1.

The input power to a three-phase motor was measured by two watt meter method. The readings were 10.4 KW and – 3.4 KW and the voltage was 400 V. Calculate (a) the power factor (b) the line current.

Answer»

As given in, when W2 reads negative, then we have

tanφ = √3(W1 + W2)/(W1 - W2).

Substituting numerical values of W1 and W2, we get

tanφ = √3(10.4 + 3.4)/(10.4 - 3.4) = 1.97; φ = tan-1(1.97) = 63.1°

(a) p.f. = cos cos . φ= ° 631 = 0.45 (lag)

(b) W = 10.4 – 3.4 = 7 KW = 7,000 W

7000 = √3IL x 400 x 0.45; IL = 22.4A



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