# Interpreting I²R and V²/R

• B
Hi. I’m new to the forums. I’m not a scientist, but I’m interested in learning more about some basic concepts of physics relating to forces, power, etc. I took first year calculus in university and did very well, but that was almost 40 years ago (it almost hurt to admit that!), so I’ve forgotten most of it. I understand the principles of mechanical force, work and power, and I’m looking to gain an understanding of electrical principles. This is purely for interest sake; I’ve got no objective in mind other than to expand my knowledge.

In a simple DC circuit, I understand the concept of voltage (“V” - electrical potential energy difference per charge) and current (“I” – rate of flow of charge, = V/R). I also understand Ohm’s law (V=IR), and I realize that, where Ohm’s law applies, there are 3 ways of expressing power – V·I, I²R, and V²/R. What I’m trying to do is interpret what each one of those formulas mean on a “stand-alone” basis.

The first one, P=V·I, is easy because I’ve seen the definition in numerous places. It’s (energy dissipated per unit of time) = (energy dissipated per charge passing through a resistor) × (charge passing through a resistor per unit of time). This makes perfectly good sense to me, since the charge in the denominator of voltage cancels the charge in the numerator of current, leaving energy per unit of time, which is the definition of power.

P=I²R is a bit trickier. Why is P proportional to the square of I (and proportional to R)? I haven't found an interpretation of this; the only explanation I’ve seen is basically a restatement of P=V·I using Ohm’s law. This isn’t what I’m looking for, though. So I’ve come up with my own interpretation - it makes sense to me, but I don’t know if it’s actually correct. The way I think of it is that, in a current passing through a resistor, electrons are moving around and colliding with the other particles in the resistor. These collisions result in kinetic energy being transferred to the resistor, which is essentially the energy dissipated. Since KE is proportional to the square of velocity, it makes sense that power would be proportional to the square of current, since a higher current means higher electron velocity (assuming a constant R). And if the current is kept constant, doubling (for example) the resistance would in effect double the number of particles the electrons can collide with, thereby doubling the rate of collisions and doubling the total energy dissipated. So like I said, all this makes sense to me, and I’ve done it without explicitly talking about voltage, but I’m wondering if it is in fact a reasonably accurate description. And if it isn’t correct, what is the correct interpretation?

P= V²/R is worse. Why is power proportional to the square of voltage, and inversely proportional to R? I can’t for the life of me come up with an interpretation of this that doesn’t require me to consider current, which basically defeats the purpose of what I’m trying to do. Is there a way to look at this formula that doesn’t essentially require it to be restated as P= V·I?

Thanks in advance for any insight you can provide.

• Delta2

## Answers and Replies

Grinkle
Gold Member
Power is not dissipated if effort and flow are not both greater than zero. Considering an electrical system, effort is voltage and flow is current. To me, it makes no sense to try and consider power dissipation without considering both voltage and current. A fully charged battery is presenting effort at its leads, there is a voltage between them. If the battery is open circuit, the resistance between the lead is infinite, there is no flow (no current) and no power is dissipated. R relates voltage to current, it relates effort to flow, as you note. This is not an artifact of algebra - the relation itself has physical implications and needs to be considered when thinking about power dissipation.

• berkeman
phinds
Science Advisor
Gold Member
I agree completely w/ Grinkle. I think you should Google "drift velocity", as I don't think electron velocity in a circuit is what you think it is.

• berkeman
Thanks guys. For the V²/R scenario, I kind of suspected your answer was the case since voltage without current can't give power, so there isn't really any way to talk about P=V²/R without bringing current into the analysis. I'm not sure about I²R, though - voltage is implicit if there is current, but I don't know if that invalidates my analysis.

I'm familiar with drift velocity (or at least I think I am). As I understand it, drift velocity is the velocity at which electrons move through a resistor/conductor. However, (and I stand to be corrected), this is different than the velocity of individual electrons as they "bounce around" inside the resistor/conductor. As an analogy, the speed of light is 300,000 km/s. The radius of the sun is ~700,000 km. However, a photon starting from the centre of the sun doesn't reach the surface in 2.33 seconds - it takes anywhere up to 1 million years because it keeps bouncing off the various particles in the sun. So 300,000 km divided by 1 million years is 333 metres/year - this would be the equivalent of "drift velocity". The photon still travels at 300,000 km/s, though - it just doesn't get very far very fast; this is similar to what I'm referring to when I talk about the velocity of individual electrons.

It usually doesn't help to think of circuits in terms of the mechanical motion of electrons. After all, the existence of electrons wasn't confirmed until decades after Maxwell's equations were written down.

##P = I V##, ##P = I^2 R##, and ##P = V^2 / R## are all the same thing. If you want to know where the second two come from, they're both simply a restatement of the fact the ##P = I V## under the condition that ##V = IR##. Substituting the latter into the former gives both of the other equations. Basically, power increases as the square of the voltage as well as the square of the current (times some constant), and this is more clear when you look at ##P = I V## and notice the fact that current increases as voltage does.

• CWatters and Drakkith
robphy
Science Advisor
Homework Helper
Gold Member
Of the three, P=IV is the most fundamental statement...
although if you write it as P=VI, then you can appreciate it as the analogue of the mechanical definition of Power=(Force)(velocity).
The analogue of the resistance R is some drag-coefficient b for a force proportional to the velocity v... (drag force)= -bv.

(Power supplied by a battery)=(Voltage across the battery)(Current through the battery)
(Power "supplied" by a resistor)=(Voltage across the resistor)(Current through the resistor)... taking care of sign conventions, the voltage across the resistor is negative... so the resistor dissipates [positive] power.

Then, as you say, using Ohm's Law, you can express this in terms of the resistance and either the voltage across or the power through.
The mechanical analogues of this can be constructed..
and maybe an interpretation of the mechanical analogy can be made... but I haven't seen these mechanical analogues in the literature.
Maybe it's better to think of work-done first, rather than power.

I am not sure electrons flow so much "faster" with voltage as in the op's mechanical analogy.

davenn
Science Advisor
Gold Member
Since KE is proportional to the square of velocity, it makes sense that power would be proportional to the square of current, since a higher current means higher electron velocity (assuming a constant R). And if the current is kept constant, doubling (for example) the resistance would in effect double the number of particles the electrons can collide with, thereby doubling the rate of collisions and doubling the total energy dissipated

electron velocity is extremely small ( slow) their kinetic energy as a result is also very low and therefore doesn't factor into the calculations

I agree with phinds' comment ......

I think you should Google "drift velocity", as I don't think electron velocity in a circuit is what you think it is.

and you also need to consider that the energy in the circuit is not carried by the electrons rather it is carried (transported, if you like) by the electric field

Dave

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• vanhees71
CWatters
Science Advisor
Homework Helper
Gold Member
I can’t for the life of me come up with an interpretation of this that doesn’t require me to consider current.. ?

Basically you can't because I and V are not independent variables. In the case of a resistor P=IV but if you change I (V) then V (I) also changes. You cannot change one without effecting the other.

A common problem is that people think V is constant. After all the mains voltage is constant, so is the voltage of a battery (mostly). The problem is that in P=IV the voltage is not the supply voltage it's the voltage across the component that is dissipating the power. These are sometimes the same thing but not always.

Thanks again everyone. I guess maybe I'm just trying to make sense out of a formula where it can't actually be done. A bit disappointing, but if that's the way it is, so be it. I'm not yet entirely convinced my interpretation of I²R is wrong, although davenn's comment suggests that KE is irrelevant, so I guess I'm wrong there too.

And I just realized I made a mistake in my previous post where I drew an analogy of drift speed to photons leaving the sun. I don't know how to edit the original post, so I'll just correct it here by saying that the "drift speed" (so to speak) of the photon is 700 metres/year, not 333 metres/year (I totally botched the original calculation - pretty embarrassing given how basic it is).

Svein
Science Advisor
The different formulae have their place in different situations. For example: If you want to transfer power through a cable, the power supplied at the receiving end is Ps=V⋅I. The power lost in the cable is Ploss=Rcable⋅I2. From these two you can infer:
1. If you need to transfer a given amount of power, you can use a low voltage and high current or a high voltage and a low current (as long as their product stays the same).
2. If you double the voltage (and thus halve the current), the cable loss is reduced to ¼.
3. That is why power companies prefer to use extremely high voltages for long cables (if you use 500 000V, the current will be one thousand of the current if you were to use 500V. The cable loss will be one millionth.)

davenn
Science Advisor
Gold Member
And I just realized I made a mistake in my previous post where I drew an analogy of drift speed to photons leaving the sun. I don't know how to edit the original post, so I'll just correct it here by saying that the "drift speed" (so to speak) of the photon is 700 metres/year, not 333 metres/year (I totally botched the original calculation - pretty embarrassing given how basic it is).

and this is, unfortunately, is still completely incorrect

photons don't have a drift speed, they travel at the speed of light ( round figures 300,000 metres / sec) and cover the distance from the sun to earth, ~ 150 million kilometres, in around 8.5 minutes

Electrons in a conductor, on the other hand and with a voltage applied, do have a drift velocity of a few millimetres / sec or so, depending on the magnitude of the current

Dave

EDIT

after reading your comments in post #4, I can see that you misquoted yourself in post # 10
my above response was purely to post #10. You were getting on the right path in post #4

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after reading your comments in post #4, I can see that you misquoted yourself in post # 10
my above response was purely to post #10. You were getting on the right path in post #4
I'm not sure where I misquoted myself in #10; there, I just corrected my shameful botched calculation in post #4. But I guess it doesn't really matter. Thanks for confirming that my analogy was on the right path in #4.

I think something for the OP to consider is that the equations involving P, V, I, R are on the one hand different ways of saying same the same thing and indeed must give exactly the same answers to a problem so can be seen validly as just algebraic manipulations of each other.

OTOH though the reality of designing, building and taking measurements of real circuits you should consider that in any real circuit you don't get infinite degrees of freedom in what values of P, V, I, R you choose.

In the real world you are constrained by many things in building a circuit whether the circuit be a national power grid or an audio speaker home sound system.

for example it is every much a political and economic reason that a coal fired power station generates at a specific frequency and voltage. the fact it is AC and the fact that there is a finite distance between source and destination with a fixed resistance/km determined by a whole lot of constraints eg type of cable, distance between posts supporting the cable...etc. then P=VI becomes more than just algebra as does in this example P = I^2R.

clearly for fixed R which you have no wriggle room to adjust the "I" needs to be as small as practical which requires the source voltage to be as large as practical while being constrained by the rotation speed of the generator and in fact the entire design of the entire plant including boiler room, furnace, miles of steam pipe, cooling tower size.......... the P's relate to different things that are linked through a subdivision of source, transmission and destination constraints.

hope that makes sense.