Three condensers of capacities 3 mF, 6 mF, 12mF are connected in series with a battery. If the charge on 12 mF condenser is 24 mC, the P.D. across the battery is

Options

A. 2 V
B. 4 V
C. 8 V
D. 14 V

Answer

To solve the problem step by step, we need to find the potential difference across the battery when three capacitors of capacities 3 mF, 6 mF, and 12 mF are connected in series, and the charge on the 12 mF capacitor is given as 24 mC.

### Step-by-Step Solution:

1. **Understanding Capacitors in Series**:
- In a series connection, the charge (Q) on each capacitor is the same. Therefore, the charge on the 3 mF, 6 mF, and 12 mF capacitors is also 24 mC.

2. **Calculating the Voltage across Each Capacitor**:
- The voltage across a capacitor is given by the formula:
\[ V = \frac{Q}{C} \]
- Where \( V \) is the voltage, \( Q \) is the charge, and \( C \) is the capacitance.

3. **Voltage across the 12 mF Capacitor**:
- For the 12 mF capacitor:
\[ V_1 = \frac{Q}{C_1} = \frac{24 \, \text{mC}}{12 \, \text{mF}} = \frac{24 \times 10^{-3}}{12 \times 10^{-3}} = 2 \, \text{V} \]

4. **Voltage across the 6 mF Capacitor**:
- For the 6 mF capacitor:
\[ V_2 = \frac{Q}{C_2} = \frac{24 \, \text{mC}}{6 \, \text{mF}} = \frac{24 \times 10^{-3}}{6 \times 10^{-3}} = 4 \, \text{V} \]

5. **Voltage across the 3 mF Capacitor**:
- For the 3 mF capacitor:
\[ V_3 = \frac{Q}{C_3} = \frac{24 \, \text{mC}}{3 \, \text{mF}} = \frac{24 \times 10^{-3}}{3 \times 10^{-3}} = 8 \, \text{V} \]

6. **Total Voltage across the Battery**:
- The total voltage across the battery (V) is the sum of the voltages across all capacitors:
\[ V = V_1 + V_2 + V_3 = 2 \, \text{V} + 4 \, \text{V} + 8 \, \text{V} = 14 \, \text{V} \]

### Final Answer:
The potential difference across the battery is **14 V**.

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