Question

In the given circuit, in the steady state condition, the potential drop across the capacitor must be-

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

Answer 1

The correct option is C $\frac{V}{3}$

In the steady-state condition, no current flows through the capacitor $C$ i.e., along the branch $BE$



Using KVL to the closed mesh $ACDFA$,

$2V-2IR-IR-V=0$

$V=3IR$

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

Using KVL to the mesh $BCDEB,$

$2V-2IR+\frac{q}{C}-V=0$

$\frac{q}{C}=-\frac{V}{3}(\because I=\frac{V}{3R})$

$\Rightarrow \left|\Delta V\right|=\frac{V}{3}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.