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My textbook says that these two circuits are equivalent. The first is not easily solvable, the second is. I have looked around the site at a number of questions and this situation is not addressed anywhere, the answers mostly revert you to using Kirchhoff's laws. My textbook says that this circuit can be solved through series/parallel relationships. Of the two equivalent circuits in the image, what is the general method (for solvable cases, don't revert to other methods) for getting from the first to the second?

It is solved by combining 3,5 in series, that with 4 in parallel, all of that with 2 in series, and putting all of that in one resistance in parallel with 6,7,8. What is the logic here if you were confronted with just the first diagram?

Circuit

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    $\begingroup$ I think that one thing that you have to do is to clarify your schematics by showing where there are electrical connections when two lines cross, and where there are not electrical connections when two lines cross. I see a number of crossed lines in the schematics and I guessing that some of these crossings involve connections between the lines but it's not clear which ones do involve connections and which ones do not. $\endgroup$ – Samuel Weir May 31 '17 at 19:41
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    $\begingroup$ I don't see how these two circuits can possibly be equivalent. In the 2nd schematic, it's possible for current to get from one terminal of the battery to the other by going through just one resistor (either R6, R7, or R8), while if you look at the top schematic it's obvious that current can't get from one terminal of the battery to the other by only going through R6 or R7 or R8. In fact, according to the top schematic, the current along any path has to flow through at least two resistors to make it to the opposite terminal of the battery. $\endgroup$ – Samuel Weir May 31 '17 at 19:53
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    $\begingroup$ Possible duplicate of Detecting if resistances are parallel or series in complex circuits $\endgroup$ – sammy gerbil May 31 '17 at 19:59
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    $\begingroup$ See also How to find the total current supplied to the circuit? Colour coding all wires which are connected enables you to redraw complex circuits in a simpler form. $\endgroup$ – sammy gerbil May 31 '17 at 20:01
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    $\begingroup$ Can you give a reference to the textbook (title, author, page, possibly edition)? Not every circuit can be transformed into a series-parallel circuit btw - e.g. the Wheatstone bridge cannot be so transformed. In general, you need Kirchhoff. $\endgroup$ – NickD May 31 '17 at 20:13
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What is the logic here if you were confronted with just the first diagram?

I recommend that you label all unique nodes (two junctions that are connected together by a wire are the same node) and redraw.

There are 5 unique nodes

A: the junction of the positive cell plate, R2, and R6

B: the junction of R2, R3, and R4

C: the junction of R4, R5, R6, R7, and R8

D: the junction of the negative cell plate, R7, and R8

E: the junction of R3 and R5

So, redraw the circuit in a sane way so that you can see the series and parallel connections. For example:

enter image description here

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Two two-terminal elements are in parallel if they are connected together on both sides. One pin of element A is connected to a pin of element B, and the other pin of element A is connected to the other pin of element B. This means that the potential across element A is the same as the potential across element B.

Two two-terminal elements are in series if they are connected together at one terminal, and nothing else is connected to the node that connects them. This means that the current through element A is the same as the current through element B.

Your two example circuits are not equivalent. In your first example, R7 and R8 are in parallel (because their terminals are connected together). But R6 is not in parallel with R7 and R8, since its second terminal is connected to the battery instead of to a second terminal of R7 and R8.

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  • $\begingroup$ Perhaps they are not equivalent but one thing that is guaranteed by the textbook is that the combinations in series/parallel necessary to solve the second diagram are exactly the same as the first. Your answer is good, but it doesn't really help because fundamentally what I don't understand is how to solve the first circuit without using Kirchhoff's laws. $\endgroup$ – Steve May 31 '17 at 20:36
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    $\begingroup$ Try re-drawing it with the crossing wires untangled. After you do that, it's much easier to see the parallel and series combinations. Also, label your nodes to make it easier to check that your untangled circuit is equivalent to the original. $\endgroup$ – The Photon May 31 '17 at 20:42
  • $\begingroup$ I've tried a million things, been trying to solve this for several days. How do you untangle it? $\endgroup$ – Steve May 31 '17 at 20:48
  • $\begingroup$ Like I said, label your nodes. Hal got the same equivalent circuit I did. $\endgroup$ – The Photon May 31 '17 at 21:54
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I agree with Samuel Weir : the two circuits are not equivalent. After colour-coding and re-drawing the circuit so that there is only one vertical line of each colour/potential, I get the same as Hal Hollis :
enter image description here
I made an adjustment to the first diagram to avoid wires crossing over. Note that in the 2nd diagram $R_4$ has flipped around, and the connection between $R_4, R_6$ has been stretched considerably.

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