Question

A resistance of $200\Omega$ and an inductor of $\frac{1}{2\pi}\text{ H}$ are connected in series to a.c. voltage of $40\text{ V}$ and $100\text{ Hz}$ frequency. The phase angle between the voltage and current is

• A. $\tan^{-1}(1/5)$
• B. $\tan^{-1}(1/4)$
• C. $\tan^{-1}(1/3)$
• D. $\tan^{-1}(0.5)$

Hint:

Phase angle $\phi$ tells you how much the voltage leads the current in an inductive circuit. $\tan \phi = X_L/R$.

Answer 1

The Correct Option is D

Step 1: Concept
In an RL series circuit, the phase angle $\phi$ is given by $\tan \phi = \frac{X_L}{R}$.

Step 2: Meaning
$X_L = 2\pi f L$. Given $f = 100 \text{ Hz}$, $L = 1/2\pi \text{ H}$, $R = 200\Omega$.

Step 3: Analysis
$X_L = 2\pi(100)(\frac{1}{2\pi}) = 100\Omega$. $\tan \phi = \frac{100}{200} = \frac{1}{2} = 0.5$.

Step 4: Conclusion
The phase angle is $\tan^{-1}(0.5)$. Final Answer: (D)