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Question

A \( 169.7 \, \text{V} \) (peak) AC source is connected to a \( 60 \, \Omega \) resistor. What is the
average power consumed?

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Choose one · Correct answer highlighted

Explanation

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. RMS voltage: V = (v_m/√(2)) = (169.7/1.414) ≈ 120 V . RMS current: I = (V/R) = (120/60) = 2 A . Average power: P = I² R = 2² × 60 = 240 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 240 W, consistent with phasor analysis and resonance condition X_L = X_C.

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