# Symmetry in a Triangular Circuit

1. Dec 9, 2013

### bananabandana

1. The problem statement, all variables and given/known data
Find the resistance across AB. Use symmetry to determine currents.

2. Relevant equations
Please see diagram.

3. The attempt at a solution
I really don't seem to understand symmetry in circuits. What is the link between the geometrical symmetry of the situation and the way in which the current divides at a junction? Is it the case that if a current can follow two possible paths, and the paths have the same resistances in the same order, that the current along each path is equal??

In the diagram below for example, why is I6 supposed to be zero? How do I know that points D and A are at the same potential??? I've tried expanding everything using Kirchoff's laws, but it just becomes very messy....

Thanks!

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2. Dec 9, 2013

### Kishlay

do you know a term something called as "weigh stone bridge" ??

3. Dec 9, 2013

### bananabandana

Um, I think you mean 'Wheatstone bridge'. Yes, I do know that the example i've got is a balanced Wheatstone bridge, but I'd still like to better understand how symmetry works in circuits in general.

4. Dec 9, 2013

### Kishlay

at here if
R1/R2=R3/R4
then there will be no current through the R6 i.e I6=0

5. Dec 9, 2013

### bananabandana

Yes, but why? How does the symmetry of the problem relate to this understanding? :P

6. Dec 9, 2013

### Kishlay

all the subs scripts represents the resistance of your question i.e
R1 is the resistance from where I1 current will flow and so on......

7. Dec 9, 2013

### bananabandana

Yes, I understand that! What I don't understand is why it is necessary, that, in this example, a balanced Wheatstone bridge must have no current flowing through the middle wire.. if that's not a stupid question :P

8. Dec 9, 2013

### Staff: Mentor

If a bridge is balanced, then the 2 outside voltage dividers give equal voltages at each end of the middle resistor.

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