Find maximum dissipated power

[chocolatecake]

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.

Homework Equations

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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!

 

[berkeman]

(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...

 

[berkeman]

(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?

 

[berkeman]

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...

 

[chocolatecake]

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 going to 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]

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]

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?

 

[gneill]

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

 

[berkeman]

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!

 

[haruspex]

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]

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?

 

[chocolatecake]

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]

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?

 

[chocolatecake]

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.

 

[chocolatecake]

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]

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!

 

[chocolatecake]

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

 

[OmCheeto]

...
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.

 

[haruspex]

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.

 

[cnh1995]

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.

 

[berkeman]

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]

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]

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?

 

[cnh1995]

But that's a different equation from the one in the problem, isn't it?

No, the equation for the power is the same as in the OP.
For finding dP/dR, the quotient rule is used.