P = I^2r = V^2/r (1 Viewer)

[Abtari]

power loss in a transmission line = I^2 R

that is why to minimise power loss we reduce current. but isn't power loss also given by the formula v^2/R (since V = IR).

so by reducing current, we are increasing voltage, which we are ALSO supposed to reduce... there is a contradiction. i know i have got it wrong somewhere just can't pinpoint where. help would be appreciated.



EDIT: for this, ohm's law V=IR used, P=VI formula used... do the Vs and the Is mean the same thing in all these formulae? have i got it wrong somewhere there perhaps.

 

[Xayma]

No.

P=I²R=VI

 

[Abtari]

just wondering, why isnt

P(loss) = v^2 / R

 

[tennille]

Power is equal to v^2/R.

 

[DementedDonkey]

I have the same confusion, and am still yet to work it out. In the prelim module electricity in the home we are taught that Power = VI = I^2R = V^2/R by equating the power law and ohms law. But then in hsc course we're taught that power loss = I^2R and not the rest? why's this? i don't understand how power and power loss =I^2R, but the rest do not work for both.. does my question make sense?

 

[Xayma]

The Physics of Everyday stuff

High-Voltage Transmission Lines

So we now finally come to the topic of this page: the transport of large amounts of electrical power over long distances. This is done with high-voltage transmission lines, and the question is: why high voltage? It certainly has a negative safety aspect, since a low voltage line wouldn't be harmful (you can put your hands on a 12 V car battery, for example, you won't even feel it; but make sure you don't put metal across the terminals, you'll get a huge current and a nasty spark!). Electric energy is transported across the countryside with high-voltage lines because the line losses are much smaller than with low-voltage lines.

All wires currently used have some resistance (the development of high-temperature superconductors will probably change this some day). Let's call the total resistance of the transmission line leading from a power station to your local substation R. Let's also say the local community demands a power P=IV from that substation. This means the current drawn by the substation is I=P/V and the higher the transmission line voltage, the smaller the current. The line loss is given by Ploss=I²R, or, substituting for I,

Ploss = P²R/V²

Since P is fixed by community demand, and R is as small as you can make it (using big fat copper cable, for example), line loss decreases strongly with increasing voltage. The reason is simply that you want the smallest amount of current that you can use to deliver the power P. Another important note: the loss fraction

Ploss/P = PR/V²

increases with increasing load P: power transmission is less efficient at times of higher demand. Again, this is because power is proportional to current but line loss is proportional to current squared. Line loss can be quite large over long distances, up to 30% or so. By the way, line loss power goes into heating the transmission line cable which, per meter length, isn't very much heat.

 

[wogboy]

so by reducing current, we are increasing voltage, which we are ALSO supposed to reduce... there is a contradiction. i know i have got it wrong somewhere just can't pinpoint where. help would be appreciated.

I have the same confusion, and am still yet to work it out. In the prelim module electricity in the home we are taught that Power = VI = I^2R = V^2/R by equating the power law and ohms law. But then in hsc course we're taught that power loss = I^2R and not the rest? why's this? i don't understand how power and power loss =I^2R, but the rest do not work for both.. does my question make sense?

The source of confusion here is which voltage (V) are you talking about when you say P = V^2 / R? And which current (I) are you talking about when you say P = I^2 * R?

Say you're delivering AC electrical power from Newcastle to Sydney through two wires (normally 3 wires are used for 3 phase power distribution, but for simplicity assume it's a single phase 2 wire system), and assume these wires have resistance R. The power loss in either of these wires is P_loss =I_wire^2*R (I_wire is the current through these wires) = V_wire^2/R (V_wire is the voltage drop across a single wire, measured from Newcastle to Sydney).

Suppose an ideal transformer is connected directly to the generator at Newcastle (primary end: Newcastle, secondary end: Sydney), which steps up the voltage by a factor n. The load voltage V_load (measured between the wires, at Sydney) will be ~ n times as large as that at Newcastle, but the load current (I_load) will be smaller by a factor of ~ n.

V_load = n*V_generator
I_load = I_generator/n

Now:

I_load = I_wire, but
V_load =/= V_wire
(think about why the above is true)

P_loss = I_wire^2 * R
=I_load^2 * R

P_loss = V_wire^2 / R
=/= V_load^2 / R

So you can see that to reduce power loss in the cables you need to reduce I_load, not V_load (i.e. step down the current, step up the voltage). Of course, once the wires reach Sydney, having such high voltages is impractical for domestic use, so the voltage is stepped back down (in a few stages) to 240V (largely irrelevant as far as power line loss is concerned, since most of the cable length is between Newcastle & Sydney).