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Here is an image of the circuit diagram below. I know that the current that is flowing through R1 and R8 is the same and so those 2 resistors are added in series. However my difficulty comes in determining the resistors in parallel. Parallel resistors ought to have different currents but the same potential difference across them and for this it would appear to me that R2, R3, R5 and R4 ought to be in parallel, as they all receive different currents due to the branching out that occurs. However this is not correct. Any help understanding how to determine when resistors are in parallel for diagrams like this would be much appreciated, as to the untrained eye like my own, the potential differences across those resistors appear to be identical.

enter image description here

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  • $\begingroup$ Why is a question about identifying series and parallel combinations not a question about concepts? $\endgroup$ Jun 23, 2021 at 16:08
  • $\begingroup$ I don’t think this is an out of scope question. To me it seems clearly conceptual $\endgroup$
    – Dale
    Jun 23, 2021 at 16:52
  • $\begingroup$ Forget series and parallel. Define four current loops. Write four voltage loop equation and solve them for the currents. $\endgroup$
    – R.W. Bird
    Jun 25, 2021 at 14:34

4 Answers 4

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To do this systematically first label all of the unknown voltagesCircuit with labeled voltages

Then for each resistor write the voltages that it is between:

R1 (V1,V2)

R2 (V2,V4)

R3 (V2,V4)

R4 (V2,V3)

R5 (V3,V4)

R6 (0,V3)

R7 (V4,Vs)

R8 (0,V1)

From this you can see that only R2 and R3 are in parallel. However, to be honest, once you have written down the voltages you are halfway to solving the circuit using the node voltage method anyway. So at that point there is no real reason to go through the hassle of rewriting the circuit as various series and parallel combinations. It is easier to simply solve the circuit directly using the node voltage method.

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    $\begingroup$ The things that you call "Voltages," would be called nodes by a EE. $\endgroup$ Jun 23, 2021 at 17:26
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This is an equivalent version of your circuit. You can see that $1$ and $8$ are in series, while $2$ and $3$ are in parallel. Once you substitute those with their equivalent resistors you are left with a Wheatstone Bridge-like circuit that can't be simplified anymore using series/parallel replacements.

enter image description here

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Here is a hint after going through, which any guy familiar with KCL and KVL, should be able to identify which capacitors are in parallel.

enter image description here

Just use KVL, solve for 4 currents and see which two resistors have same p.d. across them.

*even I haven't solved it completely ! It's a lot of work there ! But hey, you arrive at the answer!

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Resistors A and B are in parallel if and only if one end of A is joined by a wire of negligible resistance to one end of B, and the other end of A is similarly joined to the other end of B ! Only $R_2$ and $R_3$ meet this requirement. [You can spot this without re-drawing the diagram or labelling node potentials.]

Your circuit cannot be solved using just series and parallel resistance rules. $R_1$ and $R_8$ are indeed in series and $R_2$ and $R_3$ are in parallel, but these combinations form one 'arm' of a bridge configuration whose other arm is made up of $R_5$ and $R_6$. The 'bridging resistance' is $R_4$. The bridge configuration cannot be resolved directly into series and/or parallel combinations, but can be so resolved by first using a Delta-Star ($\Delta-\text Y$) transform on either the top 'square' or the bottom 'square' of the bridge. If you had numerical values for the resistances, this would be quick and easy. Without them, you generate some cumbersome algebraic expressions, but probably no more so than if you use Kirchhoff's laws, either with currents or with node potentials as the unknowns.

I found it easier to spot the bridge configuration by re-arranging the diagram so that, roughly, potentials decreased from top to bottom.

Addendum: How nicely this ties in with FrodCube's answer, which has just appeared on my screen!

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