In the shown network, the charge accumulated on the 4 μF capacitor in steady state is:

Question

G1 Physics · Accepted Answer

At steady state, no current flows through the capacitor branch, so the capacitor behaves as an open circuit. The conducting loop contains the 3 V source with 4 Ω and 6 Ω resistors in series. Current: I = 3/(4 + 6) = 0.3 A Potential difference across the 6 Ω resistor: V = IR = 0.3 × 6 = 1.8 V The capacitor has the same potential difference. Thus, Q = CV = 4 μF × 1.8 V = 7.2 μC Hence option C follows.