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We are asked to simplify the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I argue as follows, in order upper-left, upper-right, lower-left, lower-right:

schematic

simulate this circuit

While the above reasoning leads to a correct solution, I struggle to grasp why the step from upper-right to lower-left is justified. In particular, why are 2R/3 and 2R in parallel? I would feel comfortable with my reasoning if the parting state was

schematic

simulate this circuit

Yet this does not seem topologically equivalent to the circuit in the upper-right.

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    \$\begingroup\$ Your first pic and your last pic is the same circuit. How you draw the schematic lines or at what angle the resistors are drawn have no impact. \$\endgroup\$
    – Lundin
    Jan 8 at 14:20
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    \$\begingroup\$ what makes the bottom circuit un-equivalent in your mind? \$\endgroup\$
    – jsotola
    Jan 8 at 16:07
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    \$\begingroup\$ It would make it easier to discuss your circuit if you gave all the resistors designators (R1, R2, R3, ...). You already have good answers to this question, but keep it in mind for next time. \$\endgroup\$
    – The Photon
    Jan 8 at 16:31
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    \$\begingroup\$ I love all the use of color. Here's what I did while only looking at the question and before rolling down to see how much we think similarly! But apparently my left-right perception related to color choices is pretty much opposite everyone else's. ;) Funny. No idea why. \$\endgroup\$ Jan 8 at 19:01
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    \$\begingroup\$ @periblepsis: I suspect other answerers intentionally copied GodJihyo's colour scheme for consistency. \@Transistor's answer references it. \$\endgroup\$ Jan 9 at 3:44

6 Answers 6

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It is entirely equivalent to the other answers, but sometimes it helps visualising to compress each connected segment down into a single node, as in the following diagram. I imagine lengths of solder, being heated and turning into blobs.

enter image description here

Each colour denotes a single node. The connecting lines on the schematic are only to facilitate understanding. They don't change the number of nodes, no matter how many lines there are.

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  • \$\begingroup\$ maybe. Depends on the asker's learning style I guess. \$\endgroup\$
    – user253751
    Jan 8 at 18:53
  • \$\begingroup\$ So if the red and the green nodes are the only accessible nodes (as the original drawing suggests), the whole thing can be reduced to a single resistor, right? \$\endgroup\$
    – U. Windl
    Jan 10 at 7:35
  • \$\begingroup\$ @U.Windl I agree the original diagram suggests it, but it is drawn on purpose to be misleading (to teach about parallel/serial resistors) and it doesn't have, say, labelled red and green stubs. But of course you are absolutely correct: if we say no more connections can be made to blue node, then we have a two-node network with value r * par(1/2, ser(1, par(1, 2/3, 2))). \$\endgroup\$
    – jonathanjo
    Jan 10 at 8:16
  • \$\begingroup\$ @U.Windl I can't find a good source, but for circuits made only by resistors, given a two points it should always be possible to reduce the whole circuit to an equivalente single resistance. It probably works for all passive circuits with some adaptations, but i can't remember any reputable source to confirm it \$\endgroup\$
    – bracco23
    Jan 10 at 13:49
  • \$\begingroup\$ @bracco23 I believe it is possible to reduce do that for circuits that include capacitors and inductors as well... Its possible to calculate the equivalent resistance/reactivity for any circuit made up of the basic building blocks (R, L, C) that points A/B see... and from A/B point of view a circuit with the equivalent resistance/reactivity has the same behavior as the original circuit. (assuming a perfect world with no induced currents). \$\endgroup\$
    – Questor
    Jan 11 at 19:24
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Each of the three resistors in question in the top right image have their right ends connected together and their left ends connected together, they are all in parallel. All they've done is draw it in a way meant to make it less obvious than it could be. As long as two resistors have each end of one connected to one and only one of the ends of the other resistor they are in parallel.

This kind of problem is common in electronics courses, they want you to be able to figure out what connects to what without it being obvious.

Here I've colored the connections so it's a bit easier to see how they are all in parallel.

enter image description here

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    \$\begingroup\$ the "red" node is not completely colored, as it should be \$\endgroup\$
    – jsotola
    Jan 8 at 16:09
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    \$\begingroup\$ @jsotola I just wanted to highlight the connections between the resistors. \$\endgroup\$
    – GodJihyo
    Jan 8 at 17:00
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Another rearrangement.

Using @GodJihyo's colour scheme we can rearrange the resistors as shown in Figure 1. Note that all the connections remain the same but in the final diagram the parallelism is clear.

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schematic

While the above reasoning leads to a correct solution, I struggle to grasp why the step from upper-right to lower-left is justified. In particular, why are 2R/3 and 2R in parallel?

In circuit schematics, wires are not physical components. Instead, they only make one node to be the same as another node, similar to an equal sign in an algebraic expression. Thus, wires play no roles in lumped circuit analysis, and often, ignoring or removing them can improve the readability of a circuit schematics.

Questions like these are what made net labels and "ground" symbols so useful in circuit schematics - which are heavily used when drawing circuits in CAD software. It allows one to ignore the "ratsnest". Instead of thinking about wires, it uses nodes directly to reason about a circuit. For simple "planar" circuits found in practical applications, this style usually reduces most circuits to a very regular shape and allows electronics designers to immediately understand them by inspection.

For example, a Wheatstone bridge is usually drawn in physics and introduction circuit analysis textbooks using the following style in a diamond shape. A physicist would say that this representation is preferred due to its intuitiveness.

Wheatstone bridge

Meanwhile the "CAD" style is the following. Although many blame this style to early software limitation that only allowed 90-degree rotations, but also pay attention to how it uses a ground symbol to represent the common connection implicitly, rather than explicitly drawing out of the connection between two wires. This way, one can immediately pattern-match it to two resistor dividers without any thinking just by inspection. Due to the absence of the wires, it's also easy to think about the circuit in terms of two resistor dividers in isolation, rather than a complete network. Copy-paste repeated structures also becomes convenient. Thus, in CAD the second style is common and most CAD operators find it's more intuitive.

schematic

However, they also have some disadvantages. First, some circuits can become visually unfamiliar when ground wires are omitted because every textbook draws them with ground wires, such as a Wheatstone bridge, a bridge rectifier, or a multi-stage filters. In this case, it's better to explicitly drawing out the ground wires to match the convention used by most people (so please do draw bridges as diamond when software permits). Furthermore, using many ground nodes in the same place can sometimes be too repetitive and visually tedious. So when to use ground wires or ground node involves an "artistic" choice. Furthermore, in a real physical circuit, a ground wire, bus or plane does exist, and it has non-zero parasitic resistance or impedance, drawing the ground wires explicitly helps people to remember its existence. Menawhile, in schematics, all the ground wires are often completely omitted, everything just returns to ground via a magic node. Thus, noises of noise and interference become difficult to see.

Nevertheless, they still don't undermine the usefulness of labeling nodes.


Back to the question:

schematic

While the above reasoning leads to a correct solution, I struggle to grasp why the step from upper-right to lower-left is justified. In particular, why are 2R/3 and 2R in parallel?

To justify this transformation:

  1. Note that the left side of the topmost resistor R is connected to ground, so the left side must be a ground node no matter what. Thus, we remove the redundant wire and replace it with a ground symbol.

schematic

simulate this circuit – Schematic created using CircuitLab

  1. Note that the resistor R/2 at the bottom right is connected to ground. Thus, one can remove the redundant wire and replace it with a ground symbol.

schematic

  1. Note that the resistor 2R/3 is connected to ground, so the wire to ground can be replaced by a ground symbol.

schematic

  1. At this point, the answer is essentially already complete: when you see the resistors 2R/3 and 2R are sharing the same "input" and goes to the same "ground" node, they're obviously in parallel. To keep simplifying, we can slightly move the "ground" symbol of the topmost resistor R to improve readability.

schematic

  1. To further improve readability, we label the connection between between the left side and the resistor network R and R/2 at the right side as node A.

schematic

  1. Replace all wires to node A with multiple "node A" net label. Here you have it: if we imagine A as the input of the circuit, it's extremely obvious that we basically have 4 subcircuits sharing the same input A, and each can be treated independently.

schematic

  1. Rotate the left part of the schematic by 90 degrees. Now it's almost an industry-standard schematic - if A in really a common input such as a power rail.

schematic

  1. Finally we can rejoin the schematic by replace all nodes back to wires. This is equivalent to your final circuit.

schematic

I hope I've successfully showed that by labeling "nodes" appropriately, simplifying most planar circuits is entirely a mechanical process and argubly does not even involve any thinking at all.

The situation is more difficult when the circuit is not planar, or if the nodes are connected in a more complicated manner, one can find many in the exercise problems for Y-Δ transform. This is also why analyzing an unbalanced Wheatstone bridge with an arbitrary impedance at the center is also a standard exercise problem, since it cannot be treated separately as two voltage dividers in this case.

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We can understand this in this way:

enter image description here

The nodes A,B,C are same as they are short-circuited, similarly nodes D,E,F,G are same.

Based on above analysis, the elements between C&D(R), B&E(2R/3), B&F(2R) are in parallel.

I hope that makes this clear.

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    \$\begingroup\$ Tip: when you use the editor toolbar CircuitLab button to draw your schematic you can use the Save and Insert button to insert a PNG of the schematic into the post along with an edit link. There's no need for a CircuitLab account. No screengrab. No grid. It alsow means that others can copy and paste your schematic for further editing. Welcome to EE.SE. \$\endgroup\$
    – Transistor
    Jan 10 at 10:44
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After reading the other answers I came to the conclusion that the whole diagram is just a mathematical exercise about reducing the resistors into one (result).

So the question why two resistors are in parallel can be answered with "because it's an exercise".

The total resistance \$R_\mathrm{total}\$ of parallel resistors \$R_n\$ is: $$R_\mathrm{total} = \frac1{\sum_{n=1}^N\frac1{R_n}}$$ So from answer https://electronics.stackexchange.com/a/696839/282833 the resistance between the red and the blue blob is \$R/3\$, so "red" via "blue" to "green" is \$R/3 + R\$, and that in parralel with \$R/2\$ should be \$4R/11\$ between "red" and "green" if I did the maths right.

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