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This might seem like a simple question I'm asking too much about, a simple clarification is all I need.

circuit image

There's no details of current or potential difference provided at all, and it's known according to the mark-scheme that the circled resistor is connected in parallel with the 3 series resistors. Why can't I suppose the top and the bottom ones to be in parallel, and that parallel network to be in series with the other parallel network of the left and right resistors? I've seen a few existing questions similar to this but they didn't help. I've even come across a similar circuit before and considering the total resistance how I'm trying to now, I was correct.

I'll appreciate any help at all, thanks.


note from the future: I've deleted my second question because understanding the answers here is indeed enough to help in solving most related questions, particularly Noah's and Philip Wood's, but not that the others aren't helpful.

Something to note is that any interruptions at all (e.g even connection to a terminal) render a sequence of node connections as one separate network up to that point, this makes it easier to know when you can assume multiple combinations to be in parallel or series since you can distinguish between separate combinations easily.

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    $\begingroup$ Resistors aren’t connected in parallel just because they are along parallel lines in a diagram. $\endgroup$ – G. Smith Dec 19 '20 at 18:23
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    $\begingroup$ en.wikipedia.org/wiki/Series_and_parallel_circuits $\endgroup$ – G. Smith Dec 19 '20 at 18:31
  • $\begingroup$ I've gone long attempting several questions without properly ever bothering to understand or clarify my understanding rather, of the basic concepts involved in solving more difficult questions. $\endgroup$ – un-index Dec 19 '20 at 18:34
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    $\begingroup$ You can’t solve problems correctly if you aren’t clear on the concepts. $\endgroup$ – G. Smith Dec 19 '20 at 18:39
  • $\begingroup$ Yes thanks, I've learnt that already. The only thing I didn't understand was what really differentiated series and parallel circuits. See the edit too, please. $\endgroup$ – un-index Dec 19 '20 at 18:39
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The top and bottom resistors aren't a parallel combination. The left hand ends aren't connected straight together nor are the right hand ends. Both these conditions would have to be met for the resistors to be in parallel. For the same reason the left hand and right hand resistors are not connected in parallel.

What you can correctly say is that the top, right hand and bottom resistors constitute a series combination that is connected in parallel with the left hand resistor, across X and Y.

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  • $\begingroup$ Thank you very much, the phrase "left hand end of one isn't connected straight to the left hand end of the other" helped in particular. I'd accept your answer instantly if I could. The other answer is helpful too. $\endgroup$ – un-index Dec 19 '20 at 17:30
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    $\begingroup$ Good. Many resistor networks can be seen as made of series and parallel combinations, but some can't be. $\endgroup$ – Philip Wood Dec 19 '20 at 17:36
  • $\begingroup$ That solves the issue, however.. when I was going over it I noticed with that logic I could take the left and top resistors to be in series (let's call the total resistance across them A) and that A would be parallel to the series network of the bottom and right resistors. That would mean that the top-left and bottom-right series networks would together be combined in a parallel network. But then this would cause the total resistance of the circuit to be different from when I calculated the total resistance supposing the resistors are 3 series-1-paralle. Should I create a new post for this? $\endgroup$ – un-index Dec 19 '20 at 17:47
  • $\begingroup$ Since this question has shifted from "series or parallel" to "parallel or series" now. $\endgroup$ – un-index Dec 19 '20 at 17:52
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    $\begingroup$ That's a much more subtle question! The two points across which you measure the resistance may determine the series/parallel analysis. Your new analysis is correct if you take the top left and bottom right points of the square as the two points across which you measure the resistance. Indeed you can't work out or measure the resistance of a network of resistors with several nodes (junctions) unless you specify two particular nodes. It's quite an interesting question and it would, imo, be fine to post it as another question. $\endgroup$ – Philip Wood Dec 19 '20 at 18:13
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I'll just add my two cents to how I would approach such a problem systematically. Look at the whole circuit and look for resistors that are connected without any connection to anywhere else between them (for example a connection to a terminal or another part of the circuit, that disrupts the continuity of node connections). Group these in your mind into a single equivalent resistor. These are all the resistors which are directly serial. In your case the right three ones.

Now look at the updated network. Look for pairs of resistors which share a common start and end point. These are parallel and can be grouped together as a single resistor. In your case, that is the left resistor with the group you created in the first step.

In your case, at this point you are done. If the circuit is more complicated, repeatedly alternate these steps until you are left with only one resistor.

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If ever in doubt, just redraw: redrawn circuit

You are free to bend and stretch out empty sections of wire as you please, as these have 0 effect on anything happening in the circuits. If you do so here, we can nicely see 3 resistors in series, all in parallel with a single resistor

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The two top resistors are not in parallel because their end connections are not common due to the presence of the two side resistors. Similarly, the two side resistors are not in parallel because their end connections are not common due to the presence of the two top resistors.

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The one marked in red is in parallel with the remaining three. Those three are in series with each other.

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