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 ΩThe 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

A circuit tuned to a frequency of 1.5 MHz and having an effective capa

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. Effective resistance of the circuit is?(a) 2 Ω(b) 3 Ω(c) 5.5 Ω(d) 4.7 ΩThe 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» 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.

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