The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of:

Options

A. (a)`1 mu F`
B. (b)`2 mu F`
C. (c)`4 mu F`
D. (d)`8 mu F`

Explanation

To solve the problem of determining the capacitance required for maximum power transfer to the motor of an electric fan with a self-inductance of 10 H at a frequency of 50 Hz, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Resonance Condition**:
For maximum power transfer in an RLC circuit, the inductive reactance (X_L) must equal the capacitive reactance (X_C). This is expressed as:
\[ X_L = X_C \]

2. **Express Reactances**:
The inductive reactance (X_L) is given by:
\[ X_L = \omega L \]
where \(\omega\) is the angular frequency and \(L\) is the inductance. The capacitive reactance (X_C) is given by:
\[ X_C = \frac{1}{\omega C} \]
where \(C\) is the capacitance.

3. **Relate Inductive and Capacitive Reactance**:
Setting \(X_L\) equal to \(X_C\):
\[ \omega L = \frac{1}{\omega C} \]

4. **Solve for Capacitance (C)**:
Rearranging the equation gives:
\[ \omega^2 C = \frac{1}{L} \]
Thus,
\[ C = \frac{1}{\omega^2 L} \]

5. **Calculate Angular Frequency (\(\omega\))**:
The angular frequency \(\omega\) is related to the frequency \(f\) by:
\[ \omega = 2\pi f \]
For \(f = 50\) Hz:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

6. **Substitute Values**:
Now substituting \(\omega\) and \(L\) into the capacitance formula:
\[ C = \frac{1}{(100\pi)^2 \times 10} \]
\[ C = \frac{1}{10000\pi^2} \]

7. **Calculate the Numerical Value**:
Using \(\pi \approx 3.14\):
\[ C \approx \frac{1}{10000 \times (3.14)^2} \approx \frac{1}{10000 \times 9.86} \approx \frac{1}{98600} \approx 1.014 \times 10^{-5} \, \text{F} \]
Converting to microfarads:
\[ C \approx 10.14 \, \mu F \]

8. **Final Answer**:
The capacitance required to impart maximum power at 50 Hz is approximately **10.14 µF**.