During the transmission, the generated electric power is delivered after stepping up to hundreds of thousands or even more voltages by transformers.
In that case since P = I*V, increasing V reduces I in the secondary of the transformer.
The reason given is to reduce Ploss = I^2*R losses. Here I decreases so the power loss.
Is that the real reason to step up?
I'm asking because we can write the power loss equation as:
Ploss = V^2/R
Or if we use both I and V in the power equation:
Ploss = (V/R)*I
It seems like if we step up the voltage I decreases but V increases. How about the power loss?
Does the power loss decrease?
Or the real reason to step-up the voltage is to reduce the cross section area of the transmission lines significantly?
Best Answer
For these transmission lines, not the voltage will result in a power loss, but the voltage drop on these lines.
I thinks its easiest explained on an example: Lets say your transmission line has a resistance of R=100Ohm and you want to transfer P=1kW.
With "P=U*I" you get:
@1000V, you need to transfer 1A
@100kV you need to transfer 0.01A
By "dU=R*I" the voltage drop across your transmission line will be:
100V @1000V
1V @100kV
As the voltage drop in your transmission line is the power lost you can now calculate the Power loss by "Ploss = dU * I" which results in:
100W @1000V which is 10% of your original 1kW
10mW @100kV which is 0.001% of your original 1kW
Thus: The higher the voltage, the lower the power loss in your transmission line (which of course also has it's upper limits due to for example isolation of these lines and isolation of the transformers, but thats another topic).