Question

In the shown circuit, all three capacitors are identical and have capacitance $C\mu \text{F}$ each. Each resistor has resistance of $R\Omega$. An ideal cell of emf $V$ volts is connected as shown. Then the magnitude of potential difference across capacitor $C_3$ in steady state is:

• A. $\frac{V}{3}$
• B. $\frac{V}{2}$
• C. $\frac{2V}{9}$
• D. $\frac{3V}{4}$

Answer 1

The correct option is C $\frac{2V}{9}$

In steady state capacitor behaves as open circuit. No current passes through the capacitor.


$i=\frac{V}{3R}$

Potential at $1$ and $2$ are equal.

$V_1=V_2=V-iR=V-\left(\frac{V}{3R}\right)R$

$\Rightarrow V_1=V_2=\frac{2V}{3}$

Potential at point 3, $V_3=V-i\left(2R\right)$

$\Rightarrow V_3=V-\left(\frac{V}{3R}\right)\times 2R=\frac{V}{3}$

capacitor $C_1$ and $C_2$ are in parallel.


It can be redrawn as,


Potential difference across $C_3$

$V_3=\frac{V}{3}\times \left(\frac{2C}{2C+C}\right)$

$\Rightarrow V_3=\frac{2V}{9}$