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Question

If 100 volts of potential difference is applied between a and b in the circuit shown in the figure. The potential difference between 'c' and 'd' will be (Given Va>Vb).


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Solution

From the circuit given, we can see that the three 6 μF capacitors connected in upper branch are in series.

Now simplifying the circuit and redrawing it,


Since the combination of three 6 μF capacitors are connected in parallel with battery. So,

Sum of potential difference across 6 μF capacitors = potential difference of battery

Since, all the three capacitors are identical, hence

V1=V2=V3=V

V1+V2+V3=100V

3V=100 V

V=1003 V

From the shown figure, direction of flow of positive charge it is clear that Vc>Vd

VcVd=+1003 V

Alternative solution:

Let the three 6 μF capacitors be replaced an equivalent capacitor

1Ceq=16+16+16

Ceq=63=2μF



Charge flow through the 2 μF, q1=100×2=200 μC

The potential difference,

VcVd=(q1)6 μF

VcVd=200 μC6 μF=1003V

Hence, option (c) is correct answer.

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