# Find maximum dissipated power

## Homework Statement

A constant voltage source V with internal resistance r is connected to a load resistor R. The dissipated power by the resistor R is P=RV^2/(R+r)^2. Show that the maximum power dissipated by the resistor R is achieved when R = r. The maximum of P with respect to R is achieved when dP/dR = 0.

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P = RV^2/(R+r)^2

## The Attempt at a Solution

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Well, I think that I need to find the absolute maximum. That means I need to find the first derivative, find the critical points, evaluate the equation at the critical points and find the absolute maximum. And somehow show that R = r.

To find the first derivative, I used the chain rule and came up with this:

f(g(x)) -> f'(g(x))g'(x)

P = RV^2/(R+r)^2 f = RV^2/x^2 g = R+r
f' = (2Vx - 2xRV^2)/x^4 g' = 0

P'= (2V(R+r) - 2(R+r)RV^2)/(R+r)^4
= (2V - 2RV^2)/(R+r)^3

(2V - 2RV^2)/(R+r)^3 = 0

I don't know if this is correct and I don't know how to go on from here. The denominator can't be zero because the derivative won't exist at those points. But if I set the nominator to zero, it doesn't really help me because there's no r, only R. What am I doing wrong?

All help is appreciated, thanks!

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berkeman
Mentor
(2V - 2RV^2)/(R+r)^3 = 0

I don't know if this is correct
Welcome to the PF.

I don't think it's correct, because at least the units don't match. I also don't quite follow your approach at the derivative. Certainly your thoughts to take the derivative and set it equal to zero are correct...

Let me post my work in a couple of minutes to see if it helps...

Last edited:
chocolatecake
berkeman
Mentor
(using LaTeX to post the math -- you can find the tutorial for LaTeX here: https://www.physicsforums.com/help/latexhelp/)

$$P(R) = R\frac{V^2}{(R+r)^2} = (RV^2)(R+r)^{-2}$$
$$\frac{dP(R)}{dR} = (RV^2)(-2)(R+r)^{-3}(1) + (R+r)^{-2}(V^2) = 0$$
(Distribute, simplify, simplify...)
$$-R^3-R^2r+Rr^2+r^3 =0$$
Which you should be able to solve. By inspection, one solution is r=R:
$$-r^3-r^3+r^3+r^3= 0$$
But maybe there are other solutions?

Last edited:
chocolatecake
berkeman
Mentor
BTW, I've never liked the quotient rule for derivatives. I always prefer to convert the equation into a product, and apply that differentiation rule instead. It's simpler for my tiny brain...

Thank you so much for your help!

I thought that the derivative might be wrong. Thanks for the idea to convert the equation into a product. It's so much easier, I'm always gonna do that from now on.
Anyways, I tried it on my own, but I got something different than you have. I'm not sure if my solution is correct:

##P'(R) = V^2 (R+r)^{-2 }+ (RV^2)(-2)(R+r)^{-3} = 0##

if R = r:

##V^2 (r+r)^{-2} + (rV^2)(-2)(r+r)^{-3} = 0##

##(rV^2)(-2)(r+r)^{-3} = \frac{-V^2}{(r+r)^2}##

##(rV^2)(-2)(2r)^{-3} = \frac{-V^2}{(2r)^2}##

##\frac{(rV^2)}{(2r)^3} = \frac{V^2}{2(2r)^2}##

##rV{^2} = \frac{V^2(2r)^3}{2(2r)^2}##

##rV{^2} = \frac{V^2(2r)}{2}##

##rV{^2} = \frac{2V^2 rV^2}{2}##

##rV{^2} = V^2 rV^2##

##0 = \frac{V^2 rV^2}{rV^2}##

##0 = V^2##

##0 = V##

So technically, if R = r, the derivative is zero when V is zero. But is that enough to show that the maximum power is dissipated when R = r?

Also, thank you for the link to the LaTeX tutorial! :)

berkeman
berkeman
Mentor
P′(R)=V2(R+r)−2+(RV2)(−2)(R+r)−3=0P'(R) = V^2 (R+r)^{-2 }+ (RV^2)(-2)(R+r)^{-3} = 0

if R = r:

V2(r+r)−2+(rV2)(−2)(r+r)−3=0
I wouldn't set r=R so early. Just see if you can simplify the derivative down to the final form that I showed, and then see if you can solve for R in terms of r then...

chocolatecake
I understand now how to find the derivative but I don't understand how to simplify to ##-R^3-R^2r+Rr^2+r^3 =0##
How did you get R^2r or Rr^2? I tried using the binomial theorem for (R+r)^2 and (R+r)^3 but that made everything even more confusing.
What I ended up with is basically this:

##\frac{V^2(R+r)+2(RV^2)}{2R^3+6R^2r+6Rr^2+2r^3}=0##

But this doesn't look right.

So I decided to pick random values for R and V instead:

R = 3
V = 4

##(3)(4^2)(-2)(3+r)^{-3}+(3+r)^{-2}(4^2)=0##

##48(-2)(3+r)^{-3}+(3+r)^{-2}(16)=0##

##\frac{-96}{(3+r)^3}+\frac{16}{(3+r)^2}=0##

##96=\frac{16(3+r)^3}{(3+r)^2}##

##96=16(3+r)##

##3=r##

Therefore, r = R

Is that correct?

berkeman
Mentor
I understand now how to find the derivative but I don't understand how to simplify to −R3−R2r+Rr2+r3=0-R^3-R^2r+Rr^2+r^3 =0
How did you get R^2r or Rr^2?
$$P(R) = R\frac{V^2}{(R+r)^2} = (RV^2)(R+r)^{-2}$$
$$\frac{dP(R)}{dR} = (RV^2)(-2)(R+r)^{-3}(1) + (R+r)^{-2}(V^2) = 0$$
$$\frac{-2RV^2}{(R+r)^3} + \frac{V^2}{(R+r)^2} = 0$$
Divide both sides by V^2 to get rid of the voltage dependence (it is not needed), put both of the LHS terms over a common denominator, multiply both sides by that denominator to get rid of it, distribute terms, gather terms, and I get to:
$$-R^3-R^2r+Rr^2+r^3 =0$$
I think it can be factored, but I haven't gotten that to work right away. I'll give it another shot. Can you work through simplifying the derivative and eliminating the voltage V now?

chocolatecake
gneill
Mentor
I think it can be factored, but I haven't gotten that to work right away.
See if ##(r - R)(r + R)^2## fills the bill

chocolatecake and berkeman
berkeman
Mentor
See if ##(r - R)(r + R)^2## fills the bill
Lordy, I was off in the weeds with 6 unknowns and 4 equations in the factoring... Thanks for the help!

chocolatecake and gneill
haruspex
Homework Helper
Gold Member
But maybe there are other solutions?
An easier way is to make the denominator simpler by substituting S=R+r:
##\frac{P}{V^2}=\frac{S-r}{S^2}##
Differentiating wrt S, and multiplying by S4 to get rid of the denominator: ##S^2=2S(S-r)##.

chocolatecake and cnh1995
Divide both sides by V^2 to get rid of the voltage dependence (it is not needed), put both of the LHS terms over a common denominator, multiply both sides by that denominator to get rid of it
Ok, that makes sense to me now. If I do it, it looks like this:

##\frac{-2R(-2)V^2}{(R+r)^3}+\frac{V^2}{(R+r)^2}=0##

divide by ##V^2## :
##\frac{4R}{(R+r)^3}+\frac{1}{(R+r)^2}=0##

common denominator:
##\frac{4R(R+r)^2}{(R+r)^5}+\frac{(R+r)^3}{(R+r)^5}=0##

multiply by common denominator:
##4R(R+r)^2+(R+r)^3=0##

but after that it looks different from what you have:

expand the terms:
##4R(R^2+Rr+Rr+r^2)+R^3+3R^2r+3Rr^2+r^3=0##

and I end up with:
##5R^3+11R^2r+7Rr^2+r^3=0##

Where did I go wrong? Where are the negative signs in your equation coming from?

An easier way is to make the denominator simpler by substituting S=R+r:
PV2=S−rS2\frac{P}{V^2}=\frac{S-r}{S^2}
Differentiating wrt S, and multiplying by S4 to get rid of the denominator: S2=2S(S−r)S^2=2S(S-r).
But if I substitute R+r with S, then I have three variables. And I don't understand where the ##\frac{P}{V^2}## is coming from. If we have a P there, then there are four variables (S, V, P and R).

berkeman
Mentor
Ok, that makes sense to me now. If I do it, it looks like this:

−2R(−2)V2(R+r)3+V2(R+r)2=0\frac{-2R(-2)V^2}{(R+r)^3}+\frac{V^2}{(R+r)^2}=0
Where did the extra -2 come from in the numerator of the first term?

Where did the extra -2 come from in the numerator of the first term?
Well, the numerator was ##-2RV^2##, so I rewrote it as ##-2R(-2)V^2##.
But I just realized that this was nonsense because it would only work if the numerator was ##-2(R+V^2)##.
I'll try it again, give me a few minutes.

berkeman
Now I got it!

##-2R(R^2+Rr+Rr+r^2)+R^3+3R^2r+3Rr^2+r^3=0##
##-2R^3-2R^2r+(-2R^2r)-2Rr^2+R^3+3R^2r+3Rr^2+r^3=0##
##-R^3-R^2r+Rr^2+r^3=0##

and then if I substitute R=r, I get the same as you did: ##-R^3-R^3+R^3+R^3=0##
Thank you!!!

But now I still have to find the absolute maximum, or is this sufficient to show that R = r at the max. dissipated power?

berkeman
Mentor
But now I still have to find the absolute maximum, or is this sufficient to show that R = r at the max. dissipated power?
Given the help we got from @gneill in factoring, it looks like there is only one value of R that makes that equation equal to zero, right?
See if ##(r - R)(r + R)^2## fills the bill
And to show that it maximizes power, you could either evaluate the 2nd derivative at that point to verify that it is negative, or you could plug R = 1.1r and R = 0.9r into the original power equation to show that you get less power delivered to R when it's slightly more or less than matching r.

Good job, and way to hang in there!

chocolatecake
Great, I will try that!

berkeman
And thank you to the others who helped as well of course!

berkeman
OmCheeto
Gold Member
...
But maybe there are other solutions?
I came up with a total of 5 solutions.
Though, R = r was the only one that made sense.
The others were kind of impractical.

chocolatecake
haruspex
Homework Helper
Gold Member
But if I substitute R+r with S, then I have three variables.
No, S is instead of R, so still only two variables.
I don't understand where the ##\frac{P}{V^2}## is coming from
You started with ##P=\frac{RV^2}{(R+r)^2}##. I just divided through by V2 then replaced R by S-r everywhere.
In physical terms, I am keeping r fixed and varying the total resistance. The maximum P occurs when the sum, S, is 2r.

chocolatecake
cnh1995
Homework Helper
Gold Member
P=I2R=E2R/(r+R)2
∴dP/dR=E2[(r+R)2-2R(r+R)]/(r+R)4.

For dP/dR=0,
the numerator of the above equation should be 0, which gives r=R.

chocolatecake
berkeman
Mentor
P=I2R=E2R/(r+R)2
∴dP/dR=E2[(r+R)2-2R(r+R)]/(r+R)4.

For dP/dR=0,
the numerator of the above equation should be 0, which gives r=R.
That looks like a nice reason to use the quotient rule for the differentiation in this case, I guess (I still dislike it in general).

But you still have to distribute and factor the numerator to be sure r=R is the only solution, don't you?

cnh1995
cnh1995
Homework Helper
Gold Member
But you still have to distribute and factor the numerator to be sure r=R is the only solution, don't you?
Yeah, the numerator beomes
r2+2rR+R2-2rR-2R2=0, which simplifies to
r2-R2=0.
So r=R is the only sensible solution.

chocolatecake and berkeman
P=I2R=E2R/(r+R)2
∴dP/dR=E2[(r+R)2-2R(r+R)]/(r+R)4.
But that's a different equation from the one in the problem, isn't it?