# Electrical heating transmission wires

• Glenn G
In summary, the conversation discusses the apparent contradiction between using P = I^2 X R and P = V^2 / R to explain power losses in electrical heating. The conversation concludes that while both equations are correct, they are applied to different parts of the circuit and taking into account the voltage drop across the transmission lines can help understand this concept better.
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

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

Glenn G
Glenn G said:
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.
Glenn G said:
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?

Glenn G
Glenn G said:
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.

Glenn G
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.

NTL2009 and 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

## 1. What is the purpose of electrical heating transmission wires?

Electrical heating transmission wires are used to transfer electrical energy from a power source to a heating element, which then converts the electrical energy into heat. This allows for the heating of a specific area or object through the use of electricity.

## 2. How do electrical heating transmission wires work?

Electrical heating transmission wires work by conducting electricity through a material with high electrical resistance, such as copper or aluminum. This resistance causes the wire to heat up, which in turn heats up the surrounding environment or object.

## 3. What are the advantages of using electrical heating transmission wires?

One advantage of using electrical heating transmission wires is their efficiency, as they can convert almost all of the electrical energy into heat. They are also easy to install and can be used in a variety of settings, such as homes, businesses, and industrial facilities.

## 4. Are there any safety concerns with electrical heating transmission wires?

While electrical heating transmission wires can be a safe and efficient way to heat a space, there are some safety concerns to be aware of. These wires can become hot to the touch, so care should be taken when handling them. It is also important to ensure that the wires are properly installed and maintained to prevent any potential hazards.

## 5. Can electrical heating transmission wires be used for outdoor heating?

Yes, electrical heating transmission wires can be used for outdoor heating. However, they should be installed and protected in a way that prevents any potential damage from exposure to the elements, such as rain or snow. Additionally, proper safety precautions should be taken when using these wires for outdoor heating to prevent any potential hazards.

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