Math for why voltage is stepped-up in power lines?

Click For Summary

Discussion Overview

The discussion centers around the mathematical reasoning behind why voltage is stepped-up in power transmission lines, particularly focusing on the implications for power loss and efficiency in electrical systems. Participants explore concepts related to power loss, voltage, current, and resistance in both direct current (DC) and alternating current (AC) contexts.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that while power loss can be expressed as P = V^2/R or P = I^2*R, they seek clarification on how increasing voltage specifically reduces power loss in transmission lines.
  • Another participant emphasizes that the relevant voltage for power loss is the difference between the start and end of the line, which depends on current and line impedance.
  • Some participants argue that stepping up voltage allows for a reduction in current needed for the same energy transfer, which can lead to cost savings by using smaller gauge wire.
  • There is a mention of corona discharge losses occurring at high line-to-ground voltages, suggesting additional factors that may complicate the discussion.
  • A participant provides a circuit analysis, illustrating the relationship between generator voltage, load voltage, and efficiency, indicating that higher voltage and lower current lead to improved efficiency.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the relationship between voltage, current, and power loss. While some points are clarified, there remains no consensus on the fundamental reasons why increasing voltage decreases power loss, as participants continue to seek deeper explanations.

Contextual Notes

Some participants highlight that misunderstandings often arise from conflating power line voltage with voltage drop across the line. The discussion also touches on the complexities introduced by different operating conditions, such as DC versus AC systems.

Who May Find This Useful

This discussion may be of interest to students and professionals in electrical engineering, physics, or related fields who are exploring the principles of power transmission and efficiency in electrical systems.

HydroGuy
Messages
29
Reaction score
0
I've long understood that voltage is stepped-up in power t-lines to decrease losses, however, I've never really understood the math reasoning behind it.

Isn't P = V^2/R the same as P = I^2*R? If so, it seems that either increasing V or increasing I would invoke the same losses... Could someone explain?

Thanks
 
Engineering news on Phys.org
When you are talking about the power lost over the transmission line, it's the voltage difference between the start and the end of the line, not between operating voltage and earth. This voltage is dependent on current and impedance of the line only.
 
OK, thanks but that isn't really answering my question. Why does increasing the voltage decrease the power loss in the line?
 
Snoogans said:
When you are talking about the power lost over the transmission line, it's the voltage difference between the start and the end of the line, not between operating voltage and earth.
Good answer - this is the thing most students misunderstand when first seeing this
 
Another good reason for stepping up the voltage is that you need less current to get the same amount of energy transfer. Hence you can use a smaller gauge wire to carry the load saving wire cost.
 
Yes that's the point - but it confuses people who think of power = V^2/r and v as the powerline voltage rather than the voltage drop.
 
There are losses due to corona discharge also when the line to ground voltage gets very high. Not trying to confuse the issue; just noting.
 
It helps if you draw a circuit with a load and consider that you have two voltages, the voltage at the generator and a lowered voltage at the load due to line resistance.

Generator:============Load
(= is wire pair across which we measure voltage).

You have V at the power source and V' at the load with voltage drop: V-V' = RI.
Where R is the line resistance (...for both legs. You can put the load anywhere in the current loop and get the same numbers.)

The load power is V'I and of course current is constant.
(We're doing this in DC for simplicity but you can see the reasoning will apply to AC as well.)

Efficiency is power received at load over power leaving the source:
P at source = VI
P at load = V'I
Efficiency is V'I/VI =V'/V= (V-RI)/V = 1-RI/V. So the less current and more voltage, the close to 100% efficiency.
 

Similar threads

How transmission lines deliver (established?) power from generator
  • · Replies 8 ·
Replies
8
Views
3K
Copper Losses in an Electric Generator
  • · Replies 5 ·
Replies
5
Views
3K
Sizing an Isolation Transformer for a DC Power Supply
  • · Replies 8 ·
Replies
8
Views
3K
When to have only voltage gain in an oscillator
  • · Replies 40 ·
2
Replies
40
Views
3K
EMF below Power Transmission Lines
  • · Replies 2 ·
Replies
2
Views
2K
Power Dissipation along power lines and Resistance value
  • · Replies 9 ·
Replies
9
Views
5K
Question about Transmission Line Losses
  • · Replies 8 ·
Replies
8
Views
3K
Is it current or voltage that is fixed in transformer output
  • · Replies 10 ·
Replies
10
Views
3K
Identifying overhead power lines (esp. UK)
  • · Replies 20 ·
Replies
20
Views
5K
Introductory Circuit Analysis—power ratings
  • · Replies 6 ·
Replies
6
Views
2K