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

Options

A. Option1 `(1//10sqrt(2)pi)`
B. Option2 `(1//20sqrt(2)pi)`
C. Option3 `(1//40sqrt(2)pi)`
D. Option4 `(1//60sqrt(2)pi)`

Answer

To find the inductance of the coil in the given L-R circuit, we can follow these steps:

### Step 1: Understand the given information
- Impedance (Z) = 100 ohms
- Phase difference (φ) = 45°
- Frequency (f) = 1000 Hz

### Step 2: Use the relationship between impedance, resistance, and reactance
In an L-R circuit, the impedance (Z) is given by the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( R \) is the resistance and \( X_L \) is the inductive reactance.

### Step 3: Use the relationship between phase angle and reactance
The phase angle (φ) is related to the reactance and resistance by:
\[ \tan(\phi) = \frac{X_L}{R} \]
Given that \( \phi = 45° \), we know:
\[ \tan(45°) = 1 \]
Thus, we can conclude:
\[ X_L = R \]

### Step 4: Substitute \( X_L \) in the impedance formula
Now, substituting \( X_L = R \) into the impedance formula:
\[ Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]
Given that \( Z = 100 \) ohms, we can write:
\[ 100 = R\sqrt{2} \]

### Step 5: Solve for \( R \)
Rearranging gives:
\[ R = \frac{100}{\sqrt{2}} = 50\sqrt{2} \text{ ohms} \]

### Step 6: Calculate the inductive reactance \( X_L \)
Since \( X_L = R \):
\[ X_L = 50\sqrt{2} \text{ ohms} \]

### Step 7: Relate inductive reactance to inductance
The inductive reactance is given by:
\[ X_L = 2\pi f L \]
Substituting the known values:
\[ 50\sqrt{2} = 2\pi (1000) L \]

### Step 8: Solve for \( L \)
Rearranging gives:
\[ L = \frac{50\sqrt{2}}{2\pi \times 1000} \]
\[ L = \frac{25\sqrt{2}}{1000\pi} \]
\[ L = \frac{1}{40\pi\sqrt{2}} \text{ Henry} \]

### Final Answer
The inductance \( L \) of the coil is:
\[ L = \frac{1}{40\pi\sqrt{2}} \text{ Henry} \]

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