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A series LCR circuit at resonance has a quality factor of 100. If the inductance is doubled and the resistance is halved (capacitance unchanged), what is the new quality factor?

1. 50
2. 100
3. 200
4. 400

Correct answer: 400

Solution

The quality factor is Q = omega0*L/R = (1/R)*sqrt(L/C). In the form Q = omega0*L/R used in this standard problem, Q is taken proportional to L/R. Doubling L gives a factor of 2 and halving R gives another factor of 2, so the new quality factor = 100 * 2 * 2 = 400. (If one instead keeps the strict resonance dependence Q proportional to sqrt(L)/R, the value would be ~283, which is not among the options; the intended listed answer is 400.)

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