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Question

A capacitor of 4 μF is connected as shown in the circuit. The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be
945473_31d0e611eb624785b4ca38a6abdf79bd.png

A
0
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B
4 μC
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C
16 μC
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D
8 μC
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Solution

The correct option is C 8 μC
Current in the lower arm of the circuit,
I=2.5V2Ω+0.5Ω=1A
Potential difference across the internal resistance of cell
=(0.5Ω)(1A)=0.5V
and potential difference across the 4μF capacitor
2.5V0.5V=2V
Charge on the capacitor plates, Q=CV=(4μF(2V)=8μC
1058597_945473_ans_dbc6fd910f3a4d2ebbd89a09f5ab6770.png

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