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Topic: electric cells (Read 579 times) |
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Ronno
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electric cells
« on: Mar 9th, 2009, 10:02am » |
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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?
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teekyman
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Re: electric cells
« Reply #1 on: Mar 9th, 2009, 5:00pm » |
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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.
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Immanuel_Bonfils
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Re: electric cells
« Reply #2 on: Mar 9th, 2009, 5:18pm » |
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Shouldn't it depend of which of (r,R) is greater?
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Ronno
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Re: electric cells
« Reply #3 on: Mar 9th, 2009, 7:59pm » |
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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.
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teekyman
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Re: electric cells
« Reply #4 on: Mar 9th, 2009, 11:53pm » |
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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.
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towr
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Re: electric cells
« Reply #5 on: Mar 10th, 2009, 1:23am » |
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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.
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| « Last Edit: Mar 10th, 2009, 11:31am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Immanuel_Bonfils
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Re: electric cells
« Reply #6 on: Mar 10th, 2009, 8:05am » |
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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.
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Ronno
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Re: electric cells
« Reply #7 on: Mar 10th, 2009, 8:05am » |
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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  .
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Immanuel_Bonfils
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Re: electric cells
« Reply #8 on: Mar 10th, 2009, 8:14am » |
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So, it isn't for any (n, R>r) as the first post suggests?
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Ronno
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Re: electric cells
« Reply #9 on: Mar 10th, 2009, 8:51am » |
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If the general case is solvable, that's very good. But I don't think it is so I provided the exact problem.
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Leo Broukhis
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Re: electric cells
« Reply #10 on: Mar 15th, 2009, 11:50pm » |
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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.
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Ronno
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Re: electric cells
« Reply #11 on: Mar 16th, 2009, 1:31am » |
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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.
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Leo Broukhis
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Re: electric cells
« Reply #12 on: Mar 16th, 2009, 8:07am » |
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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.
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| « Last Edit: Mar 16th, 2009, 8:10am by Leo Broukhis » |
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Ronno
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Re: electric cells
« Reply #13 on: Mar 17th, 2009, 5:38am » |
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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?
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Leo Broukhis
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Re: electric cells
« Reply #14 on: Mar 17th, 2009, 8:37am » |
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x - n mod x elements of the chain will have floor(n/x) cells connected in parallel, and the rest will have one more.
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