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Could somebody please explain why we step up the voltage when transmitting electricity over long distances? I have read it is to reduce energy losses. Why does a high voltage not result in high energy losses? Could you also show me it using the p=v^2/R equation as thats where my difficulties lie. If the voltage is really high would the power losses not me bigger?

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  • $\begingroup$ related: physics.stackexchange.com/q/65147 $\endgroup$
    – CDCM
    Commented Aug 24, 2017 at 0:51
  • $\begingroup$ I agree with @CDCM ... several people have already answered this question, including myself. $\endgroup$ Commented Aug 24, 2017 at 0:56
  • $\begingroup$ Sorry guys. They dont really answer it and show how I am using the p=v^2/r equation wrong which is the real flaw Im having. $\endgroup$
    – Jake
    Commented Aug 24, 2017 at 1:08
  • $\begingroup$ The voltage isn't the total voltage between the transmission line and ground, but the voltage drop from the start to the end of the line. That's $V_L$ in the duplicate question. $\endgroup$
    – Nick T
    Commented Aug 24, 2017 at 3:01

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In your expression $v$ is the voltage drop across the cables which will decrease as the resistance of the cables $R$ is decreased.

In your expression for the power loss the voltage term is squared and so the numerator $v^2$ decreases faster than the denominator R as the resistance decreases, thus the power loss in the cables decrease as the resistance of the cables decreases.

If the current is $I$ then the voltage drop across the cables $v=RI$ and so the power loss is $\dfrac{\left ( RI\right )^2}{R}=I^2R$ which is perhaps a more familiar form?

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