To minimize the power loss in long-distance power cables it is best to minimize the current and maximize the voltage. This is because the power loss in the cable is calculated by $P=VI$, which we can also express as $I^2R$ by use of Ohm’s law. But by the same reasoning the power loss can also be expressed as $V^2/R$, which clearly shows that the power loss is minimized when the voltage is minimized. How can I explain the contradiction? Why can’t I use $V^2/R$?


1 Answer 1


Power loss in cables (Joule heating) cannot be calculated as $P=VI$ where $V$ is voltage between the lines. The proper voltage to use for power loss over some wire segment is voltage drop over this segment, which is much smaller than $V$ and is given by Ohm's law (the wire used to transfer current over long distances is usually made of aluminium):

$$ V_{drop} = RI. $$

Here $R$ is ohmic resistance of the wire segment and $I$ is current flowing through the wire.

Now we can use the formula

$$ P_{loss} = V_{drop}I $$ and we get $$ P_{loss} = RI^2. $$

So the lower the current used, the lower the power loss in the wires.

We could express this loss in terms of $V_{drop}$, but this is not common in engineering practice.

The moral of the usual story is that to minimizes losses, we need to minimize current (or, in the alternative expression, the voltage drop) and the usual way to do that while maintaining net transmitted power $VI$ at hand is to crank up voltage between the power lines $V$.

  • $\begingroup$ I have always had difficulty in understanding this. How come the voltage is high but the current is low? I get that it is because of conservation of energy but what is the reason in more detail? I would have thought that current depends only on the voltage and the resistance. So if you told me the voltage across a resistor and the resistance I would calculate the current through ohm's law. Why isn't ohms law applicable in this case? $\endgroup$ Apr 16, 2020 at 1:56
  • $\begingroup$ Ohm's law is applicable to a power line, but only to voltage drop along a single wire $V_{drop}$, see the formula above. But the voltage $V$ that appears in the usual story about high voltage transmission lines is different voltage - it is the voltage between different wires of the bunch. There is no useful way to apply Ohm's law to this voltage, as the wires are galvanically isolated from each other by air, so there is almost no current flowing from one to another laterally. $\endgroup$ Apr 16, 2020 at 17:14

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