- #1
sheldon
- 150
- 0
If you have 12 resistors, each rated for 600ohms, how could you arrange them to make a total resistance of 500ohms?
Creating 500Ω Resistance with 12 600Ω Resistors
- Context: Undergrad
- Thread starter sheldon
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- Resistance Resistors
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Discussion Overview
The discussion revolves around how to configure 12 resistors, each with a resistance of 600 ohms, to achieve a total resistance of 500 ohms. The conversation includes various methods of arrangement, including series and parallel configurations, and explores both algebraic and conceptual approaches.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant proposes using a combination of series and parallel arrangements to achieve 500 ohms, suggesting a specific algebraic method involving the equation 1/a + 1/b = 1/c.
- Another participant confirms that 6 resistors can achieve 500 ohms but expresses confusion about how to use all 12 resistors.
- A different configuration is suggested involving 3 resistors in parallel (yielding 200 ohms) in series with 2 resistors in parallel (yielding 300 ohms) to reach 500 ohms.
- One participant later corrects themselves, stating that for 12 resistors, the solution involves 3 in parallel, another 3 in parallel, and 6 in parallel, all joined in series.
- Another participant introduces a conceptual idea of arranging resistors on a cube, questioning the feasibility of such a configuration to achieve the desired resistance.
- A follow-up response humorously critiques the cube idea, but then elaborates on the current flow through the proposed arrangement, indicating a split and combination of currents at various points.
Areas of Agreement / Disagreement
Participants express differing views on how to arrange the resistors to achieve 500 ohms, with no consensus on a single method. Multiple configurations are proposed, and some participants express confusion or corrections regarding earlier claims.
Contextual Notes
Some assumptions about the configurations and the nature of the resistors may be implicit, and the discussion includes unresolved mathematical steps and varying interpretations of resistor arrangements.
- #2
FZ+
- 1,599
- 3
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.
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
sheldon
- 150
- 0
I see how you get 500 ohms with 6 resistors but don't understand how you got it with 12?
- #4
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.
3 600 Ohm resistors in parallel (200 Ohms)
In series with:
2 600 Ohm resistors in parallel (300 Ohms)
= 500 Ohms.
- #5
FZ+
- 1,599
- 3
Oops.
For 12 resistors, the solution is:
3 in parallel, 3 in parallel and 6 in parallel joined in series.
For 12 resistors, the solution is:
3 in parallel, 3 in parallel and 6 in parallel joined in series.
- #6
sheldon
- 150
- 0
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.
- #7
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?
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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