1.

A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Q factor is?(a) 50(b) 100(c) 150(d) 200I had been asked this question in an online interview.This interesting question is from Problems of Series Resonance Involving Quality Factor in section Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

The CORRECT option is (c) 150

Easiest EXPLANATION: Q = \(\FRAC{ω}{ω1 – ω2} = \frac{f}{F2-f1}\)

Here, f = 1.5 × 10^6 Hz

f1 = (1.5 × 10^6 – 5 × 10^3)

f2 = (1.5 × 10^6 + 5 × 10^3)

So, f2 –f1 = 10 × 10^3 Hz

∴ Q = \(\frac{1.5 × 10^6}{10 × 10^3}\) = 150.



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