The charge on the 4 μF capacitor in the steady state is

Step-by-step 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:- ∴ Steady State current is I = 12 2 + 4 = 2 A ∴ Potential difference across AB is V = 2 × 4 = 8 V ∴ 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, q 2 + q 4 = 8 ⇒ q = 32 3 μC