Impedence of `L-R Circuit is `100ohm` and phase difference between source voltage and source current is `45^@` of frequency of source `1000Hz` then find inductance of coil.

Options

A. `25Sqrt2`
B. `(50sqrt2/pie)`
C. `(25sqrt2/pie)`
D. `(20sqrt2/pie)`

Answer

To solve the problem of finding the inductance \( L \) of an L-R circuit given the impedance \( Z \), phase difference \( \phi \), and frequency \( f \), we can follow these steps:

### Step 1: Understand the given parameters
We have:
- Impedance \( Z = 100 \, \Omega \)
- Phase difference \( \phi = 45^\circ \)
- Frequency \( f = 1000 \, \text{Hz} \)

### Step 2: Convert phase difference to tangent form
The phase difference \( \phi \) in an L-R circuit can be expressed as:
\[ \tan(\phi) = \frac{X_L}{R} \]
where \( X_L \) is the inductive reactance and \( R \) is the resistance. Given that \( \phi = 45^\circ \):
\[ \tan(45^\circ) = 1 \implies X_L = R \]

### Step 3: Relate impedance to resistance and reactance
The impedance \( Z \) of an L-R circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Since we have established that \( X_L = R \), we can substitute this into the impedance equation:
\[ Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

### Step 4: Solve for resistance \( R \)
From the impedance equation:
\[ Z = R\sqrt{2} \]
Substituting \( Z = 100 \, \Omega \):
\[ 100 = R\sqrt{2} \implies R = \frac{100}{\sqrt{2}} = 50\sqrt{2} \, \Omega \]

### Step 5: Calculate inductive reactance \( X_L \)
Since \( X_L = R \):
\[ X_L = 50\sqrt{2} \, \Omega \]

### Step 6: Relate inductive reactance to inductance
The inductive reactance \( X_L \) is also given by:
\[ X_L = \omega L \]
where \( \omega = 2\pi f \). First, calculate \( \omega \):
\[ \omega = 2\pi \times 1000 = 2000\pi \, \text{rad/s} \]
Now substituting for \( X_L \):
\[ 50\sqrt{2} = (2000\pi)L \]

### Step 7: Solve for inductance \( L \)
Rearranging the equation gives:
\[ L = \frac{50\sqrt{2}}{2000\pi} \]
Simplifying:
\[ L = \frac{\sqrt{2}}{40\pi} \, \text{H} \]

### Step 8: Convert to millihenries
To convert \( L \) from henries to millihenries:
\[ L = \frac{\sqrt{2}}{40\pi} \times 1000 \, \text{mH} = \frac{1000\sqrt{2}}{40\pi} \, \text{mH} = \frac{25\sqrt{2}}{\pi} \, \text{mH} \]

### Final Answer
Thus, the inductance \( L \) of the coil is:
\[ L = \frac{25\sqrt{2}}{\pi} \, \text{mH} \]