# High voltage power transmission and joule's law

Hi,

Power transmission over long distances from power plants to local power grids is often done over DC with relatively high voltages/tensions. The argument for this is that since, by Joule's law, power dissipation due to heat by a resistive element is given by I^2 * R, lowering the current as a result of upping the voltage results in less power loss. But what is really puzzling me is that, by Ohm's law, I^2 * R is equal to (V^2)/R (since I = V/R), in which case upping the voltage will result in *increased* loss of power through heat generation.

I must be missing something, but I feel this isn't as clear as most people make it out to be with repeated quotes of I^2 * R without considering other equivalent forms..

Thanks a lot!

berkeman
Mentor
Hi,

Power transmission over long distances from power plants to local power grids is often done over DC with relatively high voltages/tensions. The argument for this is that since, by Joule's law, power dissipation due to heat by a resistive element is given by I^2 * R, lowering the current as a result of upping the voltage results in less power loss. But what is really puzzling me is that, by Ohm's law, I^2 * R is equal to (V^2)/R (since I = V/R), in which case upping the voltage will result in *increased* loss of power through heat generation.

I must be missing something, but I feel this isn't as clear as most people make it out to be with repeated quotes of I^2 * R without considering other equivalent forms..

Thanks a lot!

Power transmission lines use AC, not DC. Transformers need AC in order to work, and transformers are used to raise and lower the voltages for coupling to the HV power transmission lines.

The error in your equation is that the V you are listing is for the (small) voltage *drop* across the wires, not the (high) differential power voltage. Raising the differential power voltage results in lower current I, which also results in lower voltage drop along the wires.

The error in your equation is that the V you are listing is for the (small) voltage *drop* across the wires, not the (high) differential power voltage. Raising the differential power voltage results in lower current I, which also results in lower voltage drop along the wires.