Question

An ideal cell of emf $10\text{V}$ is connected in circuit shown in figure. Each resistance is $2\Omega$. The potential difference (in $\text{V}$ ) across the capacitor when it is fully charged is

Answer 1

As capacitor is fully charged no current will flow through it.

We have the current distribution as shown in the figure.
Equivalent resistance,$R_{eq}=\left(\frac{4\times 2}{4+2}\right)+2$
Net current,$i=\frac{10}{\frac{4}{3}+2}=\frac{10\times 3}{3}=3\text{A}$
Current division among resistors can be considered as
$i_1=2\text{A and}{i}_2=1\text{A}$
Potential difference across capacitor is
$V_{AEB}=1\times 2+3\times 2=8\text{V}$.