Question

While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are 100 W and 250 W. The power factor of the load is

• A. 0.8

Answer 1

The correct option is A 0.8
Given,
$W_1=250W,W_2=100W$

Power factor angle is given by

$\varphi =ta{n}^{-1}\left[\frac{\sqrt{3}(W_1-W_2)}{(W_1+W_2)}\right]$

$ta{n}^{-1}\left[\frac{\sqrt{3}(250-100)}{(250+100)}\right]$

$or,\varphi =ta{n}^{-1}\left[\frac{\sqrt{3}\times 150}{350}\right]$

$=ta{n}^{-1}\left[\frac{3\sqrt{3}}{7}\right]={36.58}^{\circ }$
$\therefore Powerfactorofload,$

$cos\varphi =cos{36.58}^{\circ }$

$=0.802\approx 0.80$