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120 volts x 20 amps = 2,400 Watts

However, if I increased the voltage and lowered the current, you can also use a smaller wire size (more inexpensive), also have less heat and achieve the same watt Power.

1,000 volts x 2.4 amps = 2,400 Watts

  1. Why doesn't it heat up like current?
  2. To me this approach seems more efficient and less costly because you don't use as much material, so why isn't this common?
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    $\begingroup$ It will heat up just the same! 2,400 Watts is the rate at which heat is given off in both cases. $\endgroup$
    – N. Virgo
    Commented Jan 6, 2014 at 5:00

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The term you're looking for is Joule Heating.

The heat dissipated in a conductor is proportional to $I^2 R$ where $I$ is the current and $R$ is the resistance. Heating happens when moving charge (electrons) collide with the molecules in the conductor inelastically (that is, they transfer some kinetic energy to the molecule).

Remember, current is defined as the amount of charge passing a given spot per unit of time. It makes sense then that the more charge passing a spot, the more collisions occur and therefor the more heat is dissipated.

This is why power lines use high voltages for transmission so that they can provide more power with less current.

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  • $\begingroup$ So why don't homes, business and electronics, etc. opt for more higher voltages than amps if it keeps costs down, and is more efficient? Ex: why bother with a 12v car battery with a high amp draw? $\endgroup$
    – user36823
    Commented Jan 5, 2014 at 22:48
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    $\begingroup$ @user36823 most countries did that and standardized on $240\: \mathrm{V}$. The trouble is that (via Ohm's law) higher voltage can overcome resistance more easily so higher voltage has the potential to be a lot more dangerous. $\endgroup$
    – Brandon Enright
    Commented Jan 5, 2014 at 22:50
  • $\begingroup$ @user36823 regarding your $12\: \mathrm{V}$ comment, we don't have a lot of choice. Batteries are electrochemical and most battery cells only generate about $1.5\: \mathrm{V}$. We did standardize on something higher for car batteries. We went with $12\: \mathrm{V}$ by putting a bunch of cells in series instead of $1.5\: \mathrm{V}$. $\endgroup$
    – Brandon Enright
    Commented Jan 5, 2014 at 22:52
  • $\begingroup$ @user36823 another thought that comes to mind is that lights (light bulbs) were the main use of electricity initially and they work by generating heat. No need to go with something terribly efficient in the case of light bulbs. $\endgroup$
    – Brandon Enright
    Commented Jan 5, 2014 at 22:57
  • $\begingroup$ RF semiconductor electronics try to use as high voltage as possible exactly for reasons of of improving efficiency but are always limited by the break-down voltage inherent to the material used. Recent developments in GaN material processing have effectively revolutionized power amplifier technology compared to the older ones using lower break-down voltage Si or GaAs based materials. $\endgroup$
    – hyportnex
    Commented Jan 6, 2014 at 15:16
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In short, because the power drop across a circuit element is $P=IV$, and a resistor (or a piece of wire) experiences a voltage drop $V = IR$, which leads to $P = I^2 R$, or $P = \frac{V^2}{R}$. The first equation is relevant, which shows that for a fixed resistance, doubling the current quadruples the power loss through the conductor. So it is best to use a smaller current.

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  1. The point is that the wire ends are not exposed to a potential difference equal to the full voltage. The wire ends differ in potential only by a (small) voltage drop $\Delta V = R\,I$, where $R$ is the resistance of the wire. Dissipation power equals $P = I\,\Delta V = I^2 R = \frac{(\Delta V)^2}{R}$. Now suppose you can transform $(I,V)$ while keeping the product $P_1 = V I$ roughly constant. Then dissipation power is $P = I^2 R = \frac{P_1^2}{V^2} R$, so higher voltage $V$ and, correspondingly, lower current $I$ results in less power dissipation $P$.
  2. Actually it is common to use high-voltage AC lines for high-power transmissions over long distances. In your home, you will not want to have kilovolt lines within millimeters from your fingertips however, cf. Brandon's comment.
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Amps travel in a straight line and so must travel inside the wire. Volts travel around the amps and usually outside the wire. So amps will generate heat - because of the atoms and valence electrons create degrees of resistance - while volts , generally, will not. But if you use thick enough wire you will not notice the heat increase.

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    $\begingroup$ -1: Either this is usual electrical engineering lingo or it is just nonsense. "Amps" and "Volts" are units, not things, so they can't do anything and I can't see the physics here. $\endgroup$
    – Martin
    Commented Apr 19, 2015 at 0:27

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