Question

A coil of resistance $450\Omega$ and self-inductance $1.5 \text{ henry}$ is connected to an a.c. source of frequency $\frac{150}{\pi} \text{ Hz}$. The phase difference between voltage and current is

• A. $\tan^{-1}(0.5)$
• B. $\tan^{-1}(1)$
• C. $\tan^{-1}(1.5)$
• D. $\tan^{-1}(2.0)$

Hint:

When $X_L = R$, the phase difference is always 45° or $\pi/4$ radians.

Answer 1

The Correct Option is B

Step 1: Concept
In an RL circuit, the phase difference $\phi$ is given by $\tan \phi = \frac{X_L}{R}$, where $X_L = 2\pi f L$.

Step 2: Meaning
We need to calculate the inductive reactance $X_L$ using the given frequency and inductance.

Step 3: Analysis
$X_L = 2\pi \left(\frac{150}{\pi}\right) (1.5) = 2 \times 150 \times 1.5 = 300 \times 1.5 = 450\Omega$.
$\tan \phi = \frac{X_L}{R} = \frac{450}{450} = 1$.

Step 4: Conclusion
The phase difference $\phi$ is $\tan^{-1}(1)$ (which is 45°). Final Answer: (B)