The reactance of a coil when used in the domestic AC power supply `(220V,50cycl es)` is `50ohm`. The inductance of the coil is nearly

Options

A. 2.2 H
B. 1.6 H
C. 0.22 H
D. 0.16 H

Explanation

To find the inductance of the coil, we can use the formula for inductive reactance, which is given by:

\[ X_L = \omega L \]

Where:
- \( X_L \) is the inductive reactance (in ohms),
- \( \omega \) is the angular frequency (in radians per second),
- \( L \) is the inductance (in henries).

### Step 1: Identify the given values
From the problem statement, we have:
- \( X_L = 50 \, \text{ohm} \)
- Frequency \( f = 50 \, \text{cycles/second} \) (or Hz)

### Step 2: Calculate the angular frequency
The angular frequency \( \omega \) is calculated using the formula:

\[ \omega = 2 \pi f \]

Substituting the value of \( f \):

\[ \omega = 2 \pi \times 50 \]

Calculating this gives:

\[ \omega = 2 \times 3.14 \times 50 = 314 \, \text{rad/s} \]

### Step 3: Rearranging the formula for inductance
We need to find \( L \). Rearranging the formula for inductive reactance gives:

\[ L = \frac{X_L}{\omega} \]

### Step 4: Substitute the values into the formula
Now, substitute \( X_L \) and \( \omega \) into the equation:

\[ L = \frac{50}{314} \]

### Step 5: Calculate the inductance
Now, performing the division:

\[ L \approx 0.159 \, \text{H} \]

Rounding this to two decimal places, we get:

\[ L \approx 0.16 \, \text{H} \]

### Conclusion
The inductance of the coil is nearly \( 0.16 \, \text{H} \).

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