When an `AC` signal of frequency `1 kHz` is applied across a coil of resistance `100 Omega`, then the applied voltage leads the current by `45^@`. The inductance of the coil is

Options

A. `16 mH`
B. `12mH`
C. `8mH`
D. `4mH`

Explanation

To solve the problem of finding the inductance of a coil when an AC signal is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Frequency \( f = 1 \, \text{kHz} = 1000 \, \text{Hz} \)
- Resistance \( R = 100 \, \Omega \)
- Phase angle \( \phi = 45^\circ \)

2. **Use the Relationship Between Phase Angle and Reactance**:
The phase angle \( \phi \) is related to the inductive reactance \( X_L \) and the resistance \( R \) by the formula:
\[ \tan(\phi) = \frac{X_L}{R} \]
Since \( \phi = 45^\circ \), we know that:
\[ \tan(45^\circ) = 1 \]
Therefore, we can write:
\[ 1 = \frac{X_L}{R} \]
This implies:
\[ X_L = R \]
Substituting the value of \( R \):
\[ X_L = 100 \, \Omega \]

3. **Relate Inductive Reactance to Inductance**:
The inductive reactance \( X_L \) is given by the formula:
\[ X_L = 2 \pi f L \]
Substituting the known values:
\[ 100 = 2 \pi (1000) L \]

4. **Solve for Inductance \( L \)**:
Rearranging the equation to solve for \( L \):
\[ L = \frac{100}{2 \pi (1000)} \]
Simplifying further:
\[ L = \frac{100}{2000 \pi} = \frac{1}{20 \pi} \]
Now, calculating the value:
\[ L \approx \frac{1}{62.83} \approx 0.0159 \, \text{H} = 15.9 \, \text{mH} \]

5. **Final Answer**:
Therefore, the inductance of the coil is approximately:
\[ L \approx 16 \, \text{mH} \]

### Summary:
The inductance of the coil is \( 16 \, \text{mH} \).