Resistance Association - How to solve this exercise in the proper way?

[jaumzaum]

Homework Statement

See below

Relevant Equations

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This was an exercise from the 2011 admission test of an university from Brazil:
40 identical generators with electromotive force 1,5V and resistance 0,25 Ohms and one resistor with resistance 2,5 Ohm are associated in such a way that the 2.5- resistor dissipates the greatest energy per second, that is? Answer: 90W.

My solution:
If we make m groups of n generators we would have:
$$m\epsilon=(\frac{mr}{n}+R)i$$
$$i=\frac{1,5mn}{0,25m+2,5n}=\frac{6mn}{m+10n}=\frac{240}{40/n+10n}=\frac{24}{4/n+n}$$
We know by the mean-inequality that
$$4/n+n>=2\sqrt{4} =4 $$
This is possible because it occurs when:
$$4/n=n \rightarrow n=2, m=20$$
$$i=6A \rightarrow P=90W$$

But why do all the groups have the same number of resistors? Why does the circuit have only series or parallel associations? Why isn't there any Delta/Star or any other more complex associations? Is there any way to solve this exercise in the right way?

P.S. I'm really not sure if the best way to post this is in the Physics or Math Forum. I'm posting in the Physics one. Sorry if I'm wrong by that/

 

[cnh1995]

I didn't understand your method of using mean-inequality, but I got 90W using Maximum Power Transfer theorem.
I believe you have interchanged the values of m and n. As per my calculations, you need to have 2 groups of 20 cells.

But why do all the groups have the same number of resistors?

To ensure equal current and power sharing among the cells, thereby avoiding under/overloading of any cell.
If the power sources are identical, they should practically share equal power.

 

[cnh1995]

I didn't understand your method of using mean-inequality,

Ok I got your method. It is almost the same as Maximum Power Transfer theorem, except that you are not using Thevenin resistance in your calculations.

My solution:
If we make m groups of n generators we would have:

mϵ=(mrn+R)i​

I believe this equation is incorrect. Put m=1, n=40 and see if you get the correct current for 1 group of 40 cells.