An inductive coil has a resistance 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. `1/(40 pi)H`
B. `1/(20 pi)H`
C. `1/(60 pi)H`
D. `1/(10 pi)H`

Explanation

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

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

### Step 2: Substitute the given values into the equation.
We know that φ = 45 degrees, R = 100 Ω, and we need to find X_L:
\[ \tan(45^\circ) = \frac{X_L}{100} \]
Since \(\tan(45^\circ) = 1\), we can write:
\[ 1 = \frac{X_L}{100} \]

### Step 3: Solve for the inductive reactance (X_L).
From the equation above, we can find X_L:
\[ X_L = 100 \, \Omega \]

### Step 4: Relate inductive reactance to inductance.
The inductive reactance (X_L) is also related to the inductance (L) and the frequency (f) by the formula:
\[ X_L = \omega L \]
where \(\omega = 2 \pi f\).

### Step 5: Calculate the angular frequency (ω).
Given that the frequency f = 1000 Hz:
\[ \omega = 2 \pi \times 1000 = 2000 \pi \, \text{rad/s} \]

### Step 6: Substitute X_L and ω into the equation to find L.
Now we can substitute X_L and ω into the equation:
\[ 100 = (2000 \pi) L \]
Solving for L gives:
\[ L = \frac{100}{2000 \pi} = \frac{1}{20 \pi} \, \text{H} \]

### Final Answer:
The inductance of the coil is:
\[ L = \frac{1}{20 \pi} \, \text{H} \]