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In a series RLC circuit connected to an AC voltage source, when L is removed the phase difference between voltage and current is pi/3. When C is removed instead, the phase difference is also pi/3. What is the power factor of the complete RLC circuit?

1. 1
2. sqrt(3)/2
3. 1/2
4. 1/sqrt(2)

Correct answer: 1

Solution

Without L (RC series): phase difference = pi/3 means 1/(omega*C*R) = tan(pi/3) = sqrt(3). Without C (RL series): phase difference = pi/3 means omega*L/R = tan(pi/3) = sqrt(3). From both equations: omega*L/R = 1/(omega*C*R), so omega*L = 1/(omega*C). This is exactly the resonance condition for the RLC circuit. At resonance, inductive reactance equals capacitive reactance, net reactance = 0, impedance = R, and power factor = cos(0) = 1.

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