Question:

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

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When $X_L = X_C$, the circuit is in resonance and the phase angle is $0$.
Updated On: Apr 30, 2026
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The Correct Option is D

Solution and Explanation


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)
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