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  • A capacitor of 4 mu F is connected as shown in the circuit. The internal resistance of the battery is 0.5\Omega. The amount of charge on the capacitor plates will be a) 0 b) 4 mu C c) 16 mu C d) 8 mu C

A capacitor of 4\mu F is connected as shown in the circuit. The internal resistance of the battery is 0.5\Omega. The amount of charge on the capacitor plates will be

a) 0

b) 4\mu C

c) 16\; \mu C

d) 8\; \mu C

 

Answers (1)

The answer is the option (d) 8\mu C

Current through the 2\Omega resistance considering internal resistance of the battery 1\Omega,

I=\frac{2.5V}{\left ( 2\Omega +0.5\Omega \right )}=1A

The voltage across the internal resistance of the battery

=\left ( 0.5\Omega \right )\left ( 1A \right )=0.5V

The voltage across the 4\mu F capacitor

2.5V-0.5V=2V

Therefore, charge on the capacitor plates

, Q=CV=\left ( 4\mu F \right )\left ( 2V \right )=8\mu C

Posted by

infoexpert23

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