A \( 226.3 \, \text{V} \) (peak) AC source is connected to a \( 80 \, \Omega \) resistor. What is the average power cons
**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. RMS voltage: V = (v_m/√(2)) = (226.3/1.414) ≈ 160 V . RMS current: I = (V/R) = (160/80) = 2 A . Average power: P = I² R = 2² × 80 = 320 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 320 W, consistent
Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values
