noobie question, but wondering if someone can clarify this for me.
i know according to Ohms law V =IR that the voltage and current are directly proportional that if one goes up the other does as well.
but when we use the power equation P=IV it is the opposite.
am i missing something here? I know in distrubtion and transmission they up the voltage so they can use smaller conductors as the current is alot smaller. but doesnt V=IR contradict this?
a bit confused here.
thanks
The bold part is a common misunderstanding, which Charlie touched on in his response. What you have to remember is that V in these equations is the voltage
difference between two points in the circuit. When dealing with a device, we just call this the input voltage -- one screw connects to L1 @ 120V, and the other screw connects to Neutral @ 0V, so the voltage difference across the device terminals is 120V. However, when dealing with transmission line losses we're not concerned with the voltage difference between Line and Neutral; we're concerned with the voltage at Point A on Line 1 compared to the voltage at Point B several miles away, also on Line 1. We commonly refer to this as voltage
drop.
Once we realize that we're talking about voltage drop, rather than line voltage, it becomes clearer why there is no contradiction. As David pointed out, if you have, say, a step-down transformer that feeds several homes, the amount of power required will remain constant regardless of what voltage is on the transformer's primary winding (as long as the same loads are turned on, anyway). Perhaps an example is in order:
Let's say we have a step-down transformer that is supplying a constant 100A, 240V, single-phase to a building. The feeder on the primary side of the transformer has a resistance of 1 Ohm. We can calculate the power being fed to the building as:
P = IV = 100A x 240V = 24,000VA, or 24kVA
If the primary voltage is 480V, single-phase, then we can calculate the primary current with:
I = P/V = 24kVA / 480V = 50A
The voltage dropped across the feeder to the transformer can then be calculated with:
V = IR = 50A x 1 Ohm = 50V
So, you would actually need 530V at the source of your feeder in order to have 480V at the transformer primary. This is obviously way too high to be an acceptable voltage drop, but let's continue with the example. Our power loss in the feeder is:
P = IV = 50A x 50V = 2500W (notice that V is the voltage drop across the feeder, NOT the nominal system voltage)
This can also be calculated by:
P = I
2R = (50A)
2 x 1 Ohm = 2500W (this is the form of the equation that we would normally use for this calculation)
P = V
2/R = (50V)
2 / 1 Ohm = 2500W
If, however, the primary voltage is 12kV, then our calculations become:
Primary Current: I = P/V = 24kVA / 12kV = 2A
Feeder Voltage Drop: V = IR = 2A x 1 Ohm = 2V
Feeder Power Loss: P = I
2R = (2A)
2 x 1 Ohm = 4W
This is why we use high voltages to traverse long distances -- it lowers the amount of current that must be delivered to the step-down transformer at the far end, which reduces the power lost as heat in the transmission lines. There are other factors that get considered in real life, such as skin effect, AC reactance, etc., but this gives the basic idea.
Sorry for being so long-winded.