Compute the resonant frequency for a series RLC circuit with L = 8 mH and C = 40 µF (standard formula f0 = 1 / (2π√(LC))).
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A2,810 Hz
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B281 Hz
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C28.1 Hz
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D10 kHz
Answer
Correct Answer: 281 Hz
Explanation
Introduction / Context:Resonant frequency is the cornerstone of tuned-circuit design. For a given L and C, f0 indicates where series reactances cancel and current peaks, which is vital for filters, oscillators, and matching networks.
Given Data / Assumptions:
- L = 8 mH = 8 × 10^−3 H.
- C = 40 µF = 40 × 10^−6 F.
- Ideal, linear, time-invariant components.
Concept / Approach:
Use the standard formula: f0 = 1 / (2 * π * √(L * C)). Carefully handle unit conversions to avoid decade and decimal errors.
Step-by-Step Solution:
LC = 8e−3 * 40e−6 = 3.2e−7.√(LC) ≈ √(3.2e−7) ≈ 5.654e−4.f0 = 1 / (2 * π * 5.654e−4) ≈ 1 / 0.003552 ≈ 281.6 Hz.Rounded to choices: 281 Hz.Verification / Alternative check:
Back-of-envelope: √(LC) ~ 5.65e−4 s ⇒ period ~ 3.55 ms ⇒ frequency ~ 1/3.55 ms ≈ 281 Hz—consistent.
Why Other Options Are Wrong:
2,810 Hz is an order-of-magnitude (×10) error. 28.1 Hz is a ÷10 error. 10 kHz is unrelated to LC values given.
Common Pitfalls:
Forgetting milli and micro conversions; misplacing decimal points; swapping series vs. parallel intuition (resonant frequency formula is the same).
Final Answer:
281 Hz