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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) 

Explanation:-

In the steady state (after a long time has passed), the capacitor will be fully charged, and the voltage across the capacitor will be equal to the battery voltage V. At this point, the current in the circuit will be zero because the capacitor will act like an open circuit.

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, a 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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