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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.

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  • $\begingroup$ Note that when the water leaves the hose you are leaving the (wired) electric current metaphor.-- It's also helpful to realize that the equations are valid for a system which has come to an equilibrium; the observed current is the consequence of the applied voltage and given resistance. This means that if you have to violently "press" the electrons with high voltage through a high resistance in order to cause a certain current, then a lot of work is done; much more than if the same current flows already with little "pushing" (because there is less resistance). $\endgroup$
    – Peter - Reinstate Monica
    Nov 29, 2018 at 11:07
  • $\begingroup$ Probably not enough for a answer - the windmill as a fixed resistance so power across the windmill with the same current in both cases is the same. It is the Hose resistor which has the higher power dissipation.So if you imagine a simple potential divider with a windmill and hose resistors, it may more sense. $\endgroup$
    – R.Joshi
    Nov 29, 2018 at 14:57
  • $\begingroup$ Spinning an ideal windmill (one with no friction and blades with no inertial mass) requires just an infinitesimal touch and then it will be spinning infinitely fast. It's best not to imagine the water squirting against the windmill -- it's actually a tricky problem when you think about real world air friction, etc. (Reminds me of the Feynman's infamous sprinkler problem actually!) $\endgroup$
    – Mark Lakata
    Nov 29, 2018 at 22:14
  • $\begingroup$ I also find the water analogy problematic. Instead, imagine bricks sliding down a slope. Each brick is a unit of charge. The number of bricks that pass a fixed point in one second is the current. The height of the hill is the voltage, and the friction that slows the bricks down is the resistance. This analogy also has significant problems, but perhaps it is different enough from the water analogy to give you some insights into your question. $\endgroup$
    – Eric Lippert
    Nov 29, 2018 at 22:42
  • $\begingroup$ <rant> I wholeheartedly hate the water analogy for the electric phenomena. It just gives a faint glimpse of understanding of the basics (e.g. the current "flows like water"), but it gives you huge headaches when trying to understand even slightly more advanced topics. "OK, that wire is like a pipe; you cut the pipe and water comes out; you cut the wire and electrons start falling out, of course!". Right. Indeed In my school lab we have a janitor especially trained to clean all the electron mess left on the floor by students during lab hours.</rant> Sad reality: Maxwell's equations are hard. $\endgroup$
    – LorenzoDonati4Ukraine-OnStrike
    Nov 30, 2018 at 18:27

8 Answers 8

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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.

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  • $\begingroup$ Just like in Thermodynamics T = P/V :) $\endgroup$
    – Jay
    Nov 29, 2018 at 19:24
  • $\begingroup$ E.g. amperage is like pressure if you apply for fluid dynamics into the aforementioned 'holes' $\endgroup$
    – Jay
    Dec 2, 2018 at 17:03
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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).

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    $\begingroup$ You could still have a restriction upstream and then have the pipe flare back out to the same diameter as the other. This would give the same current and velocity to the flow, while still increasing the resistance and therefore power consumed in the pipe with the restriction. Although his analogy did have some issues for this reason, it's not the main problem he is running into. $\endgroup$
    – JMac
    Nov 29, 2018 at 19:00
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    $\begingroup$ This answer feels unsound to me. We know that in the water pipe analogy, water pressure corresponds to electric voltage, and water current corresponds to electric current. It sounds like you're comparing electric power to the linear speed of the water, which is fallacious. The correct analogy of electric power is water power, not water speed. You're writing as if the correspondence between speed and power were significant, but as far as I know, it's just a coincidence. $\endgroup$
    – Tanner Swett
    Nov 30, 2018 at 5:44
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    $\begingroup$ Most importantly, your answer suggests that the presence of a resistor causes something else to receive a large amount of power (because, in your scenario, the presence of a narrow pipe causes something other than the pipe to receive a large amount of power). That's wrong. $\endgroup$
    – Tanner Swett
    Nov 30, 2018 at 5:48
  • $\begingroup$ Guys; The OP is asking about electrical circuits. The OP then brings up the inexact water-in-a-pipe analogy. I'm trying to point out a fundamental flaw in the OP's reasoning (that equal current = equal speed). @TannerSwett: the extra power comes from the prime mover (battery/pump): we have defined a constant current/flow so if resistance increases, the prime mover has to do more work, hence more power. It's not wrong. $\endgroup$
    – Oscar Bravo
    Nov 30, 2018 at 7:12
  • $\begingroup$ @OscarBravo OP talks about water current not speed. He then relates that current delivered to the fan to the speed that the windmill would spin. He was not trying to say equal current is equal speed of flow. He was relating the speed of the windmill to the water current, which is pretty much equivalent to the amount of power delivered to a load relative to it's current. Speed of the windmill is analogous to the intensity of a lightbulb for example in this analogy. $\endgroup$
    – JMac
    Nov 30, 2018 at 15:20
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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: $$ \text{Power} = \frac{ \text{[Charge]} }{ \text{[Time]}}\frac{ \text{[Energy]} }{ \text{[Charge]}} = \frac{ \text{[Energy]} }{ \text{[Time]}} $$

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    $\begingroup$ Dimensional analysis is often a nice tool, but it's not really layman's terms. $\endgroup$
    – JiK
    Nov 30, 2018 at 13:40
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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.

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    $\begingroup$ That's not right. A hose with greater resistance creates a bigger pressure drop across it. That's not the same as the pressure that comes out the other end, which is what makes the windmill spin. $\endgroup$
    – knzhou
    Nov 29, 2018 at 9:57
  • $\begingroup$ I am a strong believer that you do not truly understand something until you can explain it to someone else in layman's terms. If this is true, then I haven't truly understood electric current and so on. In fact I couldn't follow all answers you got, and generally speaking I find the water analogy useless and misleading. Just some examples of its inadequacy. $\endgroup$
    – Elio Fabri
    Nov 29, 2018 at 11:23
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    $\begingroup$ In your hose there is a water flow (= current). There is a pressure (= voltage, but this is questionable). But there is also a water speed, with no electrical analogy. I have read of pressure of the water coming out of the hose as a relevant quantity. Well this pressure is nothing but atmospheric pressure, the same in all cases. The work water can do on the windmill depends on its kinetic energy, a quantity nobody has mentioned. $\endgroup$
    – Elio Fabri
    Nov 29, 2018 at 11:24
  • $\begingroup$ @ElioFabri Of course it's not a perfect analogy. When the water exits the hose, the pressure is converted to kinetic energy according to Bernoulli's principle. But the two can be regraded as the same kind of thing -- pressure is really just a measure of potential energy. $\endgroup$
    – knzhou
    Nov 29, 2018 at 11:31
  • $\begingroup$ @ElioFabri You get the same thing in electricity for very strong fields; electrons will get ripped off a material, with their potential converting into the usual potential energy. $\endgroup$
    – knzhou
    Nov 29, 2018 at 11:31
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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.

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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.

enter image description here

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.

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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!

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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.

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    $\begingroup$ His example was using a fixed current through a variable voltage, not a fixed voltage and variable current. $\endgroup$
    – JMac
    Nov 29, 2018 at 15:09
  • $\begingroup$ Halving or doubling the cross sectional area increases or reduces the power output by a factor of 4 or 1/4 ie. 2^2 or 1/2^2. This can't, as far as I understand, be equated with double or half the voltage. $\endgroup$
    – Charles Russell
    Nov 29, 2018 at 23:38
  • $\begingroup$ Who said anything about halving or doubling the area? He's using a hydraulic analogy with a fixed flowrate, which corresponds to a fixed current. He then said to compare with a pipe of twice as much resistance, which in the analogy is easily done by varying the diameter. To get a fixed flowrate though different resistance you need to vary the pressure (analog of voltage). In his example the second hose would have double the pressure for the double resistance, as stated in the question. Halving or doubling diameter is irrelevant. $\endgroup$
    – JMac
    Nov 29, 2018 at 23:53
  • $\begingroup$ 'Now lets imagine two hoses... one has twice the resistance (meaning it has smaller physical width than the other). ' As I understand it, this smaller width equates to 1/4 of the cross sectional area. ce, which in the analogy is easily done by varying the diameter. To get a fixed flowrate though different resistance $\endgroup$
    – Charles Russell
    Nov 30, 2018 at 0:43
  • $\begingroup$ That still doesn't explain why you're using a fixed voltage and varying current in your example. $\endgroup$
    – JMac
    Nov 30, 2018 at 0:54

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