Electrical heating transmission wires

[Glenn G]

Hello community.
I have a question about electrical heating. I am familiar with the fact that textbooks explain how power losses are reduced by transmitting at high V and low I since P = I^2 X R (equation from Ohm's law and P=I X V)

But if you use a different sub for P = I V then you get P = V^2 / R ... So this would suggest high V gives a high heating effect. How would you describe the apparent contradiction here? My take is that the power output is fixed because it depends on the input and as such you will, with a step up transformer get a low I and high V and therefore lower I^2 R, but I don't get why we are precluded from using P=V^2/R for explaining power losses as this clearly gives the opposite conclusion.

Kind regards,
Glenn

 

[oz93666]

It doesn't give a different conclusion ...I think the best way to grasp this is to calculate the losses in a given situation ...

First imagine you have a power line of 1ohm resistance then calculate the losses in the powerline when (a) transmitting 1kw of power at 10V ...(b) ...transmitting 1kw of power at 100V ... whichever equations you use the answer will come out the same , and at higher voltages the loss is less for same power delivered

 

[cnh1995]

My take is that the power output is fixed because it depends on the input and as such you will, with a step up transformer get a low I and high V and therefore lower I^2 R

Right.

but I don't get why we are precluded from using P=V^2/R for explaining power losses as this clearly gives the opposite conclusion.

It doesn't.
Draw a simple circuit diagram with a high voltage source V, a transmission line with resistance R and a connected load. What is the voltage across the transmission line resistance? What do you understand from this?

 

[OmCheeto]

Hello community.
I have a question about electrical heating. I am familiar with the fact that textbooks explain how power losses are reduced by transmitting at high V and low I since P = I^2 X R (equation from Ohm's law and P=I X V)

But if you use a different sub for P = I V then you get P = V^2 / R ... So this would suggest high V gives a high heating effect. How would you describe the apparent contradiction here? My take is that the power output is fixed because it depends on the input and as such you will, with a step up transformer get a low I and high V and therefore lower I^2 R, but I don't get why we are precluded from using P=V^2/R for explaining power losses as this clearly gives the opposite conclusion.

Kind regards,
Glenn

This would be true if the transmission lines were the only load, but they are not. They are in series with the destination load. Doubling the line voltage cuts the current delivered in half. Since the line resistance is constant, you will actually drop less voltage across the transmission lines.

 

[Cutter Ketch]

For the loss in the transmission line the V in P = V^2 / R is NOT the voltage of the transmission line. V is the voltage DROP across the length of the transmission line. You don't know what that V is. You could figure it out by V = I R, but that puts you right back at P = I^2 R. Lower current gives less loss in the transmission line. To deliver the same power to the load at lower current you need higher voltage.

 

[Glenn G]

Thanks for your help, I see now where I was getting confused of course the voltage drop across the cables are only a proportion of the output voltage with the rest going to the load whereas the same current is going through the cables and the load, hence I suppose the ease of using I^2 X R to find the energy losses in just the transmission cables