Three condensers `1 mu F,2 mu F` and `3 mu F` are connected in series to a `p.d` of `330 " volt"`. The `p.d` across the plates of `3 mu F` is.

Options

A. 180 V
B. 300 V
C. 60 V
D. 270 V

Explanation

To solve the problem, we need to find the potential difference across the 3 µF capacitor when three capacitors (1 µF, 2 µF, and 3 µF) are connected in series to a potential difference (p.d) of 330 volts.

### Step-by-step Solution:

1. **Calculate the Equivalent Capacitance (C_eq)**:
For capacitors in series, the formula for equivalent capacitance is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
Here, \(C_1 = 1 \mu F\), \(C_2 = 2 \mu F\), and \(C_3 = 3 \mu F\).
\[ \frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \]
\[ \frac{1}{C_{eq}} = 1 + 0.5 + 0.3333 = 1.8333 \]
\[ C_{eq} = \frac{1}{1.8333} \approx 0.545 \mu F \]

2. **Calculate the Total Charge (Q)**:
The total charge (Q) on the capacitors can be calculated using the formula:
\[ Q = C_{eq} \times V \]
Where \(V = 330 V\).
\[ Q = 0.545 \mu F \times 330 V = 0.180 \text{ mC} \text{ (or } 180 \mu C\text{)} \]

3. **Calculate the Potential Difference across the 3 µF Capacitor**:
The potential difference (V_3) across a capacitor can be calculated using the formula:
\[ V = \frac{Q}{C} \]
For the 3 µF capacitor:
\[ V_3 = \frac{Q}{C_3} = \frac{180 \mu C}{3 \mu F} = 60 V \]

### Final Answer:
The potential difference across the plates of the 3 µF capacitor is **60 volts**.