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Question

A series LCR circuit with \( R = 50 \, \Omega \), \( X_L = 70 \, \Omega \), \( X_C = 30 \, \Omega \)
has a \( 200 \, \text{V} \) (rms) source. What is the power dissipated?

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Explanation

**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). Z = √(R² + (X_L - X_C)²) = √(50² + (70 - 30)²) = √(2500 + 1600) = √(4100) ≈ 64 Ω . RMS current: I = (V/Z) = (200/64) ≈ 3.125 A . Power: P = I² R = (3.125)² × 50 ≈ 488.28 W . Applying X_L = ωL, X_C =

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