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

Options

A. `8 mu F`
B. `4 `mu F`
C. `2 mu F`
D. `1 mu F`

Explanation

To solve the problem of determining the capacitance required to impart maximum power to an electric fan motor 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 Condition for Maximum Power:**
For maximum power transfer in an RLC circuit (which includes a resistor, inductor, and capacitor), the inductive reactance (X_L) must equal the capacitive reactance (X_C). This can be expressed as:
\[ X_L = X_C \]

2. **Calculate Inductive Reactance (X_L):**
The inductive reactance is given by the formula:
\[ X_L = 2 \pi f L \]
where:
- \( f \) is the frequency (50 Hz)
- \( L \) is the inductance (10 H)

Substituting the values:
\[ X_L = 2 \pi (50) (10) = 1000 \pi \, \text{ohms} \]

3. **Set Up the Equation for Capacitive Reactance (X_C):**
The capacitive reactance is given by the formula:
\[ X_C = \frac{1}{2 \pi f C} \]
where \( C \) is the capacitance we need to find.

4. **Equate X_L and X_C:**
Since \( X_L = X_C \), we can set the two expressions equal to each other:
\[ 1000 \pi = \frac{1}{2 \pi (50) C} \]

5. **Solve for Capacitance (C):**
Rearranging the equation to solve for \( C \):
\[ C = \frac{1}{2 \pi (50) (1000 \pi)} \]
Simplifying this:
\[ C = \frac{1}{100000 \pi^2} \]

Using \( \pi \approx 3.14 \):
\[ C \approx \frac{1}{100000 \times 9.86} \approx \frac{1}{986000} \approx 1.014 \times 10^{-6} \, \text{F} \]

6. **Convert Capacitance to Microfarads:**
Since \( 1 \, \text{F} = 10^6 \, \mu\text{F} \):
\[ C \approx 1.014 \, \mu\text{F} \]

Rounding gives us approximately:
\[ C \approx 1 \, \mu\text{F} \]

### Final Answer:
The capacitance that should be connected to the motor for maximum power at 50 Hz is approximately **1 microfarad (μF)**.