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The figure below is the schematic of the circuit,

enter image description here

I have to admit that this is a homework question which I have stuck for a while.

  • I try to use Wheatstone Bridge theory, but here in this question I couldn't find such a loop which satisfies that theory.
  • Even I couldn't see a conventional path to solve the circuit. For me this is a mess. That's why I am asking this here.
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  • \$\begingroup\$ See electronics.stackexchange.com/questions/170280/… It's not the same, but it might give you some hints to solve your problem \$\endgroup\$ – efox29 Jul 27 '15 at 9:10
  • \$\begingroup\$ @efox29 The question was posted by the same user. :) \$\endgroup\$ – K. Rmth Jul 27 '15 at 9:46
  • \$\begingroup\$ @K.Rmth Derp... \$\endgroup\$ – efox29 Jul 27 '15 at 9:47
  • \$\begingroup\$ Hint: Assume you are applying a voltage between x and y. Now any two points that are at the same potential you can either short together or remove any resistor between them to simplify the circuit. You should be able to see which points must be at the same potential by using arguments from symmetry. \$\endgroup\$ – Warren Hill Jul 27 '15 at 9:54
  • \$\begingroup\$ @Chu I think the question is what resistance would you expect to measure between points x and y. \$\endgroup\$ – Warren Hill Jul 27 '15 at 10:16
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One way to solve it would be to connect a 1 volt source between X and Y, then use nodal analysis to determine the source current. There are six nodes (not counting X and Y), so six equations would be needed. The equivalent resistance is the voltage divided by the current.

Another way is to take advantage of symmetry, as in your previous question. Remove \$R_{PQ}\$ and \$R_{RS}\$. Now look at nodes P, Q, R, and S. Each one is halfway between X and Y in terms of resistance along their branch. This means that they all should have the same voltage, which is \$\frac {V_X - V_Y} 2\$. Since they have the same voltage, a resistor connected between any two of the points won't draw any current. Thus, \$R_{PQ}\$ and \$R_{RS}\$ can be ignored. Now the circuit is a straightforward series and parallel resistor problem.

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you can eliminate the resistors between points RS and QP because the current through them is 0. That the current through them is 0 is given by the Wheatstone theory. the voltage on S and R are equal, same goes for Q and P .

then you get a network that is very easy to calculate. Just remove (open-circuit) the two resistors I mentioned.

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  • \$\begingroup\$ Are you sure Q and P are at the same potential? \$\endgroup\$ – Warren Hill Jul 27 '15 at 9:59
  • \$\begingroup\$ @user55924-Please explain the way that you use to deduce that the current through RS and PQ is zero. I couldn't see a clear evidence for that. \$\endgroup\$ – On the way to success Jul 27 '15 at 10:49
  • \$\begingroup\$ Yes, I am sure, explanation: Say that X and Y are connected to a voltage source. at point P, by observation exactly half of the impedance is at the Y side, exactly half is at the X side. Both sides are equal, meaning that the potential at P is 1/2 of the voltage source between X and Y. The point is at a symmetrical node. Same logic for R,S and Q. The circuit could also be seen as two perfectly balanced Wheatstone bridges connected with 1 ohm resistors, both are balanced so there is no current flow between the points. Also, I simulated to test my conclusion. \$\endgroup\$ – user55924 Jul 27 '15 at 10:51
  • \$\begingroup\$ @Onthewaytosuccess Sorry about the short explanation in the answer, I thought you would go " Aha! " when I said that there was no current between the points in the answer. \$\endgroup\$ – user55924 Jul 27 '15 at 10:59
  • \$\begingroup\$ @user55924-I may go with "Aha" if I know how the method that you use to deduce that there is no current through RS and PQ. I wanna know that way. \$\endgroup\$ – On the way to success Jul 27 '15 at 11:11

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