- #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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SUMMARY
The discussion focuses on creating a total resistance of 500Ω using 12 resistors, each rated at 600Ω. The solution involves arranging the resistors in a combination of series and parallel configurations. Specifically, the optimal arrangement is 3 resistors in parallel (yielding 200Ω), combined in series with 2 resistors in parallel (yielding 300Ω), resulting in a total of 500Ω. Additionally, an algebraic method is utilized to derive the relationship between the number of resistors in each configuration.
PREREQUISITES- Understanding of Ohm's Law
- Familiarity with series and parallel resistor configurations
- Basic algebra for manipulating equations
- Knowledge of electrical resistance calculations
- Research the principles of series and parallel circuits
- Learn about resistor color codes and ratings
- Explore advanced circuit analysis techniques
- Study the impact of resistor configurations on total resistance
Electronics enthusiasts, electrical engineering students, and anyone involved in circuit design and analysis will benefit from this discussion.
- #2
FZ+
- 1,594
- 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+
- 1,594
- 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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