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: Understanding the Question:
We are given an LCR series circuit. We need to find \(\tan \phi\), where \(\phi\) is the phase difference between the source voltage and current.
Step 2: Key Formula or Approach:
1) Inductive reactance: \(X_{L} = \omega L = 2\pi f L\).
2) Capacitive reactance: \(X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi f C}\).
3) Phase angle: \(\tan \phi = \frac{|X_{L} - X_{C}|}{R}\).
Step 3: Detailed Explanation:
Given:
\(L = \frac{100}{\pi} \text{ mH} = \frac{0.1}{\pi} \text{ H}\)
\(C = \frac{10^{-3}}{2\pi} \text{ F}\)
\(R = 10 \Omega\)
\(f = 50 \text{ Hz}\)
Calculate reactances:
\[ X_{L} = 2\pi (50) \left(\frac{0.1}{\pi}\right) = 100 \times 0.1 = 10 \Omega \]
\[ X_{C} = \frac{1}{2\pi (50) \left(\frac{10^{-3}}{2\pi}\right)} = \frac{1}{50 \times 10^{-3}} = \frac{1000}{50} = 20 \Omega \]
Now find \(\tan \phi\):
\[ \tan \phi = \frac{|X_{L} - X_{C}|}{R} = \frac{|10 - 20|}{10} = \frac{10}{10} = 1 \]
Step 4: Final Answer:
The tangent of the phase angle is \(1\).