Question:
In the given LCR series circuit the quality factor is
In the given LCR series circuit the quality factor is
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The quality factor \( Q \) represents the damping of the circuit, with higher values indicating lower energy losses relative to the stored energy.
Updated On: Mar 18, 2026
- \(54.6\)
- \(64\)
- \(48\)
- \(50\)
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The Correct Option is D
Solution and Explanation
Given Values:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \] \[ \omega_0 = \frac{1}{\sqrt{(4 \, \text{H}) \times (16 \times 10^{-6} \, \text{F})}} \] \[ \omega_0 = \frac{1}{\sqrt{64 \times 10^{-6}}} \] \[ \omega_0 = \frac{1}{8 \times 10^{-3}} \] \[ \omega_0 = 125 \, \text{rad/s} \] 2. Calculate the Quality Factor (Q):
\[ Q = \frac{\omega_0 L}{R} \] \[ Q = \frac{(125 \, \text{rad/s}) \times (4 \, \text{H})}{10 \, \Omega} \] \[ Q = \frac{500}{10} \] \[ Q = 50 \] Therefore, the quality factor of the LCR series circuit is 50. The correct answer is (4) 50.
- Resistance (\( R \)) = 10 \( \Omega \)
- Inductance (\( L \)) = 4 H
- Capacitance (\( C \)) = 16 \( \mu \)F = \( 16 \times 10^{-6} \) F
\[ \omega_0 = \frac{1}{\sqrt{LC}} \] \[ \omega_0 = \frac{1}{\sqrt{(4 \, \text{H}) \times (16 \times 10^{-6} \, \text{F})}} \] \[ \omega_0 = \frac{1}{\sqrt{64 \times 10^{-6}}} \] \[ \omega_0 = \frac{1}{8 \times 10^{-3}} \] \[ \omega_0 = 125 \, \text{rad/s} \] 2. Calculate the Quality Factor (Q):
\[ Q = \frac{\omega_0 L}{R} \] \[ Q = \frac{(125 \, \text{rad/s}) \times (4 \, \text{H})}{10 \, \Omega} \] \[ Q = \frac{500}{10} \] \[ Q = 50 \] Therefore, the quality factor of the LCR series circuit is 50. The correct answer is (4) 50.
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