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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ Feb 3, 2022 at 19:35
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    $\begingroup$ @Mehmer, the point is that the "V" that determines the power lost in the line is the V between one end of the line and the other. The "V" that determines the power delivered to the load is the V between the load end of the line and ground. $\endgroup$
    – The Photon
    Feb 3, 2022 at 19:46
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    $\begingroup$ @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. $\endgroup$
    – The Photon
    Feb 4, 2022 at 16:53
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    $\begingroup$ If the voltage rises across the ENTIRE circuit, then the voltage drop should rise as well right? If the load devices aren't changed, then yes. But you design a lightbulb differently to run on a 220 V system than on a 110 V system. The lightbulb used in the 220 V system will require less current than the one in the 110 V system, if they're designed to produce the same amount of light. $\endgroup$
    – The Photon
    Feb 4, 2022 at 16:55
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    $\begingroup$ @Mehmer, The load determines the current draw (or power draw, if you want to think at a system design level). The length and diameter of the wires then determines the voltage dropped over the wires. If the wires are severely under-sized, you might have to solve the whole design rather than simply assume the load current is fixed. $\endgroup$
    – The Photon
    Feb 5, 2022 at 16:44

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