An ideal inductive coil has a resistance of `100Omega` When an ac signal of frequency `1000Hz` is applied to the coil the voltage leads the current by `45^(@)` The inductance of the coil is .

Options

A. ` 1 / 10 pi `
B. `1 / 20 pi `
C. `1 / 40 pi `
D. ` 1 / 60 pi `

Explanation

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

### Step 1: Understand the relationship between voltage, current, and inductance
In an AC circuit with an inductor, the voltage (V) leads the current (I) by a phase angle (φ). The inductive reactance (X_L) is given by the formula:
\[ X_L = \omega L \]
where:
- \( \omega = 2\pi f \) (angular frequency),
- \( L \) is the inductance.

### Step 2: Identify the given values
From the problem, we have:
- Resistance \( R = 100 \, \Omega \)
- Frequency \( f = 1000 \, \text{Hz} \)
- Phase angle \( \phi = 45^\circ \)

### Step 3: Calculate the inductive reactance
Since the voltage leads the current by \( 45^\circ \), we can use the tangent of the phase angle to relate the resistance and the inductive reactance:
\[ \tan(\phi) = \frac{X_L}{R} \]
For \( \phi = 45^\circ \):
\[ \tan(45^\circ) = 1 \]
Thus,
\[ \frac{X_L}{R} = 1 \]
This implies:
\[ X_L = R = 100 \, \Omega \]

### Step 4: Calculate the angular frequency
Now, calculate the angular frequency \( \omega \):
\[ \omega = 2\pi f = 2\pi \times 1000 = 2000\pi \, \text{rad/s} \]

### Step 5: Relate inductive reactance to inductance
Now we can relate the inductive reactance to the inductance:
\[ X_L = \omega L \]
Substituting the values we have:
\[ 100 = (2000\pi)L \]

### Step 6: Solve for inductance \( L \)
Rearranging the equation to solve for \( L \):
\[ L = \frac{100}{2000\pi} \]
\[ L = \frac{1}{20\pi} \]

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

### Final Answer
The correct option is B: \( \frac{1}{20\pi} \).

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