Question

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. $8\, \mu C$

Answer 1

Answer: (D)

Current in the lower arm of the circuit,
$I=\frac{2.5\, V}{2\,\Omega+0.5\,\Omega}=1\, A$,
Potential difference across the internal resistance of cell
$=\left(0.5\,\Omega\right)\left(1\,A\right)=0.5\,V$
and potential difference across the $4\mu F$ capacitor
$= 2.5 V - 0.5 V = 2 V$
Charge on the capacitor plates, $Q=CV=\left(4\,\mu F\right)\left(2V\right)$
$=8\,\mu C$