In the circuit shown in the figure, a 2 μF capacitor is in a branch with a 4 Ω resistor, while the other branches contain a 2 Ω resistor and a 2 V source in series with a 6 Ω resistor. After steady state is reached, what is the potential difference across the capacitor?

Question

G1 Physics · Accepted Answer

At steady state under a DC source, the capacitor branch carries no current. Therefore there is no voltage drop across the 4 Ω resistor in the capacitor branch, and the capacitor voltage equals the potential difference between the two main nodes. The conducting loop contains the 2 Ω and 6 Ω resistors in series with the 2 V source. Thus, I = 2 ÷ (2+6) = 1 ÷ 4 A. The node-to-node voltage equals the drop across the 2 Ω resistor: V(C) = I×2 = (1 ÷ 4)×2 = 0.5 V. Correct option: A.