Question

A capacitor of capacitance 5 µF is connected as shown in the figure. The internal resistance of the cell is 0.5 Ω. The amount of charge on the capacitor plate is:

• A. Zero
• B. 5 µC
• C. 10 µC
• D. 25 µC

Hint:

The charge on a capacitor is given by the product of its capacitance and the applied voltage. Ensure the units are consistent.

Answer 1

The Correct Option is C

The total resistance in the circuit is the sum of the internal resistance and the resistance of the resistor.
The time constant \( \tau \) is given by: \[ \tau = R \cdot C \] where \( R \) is the total resistance and \( C \) is the capacitance. The time constant for this circuit is: \[ \tau = 0.5 \, \Omega \times 5 \, \mu F = 2.5 \, \mu \text{s} \] The charge on the capacitor is given by the formula: \[ Q = C \cdot V \] where \( C = 5 \, \mu F \) and the voltage \( V = 2.5 \, \text{V} \). Therefore, the charge on the capacitor is: \[ Q = 5 \, \mu F \times 2.5 \, \text{V} = 10 \, \mu C \]
Thus, the correct answer is (c).