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 power-consuming 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?

  • $\begingroup$ Can you make your post a little more exact? Are you saying Power = Current? Please define all terms. And make sure units match up. $\endgroup$
    – user196418
    Commented Dec 16, 2018 at 19:02
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    $\begingroup$ en.wikipedia.org/wiki/Hydraulic_analogy#Equation_examples $\endgroup$
    – BowlOfRed
    Commented Dec 16, 2018 at 19:29
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    $\begingroup$ In the case of a water wheel, it's the amount of water flowing times how hard the water pushes the wheel. $\endgroup$ Commented Dec 16, 2018 at 23:16
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    $\begingroup$ Pressure does matter to hydraulic power, that's why water is piped from mountain dams to valley power stations. $\endgroup$
    – Aetol
    Commented Dec 17, 2018 at 10:18
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    $\begingroup$ This illustrates one of the big flaws of the hydraulic analogy - now you have to understand hydraulics, too. $\endgroup$ Commented Dec 19, 2018 at 1:00

9 Answers 9


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).


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.

  • $\begingroup$ Very small compared to... flow in pipes? In open channels the effort variable is predominately height rather than pressure, but in any flow it's always the combination (could be evaluated via bernoulli's equation) $\endgroup$
    – Eph
    Commented Dec 17, 2018 at 13:00
  • $\begingroup$ flow in pipes, or in hydraulic systems. the driving pressure in open channel flow is so small it is measured in inches of water column rather than pounds per square inch, but it is still pressure. $\endgroup$ Commented Dec 17, 2018 at 16:49
  • $\begingroup$ For a water wheel (as pictured in Cort's answer) the effort from pressure is rather irrelevant compared to the effort from the height difference. If this height effort is included in your driving pressure that wasn't clear to me. $\endgroup$
    – Eph
    Commented Dec 17, 2018 at 17:13
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    $\begingroup$ Ahh I think I see where we miscommunicated. I think you're saying that the effort variable is measured in units of pressure (which could be calculated as $P_0=\rho\,g\,h+\frac12\rho\,v^2+p$) regardless of which term is dominant in bernoulli's equation. I was just thrown off by your comment that open channels have low effort and thought you were neglecting to consider the height portion of the equation. $\endgroup$
    – Eph
    Commented Dec 17, 2018 at 18:23
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    $\begingroup$ And in this specific case, the "stuff" that flows is a charge, so the effort per unit stuff is a voltage, the stuff per unit time is a current but the dimension of (effort per unit stuff) × (stuff per unit time) is independent of what "stuff" is. $\endgroup$
    – Alexis
    Commented Dec 18, 2018 at 19:58

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.


One of the challenges with using analogies like the water analogy is making sure you use the correct objects to draw your analogy from. Many answers here argue that hydraulic power depends on volume and pressure. This is true if your water wheel looks something like this:


This is a drawing of what a modern hydroelectric turbine looks like. They are designed to be able to efficiently draw power from large volumes of water with large pressure drops, such as the pressure drop from the bottom of a lake out to atmospheric pressure. In these cases, the water analogy works rather well, as expected.

However, when I think of a "water wheel," I have a different image. I imagine something much older:

Water wheel

These operate differently, and lead to the conclusion you drew, which is that water wheels generate power from current alone. The reason you end up drawing this conclusion is because this sort of water wheel wastes any power from pressure or velocity. The only potential it is effective at generating power from is the gravitational energy of the water from a high altitude to a low altitude. If you were to spray high pressure water at one of these water wheels, most of the energy would be wasted as the water splashes off of the blades. Some energy would indeed be transferred to the wheel, but it would be tremendously wasteful.

The water wheel is most efficient at handling cases where the vast majority of the energy of the water is stored up as gravitational potential energy -- energy from being up high. And it's best at converting water which is exactly at the height of the wheel. If you drop water down from high above the wheel it will turn, but most of the energy will be wasted in splashing and sloshing.

Thus we would treat this water wheel as a "constant voltage" device, in our electrical analogy. The setup around the wheel has the effect of guaranteeing that most of the potential of the water is its fixed height as it enters the wheel. Any energy above this is wasted as heat. And, indeed, if you look at the math, when your voltage is constant, your power is indeed proportional to the current. It's the special case where this is true.

We do indeed have devices that operate this way, but you have to enter the world of semiconductors to do so. Diodes are small semiconductor junctions used to make current flow only one way. Try to flow against it, and they're like a check valve that stops the water flow.

Well, almost like a check valve. They operate like a check valve up to a point, called the "breakdown voltage." If you put a voltage higher than this across the diode in the wrong direction, it starts to let current through. It will dissipate any energy that comes from running current across this voltage drop as heat.

So what the old fashioned water wheel is most like is a motor with a reverse-biased diode on it. Any potential of the water beyond the gravitational potential energy the wheel can handle is lost to splashing and sloshing. Any potential from higher voltages applied to the diode and motor circuit is lost to heat as the current flows through the diode. The remaining gravitational potential of the water multiplied by the volume of water sent through the wheel tells you how much mechanical power is generated by the water wheel. The voltage across that motor (after the diode limits it) multiplied by the current going through the motor is how much mechanical power the motor generates. The analogy holds, you just need a more complex circuit to model the 6000 year old device!

Incidentally, we do actually design circuits like this. In modern circuitry, we often have "zener diodes" which have a carefully tuned breakdown voltage to be a "voltage reference," and we have voltage regulators which are designed to resist the flow of electricity just enough to ensure a specific voltage across the remaining circuit.

  • $\begingroup$ The water wheel actually does benefit from velocity. During the drop the water gets accelerated by gravity. This energy gets absorbed by the wheel's blades as the fluid de-accelerates. $\endgroup$
    – Muhlemmer
    Commented Dec 17, 2018 at 16:19
  • $\begingroup$ @Muhlemmer Actually, if you think about it, it doesn't really get accelerated. It does have the force of gravity pulling on it, but if it accelerated on the way down, the bottom of the wheel would have to be spinning faster than the top of the wheel. (And it does benefit a little, but its limited) There's also angular acceleration from bending around the wheel, but that's also pretty slight. $\endgroup$
    – Cort Ammon
    Commented Dec 17, 2018 at 16:26
  • $\begingroup$ +1 for emphasizing the point that while the amount of useful energy a water wheel harvests for a unit quantity of water may be roughly constant, the amount of energy removed per unit quantity is proportional to the pressure difference before and after, with increases in pressure differential leading to increases in waste. $\endgroup$
    – supercat
    Commented Dec 17, 2018 at 18:30

The person who said that is wrong. Current doesn't push. Voltage or pressure pushes. Current is how much actually flows.

Useful power is the force of the push (voltage, pressure) multiplied by the current (amps, CFM).

Obviously with no push, you have no power. But with no flow, you also have no power.

In water generation, pressure also equates to height, called "head".

For instance, a PC power supply is a lot like run-of-river hydro -- not much pressure but lots of flow.

A very tall, but low-flow dam like Oroville is a lot like an LED light on a 240V line - lots of pressure, not much flow. Not much flow needed.


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 ($\frac {m^3}{s}$) times pressure ($\frac {N}{m^2}$) equals power ($\frac {J}{s}$ = 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.


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 $$ \text{(power)} = \text{(current)}\times\text{(voltage)} $$

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.


The river flows with fixed flow rate $\dot m$, the width of the river is constant throughout, but the river is made to go over a waterfall that drops by distance $\Delta h$. The velocity at the top of the waterfall is $v_1$ and the bottom $v_2$. By energy balance, $$ \frac{v_2^2-v_1^2}{2}=g \Delta h $$ The power that is available to us is $$ \dot W = \dot m \frac{v_2^2-v_1^2}{2} $$ We have to subtract $v_1^2$ because the river must continue to flow with its steady velocity after it hits the wheel, or else we will flood. Combining with the energy balance: $$ \dot W = \dot m g \Delta h $$ Here now we have a one-to-one analogy with electric current: $\dot m$ is the current and $g\Delta h$ is the voltage. Increase the voltage and the same current will produce more power.


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 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...


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