00:01
All right.
00:04
So for this problem, we have the circuit.
00:08
So let me draw the circuit.
00:11
We have the capacitor, c1, and we have the battery v, and we have this part consisting of c2.
00:32
So this one is c2.
00:35
And there is extra one.
00:40
Okay.
00:41
And in this circuit, all of the capacity has capacitance equal to 10 microveridates.
00:48
And also, we know that v is 10 volts.
00:55
So for part a, we want to find out the charge stored in c1.
01:02
So as you can see that the voltage drop across c1 is simply equal to v, right? just this two point.
01:12
So the charge stored in c1 will be simply equal to q1 equal to c1 times v.
01:19
And so v is 10 volts and c1 is 10 micro -fadres so the result is 10 to negative false and for part b we want to find out the charge stored in c2 but before that we can look at this part we can look at this part so we have two capacitors connect in the series and the and there is x one which is connected in parallel to this part.
01:52
So basically the total capacitance of this part would be equal to c, let's say c prime, equal to, because each of them have 10 microferiorities.
02:06
So it's just equal to 10 plus 10 times 10 divide by 10 plus 10.
02:14
So this 10 times, so this this part is simply the capacitance of this of this two...