This is a question which I seem to have tackled multiple times, solved each time after reading a dodgy internet explanation, then partially forgotten about and retackled half a year later. It is time to put an end to this.

So first, at the power plant, a set power is produced. Let's call this $P_g$. Then there is a transformer that can say hypothetically transform this power into any voltage, and therefore following the equation $V = IR$, any current desired. Cool so far.


Consider the case where you simply condense all the power lines into a single resistance, R. Now we can relate the power lost from the resistance $P_l$ as $P_l = I^2R$. By inspection, as the resistance is a constant, lowering the current drastically lowers the power lost. This is great, as lowering the current means raising the voltage, which seems all handy dandy.

But wait: here's my problem. Can't the Equation for power loss be translated as thus:

$P_l = I^2R$

$P_l = IV$

$P_l = \dfrac{V^2}{R}$

BAM. Now the power seems to depend upon the voltage alone, as the resistance does not change, so it seems pretty logical that REDUCING the voltage reduces the power lost. Nuts. Evidently I have done something wrong. Lets take a simple case, $P_g = 10$. Lets say I set the voltage to 10, and the current to 1, and the resistance to 1 as well. That seems ok, as $P = VI$. Now lets use the two equations we have to determine the loss of power.

$P_l = I^2R = 1$

Seems logical, the current is 1 definitely, and the resistance is 1 definitely, so the power loss is one. Makes sense too.

$P_l = \dfrac{V^2}{R} = 100$

WHAT IS LIFE!! But this seems pretty logical to me too, the voltage across the resistance is 10 (is it not? I might be forgetting something fundamental here? not sure), and the resistance is 1.

Any help is helpful, I'm probably just being silly at one step or another.


Well I was right, I WAS being silly i think. I pretty much completely ignored the fact that $V = IR$, so in the case where $P_g$ is 10, and the resistance was 1. I seem to have run into a complication now. If I set the voltage to 10, according to $P = VI$ the current should be one, but according to $V = IR$ the current should be 10. I run into a conundrum? I am now thoroughly confused. Any help is helpful.

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    $\begingroup$ There are several duplicates of this, but the only one I can find where the role of the transformer is described in detail is Heat loss using alternating current. Joshua, does this address your question? If so I'll flag it as a duplicate. $\endgroup$ Commented Nov 7, 2014 at 8:51
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    $\begingroup$ Other duplicates with what seem to me unsatifactory answers are Why are high voltage lines “high voltage?” and Power dissipation in High Voltage Cables. $\endgroup$ Commented Nov 7, 2014 at 8:53
  • $\begingroup$ These answers help clear up the real life situation a lot, as the high voltage power lines aren't the only source of resistance in the circuit, so my logic doesn't really apply. But how about the case I have outlined here? For this simple situation, shouldn't the voltage drop over the resistor be the whole ten volts, as there is nothing else? Then how come one equation outputs a 1, and the other a 100? $\endgroup$ Commented Nov 7, 2014 at 8:58
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    $\begingroup$ In the real-life case, the resistance of the wire and the power from the plant are both constants. In your circuit, they cannot both be constant. If you up the voltage on constant resistance, you're upping the power generated (and lost) simultanously. $\endgroup$
    – BowlOfRed
    Commented Nov 7, 2014 at 9:13
  • $\begingroup$ $P_g$ was my power generated, a constant assumed, $P_l$ was the power lost, the one I wanted to minimize $\endgroup$ Commented Nov 7, 2014 at 9:19

3 Answers 3


You've begun with this:

$P_l = I^2R$

$P_l = IV$

This is correct, but the $V$ here is not the line voltage, but instead the voltage drop across the resistor under consideration. Increasing the line voltage does not increase the voltage drop.

Your diagram with a single resistance and a power station implies that the current in the line depends on that resistance and the line voltage. In reality, it does not. The resistance is just a (usually small) portion of the circuit.

sigh I'll leave the above, but I had misread your question. I thought the diagram was oversimplified, but that's actually the situation you were asking about.

Because there is only one source of loss (the resistor), then power lost through is is equal to power generated. $P_l = P_g$. You cannot now state that $P_g$ is a constant, the resistance is a constant, and voltage is variable.

This is great, as lowering the current means raising the voltage,

It does for constant power. But you want to have constant resistance instead. You cannot have both while varying voltage.

  • $\begingroup$ "sigh" Haha thanks for writing it out for me. Ok so basically you can't keep both Power generated and resistance both constant whilst varying voltage, so my whole scenario breaks down. Does that mean in real life the resistance is also a variable, and that they change it depending on the power generated? So like I can't one day just take out a section of the transmission lines and expect it to work? $\endgroup$ Commented Nov 7, 2014 at 9:30
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    $\begingroup$ It's the other way around. Your usage of electrical equipment changes the resistance. The power generation changes in response to keep the voltage constant. $\endgroup$
    – BowlOfRed
    Commented Nov 7, 2014 at 9:35
  • $\begingroup$ Ok wait so this is changing my view on the world. I thought power generators at their simplest were like "Ok I'll burn 10 coals a minute to produce some steam to drive some generators to get some set power output?" How can the power generation be changed dynamically to suite your usage of electrical equipment? $\endgroup$ Commented Nov 7, 2014 at 9:37
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    $\begingroup$ Operators are allowed some variance in voltage, so nothing happens for small load changes. But when there is a sufficient change in demand, power plants have to be brought online or offline to stay in spec. $\endgroup$
    – BowlOfRed
    Commented Nov 7, 2014 at 9:53
  • $\begingroup$ Increasing the line voltage does not increase the voltage drop In fact, increasing the power supply voltage decreases the voltage drop in the transmission line because it decreases the current. The voltage drop and the current are inversely proportional to one another by Ohm's Law. $\endgroup$ Commented Nov 9, 2015 at 3:46

High voltage, because:

The voltage that a power plant pushes out, is consumed by wires (loss) and a machine (desired). The smaller voltage the wires consume, the smaller the loss.

In order to minimize loss, you have to minimize the voltage share of the lines.

Voltage consumed by the wires is $V=RI$ ($R$ is resistance of the lines). As you can see, the lower the current ($I$), the lower gets the voltage consumed by the wires.

As $P = VI$, to get low current, you have to get high voltage to get the total power of the system to remain. In this latter equation V is the voltage pushed by the powerplant.


The purpose of a transmission line is to transmit electrical power from a power plant to a house. Let $P$ be the power, $V$ be the voltage, and $I$ be the current at the house. Note that $P=IV$ and let's assume that $P$ is a constant.

Let $R$ be the resistance of the wires that make up the transmission line. Since the current that goes through the wires equals the current that goes through the house, the power lost in the wires is $RI^2$. So the power generated at the plant needs to be $IV+RI^2$.

To minimize power generation cost, we want to maximize the amount of power delivered to the house per unit of power generated at the power plant. Mathematically, this means that we want to maximize


Multiplying the top and bottom by $V/I$ we get


Since $P=IV$ we get


To maximize this expression we should maximize $V$. That means we should transmit the power at a high voltage.


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