00:01
Hi, here in this given problem there are four different capacitors arranged in this circuit.
00:10
The first two capacitors c1 and c2, these are in series and the other two capacitors, these are in parallel.
00:21
And this parallel combination that is in series again with the first two capacitors and in parallel these are c3 and c4 and the two points a and b are mentioned here.
00:40
This is the battery providing a voltage v, switch s.
00:46
The voltage provided to this circuit that is 9 .00 volt.
00:52
Values of the capacitors out of the four, the values of two capacitors are known to us c2 that is equal to 3 .80 microfarad and c4 this is 4 .10 microfarad.
01:10
And charge passing through point a that will be the total charge of the circuit.
01:17
As we know charge remains the same in series and that is given as 10 .0 microcoulomb and this charge will be divided into two parts.
01:31
It will be q3 passing through c3 and it will be q4 that passing through c4.
01:40
So the charge passing through point b means qb, actually it will be q4, the charge passing through c4 and this is given as 8 .00 microcoulomb.
01:54
So q3 that will be given by total charge minus charge passing through c4 means this is 10 minus 8 means this is 2 .00 microcoulomb.
02:12
Now we find v4 potential drop across capacitor c4 that will be given by v is equal to q by c means q4 by c4, q4 that is 8 and divided by c4 4 .10.
02:34
So it comes out to be equal to 1 .95 volt.
02:40
So as the voltage remains the same in parallel and this we are doing for the first part of the problem.
02:55
So here we have started doing first part in which we have to find the value of capacitor c1...