# anyone can help with a calculus question?

#### jani

##### Fan Favourite
ok, some are going to say "what the hell?" etc but Im confident we got smart people on board like Lee.

If a resistor of R ohms is connected across a battery of E volts with internal resistance of r ohms, then the power (in watts) in the external resistor is

P = (E² R) / (R + r)²

If E and r are fixed but R varies, what is the maximum value of the power?

Its application of differentiation. Please   E=MCsquared

#### jani

##### Fan Favourite
btw the answer is E^2/(4r), I just want to know how to get it

#### champdave

##### SG Sheffield Authority
jani;2366842 said:
If a resistor of R ohms is connected across a battery of E volts with internal resistance of r ohms, then the power (in watts) in the external resistor is

P = (E² R) / (R + r)²

If E and r are fixed but R varies, what is the maximum value of the power?

Its application of differentiation. Please   Well, you want to find dP/dR, and then set it equal to 0 to find out the value of R for which P is a maximum. Setting that value of R back into the original equation (the P(R) function) will give you the power value, assuming you have E and r given.

E and r are constant so it's just normal differentiation. You could use either the product rule (with P(R) = (E²R)[(R+r)-²] ) or use the quotient rule of f(x)/g(x), where f(x) = E²R and g(x) is (R+r)². I'm not going to do it for you, the differential in this case is an easy one. Just remember that E and r will not be differentiated as it is with respect to R. So d/dR[E²R] = E², for example.

Hope that helps!

#### Deisler

##### Red Card [Being a douche] exp. 22/1/06
jani;2366842 said:
P = (E² R) / (R + r)²

P = E² R / (R + r) (R + r)

P = E² R / R² + R r + R r + R²

P = E² R / R (R + r + R + r)

P = E² / 2 R + 2 r

That is as far as i got i am not sure if 2 R + 2 r = 4 r though it doesn't make sense since R is variable and r is fixed. #### champdave

##### SG Sheffield Authority
OK, I've got the answer - I'll just tidy it up and post it here. Deisler;2366859 said:
That is as far as i got i am not sure if 2 R + 2 r = 4 r though it doesn't make sense since R is variable and r is fixed. It does - it's saying that the resistor is a variable one but the internal resistance of the battery is fixed. This is a pretty reasonable approximation to make in electronics with regulated resistors, and it means you are left with a simple differential rather than a partial differential, which can get messy.

#### jani

##### Fan Favourite
damn, I got my questioned answered in just about half an hour thanks a lot man. I still have to understand it but thanks alot, a huge lot.

#### champdave

##### SG Sheffield Authority
jani;2366874 said:
damn, I got my questioned answered in just about half an hour thanks a lot man. I still have to understand it but thanks alot, a huge lot.

Not a prob - it's just applying a basic calculus rule to a problem. The bit that throws most people off is treating E and r as constants - it's quite a mental block. As long as you know what the function is (in this case, f(R)) then you know what the differential should be. (in this case, dP/dR)

You can do this question with either the Quotient Rule or the Product Rule. All you need to do is do a bit of factoring out halfway as it breaks down the function and makes it easier to manage.

#### jani

##### Fan Favourite
yeah, its not really the differentiation part thats killing me. its the step after.

I dont understand the "when dP/dR = 0, r = R", and especially when you subsituted it back in, I really dont understand that.

haha basically I dont understand the whole thing after the differentiation #### champdave

##### SG Sheffield Authority
Well, the maximum or minimum of a function occurs when the first differential of the function is 0, as this implies that the function is not changing at that point.

Now, setting dP/dR equal to 0 gives: (can't be bothered to do it fancy, sorry)

0 = (E^2)(r-R) / (R+r)^3 = [(E^2) / (R+r)^3](r-R)

If we divide both sides by (E^2) / (R+r)^3 then we get:

0 = r - R

(as 0 / a = 0, where a is any number)

Adding R to both sides gives R = r, which is saying that at the maximum power, the internal resistance is equal to the total resistance in the resistor. This means that we can now write the maximum power of the function by substituting this fact in to give P not in terms of R. (and thus no unknowns for P)

#### Filipower

##### Bunburyist
too much CM, dave.

#### champdave

##### SG Sheffield Authority
Tom;2366916 said:

I wish I was your mum's derivative, so I could lie tangent to her curves.

Maths jokes, I never thought I'd stoop so low #### Stotty

##### Fan Favourite !

I'm slightly ashamed i understand that #### Filipower

##### Bunburyist
me too, that's not that advanced #### Á&#316;é&#64303;

##### Fan Favourite
Im glad school is deep in the past for me #### Mandieta6

##### Red Card - Life
Life Ban
Yo Momma's the opposite of X inversed, everyone gets to touch her....

My math teacher actually told me that... I am not taking AS maths.

#### Tom

##### That Nice Guy
champdave;2367079 said:
I wish I was your mum's derivative, so I could lie tangent to her curves.

Maths jokes, I never thought I'd stoop so low Yeah but, your mum.

Don't even try coming back #### jani

##### Fan Favourite
champdave;2367079 said:
I wish I was your mum's derivative, so I could lie tangent to her curves.

Maths jokes, I never thought I'd stoop so low haha Replies
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