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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.

photon
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1 Answers1

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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.

mike stone
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