Why does high voltage result in less power loss than high current?

Question

Physics 2 student here. I’ve just learned that:

P=I*V

Using ohms law, V=I*R you can get the equations:

P=I^(2)*R and P=V^(2)/R

In these equation, power is proportional to I^2 and V^(2) respectively. This seems to show that power is squarely proportional to both voltage difference and current. Yet it’s stated that high voltage results in less power loss than high current. How can this be true when power scales by the same amount with a change in either.

The context of this is power lines and why they use high voltage.

Please ELI5. I’ve heard inadequate explanations in my textbooks, I can’t wrap my head around it.

In: Physics

Answer 1

For the math, one is multiplied by resistance while the other is divided by it. That makes an obvious difference.

For an ELI5 explanation of power loss in a wire, let’s use the water in a hose analogy.

Imagine the wire is a hose or pipe. Voltage is now water pressure, current is how much water flows through the pipe per unit of time so liters per minute or whatever units you prefer.

A small amount of water at very high pressure can do the same amount of work (spinning a turbine for instance) as a large amount at low pressure. So that makes it equivalent to how low current at high voltage can have equal power to high current at low voltage. With me so far?

Now imagine we have to get a lot of power through a narrow pipe, which offers a lot of resistance if you try to push a high current of water through it. You cannot squirt a swimming pool’s worth of water through a common garden hose in one minute at low pressure, the resistance is too great, so there is power loss.

But, if the hose is strong enough to not burst, you can instead transmit the same amount of power by increasing pressure at a low current. The hose restricts flow, not pressure. Likewise, a wire resists current but not voltage as much.

Now, my analogy above is for transmission power loss in a wire. Things can get confusing if you haven’t grasped the fundamentals, as the same formulae are used for things like calculating how much power a given device in your circuit has; calculating the power loss in a wire is really calculating how much heat that wire outputs as a resistor, treating the wire as a component.