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)

8 $\mu C$

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