Solving for Power Loss in a Transmission Line: (P-P')=(P^2)R/(V^2)

  • Thread starter Thread starter Fisicks
  • Start date Start date
  • Tags Tags
    Physics Power
Click For Summary

Homework Help Overview

The discussion revolves around understanding power loss in a transmission line, specifically how the difference between the power produced by a power plant (P) and the power received by the user (P') can be expressed mathematically. The context involves concepts from electrical engineering, particularly relating to power, voltage, current, and resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between power, voltage, and resistance, questioning how to express the power delivered to a load in terms of the current and resistance. There is an attempt to manipulate the equation (P-P')=(P^2)R/(V^2) to understand the implications of voltage drop across the transmission line.

Discussion Status

The discussion is active, with participants providing insights and clarifications about the relationships between power, current, and voltage. Some participants have begun to articulate their understanding of the voltage drop and its effect on power delivery, while others continue to seek clarity on the underlying concepts.

Contextual Notes

Participants are navigating through the implications of fixed power output and variable voltage in the context of a transmission line with resistance. There is an emphasis on the need to consider the voltage drop across the line, which affects the power available to the load.

Fisicks
Messages
84
Reaction score
0

Homework Statement

\
A power plant can produce electricity at fixed power P but the operator is free to choose the voltage. The electricity is carried with a current I through a transmission line ( resistance R) from the plant to the user where the user receives power P'. Show the reduction in power (P-P') is equal to (P^2)R/(V^2)



Homework Equations


P=V^2/R


The Attempt at a Solution


I seem to be missing some key idea because when I mess around with (P-P')=(P^2)R/(V^2)
I get that P' is zero which is obviously wrong. Is it not true that R/V^2 = 1/P?
 
Physics news on Phys.org


Fisicks said:
I seem to be missing some key idea because when I mess around with (P-P')=(P^2)R/(V^2)
I get that P' is zero which is obviously wrong. Is it not true that R/V^2 = 1/P?

That is true if you're interested in the power being developed in the resistance R. In this case it's the power delivered to some (unspecified) load that we're interested in.

The power plant produces power P = I*V. It arrives at the far end with the same current, but at a reduced voltage due to the voltage drop across the transmission line resistance R. So the power available for delivery to the load is P' = I*V'. What's an expression for P' in terms of the current and resistance R?
 


I still don't quite understand but here it goes.

The only thing that has changed is V, so P'=I*V'=(I^2)*R
P-P'=I*V - (I^2)R=I(V-IR)
 
Last edited:


P' = I*V'. And V' is not V, since some voltage was dropped on the transmission line. How much was dropped? (Hint: you have the current I which remains the same, and the resistance R of the transmission line).
 


Ok i understand now. The voltage drop is IR, so P-P'=IV=I(IR)=(P/V)(P/V)R
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 20 ·
Replies
20
Views
12K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K