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A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 50. If R and L are doubled and C is kept same, the new Q of the circuit is?(a) 25.52(b) 35.35(c) 45.45(d) 20.02The question was posed to me in a job interview.My doubt stems from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer» RIGHT answer is (b) 35.35

Easy explanation: Quality FACTOR Q of the series RLC circuit is given by, Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)

Qnew = \(\frac{1}{2R} \sqrt{\frac{2L}{C}} = \frac{1}{2} ×\frac{1}{R} \sqrt{\frac{2L}{C}} = \frac{1}{2} × \sqrt{2} × Q\) = 35.35.


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