Question

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) 200

The question was asked in an interview for internship.

This question is from Applications of Signals on Circuits in division Signals and Systems Basics of Signals and Systems

Answer 1

Correct answer is (c) 150

The best I can explain: \(Q = \frac{ω}{ω1 – ω2} = \frac{f}{f2-f1} \)

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.