# 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!
Thank you for your help and your time! • 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?