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Q.

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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a

0

b

16 µF

c

4 µF

d

8 µF

answer is D.

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Detailed Solution

At steady state terminal potential difference between battery

V=IR=ERr+R=2.5×20.5+2=2V

Charge on capacitor.

Q=CV=4×2μC

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