$ ext{When an inductor L and a resistor R in series are connected across a 12V, 50Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from applied voltage by } \frac{\pi}{3} ext{radian. The value of R will be :}$

Q: $\text{When an inductor L and a resistor R in series are connected across a 12V, 50Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from applied voltage by } \frac{\pi}{3} \text{radian. The value of R will be :}$

$\text{When an inductor L and a resistor R in series are connected across a 12V, 50Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from applied voltage by } \frac{\pi}{3} \text{radian. The value of R will be :}$

Question

$	ext{When an inductor L and a resistor R in series are connected across a 12V, 50Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from applied voltage by } \frac{\pi}{3} 	ext{radian. The value of R will be :}$

Accepted Answer

To determine the value of the resistor $ R $ in the given circuit, we need to analyze the information provided:

1. **Given Data:**
   - Supply Voltage, $ V = 12 \, \text{V} $
   - Frequency, $ f = 50 \, \text{Hz} $
   - Current, $ I = 0.5 \, \text{A} $
   - Phase difference between current and voltage, $ \phi = \frac{\pi}{3} \, \text{radian} $

2. **Understanding the Circuit:**
   - The circuit consists of an inductor $ L $ and a resistor $ R $ in series.
   - The impedance $ Z $ of the series circuit is given by $ Z = \sqrt{R^2 + (X_L)^2} $, where $ X_L = 2\pi f L $ is the inductive reactance.

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

4. **Calculating Resistance $ R $:**
   - From the phase angle relationship, $ \frac{X_L}{R} = \sqrt{3} $, which implies $ X_L = \sqrt{3} R $.

5. **Using Ohm's Law:**
   - The total impedance $ Z $ can also be calculated using Ohm's Law: $ Z = \frac{V}{I} = \frac{12}{0.5} = 24 \, \Omega $.

6. **Substituting into the Impedance Formula:**
   - We have $ Z = \sqrt{R^2 + (X_L)^2} = \sqrt{R^2 + (3R^2)} = \sqrt{4R^2} = 2R $.
   - Setting $ 2R = 24 $, we solve for $ R $ to get $ R = 12 \, \Omega $.

7. **Conclusion:**
   - Therefore, the value of the resistor $ R $ is $ 12 \, \Omega $, which corresponds to Option 4.

**Explanation for Incorrect Options:**

- **Option 1 (4 $\Omega$):** If $ R = 4 \, \Omega $, then $ 2R = 8 \, \Omega $, which does not match the calculated total impedance of $ 24 \, \Omega $.

- **Option 2 (6 $\Omega$):** If $ R = 6 \, \Omega $, then $ 2R = 12 \, \Omega $, which again does not match the calculated total impedance of $ 24 \, \Omega $.

- **Option 3 (8 $\Omega$):** If $ R = 8 \, \Omega $, then $ 2R = 16 \, \Omega $, which still does not match the calculated total impedance of $ 24 \, \Omega $.

Thus, the only value of $ R $ that satisfies all the conditions given in the problem is $ 12 \, \Omega $, confirming that Option 4 is indeed the correct answer.