A factory motor draws 20 A (rms) from a 230 V (rms) AC supply, and the current lags the voltage by 60°. Find (a) the power factor, (b) the real power consumed, (c) the apparent power, (d) the real power if capacitors correct the power factor to 0.95 while the motor's real power stays the same — what current is then drawn? Step 1: Power factor = cosφ = cos(60°) = 0.50. Step 2: Real power P = V_rms·I_rms·cosφ = 230 × 20 × 0.50 = 2300 W. Step 3: Apparent power S = V_rms·I_rms = 230 × 20 = 4600 VA — the lines carry twice the "useful" current-worth. Step 4: With correction, P stays 2300 W but cosφ = 0.95, so I = P/(V·cosφ) = 2300/(230 × 0.95) ≈ 10.5 A. The line current nearly halves, cutting I²R line losses to about (10.5/20)² ≈ 28% of before. This is why utilities insist on power-factor correction.