Understanding voltage and power in the fluid analogy for DC circuits I am trying to understand electric circuits (ie voltage, current, power, and resistance).  For the most part, everything makes perfect sense, but for some reason I do not feel as if I understand the proper definition of power.  Yes, I understand the formula's $V=IR$ and $P=IV$, 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 (which I cannot confidently do at this point in time).
Perhaps I can explain my confusion using the water analogy (which I'm not particularly fond of, but will use anyone for illustrative purposes).  I understand that if water is flowing through a hose (or pipe), the amount of water at a specific spot per second is analogous to electrical current, the pressure to voltage, and the width of the pipe to resistance.
Now let's 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, or another way to put this might be that the work done upon them is the same.
Now this is where my confusion starts, because in this situation, the current is the same for both hoses, but the power (watts) is doubled for the hose that requires double the voltage to maintain the same current (due to it having twice the resistance).
When I think of the word "power" and something that has twice as much of it as something else, my mind instantly thinks it is twice as powerful, and thus can exert more force or do more work on external objects.  But here it seems like current is what determines the speed at which the windmills would turn, or how bright a light bulb in a closed electrical circuit would be? In fact, it seems like 'power' in this context is a requirement, or the the amount of effort required to keep the current at a constant rate given a certain voltage. This also makes me think that the the hose with twice the power is less efficient (obviously due to the resistance). But thinking of power like this seems counter intuitive, and perhaps I am not understanding something important here. Is power truly the 'effort required' by the circuit to run or is it about 'potential work' that a circuit can exert on external things.  Any clarification is greatly appreciated.
 A: 
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, or another way to put this might be that the work done upon them is the same.

But you said that the pressure in one was higher than the other.  That means the water is going to shoot out faster.  Each bit of water is moving faster and is capable of doing greater work than an equivalent bit from the one that is moving slower.  One will make the windmills spin, but the other will make them spin really fast.
And since the flow rate from each hose is the same, that means the higher pressure hose can do more work.  
A: 
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.
A: The analogy to flowing water might be more helpful if you looked at a slightly different scenario.  Instead of a hose squirting at a toy windmill, consider a hydroelectric dam.  There is stored energy in the water above the dam, there is a system of pipes that guide water to a turbine, and there is an exit pipe that lets the used water flow into the river below the dam.  If part of the system of pipes is horizontal,  there is some loss of pressure in the pipe from one end of the horizontal stretch to the other.  But it's small, and it represents lost energy.  There is a large pressure drop between the intake of the turbine and the outlet of the turbine, and that represents energy that has been taken out of the water and put into the electricity.  And there is some pressure in the outlet pipe, representing residual energy in the water that could not be harnessed into electric energy.
But in the above description,  I have subtly shifted the focus from power to energy.  And this is where you need to expand your bag of concepts.  Power and energy are closely related concepts, but they are not the same.  The hydro electric dam is measured in terms of power  (kilowatts) but the electric company sells you electricity in terms of energy (kilowatt-hours).  Power is basically energy per unit time.  Or put the other way around,  energy is power integrated over time.  If you take some time to understand both power and energy, you may find it easier to grasp both concepts better than you can grasp either one.
Energy can be measured in Joules, Ergs, or electron-volts as well as kilowatt-hours.  The same units can be used as measures of work.  In fact the concepts of work and energy are very closely related.  Work is basically energy transferred from one system to another.  
A: 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).
A: 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:
$$ \text{Power} =  \frac{ \text{[Charge]} }{ \text{[Time]}}\frac{ \text{[Energy]} }{ \text{[Charge]}} =  \frac{ \text{[Energy]} }{ \text{[Time]}} $$
A: The circuit you have described might look like this with two tubes $AB$ and $CD$ with the same internal diameter but one double the length (resistance) of the other.
To complete the circuit any water which issues from the ends of the tube is pumped back into the reservoir of water to keep a constant head.
This last bit is analogous to the chemical reaction in a cell keeping the potential difference across the terminals (head of water) constant.    

There is a pressure difference ($\equiv$ potential difference) $P \,(=h\rho g)$ across tube $AB$ and the rate of flow of water ($\equiv$ current) is $\dot q$.
If the cross-sectional area of the tube is $A$ then the net force on the water in the tube $(P - P_{\rm atmos}) A$ and the work done per second (power) to drive the water through the tube is $(P - P_{\rm atmos}) A \times v =\Delta P\,\dot q$ where $v$ is the speed of the water.
So we have $\Delta P\,\dot q\equiv VI$ for the electrical circuit.
For the lower tube the pressure difference is twice that of the upper tube and the power to drive the water through the lower tube is twice that for the upper tube as the pressure difference ($\equiv$ potential difference) is twice as large.  
Where does that work per second do?
It drives the water through the pipes and generates heat due to fluid friction (viscosity) at a rate of $\Delta P\,\dot q$ just as the potential difference drive a current though a resistor with heat being generated.  
The windmills ($\equiv$ ammeters) at the end both rotate at the same speed thus indicating that the flow of water is the same.
If the windmills are frictionless then the water loses no kinetic energy as it flows over the paddles and hence there is no energy transfer to the windmills equivalent to saying that an ammeter has zero resistance.  
A: As far as I can see, the question you are asking is "What is power in electrical circuits?" and you have tried to model this using an analogy of water in pipes.
I believe the analogy introduces many difficulties so i'd like to suggest abandoning it and look instead at..
Resistors! :-)
The thing about resistors is that when you apply a voltage to them, some current flows.
If you apply more voltage, more current flows.
If you then somehow manage to increase the resistance, less current flows.
the realtionship between current, voltage and resistance is locked together by V=IR.
When you have a voltage between two points in a circuit and a current running between them some heat will be generated.
The heat is Energy and is measured in Joules.
If you turn on an electric cooker, the ring gets hotter and hotter.
This is because more and more joules of energy are being added to the ring.
The rate that the energy is added is Power.
Power = Joules per second
P=J/s
Power also equals Current times Voltage
P=IV
It is measured in Watts which are Joules per Second.
More power would mean that more Joules of energy were added to the ring per second and the cooker ring would heat up faster.
The reason you have I squared in the formula for power in a resistor is due to substitution.
P=IV
but
V=IR
so P=I(IR)
$$ P=I^{2}R\, $$
and
$$ I=V/R\, $$
so
$$ P=V(V/R)\,. $$
$$ P=V^{2}/R\,. $$
The voltage, current, resistance and power are all locked to each other for each resistor in the circuit.
In circuits with resistors in series, the currents will all be the same and the voltages will vary depending on the idea of the "Potential divider".
The total current will be the voltage over all the series resistors divided by the sum of their resistances.
If you multipy this current by the resistance of each resistor and add them up you will find that the total equals the voltage across all of them and the voltage across each resistor is proportional to the resistance of that resistor.
This means that power will be disippated in each resistor in the circuit.
now, wires on circuit diagrams are considered to have no resistance.
Unfortunately, this is not true in the real world!
Wire has a resistance inherrent in the material.
Each strand of thin copper can be considered a resistor.
All the strands together can be considered resistors in parallel.
So the more strands and thicker the cable, the lower the resistance.
If you include this resistance in your calculations, a lower resistance means a lower voltage drop for the same current and so a lower power loss.
If you have large currents flowing you should consider the power being converted to heat in the wires.
This is why you should put in thicker wire for cookers than lights and also why you should un-roll extansion cables when running them near their maximum capacity.
If you don't, the heat lost from the cable will not be able to escape from the roll and it will het hotter and hotter, possibly melting the insulation!
A: It might be imagined that for the same voltage and two different resistance values there could not be the same 'current' output. Joule found heat dissipated in a wire was proportional to the 'square' of the current passing through it. Hence, 4v/1Ω = 4 amps and 4v/2Ω = 2 amps.
This would give 4^2 amps * 1 ohm = 8 watts in the first case and 2^2 amps * 2 ohms = 16 watts in the second case.
Multiplying both by seconds = Joules, the amount of work done or heat energy dissipated.
