00:01
In this problem, we have been given the following circuit and we have been shown here a number of capacitors as connected.
00:09
And we are given the value of this capacitor capacitance as 11 .5 micro -farrant.
00:16
And we have to determine the charge that's flowing through this capacitor of capacitance c provided that the potential difference between the two points a and b.
00:25
That is 49 .2 volts.
00:29
So this is the potential difference across the points ab.
00:34
So here we'll use the expression q is equal to cv.
00:38
And let's first figure out the total charge that's flowing through this capacitor.
00:45
So as we can see that a and b points, they are connected to the ends of this capacitor.
00:50
In that case, the charge that will be flowing through this capacitor of capacitance 1 microferrin, that will be its capacitance multiplied by the.
01:00
Potential difference and this comes out to be 49 .2 microculum and now we can observe that this series can be simplified so that we can figure out the total capacitance we can see that these two capacitors they are connected in series and has the total capacitance after combining them it will be six times 11 .5 over 11 .5 plus 6 and this value it simplifies to approximately 3 .92 micro ferret.
01:32
And now when we combine these two capacitors, we see that they are in parallel to this 5 microferret capacitor.
01:40
And when we combine them, we're going to get the total capacitance now as 3 .92 plus 5.
01:47
That's 8 .92 microferret.
01:49
And now this 8 .92 microferret, it's connected in series with this 3 microferret capacitor.
01:56
And the total capacitance now, it will be 8 .92 times 3 over 8 .92 plus 3.
02:04
And this value comes out to be approximately 3 .24 micro ferret.
02:12
So this 3 .24 micro ferret that's the total capacitance we get...