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

Question

RRB Paper · Accepted Answer

Using the two-wattmeter method formula:

W1 = 250 kW W2 = -50 kW (after reversing the current coil terminals)

The total power is given by:

W = W1 + W2 = 250 kW - 50 kW = 200 kW

The power factor can be calculated as follows:

tan(Φ) = √3(W1 - W2) / (W1 + W2) tan(Φ)

= √3(250 kW - (-50 kW)) / (250 kW + (-50 kW)) tan(Φ)

= √3(300 kW) / (200 kW) tan(Φ)

= √3(3/2) tan(Φ) = √(27) / 2 tan(Φ)

= 3√3 / 2 Φ = arctan(3√3 / 2) Φ = 68.96°

Therefore, the power factor is:

cos(Φ) = cos(68.96°) = 0.3590

Unity