Series resonant band-pass filter at resonance: For L = 2 mH (with 12 Ω winding resistance), C = 0.005 µF, and a 120 Ω series resistor driven by 12 V(rms), what is the output magnitude across the 120 Ω resistor at the center frequency f0?

Introduction / Context:
This question assesses understanding of series resonant band-pass filters. At resonance, the reactive impedances of the inductor and capacitor cancel, leaving only the series resistances to set current and voltage division.

Given Data / Assumptions:

• L = 2 mH with winding resistance RW = 12 Ω.
• C = 0.005 µF.
• External series resistor R = 120 Ω (output across this R).
• Input voltage Vin = 12 V(rms).
• Ideal capacitor; sinusoidal steady-state at f0.

Concept / Approach:
At resonance of a series RLC circuit, XL = XC and the net reactive part is zero. The loop then reduces to a purely resistive series network whose total resistance is Rtotal = R + RW. The output across R is found by simple voltage division.

Step-by-Step Solution:

Verification / Alternative check:
The series current at resonance is I = Vin / Rtotal = 12 / 132 ≈ 0.0909 A. Then Vout = I * R = 0.0909 * 120 ≈ 10.909 V, confirming the divider result.

Why Other Options Are Wrong:

• 12 V: Would be true only if RW = 0 Ω and R = Rtotal; still the output would be less than the input if any other series resistance exists.
• 1.09 V and 1.1 V: Off by a factor of 10; likely from misplacing a decimal point or using 12/132 directly as the answer.

Common Pitfalls:

• Ignoring inductor winding resistance at resonance.
• Assuming reactive voltages vanish individually; only their net cancels while large circulating currents can still exist.

Final Answer:
10.9 V