Find maximum W for a generator

[SagarPatil]

Homework Statement

So a generation with an internal resistance r delivers a voltage E. The generator is connected to a electrical circuit with a resistance R. The work W done each second in sending a current through a circuit with a resistance R is given by W =E^2*R/(R+r)^2.

For constant r and E, show that W is a maximum when R=r. Sketch a graph of W as a function of R

Homework Equations

v=ir

The Attempt at a Solution

So, I am confused. I have solved for R, which is

R= (E^2/(2W))-r

If someone can explain the question, I will try to answer to question.

P.S it may look like a question from some electrical engineering book but its from a calculus book -.- (no answer)

 

[Mark44]

Homework Statement

So a generation generator with an internal resistance r delivers a voltage E. The generator is connected to a electrical circuit with a resistance R. The work W done each second in sending a current through a circuit with a resistance R is given by W =E^2*R/(R+r)^2.

For constant r and E, show that W is a maximum when R=r. Sketch a graph of W as a function of R

Homework Equations

v=ir

The Attempt at a Solution

So, I am confused. I have solved for R, which is

R= (E^2/(2W))-r

Solving for R won't help you. To find the maximum value of a function of a single variable (R here, since r and E are assumed to be constants), find the derivative W'(R) and set it to zero.

If someone can explain the question, I will try to answer to question.

P.S it may look like a question from some electrical engineering book but its from a calculus book -.- (no answer)

 

[Mark44]

BTW, in the future, please try to choose a thread title that is more informative. I have changed the title to "Find maximum W for a generator".

 

[SagarPatil]

Solving for R won't help you. To find the maximum value of a function of a single variable (R here, since r and E are assumed to be constants), find the derivative W'(R) and set it to zero.

Okay, I have done some work. I found the derivative of W'(R) and set it to 0. Here is what I got

 

[SteamKing]

Okay, I have done some work. I found the derivative of W'(R) and set it to 0. Here is what I got
View attachment 92408

You're not interested in dW/dx, whatever x is. You want to calculate dW/dR and set that equal to zero, not set R = 0. You also want to show that W is a maximum when r = R.

BTW, if you express W as the product of two expressions, rather than the quotient, you can save yourself some calculation when finding the derivative.

 

[SagarPatil]

You're not interested in dW/dx, whatever x is. You want to calculate dW/dR and set that equal to zero, not set R = 0. You also want to show that W is a maximum when r = R.

BTW, if you express W as the product of two expressions, rather than the quotient, you can save yourself some calculation when finding the derivative.

I find product of two expression harder to do.

So I derived for dW/DR and this is what I got

If I set R = 0
I would just get 0.

 

[Mark44]

Here's what I said earlier, in post #2.

Solving for R won't help you. To find the maximum value of a function of a single variable (R here, since r and E are assumed to be constants), find the derivative W'(R) and set it to zero.

I didn't say anything about setting R to zero -- I said you should set W'(R) to zero.

 

[SteamKing]

I find product of two expression harder to do.

I find that astounding.

If f(x) = g(x) * h(x), then f'(x) = g(x) * h'(x) + g'(x)*h(x)

whereas

if f(x) = g(x) / h(x), then f'(x) = [g'(x)* h(x) - g(x)*h'(x)] / h²(x)

So I derived for dW/DR and this is what I got
View attachment 92600

Which is incorrect, BTW.

If I set R = 0
I would just get 0.

And Mark44 is right. You don't set R = 0, you set dW/dR = 0, which is different. But you must calculate the correct expression for dW/dR.

 

[SagarPatil]

I find that astounding.

If f(x) = g(x) * h(x), then f'(x) = g(x) * h'(x) + g'(x)*h(x)

whereas

if f(x) = g(x) / h(x), then f'(x) = [g'(x)* h(x) - g(x)*h'(x)] / h²(x)

Which is incorrect, BTW.And Mark44 is right. You don't set R = 0, you set dW/dR = 0, which is different. But you must calculate the correct expression for dW/dR.

So, I was wondering, When you find the expression of dW/dR, do you need to find the second derivation of it to graph it or just find the fist derivation of dW/dR and set it to 0.

 

[SteamKing]

So, I was wondering, When you find the expression of dW/dR, do you need to find the second derivation of it to graph it or just find the fist derivation of dW/dR and set it to 0.

The value of R which makes dW/dR = 0 is the value which either maximizes or minimizes W.

It should be easy to determine which is which without calculating d²W/dR²

 

[SagarPatil]

The value of R which makes dW/dR = 0 is the value which either maximizes or minimizes W.

It should be easy to determine which is which without calculating d²W/dR²

So I found the expression for dW/dR

After the deravation, if I set dW/dR to 0, and set r and E to 1, I get a L shape graph. The x and y basically start from origin and y is + and x is positive

 

[Mark44]

Please post your work as text rather than as an image. Posting an image makes it impossible for us to insert a comment at the point of a mistake. Instead, we have to give the location where you are in error, which makes more work for us.

So I found the expression for dW/dR
View attachment 92843

After the deravation, if I set dW/dR to 0, and set r and E to 1, I get a L shape graph.

There is no need to set r and E to any particular values.

"Keep your eye on the prize." -- the goal here is to show that W is at its largest when R = r.

The x and y basically start from origin and y is + and x is positive

The variables are R and W. Neither x nor y appear in this problem.

What you ended with is correct, but it can be written in a slightly more simplified form:
$\frac{dW}{dR} = \frac{E^2(r - R)}{(R + r)^3}$
What does it take for a fraction to be zero? There is absolutely nothing to gain by expanding $(R + r)^3$.

 

[SteamKing]

So I found the expression for dW/dR
View attachment 92843

After the deravation, if I set dW/dR to 0, and set r and E to 1, I get a L shape graph. The x and y basically start from origin and y is + and x is positive

The first line of your work contains an error.

W = E²*R/(R+r)² not dW/dR = E²*R/(R+r)²

You would run into less algebra if you wrote

W = E²*R * (R+r)^-2

and then used the Product Rule to find dW/dR

 

[gag1234]

if somebody has full solution to this problem pls post an image solving or a word file.

 

[cnh1995]

Homework Statement

So a generation with an internal resistance r delivers a voltage E. The generator is connected to a electrical circuit with a resistance R. The work W done each second in sending a current through a circuit with a resistance R is given by W =E^2*R/(R+r)^2.

For constant r and E, show that W is a maximum when R=r. Sketch a graph of W as a function of R

Homework Equations

v=ir

The Attempt at a Solution

So, I am confused. I have solved for R, which is

R= (E^2/(2W))-r

If someone can explain the question, I will try to answer to question.

P.S it may look like a question from some electrical engineering book but its from a calculus book -.- (no answer)

Its actually the 'Maximum power transfer theorem' in electrical engineering.
W=E²*R/(R+r)².
Find dW/dR and equate it to 0 as already explained in #13 by SteamKing.

 

[SagarPatil]

Its actually the 'maximum power transfer theorem' from electrical engineering.
W=E²*R/(R+r)².
Find dW/dR and equate it to 0 as already explained in #13 by SteamKing.

I have found the solution to the answer. Thanks for helping everyone

 

[Mark44]

if somebody has full solution to this problem pls post an image solving or a word file.

We don't do that here. From the rules for this forum:
From the rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/):

Giving Full Answers:
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