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?

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:

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