Optimizing Current Intensity: Connecting Cells in Series and Parallel

[Nader AbdlGhani]

Homework Statement

How can I get the maximum Current Intensity by connecting 72 cells to a 3 Ohm resistor each cell has a potential difference of 1.5 Volt and internal resistance of 1 Ohm

Please don't tell me to connect them all in parallel or all in series as in parallel the potential difference will be low (1.5 Volt) and the resulted current intensity will be approximately 0.49 Amperes and in series the internal resistance will be huge (72 Ohms) and the resulted current intensity will be approximately 1.44 Amperes2. The attempt at a solution
I've tried to connect 18 cells in series and therefore I will have 4 packs of batteries , let's call each pack a battery , so each "battery" has potential difference of 27 Volts and resistance of 18 Ohms , then I connected the 4 batteries in parallel with each other and in series with the resistor , the resulted current intensity was exactly 3.6 Amperes , and that was the best that I can do , anyone have any other better configuration ? Thanks.

 

[BvU]

Hello Nader, welcome to PF !

You are posting in the homework section of PF, so I propose you write an expression for the fraction of the energy a voltage source with an internal resistance delivers to an external load, and try to work that around to a function of the ratio external/internal resistance. My guess is the optimum ratio is 1

[edit] you seem to limit yourself to choices like 72 x 1, 36 x 2, 18 x 4, 9 x 8. Would it be worth while to look into configurations where you create 3 V building blocks of 2 parallel cells and one in series with those ?

 

[Nader AbdlGhani]

Hello Nader, welcome to PF !

You are posting in the homework section of PF, so I propose you write an expression for the fraction of the energy a voltage source with an internal resistance delivers to an external load, and try to work that around to a function of the ratio external/internal resistance. My guess is the optimum ratio is 1

[edit] you seem to limit yourself to choices like 72 x 1, 36 x 2, 18 x 4, 9 x 8. Would it be worth while to look into configurations where you create 3 V building blocks of 2 parallel cells and one in series with those ?

Thanks for your reply ,Firstly, I'm sorry I don't know what or how to make an expression for the fraction of the energy a voltage source with an internal resistance delivers to an external load , perhaps you can teach me how ,Secondary, fortunately you opened my eyes on a very good point which is , to obtain the maximum current the internal resistance must be equal to the external resistance thus their ratio equals 1 , Thirdly , I've tried doing that configuration that you mentioned , but still , the results aren't greater than 3.6 Amperes

 

[BvU]

What I suggested wasn't useful:
The fraction is simply $I^2 R_{\rm load} / I^2 \left ( R_i + R_{\rm load} \right )$
but you want to maximize the current in the load

And the next idea
$$ I = {V\over R_i + R_{\rm load} } \quad \Rightarrow \quad P_{\rm load} = I^2 R_{\rm load} = ...$$ and then search for a maximum by differentiation with respect to $R_{\rm load}$ isn't useful either, because you can vary $R_i$ and not $R_{\rm load}$ (it does yield $R_{\rm load} = R_i$ I think). And varying $R_i$ also changes V.

With my funny mix you can get 12 V with $R_i = 1 \Omega$ or 24 V with $R_i = 4 \Omega$ , $\ \$ but 12 V is too low and $4 \Omega$ too high.

So the 18 x 4 is the best I can make of this too.

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[Nader AbdlGhani]

What I suggested wasn't useful:
The fraction is simply $I^2 R_{\rm load} / I^2 \left ( R_i + R_{\rm load} \right )$
but you want to maximize the current in the load

And the next idea
$$ I = {V\over R_i + R_{\rm load} } \quad \Rightarrow \quad P_{\rm load} = I^2 R_{\rm load} = ...$$ and then search for a maximum by differentiation with respect to $R_{\rm load}$ isn't useful either, because you can vary $R_i$ and not $R_{\rm load}$ (it does yield $R_{\rm load} = R_i$ I think). And varying $R_i$ also changes V.

With my funny mix you can get 12 V with $R_i = 1 \Omega$ or 24 V with $R_i = 4 \Omega$ , $\ \$ but 12 V is too low and $4 \Omega$ too high.

So the 18 x 4 is the best I can make of this too.

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I greatly appreciate all of your help , currently I'm sticking to my answer which is (18x4) configuration , but I'm still thinking there is another better configuration , my senior teacher should see my answer on the next Friday so I got good time to find better solution if there is so . Thanks for your help again :D

 

[insightful]

You've probably found that (12x6) also gives 3.6A.

 

[Nader AbdlGhani]

You've probably found that (12x6) also gives 3.6A.

Yup ! :D

 

[Nader AbdlGhani]

Finally guys ! , our solution is right :D ,Thanks a lot !