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A 440 V, 50 Hz AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, the half power frequencies of the circuit are ______________

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Answer & Explanation

Correct AnswerA. 50.53 Hz, 49.57 Hz

Quick Explanation

For a high-Q series RLC circuit, bandwidth is Δf = R/(2πL). With R=0.1 Ω and L=15 mH, Δf≈1.061 Hz. Half-power frequencies lie approximately symmetrically around 50 Hz, giving about 49.47 Hz and 50.53 Hz; the nearest option is 49.57 Hz and 50.53 Hz.

Formula / Key Relation

Bandwidth:

BW = R/(2πL)

= 0.1/(2π×0.015)

≈ 1.061 Hz

For high Q:

f₁ ≈ f₀ − BW/2 ≈ 49.47 Hz

f₂ ≈ f₀ + BW/2 ≈ 50.53 Hz

Detailed Explanation

Concept & Reasoning

At series resonance the impedance is minimum and current is maximum. The half-power points occur where |Z|=√2R. For a narrow bandwidth, f₂−f₁=R/(2πL), and f₁f₂=f₀² exactly. Using these relationships gives frequencies close to the listed pair. The 440 V value does not affect the half-power frequencies.

How to Solve It in the Exam

Series RLC bandwidth depends on R/L, not supply voltage.

Important Exam Point

For higher accuracy use f₁f₂=f₀² together with f₂−f₁=BW.

Related Revision Path

This question sits in the revision path Signals and Systems → RLC Circuits → Resonance .

Half-Power Frequencies of Series RLC explanatory diagram

Quick Trick

Series RLC bandwidth depends on R/L, not supply voltage.

Common Mistake

Using 440 V in the frequency calculation.

Exam Tip

For higher accuracy use f₁f₂=f₀² together with f₂−f₁=BW.

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