A coil of resistance 300 ohm and inductance 1.0 Henry is connected across an alternating voltage source of frequency $\frac{300}{2\pi}Hz$. The value of phase difference between voltage and current will be

Question

Accepted Answer

To determine the phase difference between the voltage and current in a circuit consisting of a coil with resistance and inductance connected to an alternating voltage source, we need to consider the impedance of the coil and the relationship between the resistive and inductive components.

1. **Explanation for Option 2 ($45^{\circ}$) as the Correct Answer:**

The impedance $ Z $ of a coil with resistance $ R $ and inductance $ L $ in an AC circuit is given by:
   $$
   Z = \sqrt{R^2 + (X_L)^2}
   $$
   where $ X_L = 2\pi f L $ is the inductive reactance, and $ f $ is the frequency of the AC source.

Given:
   - $ R = 300 \, \text{ohm} $
   - $ L = 1.0 \, \text{Henry} $
   - $ f = \frac{300}{2\pi} \, \text{Hz} $

Calculate the inductive reactance $ X_L $:
   $$
   X_L = 2\pi \left(\frac{300}{2\pi}\right) \times 1.0 = 300 \, \text{ohm}
   $$

The phase difference $ \phi $ between the voltage and current is given by:
   $$
   \tan \phi = \frac{X_L}{R} = \frac{300}{300} = 1
   $$

Therefore, the phase angle $ \phi $ is:
   $$
   \phi = \tan^{-1}(1) = 45^{\circ}
   $$

Thus, the correct answer is Option 2, which is $ 45^{\circ} $.

2. **Explanation for Why Other Options are Incorrect:**

- **Option 1 ($0^{\circ}$):** This would imply that the circuit is purely resistive with no inductive reactance, which is not the case here as the coil has both resistance and inductance. Therefore, the phase difference cannot be $ 0^{\circ} $.

- **Option 3 ($60^{\circ}$):** A phase difference of $ 60^{\circ} $ would require a different ratio of inductive reactance to resistance. Specifically, it would require $ \tan \phi = \sqrt{3} $, which is not satisfied here as $ \tan \phi = 1 $.

- **Option 4 ($90^{\circ}$):** This would imply a purely inductive circuit with no resistance, where the current lags the voltage by $ 90^{\circ} $. However, since there is a resistance of 300 ohms, the phase difference cannot be $ 90^{\circ} $.

In conclusion, the correct phase difference between the voltage and current in this circuit is $ 45^{\circ} $, which supports Option 2 as the correct answer.