Calculate Power & Fraction Lost in High-Voltage Transmission Line

  • Thread starter Thread starter goWlfpack
  • Start date Start date
  • Tags Tags
    Power
Click For Summary

Homework Help Overview

The discussion revolves around calculating power loss and the fraction of power lost in a high-voltage transmission line, focusing on the resistance of the line and the current it carries.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of different equations for calculating power loss, specifically I^2*R and V^2/R, while questioning the validity of their applications in this context.

Discussion Status

Some participants have provided guidance on the relationship between current, voltage, and power loss, while others are exploring different methods to approach the second part of the problem. There is an acknowledgment of the complexity introduced by the transformer in the system.

Contextual Notes

Participants are navigating the implications of using different formulas and the assumptions related to the transmission line's resistance and the current carried. There is a mention of the provided power and the need to clarify the relationship between the variables involved.

goWlfpack
Messages
51
Reaction score
0

Homework Statement


A high-voltage transmission line with a resistance of 0.28 /km carries a current of 990 A. The line is at a potential of 740 kV at the power station and carries the current to a city located 162 km from the station.
(a) What is the power loss due to resistance in the line?
W

(b) What fraction of the transmitted power does this loss represent?
%



Homework Equations



Power= change in V^2/R

where v is potential difference and R is resistance

Power = I^2*R

where I is current



The Attempt at a Solution



I got part a correct. I used the equation p=I^2*R and got 44457336
I can't figure out how to get the second part though
I tried using p=v^2/R with 740,000 as the V and .28(162) as the R to no avail

any suggestions?
 
Physics news on Phys.org
If you notice, V/R does not give a current of 990 A. The concept gets a little more complicated with a transformer. I*V (990A*740kV) is the provided power. I^2*R is the power lost to resistance.
 
Hi goWlfpack! :smile:
goWlfpack said:
any suggestions?

p = IV? :wink:
 
got it! thanks guys! :)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 20 ·
Replies
20
Views
12K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K