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A radio receiver’s tuned circuit contains a 50 Ω resistance, a 10 mH inductor, and a variable capacitor. If a 1 MHz radio signal induces a potential difference of 0.1 mV, the capacitor value required for resonance is

1. 2.5 pF
2. 5.0 pF
3. 25 pF
4. 50 pF

Correct answer: 2.5 pF

Solution

At resonance C = 1/((2*pi*f)^2 * L) = 1/((2*pi*1e6)^2 * 0.01) = 2.5e-12 F = 2.5 pF.

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