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## Main Question or Discussion Point

If you have 12 resistors, each rated for 600ohms, how could you arrange them to make a total resistance of 500ohms?

- Thread starter sheldon
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- #1

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If you have 12 resistors, each rated for 600ohms, how could you arrange them to make a total resistance of 500ohms?

- #2

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Let 1/a + 1/b = 1/c

c = a * b / (a + b)

let a = 600 x and b = 600 y and c = 500

(We are setting up a parallel circuit, one with x resistors, and the other with y resistors in parallel)

we get then

(x * y * 360000)/(600(x+y)) = 500

Which simplifies to...

x * y / (x + y) = 5/6

Which we can turn into:

x = 5y/(6y - 5)

Now, set y = 1.

x = 5 / 1 = 5

So one easy possiblity is to have 5 resistors in series on one branch and 1 on the other.

Multiply by two if you need to use all 12 resistors.

- #3

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I see how you get 500 ohms with 6 resistors but don't understand how you got it with 12?

- #4

elephant

3 600 Ohm resistors in parallel (200 Ohms)

In series with:

2 600 Ohm resistors in parallel (300 Ohms)

= 500 Ohms.

- #5

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For 12 resistors, the solution is:

3 in parallel, 3 in parallel and 6 in parallel joined in series.

- #6

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- #7

BoulderHead

How many cubes have twelve sides?

[edit]

Cheese, ok I drew it out and get it now,haha.

I've forgotten what the question was as I edit this, but looking at my sketch I see that starting at one of the corners of such an array and assuming a current of '1', the flow would split in three equal 'parts' (each carrying 1/3). The three split again in two directions (each then carrying 1/6). Next the 1/6 branches would combine again into three (each now carrying 1/3), and these three combine again to bring us back to the original amount inserted at an opposite corner.

Does that make any sense?

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