# Reason behind not using $H=V²/R$ for explaining the usage of high voltage in electric transmission lines

The answer to this question says that since there are multiple (3) potential difference across the entire circuit of power lines, the electricity is transmitted at high voltages. Also, they considered $$H=I^2Rt$$ and not $$H=\frac {V^2}{R}.$$ And a comment on the answer says that considering heat as a function of voltage here doesn't make sense as there are 3 voltages.

This made me think, is the voltage dropped across any resistance in the entire powerline circuit always the same? Is it the "internal" resistance changing at the source itself? That's the only way I can think of where resistance across the transmission wires can be kept constant without an increase in voltage dropped across it, explaining how we can't use $$H=\frac{V^2}{R}$$ here for comparing heat loss.

H is the heat generated in the resistance of the line in a given time. $$V^2$$/R is the corresponding power. This considers only the current in one line, where the V is the voltage drop from one end of the line to the other. The line itself is a high voltage to minimize the current required to transmit a lot of power.
• But $I²R=V²/R$, so what is preventing us from using $V²/R$ from concluding that the heat lost is more in high voltage when the line itself is at high voltage? Feb 3, 2022 at 19:35
• @Mehmer, the voltage across the power line follows Ohm's law, $V=IR$ where $V$ is the voltage between the ends of the line, $I$ is the voltage flowing along the line, and $R$ is the resistance of the line. If the current (mostly determined by the load at the end of the line) increases or decreases, the voltage between the two ends will change proportionally. Feb 4, 2022 at 16:53