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        riddles >> medium >> electric cells
        
(Message started by: ronnodas on Mar 9^th, 2009, 10:02am)

Title: electric cells
Post by ronnodas on Mar 9^th, 2009, 10:02am

You have n cells, each of emf E and internal resistance r. You have to connect these to another resistance of resistance R. What configuration of the n cells will maximize the current through the resistance?

Title: Re: electric cells
Post by 1337b4k4 on Mar 9^th, 2009, 5:00pm

[hide]current is voltage/resistance, so you clearly want to maximize voltage. Thus, put the cells (or batteries as we call them in the states) in series.[/hide]

Title: Re: electric cells
Post by Immanuel_Bonfils on Mar 9^th, 2009, 5:18pm

Shouldn't it depend of which of (r,R) is greater?

Title: Re: electric cells
Post by ronnodas on Mar 9^th, 2009, 7:59pm

If you put all of them in series, the internal resistances will add up too, and will decrease the current. You're logic would work if the resistance was constant, which is not the case.

If it helps, R>r.

Title: Re: electric cells
Post by 1337b4k4 on Mar 9^th, 2009, 11:53pm

ah it has been a while since i thought about physics. after reading wikipedia i realize that i have confused emf and voltage potential difference.

Title: Re: electric cells
Post by towr on Mar 10^th, 2009, 1:23am

If it were just the choice between putting all the cells in parallel or in series, then I think putting them in parallel is better when ~~(rR-1)/(rR-r²) < n~~
But there are more configurations possible.

Title: Re: electric cells
Post by Immanuel_Bonfils on Mar 10^th, 2009, 8:05am

It depends, at least, on the ratio R/r; for instance, for n=3 it would be all the three in series for R=2r and, two in parallel, in series with the third , for R =(5/4)R.

Title: Re: electric cells
Post by ronnodas on Mar 10^th, 2009, 8:05am

No not just all in series or all in parallel. If the solution is dependent on exact values of n, R and r, which I suspect it is, then try the following.
n=48, r=.3, R=5
Towr will probably brute force it ;).

Title: Re: electric cells
Post by Immanuel_Bonfils on Mar 10^th, 2009, 8:14am

So, it isn't for any (n, R>r) as the first post suggests?

Title: Re: electric cells
Post by ronnodas on Mar 10^th, 2009, 8:51am

If the general case is solvable, that's very good. But I don't think it is so I provided the exact problem.

Title: Re: electric cells
Post by Leo Broukhis on Mar 15^th, 2009, 11:50pm

Looking at the boundary cases, the parallel connection gives emf E and internal resistance r/n, the series connection gives emf nE and internal resistance rn. I.e. for the emf multiplier x we get the resistance multiplier x²/n.
By finding the max of x/(R/r+x²/n) we will know what cell connection to strive for.

Title: Re: electric cells
Post by ronnodas on Mar 16^th, 2009, 1:31am

If I understand you correctly, you are considering n/x rows in parallel, each of which has x cells in series. This is easily solvable. I'm looking for a solution to the more general case, where each row may have different number of cells.

Title: Re: electric cells
Post by Leo Broukhis on Mar 16^th, 2009, 8:07am

No, in my solution xE is the ideal emf, so one needs to compare
floor(x)/(R/r+floor(x)²/n) and ceil(x)/(R/r+ceil(x)²/n). Whichever value is greater tells how many cells to connect in series; the rest should be spread uniformly in parallel to minimize the resistance.
E.g. if x = n-2, we should connect n-2 cells in series, then add the two remaining parallel to any two distinct cells.

Title: Re: electric cells
Post by ronnodas on Mar 17^th, 2009, 5:38am

I don't think that is the most general case. Also, if x turns out to be <n/2, how will you connect the n-x cells in parallel to n-x distinct cells?

Title: Re: electric cells
Post by Leo Broukhis on Mar 17^th, 2009, 8:37am

x - n mod x elements of the chain will have floor(n/x) cells connected in parallel, and the rest will have one more.