MOST difficult circuit ever made

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How to find different resistance out of 5 same and one different resistance in circuit.?
image of resistances are shown here
http://www.pictures.pk/viewer.php?id=rgh1210158811g.JPG

in which 5 are same and one is different. how to recognize the different resistance in the circuit through ohm meter

Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution

 
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we just have to find the different reistance in the circuit...
 
just paste the link starting with pictures.pk and see the figure
 
I think you are given circuit and the ohmeter and you are told that 5 resistances are identical, while 6th is different - and you have to find this 6th just by measuring resistance between any points in the circuit.

I also suppose you are not allowed to cut wires :wink:

impaxion: is there a preset limit to number of measurements you can make, or not?
 
The circuit is symmetric. If all resistances were equal, the measurement across any of them would show the same result (the resistance in parallel with the rest of the circuit).
Since there is a different resistance, the measurement across it will be different from the other measurements
 
Redbelly98 said:
It might not be in English.

At any rate, I think Borek has provided an accurate phrasing of what is being asked.

Possibly true. It may be too hard for english speakers. I can't figure the problem out and I speak english. But I do agree the problem is pretty clear.
 
impaxion said:
we just have to find the different reistance in the circuit...

As drawn the circuit is symmetric along an axis through the center of the drawing. If you make two resistance measurements from a point on the axis to two symmetrical points off of the axis and get the same value, then the 'different resistance' must be one of the resistors that is rotated into itself by the symmetry, otherwise it's one of the others. That narrows the search down. Now find more symmetry axes. This isn't hard. You can really picture the circuit as lying on the edges of a tetrahedron. That's a good question!
 
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