# Power is being measured by two wattmeter method in a balanced three-phase system. The wattmeter reading are

Power is being measured by two wattmeter method in a balanced three-phase system. The wattmeter reading are
W

_{1}= 250 kW W_{2}= 50 kW If the latter reading is obtained after reversing the connection to the current coil of W_{2}, the power factor of the load is### Right Answer is:

0.359

#### SOLUTION

Reading of wattmeters

W_{1} = 250 kW

W_{2} = 50 kW

When the latter reading is obtained after the reversal of the current coil terminals of the wattmeter

W_{1} = 250 kW

W_{2} = −50 kW

Total power W = W_{1} + W_{2}

W = 250 + (−50) = 200 W

The power factor of the two wattmeters is

tanØ = √3[(W_{1} – W_{2}) / (W_{1} + W_{2})]

tanØ = √3[(200 – (−50)) ⁄ [250 + (−50)]

tanØ = √3[300/200]

Φ = 68.96°

**Power factor = cosΦ** = cos(68.96°) = **0.3590**