Question

In the circuit shown, the ratio of potential difference across the capacitors $C_2$ and $C_1$, when they are in steady state is

Answer 1

When the capacitors are in steady state, no current passes through them. Hence current passes through the two resistors and the galvanometer.
Hence $i=\frac{2}{400+100+500}=2\times {10}^{-3}A$.
Using KVL, Pd across $C_1=2\times {10}^{-3}\times (400+100)=1V$
and
Using KVL, Pd across $C_2=2\times {10}^{-3}\times (500+100)=1.2V$

Hence, ratio of potential difference across $C_2$ to $C_1$ is $\frac{1.2}{1}=1.2$