Krdytkyu

Many people think of the water analogy to try to explain how electromagnetic energy is delivered to a device in a circuit. Using that analogy, in a DC circuit, one could imagine the device is like a water wheel being pushed by the current.

In the case of an actual water wheel, the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time: power = current, but in electric circuits power = voltage x current.

Why is this?

In your water example power cannot be equal to current because they have different units (power is an energy per unit time, while current would be something like a number of particles passing through a surface per unit time).

...the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time

What you have noticed here through your analogy is that power is proportional to current (as an example, the more force you apply to an object, the larger its acceleration, but this does not mean that force and acceleration are equal). In a circuit element this proportionality is the voltage, since it tells you how much energy is associated with a "unit of current". You would need a similar way to convert your water current to the power generated by that current (although this might be a simplistic model of how power is generated using a water wheel).

You also have to keep in mind that it is an analogy, and all analogies have imperfections. With the water analogy, power is generated by water actually pushing on a wheel. In circuits, $P=IV$ is much more general and applies to any charges undergoing a potential difference.

Here is a simple way to keep this stuff straight.

Power is always the product of an effort variable and a flow variable. In hydraulic systems, the effort variable is pressure and the flow variable is the flow rate.

For flow in open channels, the effort variable is typically very small (but not zero) and the flow variable is very large. BTW power exchange which occurs at low effort and large flow represents the low-impedance regime.

Power is defined as work done per unit time. So a mass of water moving from one potential to some other lower potetial can do work when it hits the wheel. How much work per unit time? It depends how much mass falls times height times gravitational constant g and all of that divided by time. Water current, on the other hand is just total volume or mass that flows per unit time. Sure, it is connected to the work done or power but is not the same thing. If you define gravitational potential difference as gH and water flow as dm/dt then to have power you have to have curent times this potential difference: gHdm/dt...
In a conductor, work is done also and energy per unit charge or potential difference is given by U (voltage) and current by dQ/dt so it seems to me that everything is same...

Water and current flow as well as other “mechanical” analogies (pipe resistance vs. electrical resistance, voltage vs. pressure, etc.) are useful for introducing electrical circuit concepts at an elementary level. This is because mechanical concepts are easier to visualize whereas electrical concepts are more abstract. The analogies can only go so far without a deeper understanding.

Current does not equal electrical power and neither does water flow equal mechanical power.

Electrical current ($frac {Coul}{s}$) times (in phase) voltage ($frac {J}{Coul}$) equals power ($frac {J}{s}$ = watts).

Current flow ($frac {ft^3}{s}$) times pressure ($frac {lb-f}{ft^2}$) equals power ($frac {ft-lb}{s}$). ($frac {550 ft-lb}{s}=1 hp = 746 watts$)

The water pressure for the water wheel can come from water dropping from a height above the wheel (potential energy) or even from a hose directed horizontally, depending on the orientation of the water wheel.

To complete the analogy between current and water flow with respect to electrical an mechanical power, you need in addition the analogy between voltage (electrical potential) and pressure (mechanical potential).

Hope this helps.

The analogy with water actually holds really nicely if you consider a water wheel, or other hydroelectric system.

But what you're missing is that the power produced does not only depend on the amount of water going past - it also depends on the speed at which it does so. (this makes sense for the hydro system because kinetic energy depends on velocity and mass)

To make the analogy better, rather than thinking of the speed of the flow, think about how far it has fallen to aquire that speed. At this point you have a volume per second of water - the current - and you have a loss of height, which is literally a potential difference.

Power to a water-wheel depends both on the current (amount of water delivered)
and the head (vertical drop of water as it turns the wheel). So, the
water analogy does have TWO variables that multiply together to make
power: current, measuring (for instance) the water flow at Niagara,
and vertical drop (like the height of Niagara Falls).

Current is NOT the same as power, in a river, because long stretches of
moving water in a channel don't dissipate energy as much as a waterfall does.
Siting a hydroelectric power plant at Niagara Falls makes sense.
In the analogy to electricity, a wire can deliver current at little voltage
drop (and has tiny power dissipation) but a resistor which has that same
current will be warmed (it has a substantial terminal-to-terminal voltage drop).

I'd like to put what others said already into equations:

A river has mass flow rate $dot m$ (kg/s) (the "current"). The water flows with velocity $v$ (m/s). Power is the kinetic energy that is carried per unit time:

$$
dot W = frac{dot m v^2}{2}
$$
and notice that this has proper units of watt. If we take "voltage" to be $v^2/2$, then we get
$$
text{(power)} = text{(current)}timestext{(voltage)}
$$

Notice that the mass flow rate is not enough to give you high power. When the river is wide the velocity is low. If you want the water wheel to run fast you should build it at a narrow passage.