Question

The potential difference between $a$ and $b$ when the switch is open is given as $\frac{200}{x}V$. Find $x$.

Answer 1

When switch is open, the capacitors (6,3) on left are in series and capacitors (3,6) on right are in series and then these groups are in parallel.
Thus, potential across $6\mu F$ on left is $V_{Pa}=V_P-V_a=\frac{3}{6+3}{V}_{PQ}=\frac{3}{9}\times 200=\frac{200}{3}V....\left(1\right)$
and potential across $3\mu F$ on right is $V_{Pb}=V_P-V_b=\frac{6}{6+3}{V}_{PQ}=\frac{6}{9}\times 200=\frac{400}{3}V....\left(2\right)$
$\left(2\right)-\left(1\right),V_a-V_b=\frac{400}{3}-\frac{200}{3}=\frac{200}{3}V$