Question

A capacitor of capacitance \( 5\,\mu\mathrm{F} \) is connected as shown in the figure. The internal resistance of the cell is \( 0.5\,\Omega \). The amount of charge on the capacitor plate is

• A. 10 \( \mu \) C
• B. 5 \( \mu \) C
• C. 6 \( \mu \) C
• D. 10.2 \( \mu \) C

Hint:

To calculate the charge on a capacitor, use the formula \( Q = C \times V \), where \( C \) is the capacitance and \( V \) is the voltage across the plates.

Answer 1

The Correct Option is A

Step 1: Understanding the circuit and the formula for charge.
The charge on a capacitor is given by the formula: \[ Q = C \cdot V \] where \( C \) is the capacitance and \( V \) is the potential difference across the capacitor.

Step 2: Determining the voltage across the capacitor.
- The total voltage provided by the cell is 2.5 V, and the internal resistance of the cell is 0.5 \( \ohm\). - Using Ohm’s Law, the current in the circuit can be determined as: \[ I = \frac{V_{\text{cell}}}{R_{\text{internal}}} = \frac{2.5}{0.5} = 5 \, \text{A} \] - The voltage across the capacitor is therefore the same as the voltage across the capacitor plates: \[ V = I \cdot R_{\text{internal}} = 5 \, \text{A} \times 0.5 \, \text{Ω} = 2.5 \, \text{V} \]

Step 3: Conclusion.
The charge on the capacitor is: \[ Q = 5 \, \mu \text{F} \times 2.5 \, \text{V} = 10 \, \mu \text{C} \] Thus, the correct answer is (1).