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

Answer 2

: 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

$= (0.5\Omega)(1A) = 0.5A$

and potential difference across the $4\mu F$capacitor $= 2.5V - 0.5V = 2V.$

Charge on the capacitor plates, $Q = CV = (4\mu F)(2V)$

$= 8\mu C.$