Question about the use of high voltage in power lines(LC Physics)

JC 2K3

This is wrecking my head, and I usually understand everything in physics.

Basically, I've learned that voltage is the amount of energy "lost" per coulomb in a circuit.

And I know that V = IR.

Now, my book tells me that P = (I^2)R and therefore the amount of energy "lost" depends on I, so when transporting electricity over a long distance you need a small current.

It then says that since P = IV then V is inversely proportional to I and therefore high voltage is required over a long distance to minimise energy loss...

I simply cannot follow this logic at all, why?
-Why are they saying P is constant, are they not trying to lower P? Why does it matter if there is a high I?
-If voltage is the amount of energy lost per coulomb, then surely highering the voltage would result in higher energy losses....
-V = IR, so highering the voltage means highering the current or the resistance, and since R is supposed to be constant here(implied by the book), highering V would just higher I.
-Even if R was variable, why are they talking about voltage, surely lowering the resistance is the key here?

Aghh... I'm just so confused, can someone please explain this to me?

mathias

This is the most straightforward explanation that I can find , your getting hung up on that definition of a coulomb , and you seem to have misread into it that you must lose energy for voltage , this is not really the case !!

A coulomb is defined in terms of a voltage drop , this is just a potential difference and does not necessarily mean losing any energy.

Anyway , best to read this , its pretty clear.

http://www.bsharp.org/physics/stuff/xmission.html

JC 2K3

Thanks for that link! Ah, I was mixing up the power demand, which is a constant, with the power loss. The book explains this badly.... Doesn't explain the volt/potential difference very well either.

Why the book didn't include the formula Ploss = P²R/V² is beyond me....

However, I'm finding it difficult to grasp the concept of I being smaller if V is higher, considering V=IR...

Michael Collins

JC 2K3 wrote:

However, I'm finding it difficult to grasp the concept of I being smaller if V is higher, considering V=IR...

Ok so you're applying a fomula which doesn't apply to that situation! V=IR is Ohm's Law as I'm sure you're aware, but what it means is that if you have a voltage V across a resistance R - then the current flowing is I (or if you have a current flowing through a resistance R then the voltage across that resistance is V). But the voltage on a power line is NOT across the line. The voltage across the line is going to be the voltage at one end minus the voltage at the other - this is the voltage that is lost on the line i.e. this is the voltage that contributes to the power lost on the line.

So if we say the line is for arugument sake a 10 volt line i.e. at the sending end the line is at 10 volts with respect to ground, and if we measure the voltage at the other end and find it to be 8 volts - we know that 2 volts has been lost along the line - i.e. the line has 2 volts across it. So the power lost on a line of resistance R would be P=I²R, so we need to find the current I, but we know the voltage across is 2 volts, and the resistance we'll call 5 ohms, so the current I = 2 / 5 = 0.4 amps, and so the power lost is I²R = 0.4²(5) = 0.8 watts.

JC 2K3

Oh......

So Ohm's law doesn't apply at a point?

I'm actually gonna just have to think long and hard about this until it makes sense to me... I think you guys have helped me overcome my mental block though, so thanks.

Michael Collins

No bother, it's difficult to grasp at the start but once you have it you're grand.

JC 2K3 wrote:

Oh......
So Ohm's law doesn't apply at a point?

Well more or less yeh, it only applies to voltages across resistances. There's no such thing as a voltage at a point really, as all voltage is measured with respect to a reference. When I said a 10 volt line my previous post, that 10 volts is the voltage between the line and ground - no resistances involved there.

dingding

Ohms law would apply when you measure the voltage drop from one end of a transmission line to the other.

V - I/R

The resistance is constant so as more current passed through the line the voltage drop would increase.

In the case you describe in the original post you want to transfer a given power from one location to another.

The higher the voltage the lower the current needed to transmit the same amount of power.

w = V X I For the same power as you increase the voltage you reduce the current through the conductor.


The loss in a cable W= I squared R

AS you reduce the current by increasing the voltage you reduce the losses in the cable.

That is why the esb use high voltage power lines.

Hope this has not confused it more