Answer:
Current in the lower arm of the circuit,
I=
2Ω+0.5Ω
2.5V
=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
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Answers & Comments
Answer:
Current in the lower arm of the circuit,
I=
2Ω+0.5Ω
2.5V
=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