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Question

A series LCR circuit has \( R = 25 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 70 \, \Omega \).
What is the phase angle?

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Explanation

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. tan Φ = (X_C - X_L/R) = (70 - 50/25) = (20/25) = 0.8 . Φ = tan⁻¹(0.8) ≈ 38.66° . Since X_C > X_L , current leads voltage. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 38.66°, consistent with phasor analysis and resonance condition X_L = X_C.

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