Why Does Using High Voltages in Power Transmission Reduce Losses?

[paul_harris77]

We are constantly told at school that in order to reduce power loss in overhead cables, high voltages and low currents are used as P = I squared R. This seems to make sense until you substitute I=V/R into P=VI and get P = V squared / R. Now if voltage is increased in the cable, and resistance is decreased, power loss is at its greatest, completely opposite to the other example. Am I doing something wrong?

Many thanks

Paul Harris

 

[skeptic2]

We are constantly told at school that in order to reduce power loss in overhead cables, high voltages and low currents are used as P = I squared R. This seems to make sense until you substitute I=V/R into P=VI and get P = V squared / R. Now if voltage is increased in the cable, and resistance is decreased, power loss is at its greatest, completely opposite to the other example. Am I doing something wrong?

Many thanks

Paul Harris

What you're saying would be true if the high power line were shorted to ground and all the voltage were across the line. In practice the line feeds a transformer with an impedance much higher than the line. So in reality most of the voltage is across the transformer and the voltage across the line is very low.