0:00
Hi.
00:01
In the given problem, this is the circuit diagram having the resistance r1 here, then the two capacitors in parallel.
00:20
Here this is c1 and here this is c2.
00:28
And then another resistance r2 that is also in parallel with these two capacitors.
00:38
And finally, the battery having emf epsilon.
00:45
The value of the emf is 72 .0 volt.
00:51
Resistance r2 is 2 .00 oom.
00:57
R1 is missing here.
01:00
Capacitor.
01:03
The first capacitor is having capacity and c1 is equal to 3 .00 microferret.
01:09
And the second one is having c2 is equal to 6 .00 microfarrant.
01:17
It is said that the final charge stored by the capacitor c1 is q1 is equal to 18 .0 microculum.
01:30
And here final charge means the circuit is in steady state.
01:48
So no current will pass through the branches having capacitors.
02:02
So whole of the current will pass through these two resistors in series.
02:09
Now for the first capacitor, first of all, we consider here these two terminals as a and b.
02:19
Now, in the first part of the problem, vab, first of all, the potential drop taking place across these two terminals will be given by q1 by c1 because all these three means the two capacitors and one resistor, r2, all these three are in parallel and we know in parallel potential remains the same...