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What inductance L must be connected in series with a 5 microF capacitor, a 10 ohm resistor and a 50 Hz AC source so that the circuit power factor becomes unity (resonance)? (Construct plausible options.)
- 2.02 H
- 1.01 H
- 0.51 H
- 5.05 H
Correct answer: 2.02 H
Solution
For unity power factor the circuit must resonate: omega*L = 1/(omega*C), giving L = 1/(omega²*C). With omega = 2*pi*50 = 314.16 rad/s and C = 5e-6 F, L = 1/((314.16)² * 5e-6) = 1/(98696*5e-6) = 1/0.4935 = 2.026 H.
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