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. Effective resistance of the circuit is?

(a) 2 Ω

(b) 3 Ω

(c) 5.5 Ω

(d) 4.7 Ω

The question was asked in an online interview.

The question is from Problems of Series Resonance Involving Quality Factor in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer 1

Right option is (d) 4.7 Ω

Easy explanation: R =  \(\frac{f2-f1}{2πf^2 L}\)

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

∴ R = 4.7 Ω.