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 W1 = 250 kW W2 = 50 kW If the latter reading is obtained after reversing the connection to the current coil of W2, the power factor of the load is

Right Answer is:

0.359

SOLUTION

Reading of wattmeters

W1 = 250 kW

W2 = 50 kW

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

W1 = 250 kW

W2 = −50 kW

Total power W = W1 + W2

W = 250 + (−50) = 200 W

The power factor of the two wattmeters is

tanØ =  √3[(W1 – W2) / (W1 + W2)]

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

tanØ = √3[300/200]

Φ = 68.96°

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

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