In the two-wattmeter method of three-phase power measurement of a balanced load, if the reading of one meter is −200W, then the power factor of the load is

• A. Less than 0.5
• B. Equal to 0.5
• C. Greater than 0.5
• D. Equal to zero

Answer: (A)

The reading of two wattmeters can be expressed as
W1 =  VLILcos(30 + φ)
W2 =  VLILcos(30 − φ)
(i) When PF is unity ( φ = 0°)
W1 =  VLILcos30°
W2 =  VLILcos30°
Both wattmeters read equal and positive reading i.e upscale reading
(ii) When PF is 0.5 (φ = 60°)
W1 =  VLILcos90° = 0
W2 =  VLILcos30°
Hence total power is measured by wattmeter W2 alone
(iii) When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)
W1 = Negative
W2 = positive (since cos(−φ) = cosφ)
The wattmeter W2 reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°. However, in wattmeter W1, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives negative (i.e. downscale) reading.