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I define P is the average power. So $P=IV$ and $I=\frac{P}{V}$. $P_{loss}$ I define to be the power loss, which is equal to $I^2R$.

Substituting for $I$, $P_{loss} = \frac {P^2R}{V^2}$

So I get that the you need high voltage to minimise heat loss as R and P are constant.

However, I am confused about the following: Can I solve $P_{loss}=I^2R$ into $P_{loss}=IV$? I don't see why I can't, however, I am confident that I can't do this as then the loss in power would be equal to the average power, which doesn't make sense.

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The $P=IV$ expression includes the power delivered by the line to the users at the far end. The $I^2R$ is the part of the $IV$ power that does not get delivered to them.

To be more precise: Let $R_{\rm line} $ be the resistence of the line and $R_{\rm load}$ then $R_{\rm total}= R_{\rm line}+R_{\rm load}$ and $$ IV = I^2 R_{\rm total}=I^2 R_{\rm line}+ I^2R_{\rm load}. $$ The first term $I^2 R_{\rm line}$ is the loss and the second term $I^2R_{\rm load}$ is the useful power delivered to the user.

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  • $\begingroup$ But does $I^2R$ not equal $IV$? If not, why? $\endgroup$
    – photon
    Commented Nov 25, 2022 at 20:59
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    $\begingroup$ @photon The V for the power loss isn't the voltage delivered, but the loss of voltage between the generator and the users. That depends on the current delivered, but not on the voltage delivered. $\endgroup$
    – John Doty
    Commented Nov 25, 2022 at 21:55
  • $\begingroup$ This answer might help you. $\endgroup$
    – Cosmas Zachos
    Commented Nov 25, 2022 at 22:25
  • $\begingroup$ I'll edit my answer. $\endgroup$
    – mike stone
    Commented Nov 25, 2022 at 22:56

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