$ ext{A bulb of resistance } 10 \, \Omega ext{, connected to an inductor of inductance } L, ext{ is in series with a source marked 100 V, 50 Hz. If the phase angle between the voltage and current is } \frac{\pi}{4} ext{ radian. The value of } L ext{ will be :}$

Q: $\text{A bulb of resistance } 10 \, \Omega \text{, connected to an inductor of inductance } L, \text{ is in series with a source marked 100 V, 50 Hz. If the phase angle between the voltage and current is } \frac{\pi}{4} \text{ radian. The value of } L \text{ will be :}$

$\text{A bulb of resistance } 10 \, \Omega \text{, connected to an inductor of inductance } L, \text{ is in series with a source marked 100 V, 50 Hz. If the phase angle between the voltage and current is } \frac{\pi}{4} \text{ radian. The value of } L \text{ will be :}$

Question

$	ext{A bulb of resistance } 10 \, \Omega 	ext{, connected to an inductor of inductance } L, 	ext{ is in series with a source marked 100 V, 50 Hz. If the phase angle between the voltage and current is } \frac{\pi}{4} 	ext{ radian. The value of } L 	ext{ will be :}$

Accepted Answer

To determine the value of the inductance $ L $ that results in a phase angle of $\frac{\pi}{4}$ radians between the voltage and current in a series circuit consisting of a resistor and an inductor, we can use the concept of impedance in an AC circuit.

### Explanation:

1. **Impedance in an R-L Circuit:**
   - The impedance $ Z $ of a series R-L circuit is given by:
     $$
     Z = \sqrt{R^2 + (X_L)^2}
     $$
     where $ R $ is the resistance and $ X_L $ is the inductive reactance.

2. **Inductive Reactance:**
   - The inductive reactance $ X_L $ is given by:
     $$
     X_L = 2\pi f L
     $$
     where $ f $ is the frequency of the AC source.

3. **Phase Angle:**
   - The phase angle $ \phi $ between the voltage and current is given by:
     $$
     \tan \phi = \frac{X_L}{R}
     $$
   - Given that $ \phi = \frac{\pi}{4} $, we have:
     $$
     \tan \frac{\pi}{4} = 1 = \frac{X_L}{R}
     $$
     Therefore, $ X_L = R $.

4. **Calculating $ L $:**
   - Since $ X_L = R $, we have:
     $$
     2\pi f L = R
     $$
   - Substituting the given values $ R = 10 \, \Omega $ and $ f = 50 \, \text{Hz} $:
     $$
     2\pi \times 50 \times L = 10
     $$
   - Solving for $ L $:
     $$
     L = \frac{10}{2\pi \times 50} = \frac{1}{10\pi} \approx 0.0318 \, \text{H}
     $$

### Conclusion:

- **Option 1 (0.0318 H)** is correct because it matches the calculated value of $ L $ that satisfies the condition for the given phase angle.

### Explanation of Incorrect Options:

- **Option 2 (0.318 H):** This value is ten times larger than the correct value. It would result in a much larger inductive reactance, leading to a phase angle greater than $\frac{\pi}{4}$.

- **Option 3 (0.0787 H):** This value is more than double the correct value. It would also result in a larger inductive reactance, leading to a phase angle greater than $\frac{\pi}{4}$.

- **Option 4 (0.787 H):** This value is significantly larger than the correct value. It would result in an even larger inductive reactance, leading to a phase angle much greater than $\frac{\pi}{4}$.

Thus, the correct answer is indeed Option 1, which is $ L = 0.0318 \, \text{H} $.