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In a series LCR circuit, the current amplitude drops to 1/sqrt(2) of its peak (resonance) value at the two angular frequencies 212 rad/s and 232 rad/s. The circuit resistance is R = 5 ohm. Determine the self-inductance L of the circuit (in mH).

1. 250 mH
2. 125 mH
3. 500 mH
4. 62.5 mH

Correct answer: 250 mH

Solution

At the half-power frequencies the current amplitude is 1/sqrt(2) of the resonance value, so the power is half. The angular-frequency bandwidth between these two points equals R/L. Using the given separation, L = R/delta-omega.

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