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Jahnavi
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BvU said:Try to flatten the diagram: a top horizontal line is A, a bottom one is B.
BvU said:There are three paths from top to bottom: AB, ADB and ACB
As I indicated: a top horizontal line is A, a bottom one is B.Jahnavi said:How should I use this to simplify the circuit ?
ehild said:View attachment 223752
The red path is out of the plane, but it is equivalent with the blue path otherwise.What does that tell you about the potentials at D and C?
tnich said:Sorry. Symmetry is not going to help you on this problem.
What would happen if you interchanged the blue and red triangles, would be anything different?Jahnavi said:Thanks for replying .
This is exactly what I would like to understand .
If I draw a line perpendicular to AB , the left and right parts are mirror image , this make points D and C symmetric . But symmetric points so not necessarily have to be equipotential . Right ?
How can then we argue that points C and D are equipotential .
They don't look symmetric to me. Try it with a mirror.Jahnavi said:Thanks for replying .
If I draw a line perpendicular to AB , the left and right parts are mirror image , this make points D and C symmetric . But symmetric points so not necessarily have to be equipotential . Right ?
Oh, sorry, I see the symmetry now.tnich said:They don't look symmetric to me. Try it with a mirror.
tnich said:Oh, sorry, I see the symmetry now.
Redraw the circuit as @BvU has suggested. Then you will see it.Jahnavi said:I am not saying that the circuit is not symmmetric . All I would like to understand is that why points D and C are equipotential .
If you remember in the previous thread you had given a very nice symmetry argument . But points L and Q , although being symmetric were not equipotential .
Please help me understand why D and C are equipotential in this problem .
Remember in the previous thread, we redrew the circuit to make the symmetric obvious. We replaced one resistor with two resistors in series.Jahnavi said:If you remember in the previous thread you had given a very nice symmetry argument . But points L and Q , although being symmetric were not equipotential .
tnich said:Remember in the previous thread, we redrew the circuit to make the symmetric obvious. We replaced one resistor with two resistors in series.
And in this case, also, redrawing the circuit does not change it.Jahnavi said:But redrawing the circuit didn't change the circuit . Both were completely equivalent .
A B C D
A - 2 1 1
B 2 - 1 1
C 1 1 - 1
D 1 1 1 -
ehild said:View attachment 223752
The red path is out of the plane, but it is equivalent with the blue path otherwise.What does that tell you about the potentials at D and C?
ehild said:What would happen if you interchanged the blue and red triangles, would be anything different?
vela said:In analyzing a circuit, one of the most useful tools you have is redrawing the circuit in a way that it's easier to work with.
vela said:I'm not sure why you're so resistant (pun not intended, but I'll go with it) to the idea.
BvU said:As I indicated: a top horizontal line is A, a bottom one is B.
Have the three paths well separated, but going straight down from the top line to the bottom one
Path symmetry indicates that D and C are equipotentialJahnavi said:Thanks for replying .
This is exactly what I would like to understand [emoji2] .
Please assume this is a 2D circuit .
If I draw a line perpendicular to AB , the left and right parts are mirror image , this make points D and C symmetric . But symmetric points do not necessarily have to be equipotential . Right ?
How can then we argue that points C and D are equipotential .
Hello again,Jahnavi said:Thanks !
ehild said:The original circuit cannot be drawn in-plane
BvU said:and now I have a question for you:
with hindsight, what would have been a good hint that could have helped you effectively at that point ?
Effective resistance is a measure of the resistance between two points in an electrical circuit, taking into account all the resistors and their connections between those two points.
Effective resistance can be calculated using the symmetry method, which involves identifying symmetrical points in the circuit and using their resistances to determine the overall effective resistance.
The symmetry method is used because it simplifies the calculation process and reduces the number of equations that need to be solved, making it a more efficient method for determining effective resistance.
Yes, the symmetry method can be used for any circuit as long as there are symmetrical points present. However, it may not always be the most efficient method for calculating effective resistance in more complex circuits.
One limitation of the symmetry method is that it can only be used for circuits with symmetrical points. Additionally, it may not be accurate for circuits with non-linear components or for circuits with multiple sources of electricity.