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In my previous Phys.SE question, situated here, I asked about finding the equivalent resistance of the following circuit : Circuit 1

I got some very good answers and some tips. Now what if the same circuit is modified by adding two more resistors on the free connecting wires like this :
Circuit 2

How should I find the equivalent resistance now ?
I probably can't redraw the circuit as mentioned in answers to my previous question since every wire has a resistor on it.

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  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/22252/2451 $\endgroup$ – Qmechanic Oct 2 '13 at 14:31
  • $\begingroup$ @Qmechanic I don't think it is related. Also most of the answers their seem difficult to me. $\endgroup$ – A Googler Oct 2 '13 at 14:41
  • $\begingroup$ The circuit you've drawn can't be represented as a simple set of resistors in series and/or parallel, so redrawing it doesn't help. That means you have resort to the usual tools for analysing circuits, which unfortunately are much more complicated than the simple redrawing I did for your first question. The link Qmechanic posted gives you examples of how to do the analysis. The standard technique is to use Kirchoff's laws (en.wikipedia.org/wiki/Kirchhoff's_circuit_laws). $\endgroup$ – John Rennie Oct 2 '13 at 14:42
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If you have a circuit like this, that cannot be simplified by the serial circuit or parallel circuit law, you should check out the star-delta transform (http://en.wikipedia.org/wiki/Y-%CE%94_transform).

General Kirchhoff laws will also work of course.

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    $\begingroup$ I don't know if it is a proven result, but I'm fairly certain that any network of resistors can be solved with the parallel, series, and star-delta rules. $\endgroup$ – Robert Stiffler Nov 20 '15 at 11:23
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    $\begingroup$ @RobertStiffler: According to Wikipedia, this is only true for a planar circuit, i.e. when it can be drawn without any connections crossing over over another connection. For non-planar circuits one has to use the Kirchhoff laws. $\endgroup$ – Andreas H. Nov 22 '15 at 9:13
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The way you have the circuit drawn is obscuring the topology. Schematics should clarify, not obfuscate. It also helps to draw them neatly.

Here is a schematic drawn to make things more obvious:

Hopefully you can now see a strategy to solve this.

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  • $\begingroup$ How did you convert the circuit into this? Are resistance R2 and R3 in parallel? $\endgroup$ – A Googler Oct 3 '13 at 6:25
  • $\begingroup$ Is it wheatstone bridge? $\endgroup$ – A Googler Oct 3 '13 at 6:51
  • $\begingroup$ @AGoogler you don't have to convert at all to get to this, it's just redrawing your diagram more clearly. It's exactly the same as yours. And no they are not strictly in parallel. $\endgroup$ – Dan Oct 3 '13 at 7:13
  • $\begingroup$ @Dan then how should I solve this? $\endgroup$ – A Googler Oct 3 '13 at 13:43
  • $\begingroup$ @AGoogler exactly as I described in my answer... use a star delta transformation! It's right there in the wikipedia article, exactly this question. At least give it a try. $\endgroup$ – Dan Oct 3 '13 at 14:48
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You diagram is exactly the same as the bridge example on Wikipedia's Y-Δ transform page. After the transforms (shown in that example) it becomes trivial.

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  • $\begingroup$ I saw the page but I think that example is different . Which example are you referring to? $\endgroup$ – A Googler Oct 3 '13 at 6:53
  • $\begingroup$ @AGoogler: No it's not different, see Olin's answer as he redrew the circuit in the same form as these examples. After a Y-Δ transform your circuit becomes just normal series and parallel. $\endgroup$ – Dan Oct 3 '13 at 7:12
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The diagram is a network of wheatstone bridge, but it should satisfy the condition after which it is easy to calculate the net resistance between the points

The R2 resistance will be in the middle and will be neglected if the condition of wheatstone bridge is satisfied, i.e. R1.R4=R5.R3

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    $\begingroup$ This statement is true but it is not an answer to the question. $\endgroup$ – dmckee Apr 8 '15 at 20:35
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This circuit can be redrawn into a simpler version. The circuit so formed will exactly resemble as the "Wheatstone's bridge" with a resistor in place of a galvanometer(which is commonly used for checking slightest of the current passing through it).

First mark the points at the junctions as "1","2","3","4" respectively from left to right. Then you can observe that point 1 to point 2 has resistor no.1.so draw it first. Then we can observe that point 1 to point 3 has resistor no. 4, so draw Resistor no. 4 from the point1 (marked before) to point3. Now we see that point 2 to point 3 has resistor no. 2 in between, so draw it between the points 2 and 3 (both points are marked in the earlier steps so just draw a resistor between them). Now draw resistor 3 and resistor 5 from points 3 and 4 respectively to point 5. Now point 1 resembles point "A" and point 5 resembles point "B"(as no resistor is placed between point "A" and point 1 and between point "B" and point 5). Hint: Draw the straight i.e. without any curve between two points.

You'll observe that a Rhombus resembling Quadrilateral is formed with a resistor on all its side and a diagonal. Prefer to draw same size of the wire. For solving and knowing more about the Wheatstone's bridge thus formed.

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