A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter.

• A. W1 = 5 kW and W2 = 25 kW
• B. W1 = -5 kW and W2 = 35 kW
• C. W1 = -10 kW and W2 = 40 kW
• D. W1 = 10 kW and W2 = 20 kW

Answer: (B)

Here,
Cosφ = 0.397
P = 30 kW = √3*V_L*I_L*cosφ
V_L*I_L = 30kW/(√3*0.397) = 43628.48 VA

Readings of wattmeter W1 = V_L*I_L*cos(30+φ)

Readings of wattmeter W2 = V_L*I_L*cos(30-φ)