The inductance of the motor of a fan is 1.0 H. To run the fan at 50 Hz the capacitance of the capacitor that will cancel its inductive reactance, will be -

Options

A. `10 mu` F
B. `40 mu` F
C. `0.4 mu` F
D. `0.04 mu` F

Answer

To find the capacitance that will cancel the inductive reactance of a fan motor with an inductance of 1.0 H at a frequency of 50 Hz, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between inductive reactance (XL) and capacitive reactance (XC)**:
- The inductive reactance \(X_L\) is given by the formula:
\[ X_L = \omega L \]
- The capacitive reactance \(X_C\) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
- To cancel the inductive reactance, we set \(X_L = X_C\).

2. **Express angular frequency (\(\omega\)) in terms of frequency (f)**:
- The angular frequency \(\omega\) is related to the frequency \(f\) by:
\[ \omega = 2\pi f \]

3. **Substitute \(\omega\) into the reactance equations**:
- Substitute \(\omega\) into the equation for \(X_L\):
\[ X_L = (2\pi f) L \]
- Substitute \(\omega\) into the equation for \(X_C\):
\[ X_C = \frac{1}{(2\pi f) C} \]

4. **Set the two reactances equal to each other**:
- From \(X_L = X_C\):
\[ (2\pi f) L = \frac{1}{(2\pi f) C} \]

5. **Rearrange the equation to solve for capacitance (C)**:
- Multiply both sides by \(C\) and rearrange:
\[ C = \frac{1}{(2\pi f)^2 L} \]

6. **Substitute the known values**:
- Given \(L = 1.0 \, H\) and \(f = 50 \, Hz\):
\[ C = \frac{1}{(2\pi \cdot 50)^2 \cdot 1} \]

7. **Calculate the value**:
- First, calculate \(2\pi \cdot 50\):
\[ 2\pi \cdot 50 \approx 314.16 \]
- Now square this value:
\[ (314.16)^2 \approx 98706.5 \]
- Finally, calculate \(C\):
\[ C \approx \frac{1}{98706.5} \approx 1.014 \times 10^{-5} \, F \]

8. **Convert to microfarads**:
- Since \(1 \, F = 10^6 \, \mu F\):
\[ C \approx 10.14 \, \mu F \]

### Final Answer:
The capacitance required to cancel the inductive reactance is approximately **10 microfarads**.

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