# Resistance of a Pentagon

phantomvommand
Homework Statement:
See diagram below
Relevant Equations:
Methods of Resolving resistors

The question is to find the resistance between points A and F.

I understand that the resistor between OC can be removed. From this point onwards, resistor OC has been removed. Let the far right point of the diagram, where i5 and i7 exit from, be B.

Is it possible to detach the remaining circuit at O, such that FOA and DO'B become separate, and then resolve the remaining resistors? Why is it not possible to detach the remaining circuit at O, to form FOD and AO'B? What exactly is the reason why we can detach some circuits?

Mentor
There may be a trick to solving it, but if I don't see the trick within about 30 seconds, I just use KCL to solve it...

phantomvommand
Homework Helper
Is it possible to detach the remaining circuit at O, such that FOA and DO'B become separate
Yes. From symmetry alone, one can see the potential at that point does not change by doing that.
Why is it not possible to detach the remaining circuit at O, to form FOD and AO'B?
That would interrupt currents flowing.

##\ ##

phantomvommand
It's correct that, due to mirror symmetry, ##i_9 =0## and the branch ##\mathrm{OC}## can be removed without affecting the circuit. You can go further! When you reverse the orientation of the battery all of the currents will change direction except again - by mirror symmetry - you have an identical circuit to what you had before (imagine looking at the circuit from the other side of the paper)!

Hence ##i_1 = i_4##, then ##i_2 = i_3##, then ##i_5 = i_6## and ##i_7 = i_8##.

Now we would like to detach the circuit at ##\mathrm{O}## to simplify it further. We must do this in such a way that we don't alter anything about the circuit (currents, voltages, etc.). Hopefully you can see that it's acceptable to detach it into ##\mathrm{FOA}## and ##\mathrm{DO'B}##, because the currents in ##\mathrm{FO}## and ##\mathrm{OA}## are both ##i_1## and the currents in ##\mathrm{DO'}## and ##\mathrm{O'B}## are both ##i_5##. Detatching it into ##\mathrm{FOD}## and ##\mathrm{AO'B}## is not possible, because the currents in ##\mathrm{FO}## and ##\mathrm{OD}## are not equal, nor are those in ##\mathrm{AO}## and ##\mathrm{OB}##!

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phantomvommand
phantomvommand
Yes. From symmetry alone, one can see the potential at that point does not change by doing that.

That would interrupt currents flowing.

##\ ##
Thanks for this! Yes, I realize that the potential is the same. But even if the potential is the same, there can be current flowing through? How do we know for sure there is no current flowing through?

Homework Helper
Not necessary at all. The steps described and a quick redraw of the circuit shows two independent branches (AOF and ABDF) with very simple resistance in terms of R

phantomvommand
It's correct that, due to mirror symmetry, ##i_9 =0## and the branch ##\mathrm{OC}## can be removed without affecting the circuit. You can go further! When you reverse the orientation of the battery all of the currents will change direction except again - by mirror symmetry - you have an identical circuit to what you had before (imagine looking at the circuit from the other side of the paper)!

Hence ##i_1 = i_4##, then ##i_2 = i_3##, then ##i_5 = i_6## and ##i_7 = i_8##.

Now we would like to detach the circuit at ##\mathrm{O}## to simplify it further. We must do this in such a way that we don't alter anything about the circuit (currents, voltages, etc.). Hopefully you can see that it's acceptable to detach it into ##\mathrm{FOA}## and ##\mathrm{DO'B}##, because the currents in ##\mathrm{FO}## and ##\mathrm{OA}## are ##i_1## and the currents in all of ##\mathrm{DO'}## and ##\mathrm{O'B}## are ##i_5##. Detatching it into ##\mathrm{FOD}## and ##\mathrm{AO'B}## is not possible, because the currents in ##\mathrm{FO}## and ##\mathrm{OD}## are not equal!
Thanks, this has been very helpful!

etotheipi
Gold Member
If you redraw the schematic so that each node has only 3 connections, perhaps with a wire connecting nodes that would have more than three connections. Any branch that can be proven to ALWAYS have zero current can be removed without changing the solution.

One way to think about this is that you can add a resistor in that branch without changing the voltage drop. Then you are free to use any value for that resistor, including ∞. Of course the two new nodes you create will ALWAYS have the same voltage too.

There is a similar argument for connecting nodes that ALWAYS have the same voltage.

No, it's not the same network, but it has the same solution.

phantomvommand
phantomvommand
If you redraw the schematic so that each node has only 3 connections, perhaps with a wire connecting nodes that would have more than three connections. Any branch that can be proven to ALWAYS have zero current can be removed without changing the solution.

One way to think about this is that you can add a resistor in that branch without changing the voltage drop. Then you are free to use any value for that resistor, including ∞. Of course the two new nodes you create will ALWAYS have the same voltage too.

There is a similar argument for connecting nodes that ALWAYS have the same voltage.

No, it's not the same network, but it has the same solution.
Hi, thanks for this. I would like to clarify something:
- Why can’t a current flow between 2 same voltage nodes, when u connect them together? Even though V = 0, R =0, and since V = RI, I can be > 0, because R = 0.

in a simple parallel circuit, with 2 branches, a current flows between same voltage nodes.

there is probably some simple resolution to this, but I can’t see it!