00:01
In this problem, we have been given that this is the capacitors connected in the circuit, and we need to determine the equivalent capacitance across the point a and b.
00:11
So as we can observe that the capacitors, c2 here and this c2 capacitor, they are short -circuited, because the potential is same as they are connected across the wires.
00:26
So in this case now, to determine the potential across a and b, we can say that at this point the potential is a and even at this point the potential is b because it's connected by a straight wire.
00:39
And now we can just find out the equivalent capacitance of this circuit.
00:44
So for that, we see that c1 and c2, they are connected in series.
00:49
And here also c1 and c2 are connected in series.
00:51
So as a result, we're going to apply the combination formula for capacity.
00:55
Connected in series, so that will be c1 into c2 divided by c1 plus c2.
01:01
So when we multiply c1 and c2, we get 50.
01:04
When we add them, we get 15.
01:06
So that comes out to be 10 by 3 microferret.
01:10
So the circuit here will simplify as it contains c3, that's 20 microferred.
01:16
And there is one combined capacitors having capacity 10 by 3 microferred.
01:22
And here also 10 by 3 microferred, because on both the ends, the capacitors have the same value...