Question

3 capacitors of capacitance \( 12 \mu \text{F} \) each is connected in series. If the voltage is 12V, what charge is drawn from the cell?

• A. \( 12 \mu \text{C} \)
• B. \( 36 \mu \text{C} \)
• C. \( 4 \mu \text{C} \)
• D. \( 24 \mu \text{C} \)

Hint:

In series connection, the total capacitance is always smaller than the smallest individual capacitance, and the charge on each capacitor is the same. Use the formula for series capacitance to calculate the total capacitance, then apply \( Q = C_{\text{total}} \times V \).

Answer 1

The Correct Option is C

When capacitors are connected in series, the total capacitance \( C_{\text{total}} \) is given by: \[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] For three identical capacitors of \( 12 \mu \text{F} \): \[ \frac{1}{C_{\text{total}}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \] Thus, \( C_{\text{total}} = 4 \mu \text{F} \). The charge \( Q \) on a capacitor is given by: \[ Q = C_{\text{total}} \times V \] Substituting \( C_{\text{total}} = 4 \mu \text{F} \) and \( V = 12 \, \text{V} \): \[ Q = 4 \mu \text{F} \times 12 \, \text{V} = 48 \mu \text{C} \] So, the charge drawn from the cell is \( 4 \mu \text{C} \).