},{"Start":"13:18.565 ","End":"13:21.670","Text":"both charging and discharging acts like"},{"Start":"13:21.670 ","End":"13:28.255","Text":"the charge when charging and discharging a capacitor."},{"Start":"13:28.255 ","End":"13:30.790","Text":"When we\u0027re dealing with an inductor,"},{"Start":"13:30.790 ","End":"13:34.240","Text":"we have a charge and discharge of current,"},{"Start":"13:34.240 ","End":"13:39.130","Text":"which is identical to dealing with a capacitor when we have the charge"},{"Start":"13:39.130 ","End":"13:44.900","Text":"and discharge of the charge itself."},{"Start":"13:45.420 ","End":"13:49.195","Text":"Now as promised, I\u0027m going to solve"},{"Start":"13:49.195 ","End":"13:53.875","Text":"the differential equation to see how we got from here to this solution."},{"Start":"13:53.875 ","End":"13:55.450","Text":"If you already know how to do that,"},{"Start":"13:55.450 ","End":"13:58.285","Text":"you can end the lesson now and if not,"},{"Start":"13:58.285 ","End":"14:00.830","Text":"you can carry on watching."},{"Start":"14:01.770 ","End":"14:05.245","Text":"Now we\u0027re solving this differential equation."},{"Start":"14:05.245 ","End":"14:06.940","Text":"Let\u0027s write it out."},{"Start":"14:06.940 ","End":"14:09.925","Text":"Now, when we\u0027re doing the differential equation,"},{"Start":"14:09.925 ","End":"14:11.935","Text":"what we want is,"},{"Start":"14:11.935 ","End":"14:17.650","Text":"as we can see we have I and then over here we have dI by dt."},{"Start":"14:17.650 ","End":"14:19.960","Text":"First of all, actually, let\u0027s just write out"},{"Start":"14:19.960 ","End":"14:23.905","Text":"the equation making the I dot into dI by dt."},{"Start":"14:23.905 ","End":"14:27.520","Text":"This goes for everything if you have x dot,"},{"Start":"14:27.520 ","End":"14:30.070","Text":"then do it dx by dt and so on."},{"Start":"14:30.070 ","End":"14:32.080","Text":"It will make it much easier to solve."},{"Start":"14:32.080 ","End":"14:38.290","Text":"You have V_naught minus IR minus"},{"Start":"14:38.290 ","End":"14:45.355","Text":"L multiplied by dI by dt and this is equal to 0."},{"Start":"14:45.355 ","End":"14:48.235","Text":"As we can see, we have dI by dt."},{"Start":"14:48.235 ","End":"14:51.730","Text":"That means that we\u0027re going to want an equation of"},{"Start":"14:51.730 ","End":"14:57.040","Text":"f with respect to t because we\u0027re looking at with respect to the change in time."},{"Start":"14:57.040 ","End":"15:03.505","Text":"This is going to have to be equal to dt on one side,"},{"Start":"15:03.505 ","End":"15:07.180","Text":"which is going to be equal to f with respect to I,"},{"Start":"15:07.180 ","End":"15:10.690","Text":"so we\u0027re looking also at the change in I, dI."},{"Start":"15:10.690 ","End":"15:18.130","Text":"This is the pattern or the shape of the equation that we\u0027re looking for."},{"Start":"15:18.130 ","End":"15:22.660","Text":"In that case, we can learn from here that I want my dt on one side and"},{"Start":"15:22.660 ","End":"15:28.195","Text":"my dI with all the I variables on the other."},{"Start":"15:28.195 ","End":"15:32.320","Text":"All my t\u0027s on one side and all my I\u0027s on the other."},{"Start":"15:32.320 ","End":"15:38.305","Text":"Let\u0027s first move this expression over to the other side of the equation."},{"Start":"15:38.305 ","End":"15:48.170","Text":"I have V_naught minus IR is equal to L dI by dt."},{"Start":"15:48.540 ","End":"15:54.370","Text":"Now what I want to do is I want to get rid of my t from this side of the equation."},{"Start":"15:54.370 ","End":"15:56.740","Text":"I\u0027ll multiply both sides by dt."},{"Start":"15:56.740 ","End":"16:05.650","Text":"That means I\u0027ll have V_naught minus IR multiplied by dt and this is equal to LdI."},{"Start":"16:05.650 ","End":"16:08.095","Text":"This side is great,"},{"Start":"16:08.095 ","End":"16:09.400","Text":"but on this side,"},{"Start":"16:09.400 ","End":"16:12.340","Text":"my dt side, I have this I."},{"Start":"16:12.340 ","End":"16:17.140","Text":"Because my I is also a variable,"},{"Start":"16:17.140 ","End":"16:20.215","Text":"I\u0027m looking at how I changes with respect to time."},{"Start":"16:20.215 ","End":"16:22.510","Text":"I need to get rid of my I over here."},{"Start":"16:22.510 ","End":"16:25.760","Text":"I need my I on this side of the equation."},{"Start":"16:25.830 ","End":"16:30.955","Text":"In that case, I\u0027m going to divide both sides by what\u0027s in the brackets."},{"Start":"16:30.955 ","End":"16:36.655","Text":"What I\u0027m going to get is I\u0027m going to get the dt is equal to L divided by"},{"Start":"16:36.655 ","End":"16:43.240","Text":"V_naught minus IR dI."},{"Start":"16:43.240 ","End":"16:45.790","Text":"This is great. Now because I have dt and dI,"},{"Start":"16:45.790 ","End":"16:48.550","Text":"I of course I have to integrate."},{"Start":"16:48.550 ","End":"16:50.095","Text":"I\u0027ll add in my integrals."},{"Start":"16:50.095 ","End":"16:55.885","Text":"We can do this integral as bound or non-bound integrals."},{"Start":"16:55.885 ","End":"16:58.165","Text":"Let\u0027s put in the bounds."},{"Start":"16:58.165 ","End":"17:07.160","Text":"For t, I\u0027m integrating from t is equal to naught until some general time t. Then I\u0027m"},{"Start":"17:07.160 ","End":"17:11.210","Text":"looking at my current at the time t is equal to naught"},{"Start":"17:11.210 ","End":"17:16.550","Text":"and as I saw that current at the beginning is equal to 0."},{"Start":"17:16.550 ","End":"17:18.796","Text":"That\u0027s my initial condition."},{"Start":"17:18.796 ","End":"17:24.365","Text":"Then I\u0027m integrating until I get I at some general time t,"},{"Start":"17:24.365 ","End":"17:30.073","Text":"where this t and this t obviously correspond."},{"Start":"17:30.073 ","End":"17:32.080","Text":"Let\u0027s integrate."},{"Start":"17:32.080 ","End":"17:37.615","Text":"Here I have dt and if I integrate and then plug in t and 0,"},{"Start":"17:37.615 ","End":"17:40.050","Text":"I\u0027ll get t minus 0,"},{"Start":"17:40.050 ","End":"17:45.335","Text":"which is just t. So I get t on this side and this is equal to."},{"Start":"17:45.335 ","End":"17:49.840","Text":"Over here, what I have is I\u0027m integrating with respect"},{"Start":"17:49.840 ","End":"17:54.175","Text":"to I and I have over here I in my denominator."},{"Start":"17:54.175 ","End":"18:02.080","Text":"First of all, I\u0027m going to have a minus and where does the minus come from?"},{"Start":"18:02.080 ","End":"18:05.080","Text":"It comes from this over here."},{"Start":"18:05.080 ","End":"18:08.365","Text":"I have V_naught minus IR,"},{"Start":"18:08.365 ","End":"18:12.340","Text":"or in other words, V_naught plus negative IR."},{"Start":"18:12.340 ","End":"18:14.964","Text":"My coefficient for I is a negative,"},{"Start":"18:14.964 ","End":"18:18.025","Text":"so I have a negative over here."},{"Start":"18:18.025 ","End":"18:25.629","Text":"Then of course, because I have my variable in the denominator,"},{"Start":"18:25.629 ","End":"18:29.990","Text":"I know I\u0027m going to have to be dealing with ln."},{"Start":"18:30.570 ","End":"18:32.935","Text":"When it\u0027s in this shape,"},{"Start":"18:32.935 ","End":"18:37.795","Text":"unless it was I^2 or something like that but if it\u0027s just I,"},{"Start":"18:37.795 ","End":"18:40.555","Text":"then it\u0027s going to be ln."},{"Start":"18:40.555 ","End":"18:42.880","Text":"Then my other coefficient of I is,"},{"Start":"18:42.880 ","End":"18:44.500","Text":"of course, R where R,"},{"Start":"18:44.500 ","End":"18:46.285","Text":"of course, goes in the denominator,"},{"Start":"18:46.285 ","End":"18:48.655","Text":"and here I have my L in the numerator."},{"Start":"18:48.655 ","End":"18:52.465","Text":"I have negative L divided by R with a negative 1 divided by"},{"Start":"18:52.465 ","End":"18:57.650","Text":"I comes from being the coefficient of I in the denominator over here."},{"Start":"18:57.900 ","End":"19:07.750","Text":"Then this is multiplied by ln so that I have V_naught minus IR,"},{"Start":"19:07.750 ","End":"19:11.830","Text":"and then I plug in I(t) minus naught."},{"Start":"19:11.830 ","End":"19:14.680","Text":"What I\u0027ll have is V_naught"},{"Start":"19:14.680 ","End":"19:23.800","Text":"minus I(t)R and then"},{"Start":"19:23.800 ","End":"19:25.810","Text":"divided from law of ln,"},{"Start":"19:25.810 ","End":"19:30.145","Text":"divided by V_naught minus 0 times I,"},{"Start":"19:30.145 ","End":"19:31.765","Text":"which is just V_naught."},{"Start":"19:31.765 ","End":"19:38.680","Text":"What I could\u0027ve done is either ln of V_naught and then plug"},{"Start":"19:38.680 ","End":"19:42.489","Text":"in K minus ItR"},{"Start":"19:42.489 ","End":"19:49.600","Text":"minus ln of V_naught minus 0 or through law of ln,"},{"Start":"19:49.600 ","End":"19:54.935","Text":"I can just write it like so as a numerator and denominator."},{"Start":"19:54.935 ","End":"19:57.255","Text":"If you\u0027re not familiar with the law of ln\u0027s,"},{"Start":"19:57.255 ","End":"19:59.235","Text":"please look that up."},{"Start":"19:59.235 ","End":"20:02.295","Text":"Now what I\u0027m going to do is I\u0027m going to divide"},{"Start":"20:02.295 ","End":"20:05.940","Text":"both sides by this negative L divided by R,"},{"Start":"20:05.940 ","End":"20:14.620","Text":"so I\u0027m going to have negative Rt divided by L. This is of course equal to"},{"Start":"20:14.620 ","End":"20:23.365","Text":"ln of V_naught minus I as a function (t) R divided by V_naught."},{"Start":"20:23.365 ","End":"20:26.290","Text":"Now, in order to get rid of this ln,"},{"Start":"20:26.290 ","End":"20:31.090","Text":"I\u0027m going to raise both sides by e. I\u0027m going"},{"Start":"20:31.090 ","End":"20:37.420","Text":"to have e to the power of negative Rt divided by L,"},{"Start":"20:37.420 ","End":"20:43.030","Text":"which is equal to E of ln of V_naught minus"},{"Start":"20:43.030 ","End":"20:48.772","Text":"I as a function (t)R divided by V_naught."},{"Start":"20:48.772 ","End":"20:51.820","Text":"Of course, if I have e to the power of ln,"},{"Start":"20:51.820 ","End":"20:55.930","Text":"so it just gets rid of both of these and then I\u0027m just left with"},{"Start":"20:55.930 ","End":"21:03.010","Text":"V_naught minus I as a function (t) R divided by V_naught."},{"Start":"21:03.010 ","End":"21:06.190","Text":"Now, of course, what I want to do is I want to isolate my"},{"Start":"21:06.190 ","End":"21:12.820","Text":"I(t) because I want to see how my current changes as a function of time."},{"Start":"21:12.820 ","End":"21:15.820","Text":"I can multiply both sides by V_naught,"},{"Start":"21:15.820 ","End":"21:22.495","Text":"so I\u0027ll have V_naught e to the power of negative Rt divided by L,"},{"Start":"21:22.495 ","End":"21:26.140","Text":"which is equal to V_naught minus"},{"Start":"21:26.140 ","End":"21:33.805","Text":"I(t)R. Now I can subtract V_naught from both sides or rather,"},{"Start":"21:33.805 ","End":"21:38.680","Text":"I can add I as a function of I to both sides and subtract V_naught e to the power of"},{"Start":"21:38.680 ","End":"21:45.085","Text":"negative Rt divided by L. What I\u0027ll have is I(t),"},{"Start":"21:45.085 ","End":"21:50.320","Text":"is equal to V_naught minus V_naught e to"},{"Start":"21:50.320 ","End":"21:58.240","Text":"the negative Rt divided by L. This is,"},{"Start":"21:58.240 ","End":"21:59.905","Text":"of course, multiplied by R,"},{"Start":"21:59.905 ","End":"22:06.835","Text":"so now I can divide both sides by R. What I\u0027ll have is I(t),"},{"Start":"22:06.835 ","End":"22:10.165","Text":"is equal to V_naught divided by R,"},{"Start":"22:10.165 ","End":"22:13.150","Text":"and then we can do brackets because we have like terms."},{"Start":"22:13.150 ","End":"22:19.630","Text":"We have 1 minus e to the negative R divided by L t,"},{"Start":"22:19.630 ","End":"22:21.730","Text":"so let\u0027s just write it like that."},{"Start":"22:21.730 ","End":"22:32.095","Text":"Then R divided by L. Let\u0027s say that R divided by L is equal to 1 divided by Tau,"},{"Start":"22:32.095 ","End":"22:40.480","Text":"therefore, we get the Tau is equal to L divided by R. Then we just plug that in,"},{"Start":"22:40.480 ","End":"22:43.360","Text":"so we get that I(t),"},{"Start":"22:43.360 ","End":"22:49.570","Text":"is equal to V_naught divided by R multiplied by 1 minus e to"},{"Start":"22:49.570 ","End":"22:56.005","Text":"the power of negative R divided by L. That\u0027s 1 over Tau,"},{"Start":"22:56.005 ","End":"23:00.115","Text":"so t divided by Tau,"},{"Start":"23:00.115 ","End":"23:06.385","Text":"we can leave that like that and there we have it."},{"Start":"23:06.385 ","End":"23:12.310","Text":"We got to the solution I(t) is equal to V_naught divided by"},{"Start":"23:12.310 ","End":"23:18.715","Text":"R multiplied by 1 minus e to the power of negative t divided by Tau,"},{"Start":"23:18.715 ","End":"23:25.225","Text":"where Tau is L divided by R. Now we saw how to get to the answer."},{"Start":"23:25.225 ","End":"23:27.430","Text":"I just want to speak a little bit about this Tau,"},{"Start":"23:27.430 ","End":"23:29.155","Text":"which is L divided by R,"},{"Start":"23:29.155 ","End":"23:32.515","Text":"the inductance divided by the resistance."},{"Start":"23:32.515 ","End":"23:36.970","Text":"Generally, this number speaks about the time"},{"Start":"23:36.970 ","End":"23:43.885","Text":"taken for the inductor to fully charge."},{"Start":"23:43.885 ","End":"23:47.050","Text":"It\u0027s every L divided by R,"},{"Start":"23:47.050 ","End":"23:51.460","Text":"we\u0027re going to get some significant charging"},{"Start":"23:51.460 ","End":"23:54.160","Text":"and we generally say that 5 times this number."},{"Start":"23:54.160 ","End":"24:01.870","Text":"So 5 Tau is the amount of time that it will take to fully charge the inductor."},{"Start":"24:01.870 ","End":"24:05.425","Text":"Of course, when the inductor is fully charged,"},{"Start":"24:05.425 ","End":"24:12.640","Text":"then the maximum current is V_naught divided by R. Of course,"},{"Start":"24:12.640 ","End":"24:16.270","Text":"in the same way, 5 times Tau is also the amount"},{"Start":"24:16.270 ","End":"24:20.365","Text":"of time it will take for the inductor to fully discharge."},{"Start":"24:20.365 ","End":"24:23.875","Text":"Tau is just some value that gives us"},{"Start":"24:23.875 ","End":"24:30.310","Text":"an idea of how long it will take to fully charge and fully discharge the inductor,"},{"Start":"24:30.310 ","End":"24:35.665","Text":"so 5 times Tau is generally the amount of time it will take."},{"Start":"24:35.665 ","End":"24:39.140","Text":"That is the end of this lesson."}],"ID":21435}],"Thumbnail":null,"ID":99483}]

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