Question

A series LCR circuit with \( L = \frac{100}{\pi} \) mH, \( C = \frac{10^{-3}}{\pi} \) F and \( R = 10 \, \Omega \) is connected across an AC source of 220 V, 50 Hz supply. The power factor of the circuit would be _____.

Answer 1

Correct Answer: 1

The solution involves calculating the power factor of the LCR circuit using given values. First, calculate the inductive reactance \( X_L \) and capacitive reactance \( X_C \) using the given formulas:

• \( X_L = 2\pi f L \) 
• \( X_C = \frac{1}{2\pi f C} \)

With \( f = 50 \) Hz, \( L = \frac{100}{\pi} \) mH, and \( C = \frac{10^{-3}}{\pi} \) F, the reactances are:

• \( X_L = 2\pi \times 50 \times \frac{100}{\pi} \times 10^{-3} = 10 \, \Omega \)
• \( X_C = \frac{1}{2\pi \times 50 \times \frac{10^{-3}}{\pi}} = 10 \, \Omega \)

Next, compute the impedance \( Z \) of the circuit where \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Since \( X_L = X_C \), the impedance becomes:

• \( Z = R = 10 \, \Omega \)

The power factor \( \cos\phi \) is the ratio of resistance to impedance:

• \( \cos\phi = \frac{R}{Z} = \frac{10}{10} = 1 \)

Thus, the power factor of the circuit is 1, which is within the expected range.

Answer 2

Given:
\(L = \frac{100}{\pi} \, \text{mH} = \frac{100}{\pi} \times 10^{-3} \, \text{H}, \, C = 10^{-3} \, \text{F}, \, R = 10 \, \Omega, \, f = 50 \, \text{Hz}.\)

The inductive reactance is given by:
\(X_L = 2\pi f L = 2\pi \times 50 \times \frac{100}{\pi} \times 10^{-3} = 10 \, \Omega.\)

The capacitive reactance is given by:
\(X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 10^{-3}} = 10 \, \Omega.\)

Since \(X_L = X_C\), the circuit is in resonance. Therefore, the impedance is:
\(Z = R = 10 \, \Omega.\)

\(\text{Power Factor} = \frac{R}{Z} = 1.\)

The Correct answer is: 1