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Net Resistance

  • Thread starter andyrk
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  • #1
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Find the Net Resistance between A and B in attachment 1.
The circuit in attachment 1 simplifies to attachment 2 from the fact that the points C and E are equipotential. Same goes for D and F. So we remove CE and F as no current passes through equipotential points. But why? How are these 2 pairs of points equipotential?
 

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  • #2
haruspex
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Would you agree there must be a unique solution to the potentials and currents?
Because of the symmetry in the diagram, all equations which you obtain must be similarly symmetric: swapping the potentials and flows around a horizontal axis must also be a solution. As the solution is unique, potential at C = that at E, etc.
 
  • #3
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You might think of the problem this way. Start with diagram 2. Put a battery or any voltage between A and B and call B Ground. Then calculate the voltage at points C and E with respect to Ground. You will find they are the same. So no matter what resistance you place between C and E, no current will flow because there is no voltage difference between C and E. This is why you can remove the 5 ohm resistors. If the circuit was not symmetrical, then there is a different problem.
 
  • #4
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Symmetry is a shortcut answer. There must be some logical reasoning to it? Like if we apply Kirchoff's Voltage Law, and form the equations then by solving them we get VC=VD? But I am not able to form the equations because in the CE branch I am not able to determine what current would be passing through it. I know this is because they are equipotential, but why are they? Reply Soon!
 
  • #5
phinds
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Symmetry is a shortcut answer. There must be some logical reasoning to it?
In what way is symmetry not logical reasoning?

"reply soon" would seem to imply that this is a homework problem that you want US to solve for you. You need to show some attempt on your own. We are not here to DO your work, only to help you figure out how to do it yourself.
 
  • #6
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In what way is symmetry not logical reasoning?

"reply soon" would seem to imply that this is a homework problem that you want US to solve for you. You need to show some attempt on your own. We are not here to DO your work, only to help you figure out how to do it yourself.
It is nowhere written that for symmetric circuits simply remove the branch that doesn't effect symmetric behavior of the circuit. No this is not for Homework. I just randomly turned it up while doing this topic. The thing is the only way this can be done is by saying that we do it by symmetry. This is what bugs me. There has got to be some explanation to symmetry?? Why do we use symmetry? Explain symmetry? That's why I cannot show you my attempt because the only attempt that's worth is the symmetrical one and i didn't get. As for my attempt, as I mentioned I applied Kirchoff's Voltage Law in the loops but didn't get the answer as I am unabe to get the current in CE and DF branch.
 
  • #7
phinds
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It is nowhere written that for symmetric circuits simply remove the branch that doesn't effect symmetric behavior of the circuit. No this is not for Homework. I just randomly turned it up while doing this topic. The thing is the only way this can be done is by saying that we do it by symmetry. This is what bugs me. There has got to be some explanation to symmetry?? Why do we use symmetry? Explain symmetry? That's why I cannot show you my attempt because the only attempt that's worth is the symmetrical one and i didn't get. As for my attempt, as I mentioned I applied Kirchoff's Voltage Law in the loops but didn't get the answer as I am unabe to get the current in CE and DF branch.
So look up delta-Y transforms and do the whole circuit analysis.
 
  • #8
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If you put a battery across A and B, then set up the 4 current loops, and then solve the equations, you will find that there is no current through the 5 ohm resistors. and that the potential at C and E are equal. After working a few of these problem, you will begin to see where symmetry will save a lot of work.
 

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