The charge stored by the capacitor C in the given circuit in the steady state is ………. μC.

Question

G1 Physics · Accepted Answer

In steady state, the capacitor behaves as an open circuit. From the diode orientations shown in the figure, the conducting path contains the 2.5 V source, the 1 Ω resistor and the 4 Ω branch. The 3 Ω branch is blocked by its diode. Hence the steady current is i = 2.5/(1 + 4) = 2.5/5 = 0.5 A. The potential difference across the 4 Ω branch is V(C) = 4 × 0.5 = 2 V. The capacitor has capacitance C = 5 μF. Therefore the stored charge is Q = CV = 5 × 2 = 10 μC. Correct option: B.