Question

In the circuit given below, the charge on the capacitor $ C_1 $ is

• A. 6 \( \mu \text{C} \)
• B. 12 \( \mu \text{C} \)
• C. 18 \( \mu \text{C} \)
• D. 24 \( \mu \text{C} \)

Hint:

For capacitors in parallel, the total charge is the sum of the charges on each capacitor, and the charge on each capacitor is proportional to its capacitance.

Answer 1

The Correct Option is B

In the given circuit, we have two capacitors \( C_1 \) and \( C_2 \) connected in parallel with capacitance values of 2 \( \mu \text{F} \) each. The total capacitance \( C_{\text{total}} \) is: \[ C_{\text{total}} = C_1 + C_2 = 2 \, \mu \text{F} + 2 \, \mu \text{F} = 4 \, \mu \text{F} \] The total charge stored in the capacitors is: \[ Q_{\text{total}} = C_{\text{total}} \times V = 4 \, \mu \text{F} \times 6 \, \text{V} = 24 \, \mu \text{C} \] Since the capacitors are in parallel, the charge on \( C_1 \) is: \[ Q_1 = \frac{C_1}{C_{\text{total}}} \times Q_{\text{total}} = \frac{2}{4} \times 24 = 12 \, \mu \text{C} \]
Thus, the charge on \( C_1 \) is 12 \( \mu \text{C} \).