00:01Hello students, we have a problem which is switching problem of capacitor and how to deal with switching problem.
00:11Yes, we have come in this example, we understand
00:14You have three identical capacitors C1, C2, C3 and the capacitance of each is given as how much is one microfarad
00:22Each one is uncharged. It is very important that the capacitors that we are installing were initially uncharged, so now tell me.
00:34It is being
00:34that they are connected in the circuit as well in the figure and C1 is filled completely with the dielectric
00:39material of relative permittivity epsilon r
00:42Meaning we put C1, C2, C2, C3 in the circuit but we put a dielectric material in C1.
00:49whose dielectric permittivity, dielectric constant epsilon r is given by
00:53And battery electromotive force EMF of the cell is given how much V note value we have is given 8
01:00volt 8 volt is ok in question
01:02very important first the switch S1 is closed very important line first we close S1 while S2 is
01:12kept open and S2 was open at that time
01:17when the capacitor C3 is fully charged S1 is opened and S2 is closed simultaneously meaning
01:26When connected, S3 got fully charged.
01:28When capacitor C3 is fully charged, you do not want to charge it, so what do you do, open switch S1?
01:34I did it and as soon as I opened it, I immediately connected S2.
01:39So the question being asked is when all the capacitor reaches equilibrium the charge on C3
01:46is found to be 5 microgram
01:48So what will be the value of epsilon R, equilibrium means that brother everything has reached its steady state.
01:55At that point, the final charge on capacitor C3, which is four times larger, is pipe micrograms.
02:02So what will be the value of epsilon R, what will be the value of the dielectric material that you put in C1, this is the question
02:08Bocha is being
02:09So let us understand this, information or this is for you that C1, C2, C3 which we get from the market
02:16Bring it, friends, what is the capacity of each one of them?
02:19is 1 microgram, and the capacitors are initially uncharged, so one by one we deal with the case that
02:27When switch S1 is close, you have a diagram like this
02:29This capacitor is C3, C3, and this is our C2, C2, and we filled dielectric in C1, now C1,
02:37Take C1 dash value, this will become how much, epsilon r times, the initial value which was C1, what was C1
02:45We have, that is one microfarad, C2 we also have, friends, is also one microfarad, and C3 we have how much
02:53C3 you also have one microfarad
02:56So now initially we have this switch S1, and our S2 is open, S1 is closed, and this 8
03:02volt of battery is charged here, so when S1 is close, capacitor C3 will charge, and
03:12Take the charge on capacitor C3, let's say Q1 charge, on this plate plus Q1, and
03:17Pay minus Q1 is the charge, study statement, and Q1 is equal to what I would write, that is C into B,
03:23C3 into B, applied voltage,
03:25which is B0, C3 into B0, B0, B0, and the value of C3 is 1, and the value of B0 is 8 so
03:33How much does it cost, 8 microclam, this is the charge on capacitor C1, where, at steady state, maximum charge, capacitor
03:42C3 will be fully charged, now capacitor C3 does not need any charge at all, so what did you do at that instant?
03:49what what what what what what what what what what what what what
03:53It is given, so let us understand that now the second case, this is the situation of the first case.
03:58That was the first case, now for the second case we have to draw the diagram again, so I will create the diagram for you.
04:04I have created it, now look here, so what has happened now is that S1 is open and S2 which is here
04:10Pay �
04:10The incoming switch S2 is closed, now the charge on this capacitor which was earlier 8, is now
04:18Suppose it is 8-Q, meaning it has supplied Q charge, because now this source will work.
04:25This will charge these two uncharged capacitors, this one has a charge of plus Q, this one has a charge of minus Q, here also
04:32plus Q, minus Q, so it's clear, which capacitor is this, C
04:37C1 is the dash, which you said is epsilon r times of C1, and this is the C2 capacitor, and this is C3
04:42The capacitor is C3, and this positive plate, its negative plate, its positive, if I write C2, C3 here
04:50Because I have to use KVL, it would be better if you write here, its capacitance is epsilon r times of C1,
04:56And this we have C2, and this is positive plate, negative plate,
05:00Positive plate, negative plate, now friends let's do this, we apply KVL in this loop, KVL, Krichoff's
05:09If you apply voltage law in this loop, what will happen, let us understand this situation,
05:17That when we moved in a loop, applied KVL, using KVL in loop 1, in loop 1 if
05:23I will put KVL
05:24So, starting from here, finally we have, what is Q equal to formula
05:29It happens, CV, what happens if you want to write more, I have been a Capster, I have been a Capster, because
05:33So, let's use this same concept, so here if I'm moving, I have this, last
05:38I stayed in
05:52I can write Q upon epsilon r times of C1, here the last one has been negative, so minus
06:00Q upon C2 equal to 0, reached the same point from where we started, now given in C1 question
06:07If the value of C1, C2, C3 is 1 microfarad, then replace it with 1, 1, and also add it.
06:12You can, it is clear, and if you read the question, it is saying in the question that,
06:17When all the capacitors reach their equilibrium, at that time, the charge on C3 will be the final charge.
06:25on capacitor, C3 is 5 microfarad, the question is saying, 5 microfarad, that is to say, when we
06:34We're closing S2, S1 is open, at that time, at that time, this 8 minus Q
06:40Yes friends, 8 minus Q, its value is 5 microfarad, meaning the value of Q is, �
06:463 microfarad, means 3 microfarad charge is supplied by the capacitor, C3, and if you want to write it here
06:54equation, so this is what we have: 8 minus Q divided by 5, 5 by C3
06:58The value of 1, minus Q is 3, 3 by epsilon r into 1, then epsilon r will be, minus 3 upon 1.
07:05equal to 0,
07:08So 5 minus 3, 2 equals 3 by epsilon r you can write, to be clear, so epsilon r
07:16If I want to write the value of 3 by 2, the unit of all is kept in microfarad, microfarad, the microfarad will be cancelled out,
07:21From all over,
07:22Here the microclaim, this is in your microclaim, the capacitor is in macro farad, so microfarad, the microclaim is getting cancelled
07:28Yes, it doesn't matter, we have the answer 1.5, and this is our correct answer, so I'll hope
07:36Friends, you must have understood the concept of capacitor switching problem with dielectric.
07:43See you with the next session, till then thank you very much, she is happy, she keeps smiling.
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