Question

A coil having an inductance of $\frac{1}{\pi} \text{ H}$ is connected in series with a resistance of $300 \Omega$. If $20 \text{ V}$ from a $200 \text{ Hz}$ source are impressed across the combination, the value of the phase angle between the voltage and the current is

• A. $\tan^{-1} \left(\frac{5}{4}\right)$
• B. $\tan^{-1} \left(\frac{4}{5}\right)$
• C. $\tan^{-1} \left(\frac{3}{4}\right)$
• D. $\tan^{-1} \left(\frac{4}{3}\right)$

Hint:

In an L-R circuit, the current lags the voltage by an angle $\phi = \tan^{-1}(X_L/R)$.

Answer 1

The Correct Option is A

Step 1: Calculate the inductive reactance ($X_L$) using the frequency ($f = 200 \text{ Hz}$) and inductance ($L = \frac{1}{\pi} \text{ H}$). \[ X_L = 2 \pi f L = 2 \pi \times 200 \times \frac{1}{\pi} \] \[ X_L = 400 \Omega \]
Step 2: In a series L-R circuit, the phase angle ($\phi$) between voltage and current is given by: \[ \tan \phi = \frac{X_L}{R} \]
Step 3: Substitute the resistance ($R = 300 \Omega$) and the calculated $X_L$. \[ \tan \phi = \frac{400}{300} = \frac{4}{3} \]
Step 4: Solve for the phase angle. \[ \phi = \tan^{-1} \left(\frac{4}{3}\right) \]