Study Guide For PE 4.1

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

yaoyaodes

Well-known member
Joined
Apr 22, 2021
Messages
69
Reaction score
1
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.
 

Attachments

  • IMG_2971.jpg
    IMG_2971.jpg
    517 KB
  • IMG_2972.jpg
    IMG_2972.jpg
    445.2 KB
  • IMG_2973.jpg
    IMG_2973.jpg
    480.5 KB
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.
 
Last edited:

Latest posts

Back
Top