A 3-phase balanced load which has a power factor of 0.707 is connected to a balanced supply. The power consumed by the load is 5 kW. The power is measured by the two-wattmeter method. The readings of the two wattmeters are
A
3.94 kW and 1.06 kW
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B
2.50 kW and 2.50 kW
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C
5.00 kW and 0.00 kW
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D
2.96 kW and 2.04 kW
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Solution
The correct option is A 3.94 kW and 1.06 kW p.f.=0.707,ϕ=cos−1(0.707)=45∘ P=5kW W1+W2=5...(i) tanϕ=√3(W1−W2)(W1+W2) (W1+W2)×1=√3(W1−W2) W1−W2=5√3...(ii)
Form equation (i) and (ii), W1=3.94kW W2=1.06kW