Question

An inductor of ((\frac{100}{\pi}) mH), capacitor of capacitance ((\frac{10^{-3}}{2\pi}) F) and resistance of (10 \Omega) are connected in series with an AC voltage source of (110 V), (50 Hz) supply. The tangent of the phase angle ' (\phi) ' between voltage and current is

• A. (4)
• B. (3)
• C. (2)
• D. [suspicious link removed]

Hint:

When $X_L = X_C$, the circuit is in resonance and the phase angle is $0$.

Answer 1

The Correct Option is D

Step 1: Calculate Reactances
$X_L = 2\pi fL = 2\pi(50)\left(\frac{100 \times 10^{-3}}{\pi}\right) = 10 \Omega$.
$X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi(50)\left(\frac{10^{-3}}{2\pi}\right)} = \frac{1}{0.05} = 20 \Omega$.

Step 2: Phase Angle Formula
$\tan \phi = \frac{|X_L - X_C|}{R} = \frac{|10 - 20|}{10} = \frac{10}{10} = 1$.

Step 3: Conclusion
The tangent of the phase angle is 1.
Final Answer: (D)