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The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of lower reading wattmeter.(a) 1.835(b) 2.835(c) 3.835(d) 4.835This question was addressed to me in class test.Question is taken from Power Measurement in Three-Phase Circuits in chapter Polyphase Circuits of Network Theory

Answer»

Right option is (B) 2.835

To elaborate: WR + WY = 10KW. Ø = cos^-10.8=36.8^o => tanØ = 0.75 = √3 (WR-WY)/(WR+WY)=(WR-WY)/10. WR-WY=4.33kW. WR+WY=10kW. WY=2.835kW.



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