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I am high school boy, our teacher gave us circuit and we have to find out the equivalent resistance. enter image description hereenter image description here.

Our professor gave us idea that we can solve this question either by either superposition principle or symmetry I do not what he want to say by this, symmetry for the same current carrying resistance. Am I right is there any false in my thinking. Till now I have only study the kvl.

Edit:) I am not asking the answer of the question, I just want to know what is this superposition, or symmetry concept.

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    \$\begingroup\$ What, exactly, is your question? Be aware, this is not a homework site. \$\endgroup\$ – Chu Oct 17 '19 at 7:14
  • \$\begingroup\$ I am not asking the answer, I actually what is superposition, and symmetry concept how do they apply. \$\endgroup\$ – Jack Rod Oct 17 '19 at 7:15
  • \$\begingroup\$ Cube of resistors \$\endgroup\$ – Andy aka Oct 17 '19 at 7:19
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Let's take a symmetrical view of your first schematic:

enter image description here

It shouldn't be very difficult for you to see that the above nodes are labeled correctly and that \$V_\text{out}\$ is accurately stated. That will be the Thevenin voltage for the circuit.

You can probably easily also see that the current flowing in from adjacent \$V\$ nodes, via \$R\$ into the \$V_x\$ node, must be the same current. The sum of those two currents must also flow through the remaining \$R\$. So it follows that the voltage drop across the \$R\$ heading back to the \$0\:\text{V}\$ labeled node must be twice the voltage drop for either of the other two \$R\$, flowing from \$V\$ to \$V_x\$. Therefore, \$V_x=\frac23 V\$ from symmetrical arguments. From this, you find that \$V_\text{out}=V_\text{TH}=V-\frac23 V=\frac13 V\$.

Now, imagine what might happen if you shorted out those external terminals:

enter image description here

From this, you can infer that the Thevenin resistance is \$R_\text{TH}=\frac{\frac13 V}{\frac{V}{R}}=\frac{R}{3}\$.

So the use of symmetrical ideas can make this pretty easy to test out.

Your second problem can be solved using KVL by forming loops and simultaneously solving the equations for the voltages after you "inject" \$1\:\text{A}\$ into the two given nodes (either direction is okay, your choice) and see how that affects those two node voltages. (Obviously, they are the same before you inject the current since there is no power source in the 2nd circuit, to start.) Or you can apply a voltage at those two nodes and again solve your KVL current loops simultaneously and then find the voltage supply's resulting current. Either works.

But, as a comment from Andy suggests, the 2nd circuit is merely a cube where each of the vertices are connected along the edges with resistor \$R\$. If you think about it that way, do you see that node \$B\$ and node \$A\$ are at diametrically opposite corners of the cube? Does this suggest to you that there is a very simple symmetry to be had, given that fact? Think about it?

(I think your teacher enjoys symmetries and has a more geometrical way of thinking, than an algebraic one. A teacher who is more algebraic wouldn't care so much to create such self-simplifying geometric puzzles and would just push your algebraic manipulation skills, instead.)

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  • \$\begingroup\$ Thank you, i got the answer. \$\endgroup\$ – Jack Rod Oct 17 '19 at 12:23
  • \$\begingroup\$ Can I fold the circuit about point 0. \$\endgroup\$ – Jack Rod Oct 17 '19 at 12:23
  • \$\begingroup\$ @yuvraj I'm not exactly sure what kind of folding you have in mind. But, yes, if you do the folds correctly, you can also get the same result. I recommend you try out your own ideas about folding and see if it works for you. Just do it. If it falls, then of course you need to modify your ideas. (But folding isn't something usually found in circuit analysis.) \$\endgroup\$ – jonk Oct 17 '19 at 13:16
  • \$\begingroup\$ Folding I mean vertical, folding such two resistance who have same potential difference, come in parallel, I want to fold the circuit about point o. \$\endgroup\$ – Jack Rod Oct 17 '19 at 13:46
  • \$\begingroup\$ @yuvrajsingh Then do so as you see it and see if you get to the same place. Don't ask. Just do it. Then see how it works out for you. If it does, you are golden. If not, you need to adjust your viewpoint. This is exactly how you expand your thinking capabilities, internal mental models, and breadth. (If someone just gives you everything, you never develop those skills as fully as you should. So try it out and tell me how it works for you.) \$\endgroup\$ – jonk Oct 18 '19 at 5:57
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  • Superposition is one of the properties of linear systems. It's easier giving you an example with only two inputs: say that input A produces output X and input B produces output Y. Then the input A+B produces the output X+Y. In your case the inputs are the voltage generators while the output is Vout (n the first example). To apply superposition you simply calculate Vout only with one voltage generator at the time while the others are short circuited. The output will be the sum of the individual Vout you've calculated.

Edit:

schematic

simulate this circuit – Schematic created using CircuitLab

This is an example of a basic circuit. Say you want to calculate the current flowing through R3. You can apply Kirchhoff's (or simply simulate the circuit) to find out that the current through R3 will be 16mA. Let's apply the superposition now: You calculate I(R3) only when one of the two voltage generators is acting, while the other is substituted with a short circuit.

schematic

simulate this circuit

In this case the current flowing through R3 is 6mA. Now you proceed substituting V1 with a short circuit and calculate I(R3).

schematic

simulate this circuit

In this case I(R3) is 6mA. Now, when V1 is in the system and V2 is not I(R3)_V1 = 10mA, while when V2 is in the system and V1 is not I(R3)_V2 = 6mA. Therefore the total current for the first circuit flowing through R3 will be I(R3) = I(R3)_V1 + I(R3)_V2 = 16mA (which is the result you'd obtain when you try to apply Kirchhoff's laws and solve the system without superposition.

  • Symmetry is a property that can be used to simplify the circuit you are analyzing (like the second in your post) and it mainly derives from symmetric properties when you represent your system with matrices. It involves a little bit of linear algebra, but as a general rule you can identify the symmetry lines (geometrical) in your system and assume that the potential along those lines is the same (you could in principle draw a short circuit along the symmetry lines since there is no potential difference). That allows you to simplify the analysis of your circuit

Best of luck with your homework ;)

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  • \$\begingroup\$ Maybe :P what does sound absurd or what is that you don't understand? \$\endgroup\$ – vinnie90 Oct 17 '19 at 11:12
  • \$\begingroup\$ If you can take example of any circuit, sound like reading something without knowing how it actually happen. \$\endgroup\$ – Jack Rod Oct 17 '19 at 11:23
  • \$\begingroup\$ I've added an example for the superposition. The symmetry has already been explained very well by jonk. Ps. for the future do not call an answer absurd when you don't understand it ;) \$\endgroup\$ – vinnie90 Oct 17 '19 at 12:42
  • \$\begingroup\$ Sorry if you feel it rude, I do not what to sound like that. \$\endgroup\$ – Jack Rod Oct 17 '19 at 12:46

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