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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:

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4 µF

16 µF

8 µF

answer is D.

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Detailed Solution

At steady state terminal potential difference between battery

$V = IR = \frac{ER}{r + R} = \frac{2 .5 \times 2}{0 .5 + 2} = 2 V$

Charge on capacitor.

$Q = CV = 4 \times 2 μC$

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