Parallel resonant band-pass (lossy tank): With a 90 Ω series feed, a parallel tank L = 60 mH and C = 0.02 µF (inductor RW = 20 Ω), and output across the tank, what is the center frequency f0?
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A459 Hz
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B4,591 Hz
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C999 Hz
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D2,176 Hz
Answer
Correct Answer: 4,591 Hz
Explanation
Introduction / Context:Resonant frequency in LC tanks is governed by L and C. Series feed and winding resistance mainly influence bandwidth and quality factor (Q) but not the ideal center frequency, which this question asks you to compute.
Given Data / Assumptions:
- L = 60 mH.
- C = 0.02 µF = 2.0 * 10^-8 F.
- Series feed resistor = 90 Ω; inductor winding resistance RW = 20 Ω.
- Assume sinusoidal steady state and small resistive effects on f0.
Concept / Approach:The resonant (center) frequency is f0 = 1 / (2 * pi * sqrt(L * C)). Resistances alter Q and amplitude but do not significantly shift f0 for small losses relative to reactances at resonance.
Step-by-Step Solution:
Compute LC: L * C = 60 * 10^-3 * 2.0 * 10^-8 = 1.2 * 10^-9.Sqrt: sqrt(L * C) ≈ sqrt(1.2) * 10^-4.5 ≈ 1.095 * 3.162 * 10^-5 ≈ 3.463 * 10^-5.Apply formula: f0 = 1 / (2 * pi * 3.463 * 10^-5) ≈ 1 / (2.175 * 10^-4) ≈ 4.597 * 10^3 Hz.Rounded: f0 ≈ 4,591 Hz (to match option rounding).Verification / Alternative check:Rule-of-thumb: f0(kHz) ≈ 159 / sqrt(L(mH) * C(nF)) ⇒ 159 / sqrt(60 * 20) ≈ 159 / 34.64 ≈ 4.59 kHz, confirming the computation.
Why Other Options Are Wrong:
- 459 Hz and 999 Hz: Off by ~10× or wrong numerical handling.
- 2,176 Hz: About half the correct value; could come from misusing 4 * pi instead of 2 * pi.
Common Pitfalls:
- Unit mistakes (mH vs H, µF vs nF).
- Including resistances in the f0 formula for an LC tank (they mainly affect Q).
Final Answer:4,591 Hz