Question

In the circuit shown in figure, a potential difference of $60\text{V}$ is applied between a and b. The potential difference between the points c and d ( in Volts) is

Answer 1

The equivalent capacitance for series combination of (2C, C, 2C) is $C_{eq}$
$={(\frac{1}{2C}+\frac{1}{C}+\frac{1}{2C})}^{-1}=\frac{C}{2}$
As the capacitor C and $C_{eq}$ are in parallell, potential across each is equal to $V_{AB}=60V$
The equivalent charge $Q_{eq}=C_{eq}{V}_{AB}=(\frac{C}{2}\times 60)=30C$
As, 2C, C and 2C are in series so charge on each is equal to $Q_{eq}$
Potential across c and d = potential across C
$=\frac{Q_{eq}}{C}=\frac{30C}{C}=30V$