# Understanding voltage and power in the fluid analogy for DC circuits

Now lets imagine two hoses... one has twice the resistance (meaning it has smaller physical width than the other). But we also make sure that both hoses have the same current (meaning that the smaller hose has twice the voltage (water pressure)). If we were to hit some toy windmills with the water coming out of each of these hoses (from the same distance of course), it is my understanding that they would start spinning at the same speed.

That's where the confusion comes from -- you're not interpreting the equation $P = IV$ correctly. The equation states that the power dissipated in an object is equal to the current through that object, times the voltage drop across that object.

When we apply $P = IV$ to a resistor, which you've corresponded to a hose, $P$ is the power dissipated *in that resistor*, while $V$ is the *difference* in voltage between the two ends of the resistor. For a fixed current, the power dissipated in a resistor with higher resistance is greater, because the voltage drop is larger.

This is independent of how much energy is dissipated in the windmill, which is $I V$ where $V$ is the voltage drop across the *windmill*. In other words, the amount of power you lose in the hose depends on the pressure drop across the hose, while the amount of power you deliver to the windmill depends on the pressure of the water as it comes out of the hose.

To expand on BowlOfRed's answer, your original premise is mistaken. I read your original analysis and, while it is mostly correct, I think you think equal current = equal speed. That's not correct. Equal current = equal flow (i.e. litres/second).

In the water analogy, resistance corresponds to area of pipe, so a high resistance pipe has a smaller area. For it to have the same current, and hence flow, then the water must have higher speed.

In one pipe the water is lazily looping out, while in the other it is squirting out with a fizz. Directed into a bucket, they'd fill it in the same time (same current/flow), but the squirty water can do more work (higher pressure/voltage).

## Forget the analogies and just look at the units

You said it yourself:

but I am a strong believer that you do not truly understand something until you can explain it to someone else in layman's terms

There’s nothing that says layman terms have to be tired old analogues. Just take the thing you are studying at face value:

### Voltage

Voltage is measured in (surprise) Volts. But a Volt is a Joule per Coulomb, or: $$ \text{Voltage} = \frac{ \text{[Energy]} }{ \text{[Charge]}} $$

### Current

Current is measured in Amps and an Amp is a Coulomb per second, or: $$ \text{Current} = \frac{ \text{[Charge]} }{ \text{[Time]}} $$

### Power

Power is calculated as $P=IV$ simply because that’s how the units work out: $$ \text{Power} = \frac{ \text{[Charge]} }{ \text{[Time]}}\frac{ \text{[Energy]} }{ \text{[Charge]}} = \frac{ \text{[Energy]} }{ \text{[Time]}} $$