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Q.
The charge on the $4\,μF$ capacitor in the steady state is
NTA AbhyasNTA Abhyas 2022
Solution:
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 $V=2\times 4=8 \, V$
$\therefore $ Sum of potential difference across $2\,μF$ and $4\,μF$ capacitors is $8\,V$ . As these are in series, so charges on them would be same. Let $q$ be the charge on them, then:-
From KVL,
$\frac{q}{2}+\frac{q}{4}=8\Rightarrow q=\frac{32}{3}μC$
