An inductive coil has a reactance of `100 Omega`. When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by `45^(@)`. What is the inductance of the coil?

Options

A. `6.7 xx10^(-7)` H
B. `5.5 xx10^(-5)`H
C. `1.1 xx10^(-1)` H
D. `1.1 xx10^(-2)` H

Explanation

To find the inductance of the coil, we can follow these steps:

### Step 1: Understand the relationship between reactance and phase angle
In an LR circuit, the phase difference (φ) between the voltage and current is given by the formula:
\[ \tan(\phi) = \frac{X_L}{R} \]
where \(X_L\) is the inductive reactance and \(R\) is the resistance.

### Step 2: Use the given phase angle
The problem states that the voltage leads the current by \(45^\circ\). Therefore:
\[ \tan(45^\circ) = 1 \]
This implies:
\[ \frac{X_L}{R} = 1 \implies X_L = R \]

### Step 3: Use the given reactance value
We know that the reactance \(X_L\) is given as \(100 \, \Omega\). Therefore:
\[ X_L = 100 \, \Omega \implies R = 100 \, \Omega \]

### Step 4: Calculate the impedance
The impedance \(Z\) in an LR circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values we have:
\[ Z = \sqrt{100^2 + 100^2} = \sqrt{10000 + 10000} = \sqrt{20000} = 100\sqrt{2} \, \Omega \]

### Step 5: Relate reactance to inductance
The inductive reactance \(X_L\) is also related to the inductance \(L\) and frequency \(f\) by the formula:
\[ X_L = 2\pi f L \]
Substituting \(X_L = 100 \, \Omega\) and \(f = 1000 \, \text{Hz}\):
\[ 100 = 2\pi (1000) L \]

### Step 6: Solve for inductance \(L\)
Rearranging the equation to solve for \(L\):
\[ L = \frac{100}{2\pi \times 1000} \]
Calculating this gives:
\[ L = \frac{100}{2000\pi} = \frac{1}{20\pi} \, \text{H} \]

### Step 7: Convert to a numerical value
Using \(\pi \approx 3.14\):
\[ L \approx \frac{1}{20 \times 3.14} \approx \frac{1}{62.8} \approx 0.0159 \, \text{H} \approx 1.59 \times 10^{-2} \, \text{H} \]

### Final Answer
Thus, the inductance of the coil is approximately:
\[ L \approx 1.59 \times 10^{-2} \, \text{H} \text{ or } 15.9 \, \text{mH} \]