A series LCR circuit is connected to an AC voltage source. When the capacitor is removed, the phase difference between current and voltage is pi/3. When the inductor is removed, the phase difference is again pi/3. The power factor of the original circuit is:

B. 1 — Removing the capacitor gives tan(phi) = XL/R = sqrt(3), and removing the inductor gives tan(phi) = XC/R = sqrt(3). Since XL = XC, the circuit is at resonance, making the power factor equal to 1.

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$A series LCR circuit is connected to an AC voltage source. When the capacitor is removed, the phase difference between current and voltage is π /3 . When the inductor is removed, the phase difference is again π /3 . The power factor of the original circuit is:$

A. $0$
B. $1$
C. $0.5$
D. $1/ 2$

Show correct answer & step-by-step solution

Correct answer: B — $1$

Solution

Removing the capacitor gives tan (ϕ) = X_{L} / R = 3, and removing the inductor gives tan (ϕ) = X_{C} / R = 3 . Since X_{L} = X_{C}, the circuit is at resonance, making the power factor cos (0) = 1 .

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A series LCR circuit is connected to an AC voltage source. When the capacitor is removed, the phase difference between current and voltage is π/3. When the inductor is removed, the phase difference is again π/3. The power factor of the original circuit is:

Solution

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More AC Circuits practice questions

$A series LCR circuit is connected to an AC voltage source. When the capacitor is removed, the phase difference between current and voltage is π /3 . When the inductor is removed, the phase difference is again π /3 . The power factor of the original circuit is:$