Question

The charge on the $4\mu F$ capacitor in the steady state is

• A. $10/3\mu C$
• B. $32/3\mu C$
• C. $4/3\mu C$
• D. $8/3\mu C$

Answer 1

Answer: (B)

In steady state, the capacitors are fully charged and act as open circuit, so the equivalent circuit in steady state would be as shown:-

$\therefore$ Steady State current is $I=\frac{12}{2+4}=2A$
$\therefore$ Potential difference across $AB$ is $V=2\times 4=8V$
$\therefore$ Sum of potential difference across $2\mu F$ and $4\mu F$ capacitors is $8V$ . As these are in series, so charges on them would be same. Let $q$ be the charge on them, then:-
From KVL,
$\frac{q}{2}+\frac{q}{4}=8\Rightarrow q=\frac{32}{3}\mu C$