# Why the current is maximum When Rᵉˣᵗᵉʳⁿᵃˡ=Rᴵⁿᵗᵉʳⁿᵃˡ?

• B

## Main Question or Discussion Point

My teacher proved it like this
I= ##\frac {E}{R+r}##
Where R is external resistance
And r is internal resistance
E is emf of the cell,
Now we can write it like this
I= ##\frac {E}{(√R-√r)²+2√R√r}##
if R=r then the denominator value will be minimum but i am unable to understand that why this is the case as for any value of R their will be some less value of r and vice versa which means current is not maximum when R=r.
Thanks .

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etotheipi
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For a fixed ##R+r##, the current doesn't depend on the relative proportions of ##R## and ##r##. Since the two resistances are in series, it's only their total that matters.

For maximum power dissipated by the load, you require ##R = r##. Is that what you were thinking of?

• Hemant and berkeman
berkeman
Mentor
For maximum power dissipated by the load, you require . Is that what you were thinking of?
Yeah, that's the condition for maximum power transfer to the load, not any max current. I suppose maximum current out would be into a short circuit R=0.

• Hemant and etotheipi
kuruman
Homework Helper
Gold Member
##\dots## but i am unable to understand that why this is the case as for any value of R their will be some less value of r ##\dots##
You misunderstood what is being done here. The internal resistance ##r## is fixed at whatever value it has. You are varying the load resistance ##R## to maximize ##\dots~## what? It can't be the current because that is maximized for zero load. It could be the power as others have remarked, but then you need an expression for the power dissipated in the load, not the current through it. What would that expression be?

• Hemant, jbriggs444 and etotheipi
I was trying to understand this,
Page-I Page-II It feels that I have done something stupendous in writing the proof,
I just thought that for a single battery the case is same as for multiple batteries but from posts it seems to me that I have done something wrong in converting all batteries into one single battery.
You misunderstood what is being done here. The internal resistance r is fixed at whatever value it has. You are varying the load resistance R to maximize … what? It can't be the current because that is maximized for zero load. It could be the power as others have remarked, but then you need an expression for the power dissipated in the load, not the current through it. What would that expression be?
as ##P##=##I##²##R##--(a)
And we know that ##I##= ##\frac{E}{R+r}##so if we put ##I## value in equation 'a' then we get
##P##= (##\frac{E}{R+r}##)²##R##
So power would be maximum when denominator is minimum and denominator can't be equal to ##R## only as we have taken situation in which their is some internal resistance in battery.
So problem is again that is why ##R##+r is minimum when ##R##=r?

kuruman
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When ##R=r##, $$den=(\sqrt{r}-\sqrt{r})^2+2\sqrt{r}\sqrt{r}=2r$$
When ##R=\frac{1}{4}r##,
$$den=\left(\sqrt{\frac{1}{4}r}-\sqrt{r}\right)^2+2\sqrt{\frac{1}{4}r}\sqrt{r}=\left(-\frac{\sqrt{r}}{2}\right)^2+r=\frac{5}{4}r$$The claim that the denominator is a minimum when ##r=R## is simply not true. How come?

• etotheipi
etotheipi
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I don't follow their reasoning. The condition for maximum current through the external resistor is certainly still ##R = 0##, for any combination of ##m## and ##n##. Specifically, I think their mistake is taking the term ##2\sqrt{mRnr}## as constant... given that it definitely still depends on ##R##!

It should just be a case of saying that ##R + r## is minimised with ##R=0## for fixed ##r##, and not bothering with whatever they're doing with those square roots!

as ##P##=##I##²##R##--(a)
And we know that ##I##= ##\frac{E}{R+r}##so if we put ##I## value in equation 'a' then we get
##P##= (##\frac{E}{R+r}##)²##R##
So power would be maximum when denominator is minimum and denominator can't be equal to ##R## only as we have taken situation in which their is some internal resistance in battery.
So problem is again that is why ##R##+r is minimum when ##R##=r?
$$I = \frac{\mathcal{E}}{R+r} \implies P_L = \frac{\mathcal{E}^2 R}{(R+r)^2} = \frac{\mathcal{E}^2}{R + 2r + \frac{r^2}{R}}$$To maximise ##P_L## is to minimise ##R + 2r + r^2/R## by wiggling ##R##,$$\frac{d}{dR} \left( R + 2r + \frac{1}{R}r^2 \right) = 1 - \frac{r^2}{R^2} = 0 \implies R = \pm r$$Since ##R## is strictly positive we must take ##R = r##.

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jbriggs444
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As I read the text, R is not a variable. We are not optimizing for current as a function of external resistance R. We are optimizing for current as a function of the available parameters, m and n while holding:

R fixed
r fixed
m*n fixed

To put that in a physical context, we have a load device. And we have a shoe box full of C cells. We want to connect the cells in a rectangular matrix to maximize the current we can pump through this device. We do not control the load device (R is fixed). We do not control the nature of the C cells (r is fixed). And the number of cells in the box is also a limitation (m*n fixed).

• Hemant, DaveE and etotheipi
etotheipi
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To put that in a physical context, we have a load device. And we have a shoe box full of C cells. We want to connect the cells in a rectangular matrix to maximize the current we can pump through this device. We do not control the load device (R is fixed). We do not control the nature of the C cells (r is fixed). And the number of cells in the box is also a limitation (m*n fixed).
Actually you are right, nice insight They will get ##R = \frac{n^2 r}{C} \implies n = \frac{\sqrt{RC}}{\sqrt{r}}## which means$$I_{max} = \frac{E \sqrt{RC}}{2R \sqrt{r}} = \frac{E\sqrt{C}}{2\sqrt{Rr}}$$in terms of those fixed parameters

• Hemant
sophiecentaur
Gold Member
@Hemant Perhaps you should look at your notes again - or better still, any text book you are using. The Maximum Power Theorem is something we all go through and, even in an arm waving sense, it's pretty easy to accept and understand. This "Maximum Current' looks like a Typo because (and I think you understand it now) it's highest when the total resistance in the circuit is zero. Maximum Power ( VI ) into the load is when R (a variable quantity) = r ( a fixed quantity) and that's very simple book work.
If you have 'actual hard copy' of the teacher's apparent statement then challenge the teacher with the Maximum Power Theorem. Otherwise you can put it down to a typo on the board or into your notes. All part of the learning process.

I think people on this thread are being a bit too polite in their reactions to a bit of nonsense. This is not an attack on you, in any way. It's mainly sympathetic to you if you really have been told something so wrong.

• • Hemant and jbriggs444
DaveE
Gold Member
Actually you are right, nice insight They will get ##R = \frac{n^2 r}{C} \implies n = \frac{\sqrt{RC}}{\sqrt{r}}## which means$$I_{max} = \frac{E \sqrt{RC}}{2R \sqrt{r}} = \frac{E\sqrt{C}}{2\sqrt{Rr}}$$in terms of those fixed parameters
C?

etotheipi
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C?
I used J. Briggs' definition of ##C := mn## as the fixed number of cells available.

jbriggs444
Homework Helper
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However they derive the expression ##R = \frac{nr}{m}##, which does not hold in the general case for those parameters. What they seem to have done is optimised (incorrectly) for ##R## and then substituted this back in to find a current.
Let me go through their derivation slowly and see if I can make sense of it. I think slow, so bear with me.

They are trying to get an equivalent circuit out of this. They compute the EMF of the equivalent circuit:$$E_{\text{eq}}=nE$$Here n is clearly the number of cells in series in the matrix. They go on to compute the equivalent resistance:$$r_{\text{eq}}=\frac{nr}{m}$$Here m is clearly the number of rows in parallel. So far, so good. Trivial stuff.

Now we compute the current being pumped through the load. This is EMF divided by total resistance:$$I=\frac{nE}{R+\frac{nr}{m}} = \frac{mnE}{mR+nr}$$This is just substituting the equivalent values for the battery matrix into I=E/R and then simplifying a bit.

The next bit is where they exercise a bit of foresight. They are trying to set up to rewrite the denominator as a sum of squares. So they add zero, cleverly written as ##-2\sqrt{mR}\sqrt{nr} + 2 \sqrt{mR}\sqrt{nr}## to the denominator. This lets them then regroup and rewrite the denominator as ##(\sqrt{mR} - \sqrt{nr})^2 + 2\sqrt{mR}\sqrt{nr}##. That was the hard part. It is all downhill from there.

What we have at this point is:$$I=\frac{mnE}{(\sqrt{mR} - \sqrt{nr})^2 + 2\sqrt{mR}\sqrt{nr}}$$We are trying to maximize this.

The numerator is fixed. The denominator is positive. So we are trying to minimize the denominator:$$(\sqrt{mR} - \sqrt{nr})^2 + 2\sqrt{mR}\sqrt{nr}$$The right hand term is fixed. So we are trying to minimize the left hand term:$$(\sqrt{mR} - \sqrt{nr})^2$$That reduces to trying to make ##mR## as close to ##nr## as can be arranged.

And we are basically done.

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• • Hemant, DaveE and etotheipi
berkeman
Mentor
The thread starter @Hemant has been back to the PF and has seen your replies. I'm not liking the fact that he has not responded to all of the information you've posted for him... Let me go through their derivation slowly and see if I can make sense of it. I think slow, so bear with me.

They are trying to get an equivalent circuit out of this. They compute the EMF of the equivalent circuit:$$E_{\text{eq}}=nE$$Here n is clearly the number of cells in series in the matrix. They go on to compute the equivalent resistance:$$r_{\text{eq}}=\frac{nr}{m}$$Here m is clearly the number of rows in parallel. So far, so good. Trivial stuff.

Now we compute the current being pumped through the load. This is EMF divided by total resistance:$$I=\frac{nE}{R+\frac{nr}{m}} = \frac{mnE}{mR+nR}$$This is just substituting the equivalent values for the battery matrix into I=E/R and then simplifying a bit.

The next bit is where they exercise a bit of foresight. They are trying to set up to rewrite the denominator as a sum of squares. So they add zero, cleverly written as ##-2\sqrt{mR}\sqrt{nr} + 2 \sqrt{mR}\sqrt{nr}## to the denominator. This lets them then regroup and rewrite the denominator as ##(\sqrt{mR} - \sqrt{nr})^2 + 2\sqrt{mR}\sqrt{nr}##. That was the hard part. It is all downhill from there.

What we have at this point is:$$I=\frac{mnE}{(\sqrt{mR} - \sqrt{nr})^2 + 2\sqrt{mR}\sqrt{nr}}$$We are trying to maximize this.

The numerator is fixed. The denominator is positive. So we are trying to minimize the denominator:$$(\sqrt{mR} - \sqrt{nr})^2 + 2\sqrt{mR}\sqrt{nr}$$The right hand term is fixed. So we are trying to minimize the left hand term:$$(\sqrt{mR} - \sqrt{nr})^2$$That reduces to trying to make ##mR## as close to ##nr## as can be arranged.

And we are basically done.
Thanks for writing this long post and rectifying most of my confusions,the only confusion left was that if ##r## and ##R## are constant then why ##R## depends on r as ##R##=##\frac {mr}{n}## and now this problem is also cleared as it was stated in your post that it is internal resistance of a single battery that is constant not the equivalent internal resistance.
Thanks .

total resistance:I=nER+nrm=mnEmR+nRThis is just substituting the equivalent values for the battery matrix into I=E/R and then simplifying a bit.
Their is a small problem in this equation,
In denominator their should be ##mR+nr## but not ##mR+nR##
@Hemant Perhaps you should look at your notes again - or better still, any text book you are using. The Maximum Power Theorem is something we all go through and, even in an arm waving sense, it's pretty easy to accept and understand. This "Maximum Current' looks like a Typo because (and I think you understand it now) it's highest when the total resistance in the circuit is zero. Maximum Power ( VI ) into the load is when R (a variable quantity) = r ( a fixed quantity) and that's very simple book work.
If you have 'actual hard copy' of the teacher's apparent statement then challenge the teacher with the Maximum Power Theorem. Otherwise you can put it down to a typo on the board or into your notes. All part of the learning process.

I think people on this thread are being a bit too polite in their reactions to a bit of nonsense. This is not an attack on you, in any way. It's mainly sympathetic to you if you really have been told something so wrong.
I talked about this to my teacher again and he told me that it is not current that is maximum but power,the problem was that it was me who wanted to listen maximum current and that's why whole confusion is created .
The thread starter @Hemant has been back to the PF and has seen your replies. I'm not liking the fact that he has not responded to all of the information you've posted for him... I am really very sorry for replying too late,
The reason for this is that it was around 11 pm and I wanted to talk about this to my teacher again so as to eliminate all confusions and another reason was that I was very sleepy and if had replied at that time then probably there would be some mistake in my post.
I hope their is no mistake in this post .
Thanks to all for helping me .

• berkeman, sophiecentaur and jbriggs444
jbriggs444
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Their is a small problem in this equation,
In denominator their should be ##mR+nr## but not ##mR+nR##
Right you are. I had fixed up a number of those typos but missed that one.

• Hemant
Right you are. I had fixed up a number of those typos but missed that one.
For first time I got a like from you,
It feels like lifetime achievement .

• etotheipi
sophiecentaur
Gold Member
who wanted to listen maximum current and that's why whole confusion is created .
I've been there so many times. It happens with song titles, peoples' names and Science. (And it gets worse as you get older, too!!)

• Hemant
I've been there so many times. It happens with song titles, peoples' names and Science. (And it gets worse as you get older, too!!)
Another lifetime achievement by getting like from sophiecentaur .

• sophiecentaur
sophiecentaur
Gold Member
And there's another one. I sometimes look at a really interesting thread and realise I haven't bothered to like some pretty good posts. Perhaps I'm just a grumpy old man. My wife would certainly tell you that's true.

• Hemant and jbriggs444
And there's another one. I sometimes look at a really interesting thread and realise I haven't bothered to like some pretty good posts. Perhaps I'm just a grumpy old man. My wife would certainly tell you that's true.
Not at all!
You are not grumpy by any means. Then I am not sure.

• hutchphd
sophiecentaur
Gold Member
Nah - you're not grumpy. You're kind to cats. (Or was that one destined for the pot?)

Nah - you're not grumpy. You're kind to cats. (Or was that one destined for the pot?)
Okay,I accept .
I wanted to say that you are just trying to be grumpy like stanlee and from Stanlee face and your post I can't see any grumpiness.

sophiecentaur
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