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In a series LR circuit, the inductive reactance is X_L = 3R. A capacitor with reactance X_C = R is now added in series. Find the ratio of the new power factor to the old power factor.
- 1
- 2
- 1/sqrt(2)
- sqrt(2)
Correct answer: sqrt(2)
Solution
Old impedance Z_old = sqrt(R² + (3R)²) = R*sqrt(10); pf_old = R/(R*sqrt(10)) = 1/sqrt(10). New X_net = 2R, Z_new = sqrt(R² + (2R)²) = R*sqrt(5); pf_new = 1/sqrt(5). Ratio = (1/sqrt(5))/(1/sqrt(10)) = sqrt(10/5) = sqrt(2).
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