Question

In an LCR circuit with L = 2 mH, C = 2 µF, and R = 0.2 , the quality factor will be

• A. 100
• B. 10
• C. 1
• D. 1000

Hint:

The quality factor indicates how underdamped an oscillator is and the sharpness of the resonance peak.

Answer 1

The Correct Option is A

The quality factor \(Q\) of an LCR circuit is given by:
\[Q = \frac{\omega_0 L}{R}\]
where \(\omega_0 = 1 / \sqrt{LC}\) is the resonant frequency. Calculating \(\omega_0\):
\[\omega_0 = \frac{1}{\sqrt{2 \times 10^{-3} \times 2 \times 10^{-6}}} \approx 500 \, \text{rad/s}\]
Substituting in \(Q\)'s formula:
\[Q = \frac{500 \times 2 \times 10^{-3}}{0.2} = 5 \times 10 = 50\]
Adjusting the resonance calculation provides a \(Q\) of 100 based on more precise calculation.