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In an L-C-R A.C. series circuit, L = 9 H, R = 10 Ω and C = 100 μF. Hence the Q-factor of the circuit is ______.

Asked in GUJCET 2019 · Power and bandwidth at resonance

(1) 25
(2) 45
(3) 35
(4) 30

Answer: (4) 30

Step-by-step solution

Given: L = 9 H, R = 10 Ω, C = 100 μF = 10⁻⁴ F.

Idea: Q = (ω₀L)/R = 1/R√L/C.

L/C = 9/(10⁻⁴) = 9×10⁴ Ω², so √L/C = 300 Ω.

Q = (300)/(10) = 30.

Why the other options are wrong

(1) No single slip gives 25. Sanity check: √L/C = √9/(10⁻⁴) = 300 Ω, and (300)/(10) = 30.
(2) No single slip gives 45; it would need R ≈ 6.7 Ω with this L and C, but R = 10 Ω.
(3) No single slip gives 35. Sanity check: ω₀ = 1/(√LC) = (100)/3 rad s⁻¹, and (ω₀L)/R = (100)/3×9/(10) = 30.

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