Question

In the circuit shown, the voltage drop across the $2R$ resistor must be

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

Answer 1

Answer: (D)

$\frac{2V}{3}$

We know,

Capacitor acts as an open circuit in steady state.

Then it is clear that battery V isn't included in the circuit.

Hence potential difference 2V is dropped in R and 2R

in 1:2 respectively.

Let current in the circuit be $I$

By Kirchhoff loop law,

$V-IR-I2R=0=V-3IR$

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

Potential difference along resistance 2R,

$I\times 2R=\frac{V}{3R}\times 2R=\frac{2V}{3}$

Option $\text{C}$ is the correct answer.