Question

In the given circuit, all three capacitor are identical and have capacitance $C\mu \text{F}$ each. Each resistor has resistance $R\Omega$. An ideal cell of emf $V\text{volt}$ is connected as shown. If potential difference across capacitor $C_3$ in steady state is $\left(\frac{a}{b}\right)V$, then value of $a\times b$ is_______:

Answer 1

At steady state, no current will flow through capacitors.
Hence, current will not flow through the branches containing capacitors.

Thus effectively there are resistances $3R$ and $3R$ connected in parallel with battery $V\text{volt}$.


The potential at point $1$ and $2$ is same because both the parallel branches are identical.
So, potential difference across both $R\Omega$ resistors is same i.e $\frac{i}{2}R$

$\therefore$The potential difference across each $R\Omega$ resistance is $V^{\prime }=\frac{V}{3}$

Drawing the equivalent capacitor circuit by assigning zero potential as shown in figure.


Here, $\Delta V=\frac{2V}{3}-\frac{V}{3}=\frac{V}{3}$

In series combination of capacitors, potential difference will distribute in inverse ratio of capacitance,

$\Rightarrow (\Delta V{)}_3=\frac{(C_1+C_2)}{(C_1+C_2)+C_3}.\Delta V$

$(\Delta V{)}_3=\frac{2C}{3C}\left(\frac{V}{3}\right)$

$(\Delta V{)}_3=\frac{2}{3}\times \frac{V}{3}=\frac{2V}{9}$

On comparing with $\left(\frac{a}{b}\right)V$
we get,

$a=2$ & $b=9$

$\therefore a\times b=2\times 9=18$

Accepted answer : $18$