# Power Law Derivation for dynamic systems

1. Jul 23, 2014

### FallenLeibniz

Just a quick question. Let A and B be two points. Electrical work is defined as the amount of energy it takes to move an amount of charge Q through a potential difference VB-VA (for our purposes here, we will assume that the voltage values are measured with respect to an Earth ground) and is calculated as the product of QV.

Now to find the instantaneous power, you would take the derivative of QV with respect to time. Why is it that most texts tend to throw out the second term of this product and just keep the result of the calculation as P=VI? Are there physical grounds for so many texts falling back on this assumption?

Note to admins: Yes this question is of a mathematical nature, but I'm trying to see if there are physical grounds for most texts assuming that the Q(dV/dt) term can be thrown out other than just assumng that V is constant (which is not always the case, but doesn't stop text writers from still assuming the P=VI form of the law anyway).

2. Jul 24, 2014

### Simon Bridge

They don't.

The power dissipated in a resistor is the product of the current through it and the potential difference across it, so in the equation P=VI, V=VB-VA where A and B are points on either side of the resistor.

In general, you don't have an absolute reference point for potential energy.
Therefore you pick whatever reference point makes the maths easy.

3. Jul 24, 2014

### FallenLeibniz

That's the point I'm trying to make though. If you actual take the time derivative of the work done across the resistor, then power is no longer P=VI it's P=VI+Q(dV/dt) since W=QV. Basically I'm asking why do they call P=IV the power law when there is that extra term that's being neglected?

I'm trying to find out the physical grounds for throwing out the Q(dV/dt) term.

Last edited: Jul 24, 2014
4. Jul 24, 2014

### Simon Bridge

For a resistor, the potential difference does not change with time.
Thus dV/dt=0

5. Jul 24, 2014

### FallenLeibniz

That's only for systems in which that would be guaranteed for though, right? Such as a battery draining over a resistor. It wouldn't be valid to hold that assumption for something like sinusoidal voltage being applied over it.

Only reason why I'm on about this is because I've got a text that's trying to use p=iv as a general rule to calculate the timed rate of change in stored energy over a capacitor as well.

Last edited: Jul 24, 2014
6. Jul 24, 2014

### FallenLeibniz

I can't find an explanation for as to why assuming P=IV is valid in the case where you do have a non constant voltage. If I could get some explanation or justification, I could let it go.

7. Jul 24, 2014

### Simon Bridge

In the case of a sinusoidal voltage, I and V in P=IV are rms values.
Where the voltage is not sinusidal, you'd have to decompose the actual waveform into sinusoids.

8. Jul 24, 2014

### FallenLeibniz

But why is P=IV? How does making it an rms value balance the fact that the q(dV/dt) term is not included?

I'm not trying to be dense and I do appreciate you taking the time to help me, but I just don't see q(dV/dt) is not included in the Power Law for the cases when dV/dt is not zero.

Edit: In short my question is, why is instantaneous power defined as P(t)=I(t)V(t) and not P(t)=I(t)V(t)+q(t)[dV(t)/dt]?

UPDATE: Ok. I think I may actually have a reasoning, and wonder if it's valid. Since instantaneous power is just that, you are actually shrinking the section of circuit you are looking at. As a consequence of this, q(t) goes to zero (since as you shrink Δt, you come to have less and less charge pass through the area of interest) but i(t) does not since it is the ratio of Δq to Δt as Δt goes to zero. Does that sound plausible???????

Last edited: Jul 24, 2014
9. Jul 24, 2014

### FallenLeibniz

Given W=QV and the voltage drop over the resistor being constant gives P=IV. I'm trying to find out when the voltage drop does have a time dependence, why isn't the Q(dV/dt) included?

10. Jul 24, 2014

### Simon Bridge

It looks to me that nobody can tell you - you'll just have to do it yourself.
To figure that out - set up the circuit, and work out the power dissipated in the resistor for an AC voltage.
Go back to first principles.

11. Jul 24, 2014

### Delta²

Well the definition V=W/Q holds for a point charge Q that is accelerated between the end points on which the potential difference is V. In the case we have a continuous distribution of charge flowing between two endpoints , and such is the case when we have a current between the two points, then things are different:

We consider an ifinitesimal charge dq, then during time dt and assuming that the current is constant across the region between the two points, this charge dq moves a distance dl (which is the same for all the charges dq across the region) and the infinitesimal work done by the electric field for the charge dq is E(t)dqdl. Thus summing over all charges dq across the region we get that the infinetesimal work done for the whole charge Q inside the region is dW=E(t)Qdl.

This work dW is for time dt, hence $P=\frac{dW}{dt}=E(t)Q\frac{dl}{dt}=E(t)Qv(t)$ where v(t) is the velocity of the charge distribution among the end points which we consider constant through the region (since we assume constant current across region, also for the same reason the total charge Q among the region remains constant since the current is always the same at the two end points).But assuming also that the electric field between the two end points is homogeneous , it is E=V(t)/L, hence we have $P=V(t)\frac{Qv(t)}{L}$. It is easy to see that the quantity $\frac{Qv(t)}{L}$ is equal to the current I(t).

Update: Q/L is the charge per unit length. The charge dq that flows through a cross section S in time dt, lies within cylinder with base S and height v(t)dt. It is equal $dq=\rho Sv(t)dt$ where $\rho$ volume charge density. $\rho S=\frac{Q}{L}$ holds assuming constant volume charge density throughout the region.

I know i did some extra simplifying assumptions to simplify the derivation, i believe it can be done with just the assumption that $\rho (x,y,z,t)v(x,y,z,t)=J(t)$ is constant throught the region, that is we have constant current density (wrt spatial coordinates).

Last edited: Jul 24, 2014
12. Jul 24, 2014

### FallenLeibniz

Thank you Delta for that argument. I do see the sense in it. However for any kind of varying waveform, the assumption that i(t) is constant no longer can be taken for granted. Say you have an ohmic resistor of resistance R hooked up across a function generator that is producing a sinusoidal waveform for the voltage. i(t) would not be a constant in this case. While I do see, in the case were i(t) is constant, that p(t)=i(t)v(t), I do not see how it can hold in the cases where the current and the fields are not taken to be uniform.

I had the thought (which I put in my update) that as you begin to make Δt go to zero, the section that you are looking at gets smaller. This would make the value of q(t) approach zero as well but not necessarily i(t) since the instantaneous current can be considered to be the differential ratio dq/dt (the reason for the stress on the fact that this is a ratio is that it allows it to maintain an approach to a non-zero value since the numerator and denominator are both shrinking as the area of the circuit being observed shrinks). Does that reasoning make any sense? I am not being too convoluted am I? I do want to get this and am not trying to be dense, but I'm looking for an explanation of why the general case is still considered to be p(t)=i(t)v(t).

13. Jul 24, 2014

### Delta²

In my post I(t) is not considered to be constant in time but constant throughout space. This is a basic assumption made in any dc or ac circuits in series, that the current is everywhere equal at a given time t, regardless of what are the components consisisting the circuit.

This assumption doesnt hold if the frequency is very big such that the wavelength of the current is comparable or much smaller than the dimensions of the circuit.

And yes it is obvious that if we consider non uniform (in space) current I(t) then the formula V(t)I(t) doesnt hold. What holds in this case is $P=\int_{\Omega}\vec{E}\cdot\vec{J}dV$ where E is the electric field and J is the current density inside the region $\Omega$. We still ignore the work of the magnetic field in this formula.

Last edited: Jul 24, 2014
14. Jul 24, 2014

### FallenLeibniz

Ok. I think I get it better now. Would I be better off when working with continuous charge distributions to start with V=-∫dot(E,dl) (i.e. am I wrong in assuming that for the differential case V=(dW/dq)?)?

Last edited: Jul 24, 2014
15. Jul 24, 2014

### Delta²

Well V=dW/dq holds if V is constant in time. It turns out it holds even if V is time varying, with the extra assumption that dq/dt is constant through out a circuit.

16. Jul 24, 2014

### FallenLeibniz

Ok, so in the case with V=dW/dq, we have that dW=Vdq. Dividing by dt, we have that P=VI. I'm cool with that. However, the there is the fact that you could technically specify q=dW/dV. This implies that dW=Vdq+qdV. Dividing this by dt would give P=VI+q(dV/dt). My question is why is this dV/dt term not considered as part of the "Generalized Power Law"?

Last edited: Jul 24, 2014
17. Jul 25, 2014

### Delta²

I say again the definition V=W/q is for a point charge q that is moving between two points for which the potential diff is V. From that we can conclude if the point charge is q(t)=q=constant that W=qV and differentiating wrt time that dW/dt=qdV/dt or dW=qdV

It "just happens" when we have a current dq/dt flowing between two points with potential diff V the work dW to be dW=Vdq, it is not a direct consequence of the definition for the point charge, it holds because of an analysis like that in my first post.

Anyway, I really havent seen a term qdV/dt in any formula regarding power in ac/dc circuits and we know that this term is too large in order to be neglected. It just doesnt exit in the power formula thats all.

Last edited: Jul 25, 2014
18. Jul 27, 2014

### FallenLeibniz

So is the only way to derive the formula for instantaneous power through using the integral of the field and the current density? If not, then I am at a loss as to where this decision comes from other than the dot(E,J) integral and/or empirical observation (which I guess given a couple hundred years of EM would stand to reason that this has been tested again and again as correct).

Also, in my previous post, I stopped trying to use V=W/q for the power law since I do see that it only refers to point charge and I remember a professor saying something to the effect that you can't really carry a current off of just one charge. However when you say " definition V=W/q", I assume the addition of more charges would require the analysis as you give above. In which case I guess I'm up a river and just to have to accept the fact that undergrad texts are just going to gloss over this point without giving any rigourous justification?

19. Sep 20, 2016

### nicw

Hi, I'm not sure whether it's okay to revive old posts here, but I came across this discussion after searching (without good results) for this particular sticking point...

Well, I'd like to confirm my understanding on one particular part of this post:

I'm concerned about the part where you say that dW done on dq is E(t)dqdl.

Does this require the assumption that dt is short enough such that although E varies with t, E can be treated as almost static since dE/dt is negligible compared to the other changes?

I understand that in general, for electric fields, we're talking about extremely fast timings. However, hypothetically, if you have a time-varying E-field that varies incredibly rapidly, would this still be valid or is there some other effect that would make this still hold?