50 ALTERNATING CURRENTS 



The total current 



50,000 

 7 - 22CT>Ta6 =379 amp. 



(a) ii = 379 cos e = 379 X 0.6 = 227 amp. Ans. 



(b) ij = 379 sin 6 = 379 X 0.8 = 303 amp. Ans. 



(c) ii 2 X 0.04 = 2,070 watts. Ans. 



(d) t2 2 X 0.04 = 3,680 watts. Ans. 



(e) 7 2 X 0.04 = 5,750 watts. Ans. 



(/) If the power-factor of the load were unity, the quadrature current i 2 

 would be zero and the line current 7 = ii. 

 Therefore the loss would be 



7 2 X 0.04 = 2,070 watts. Ans. 



In this particular case, the line loss due to the quadrature 

 current is considerably in excess of that due to the energy current, 

 yet the quadrature current contributes no power to the load. 



From the foregoing it must not be inferred that the energy and 

 quadrature currents exist separately. Only one current actually 

 flows, but this current is resolved into two components, each of 

 which produces different effects in the circuit. The effect of 

 each component can then be studied, resulting in a much better 

 understanding of the circuit relations than if an attempt were 

 made to consider the current as a whole. 



'ERTY OF ELECTRICAL LABORATORY, 



FACULTY OF APPLIED SCIENCE. 



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