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When reading about power transmission one very common statement is that power is transmitted at high voltage to minimize loss in the form of heat.

What confuses me is the governing equation. Aren't $H=V^2t/R$ and $h=i^2Rt$ essentially the same equations with the common form $V=iR$?

Won't a potential difference of $22000\ \mathrm{V}$, with a resistance of $1\ \mathrm{k\Omega}$, produce a $22$ ampere current, with the heat dissipated the same as if I had used a higher current on a smaller voltage? Thanks.

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  • $\begingroup$ $i=\frac{P}{v}$ Increase voltage, decrease current for same power, current causes the heat loss. $\endgroup$ – Feyre May 14 '16 at 14:23
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Let say the power produced by a transmission company is 50 Giga_Watt(we can take any value really). What happens is, they convert this to very high voltage.When you use a 1 kilo_ohm resistor, the current draw should be 22 amps, but the power generated(50Kw) must be kept in mind! You cannot have a power source of 10 W and expect it to provide 10 Volts at 2 amps if you use a 5 ohm resistor. It is converted to high voltages, so the current draw is less and lower wastage of power occurs by heat loss.

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Power transmission lines possess a resistance which depends on the distance the energy is conveyed and the thickness and conductivity of the wire used. We can only minimize this resistance to the point where it is uneconomic or otherwise impractical to make the wires any thicker.
For any given current through a particular line, there results a power loss due to heat dissipation in the resistance of that line. By increasing the voltage through the line, the required current for any given power level (P=E.I) is reduced, thus reducing the power loss. You are correct in that a potential difference of 22000 V, with a resistance of 1 kΩ, will result in a 22 ampere current, but you cannot obtain that amount of current with a lower voltage. This would require a lower resistance to produce the same amount of heat dissipation.

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power line losses are equal to (current)^2 x (resistance). all power transmission lines are designed to minimize losses by minimizing current, which requires maximization of voltage.

However, the high-voltage limit is set by capacitive losses by coupling to the ground, and by corona discharge effects, both of which get progressively worse with increasing voltage.

minimization of both loss mechanisms yields a practical voltage optimum somewhere around 225 KeV for long-distance transmission lines.

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