A series combination of three capacitors of capacities `1 muF, 2muF` and `8 muF` is connected to a battery of e.m.f. `13 volt`. The potential difference across the plates of `2 muF` capacitor will be

Options

A. `1V`
B. `8 V`
C. `4 V`
D. `(13)/(3) V`

Explanation

To find the potential difference across the plates of the 2 µF capacitor when connected in series with a 1 µF and an 8 µF capacitor to a 13 V battery, we can follow these steps:

### Step 1: Calculate the total capacitance in series
For capacitors in series, the total capacitance \( C_{\text{total}} \) can be calculated using the formula:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
where \( C_1 = 1 \mu F \), \( C_2 = 2 \mu F \), and \( C_3 = 8 \mu F \).

Substituting the values:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{8} \]
\[ \frac{1}{C_{\text{total}}} = 1 + 0.5 + 0.125 = 1.625 \]
\[ C_{\text{total}} = \frac{1}{1.625} \approx 0.615 \mu F \]

### Step 2: Calculate the total charge in the circuit
The total charge \( Q \) stored in the series combination can be calculated using the formula:
\[ Q = C_{\text{total}} \times V \]
where \( V = 13 V \).

Substituting the values:
\[ Q = 0.615 \mu F \times 13 V = 8.0 \mu C \]

### Step 3: Calculate the potential difference across the 2 µF capacitor
The potential difference \( V_2 \) across the 2 µF capacitor can be calculated using the formula:
\[ V_2 = \frac{Q}{C_2} \]
Substituting the values:
\[ V_2 = \frac{8.0 \mu C}{2 \mu F} = 4.0 V \]

### Conclusion
The potential difference across the plates of the 2 µF capacitor is **4.0 volts**.

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