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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?
- 50
- 100
- 200
- 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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