Two watt meters are connected to measure power in a three phase circuit. The reading of one of the power factor is 7KW when load power factor is unity. If the power factor of the load is changed to 0.707 lagging without changing the total input power, calculate the readings of the two watt meters. -

Given:

$pf = 1 W1 = 7kW \\ pf = 0.707 (lagging) \\ \text{When the power factor is unity} \\ W1 = W2 \\ W2 = 7kW \\ Power = W1 + W2 = 7 + 7 = 14kW – 1 \\ \text{When the power factor is changed to 0.707 lagging} \\ Pf = \cos (ф) = 0.707 \\ ∴ ф = \cos-1 (0.707) \\ \boxed{ф = 450} \\ ∴ \tan ф = \sqrt3 W1 – W2/ W1 + W2 \\ ∴ \tan (450) = \sqrt3 W1 – W2 / 14 \\ \boxed{1= \sqrt3 W1-W2/14} \\ W1-W2 = 14/\sqrt3 = 8.082 kW -2 \\ \text{Solving eq. (1) & (2)} \\ W1 = 11.041 kW \\ W2= 2.959 kW$