The frequency at resonance for the circuit is (a) (1)/(4π√(LC)) (b) (1)/(2π√(LC)) (c) (1)/(π√(LC)) (d) (2)/(π√(LC))

Answer: (c). Two L in parallel: Leq = L/2. Two C in series: Ceq = C/2. f₀ = 12π√LeqCeq = (1)/(2π√(LC/4)) = (1)/(π√(LC)) So the answer is (c).

The frequency at resonance for the circuit is | MHT CET 2025 (26 Apr, Shift 2)

Q25MHT CET 2025 · 26 APR · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

The frequency at resonance for the circuit is

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Answer: (c)

Two $L$ in parallel: $L_{eq} = \dfrac{L}{2}$ .
Two $C$ in series: $C_{eq} = \dfrac{C}{2}$ .
$f_0 = \dfrac{1}{2\pi\sqrt{L_{eq}C_{eq}}} = \dfrac{1}{2\pi\sqrt{LC/4}} = \dfrac{1}{\pi\sqrt{LC}}$
So the answer is (c).

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