Question

In the given circuit of figure, with steady current, the potential drop across the capacitor must be

• A. $V$
• B. $V/2$
• C. $V/3$
• D. $2V/3$

Answer 1

Answer: (C)

$V/3$

When the capacitor is fully charged, no current draws from cell. So, the current will flow in the outer loop.
Now applying KVL for this loop we get, $2V-V=(2R+R)I$ or $I=\left(V/3R\right)$
Thus, $V_{DE}=2V-2RI=2V-2R\left(V/3R\right)=\left(4V/3\right)$
Potential across C is $V_{AB}=V_{DE}-V=4V/3-V=V/3$