Practice question
Question
A \( 150 \, \text{V} \) (rms) AC source supplies a \( 75 \, \Omega \) resistor. What is the average
power consumed?
Explanation
**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. RMS current: I = (V/R) = (150/75) = 2 A . Average power: P = I² R = 2² × 75 = 4 × 75 = 300 W . 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 300
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