Creating 500Ω Resistance with 12 600Ω Resistors

[sheldon]

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

 

[FZ+]

The joyous algebraic method.

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.

 

[sheldon]

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

 

[elephant]

You could do it with only 5 resistors

3 600 Ohm resistors in parallel (200 Ohms)

In series with:

2 600 Ohm resistors in parallel (300 Ohms)

= 500 Ohms.

 

[FZ+]

Oops.

For 12 resistors, the solution is:

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

 

[sheldon]

Ok I heard if you made a cube, which has 12 sides and put a 600 ohm resistor on them it would equal 500 ohms.

 

[BoulderHead]

Yeah but...

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?