Question

Figure shows a network of a capacitor, a battery and resistors. The potential drop (in $\text{V}$) across capacitor in steady state is

Answer 1

Let the potential of the junction be $V$. We know that at steady state there will be no current flowing in the capacitor.

Thus, using KCL at junction $V$:
$\frac{6-V}{2}+\frac{4-V}{4}+\frac{8-V}{4}=0$

$12-2V+4-V+8-V=0$
$24=4V$
$V=6\text{V}$

Potential drop across capacitor
$=6-(-10)$
$=16\text{V}$