Study Guide For PE 4.1

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yaoyaodes

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Question regarding 4.1k Three phase power. I guess I don't understand the solution very well. I posted my solution as well. I got the right answer because that was close to my calculation. Can someone explain the solution?

I don't understand why he ignores the reactance of the line when calculating the line current, and also, I don't understand why the pf is not taken into account when calculating the current.
 

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That's because the power loss is given in Watts, which is the real power loss. Real power loss per phase= I^2*R. Power Loss per phase= 100W. R is given. So, 100=(I^2)*3, which gives I= 5.77A. Here phase current=line current.
 
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