A capacitor of 4μF is connected as shown in the circuit. The internal resistance of the battery is 0.5Ω. The amount of charge on the capacitor plates will be a) 0 b) 4μC c) 16μC d) 8μC
Correct option is D) Current in the lower arm of the circuit, I=2Ω+0.5Ω2.5V=1A Potential difference across the internal resistance of cell =(0.5Ω)(1A)=0.5V and potential difference across the 4μF capacitor 2.5V−0.5V=2V Charge on the capacitor plates, Q=CV=(4μF(2V)=8μC
