Question

In the given circuit with the steady current, the potential drop across the capacitor is found to be $\frac{V}{x}\text{volts}$. Find the value of $x$ (Give integer value)

Answer 1

At the steady-state condition, no current will flow through the capacitor $C$.


$\therefore$ Current in the outer circuit is ,

$I=\frac{2V-V}{2R+R}=\frac{V}{3R}$

Moving along the loop from $\text{A}$ to $\text{B}$ we get,

$V_A-V+V+IR=V_B$

$\Rightarrow V_B-V_A=IR=\left(\frac{V}{3R}\right)R=\frac{V}{3}$

Accepted answer : $3$.

Note: In this problem, charge stored in the capacitor can also be asked, which is equal to [$q=\frac{CV}{3}$] with positive charge on $\text{B}$ side and negative on $\text{A}$ side because $V_B>V_A$