Series RLC at resonance with winding resistance: For L = 20 mH, C = 0.02 µF, and inductor winding resistance RW = 90 Ω, what is the series circuit impedance exactly at the resonant frequency?
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A90 Ω
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B0 Ω
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C20 kΩ
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D40 kΩ
Answer
Correct Answer: 90 Ω
Explanation
Introduction / Context:At series resonance, the inductive and capacitive reactances cancel, leaving only the real resistive part. In practical coils, winding resistance RW contributes a real component that limits current and sets the minimum achievable impedance.
Given Data / Assumptions:
- L = 20 mH, C = 0.02 µF.
- Inductor winding resistance RW = 90 Ω.
- Series configuration, sinusoidal steady state.
Concept / Approach:
Series impedance is Z = R_total + j(XL − Xc). At resonance, XL = Xc, hence the imaginary part vanishes and Z reduces to the real part only—here represented by RW (assuming other resistances are negligible).
Step-by-Step Solution:
Resonant condition: XL = Xc ⇒ XL − Xc = 0.Z_res = RW + j0 = 90 Ω.Verification / Alternative check:
Measured current at resonance equals V/RW, confirming that only RW limits current. Any additional series resistance would add directly to 90 Ω.
Why Other Options Are Wrong:
0 Ω ignores real-world winding loss. 20 kΩ and 40 kΩ are unrelated to resonance cancellation in series RLC.
Common Pitfalls:
Thinking that resonance makes impedance zero in series circuits; that is only true in an ideal, lossless case (R = 0). Real coils have resistance.
Final Answer:
90 Ω