Three capacitors `3 muF, 6 muF` and `6 muF` are connected in series to source of 120 V. The potential difference in volt, across the `3 muF` capacitor will be

Options

A. 24 V
B. 30 V
C. 40 V
D. 60 V

Answer

To solve the problem of finding the potential difference across the `3 µF` capacitor in a series circuit with three capacitors (`3 µF`, `6 µF`, and `6 µF`) connected to a `120 V` source, we can follow these steps:

### Step 1: Calculate the Equivalent Capacitance
For capacitors in series, the formula for the equivalent capacitance \( C_{eq} \) is given by:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

Substituting the values:

\[ \frac{1}{C_{eq}} = \frac{1}{3 \mu F} + \frac{1}{6 \mu F} + \frac{1}{6 \mu F} \]

Calculating each term:

\[ \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} \]

Finding a common denominator (which is 6):

\[ \frac{1}{C_{eq}} = \frac{2}{6} + \frac{1}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \]

Now, taking the reciprocal to find \( C_{eq} \):

\[ C_{eq} = \frac{3}{2} \mu F = 1.5 \mu F \]

### Step 2: Calculate the Total Charge in the Circuit
The total charge \( Q \) stored in the circuit can be calculated using the formula:

\[ Q = C_{eq} \times V \]

Substituting the values:

\[ Q = 1.5 \mu F \times 120 V = 1.5 \times 10^{-6} F \times 120 V = 180 \times 10^{-6} C \]

### Step 3: Calculate the Potential Difference Across the `3 µF` Capacitor
Using the formula for charge \( Q = C \times V \), we can find the potential difference \( V_1 \) across the `3 µF` capacitor:

\[ V_1 = \frac{Q}{C_1} \]

Substituting the values:

\[ V_1 = \frac{180 \times 10^{-6} C}{3 \mu F} = \frac{180 \times 10^{-6} C}{3 \times 10^{-6} F} = 60 V \]

### Conclusion
The potential difference across the `3 µF` capacitor is \( 60 V \).

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