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A resonant (tuned) circuit has inductance 2 x 10⁻⁴ H and resistance 6.28 ohm, and oscillates at a frequency of 10 MHz. Determine the quality factor Q of this resonator. (Take pi = 3.14.)

1. 2000
2. 200
3. 20000
4. 1000

Correct answer: 2000

Solution

The quality factor of a series resonant circuit is Q = omega*L/R = 2*pi*f*L/R. With f = 10 MHz = 1e7 Hz, L = 2e-4 H, R = 6.28 ohm: omega = 2*3.14*1e7 = 6.28e7 rad/s. Q = (6.28e7 * 2e-4)/6.28 = (1.256e4)/6.28 = 2000.

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