Question

The Charge on the capacitor of capacitance 4iF the Circuit is ?
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Answer 1

Explanation:

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

Answer 2

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