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I have just read that using high voltage results in low current, which limits the energy losses caused by the resistance of the wires.
What I don't understand is why it works this way. Does it have anythnig to do with electromagnetic induction in the wire which resists the current?

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    $\begingroup$ For the same power transmitted, high voltage means low current (P=V*I) $\endgroup$ – nasu Mar 28 '17 at 20:27
  • $\begingroup$ What is the problem with high current transmission? $\endgroup$ – ILoveChess Mar 28 '17 at 20:28
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    $\begingroup$ The heat dissipated in the transmission wires is proportional to the square of the current through wires. $\endgroup$ – nasu Mar 28 '17 at 20:29
  • $\begingroup$ But when we toy with the formula, we may well come to this form: P = U^2 / R $\endgroup$ – ILoveChess Mar 28 '17 at 20:33
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    $\begingroup$ But if you are looking at the power dissipated in the transmission line, this U is the voltage drop on the lines and not the voltage at the consumer. If the current is low, the voltage drop on the wires is also low. $\endgroup$ – nasu Mar 28 '17 at 20:38
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If the total resistance of the transmission line leading from a power station to you is R and the city/town you're in demands an average amount of power P. Then P=I*V . This makes the current drawn by the city/town is I=P/V and so the higher the transmission line voltage, the smaller the current. The line loss is given by P(loss)=I²R, or, substituting for I, P(loss) = P²R/V² Since P is fixed by demand, and R is as small as you can make it(with the big cable), line loss decreases strongly with increasing voltage(in the denominator). So smallest amount of current that you can use to deliver the power leads to the least amount of power loss. It may help to think of it as current causing 'friction/heat' which is lost on transmission.

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    $\begingroup$ This. The mental hurdle that people stumble on is the two powers (the demand and the loss). The demand is not negotiable, and you are trying to minimize the loss. $\endgroup$ – dmckee Mar 28 '17 at 20:50
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From the formula $P=VI $, $I=\frac {P}{V} $. So, if the voltage is high, current becomes low for same power. Now, $H=I^2RT $, so lower the current, lower is the heat production. Mainly to reduce heat production, the voltage is increased.

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