0
\$\begingroup\$

Two circuits are equivalent if you provide a certain voltage to the terminals the current through the terminals will be the same and and vice versa; i.e. if you let flow a certain current through the terminals, the voltage across the terminals will be the same for both circuits.

Now, I have these two circuits:

enter image description here

I solve the second circuit and I find the voltage on R23, then usually in exercises I use that voltage in first circuit to compute current in R2 and R3 in first circuit. My question is: why the voltage in the second circuit can be used for the first (I want a rigorous proof)? Is there a theorem concerning equivalent circuits (in this case R23 is equivalent of R2, R3 in parallel) embedded in other circuits?

Important: I gave that picture only for example, I want a proof that it is true in general. I want to prove that if I have two equivalent circuits A, B respect to a pair of terminals and I plug them to another circuit C so we have C+A and C+B, plug using the terminal in which A and B are equivalent, then the voltage on A is equals on B in both the circuits on the terminals.

In substance starting from my definition of equivalent circuit I would like prove that: Given A and B equivalent in my definition we have that the voltage across the terminals and current flowing through terminals will be the same, when placed in the same context.

Update To give an example for proof I would like to see, we can take the substitution theorem which similar but not equal to my question (as you can see this theorem seems obvious and has a rigorous proof)

enter image description here

enter image description here

Taken from Circuit Theory T. S. K. V. Iyer

I have insert this update only to say that my question seems obvious like substitution theorem and as you can see substitution theorem has an detailed proof and I think that my question maybe should have similiar proof in which are involved considerations on KCL, KVL and v-i equations. In my question KCL, KVL and v-i equations are not exactly equals in C+A and C+B (like in substitutions theorem) but equals in C and more we have that A is equivalent to B (in sense of my definition).

\$\endgroup\$
  • \$\begingroup\$ Voltages in parallel are the same but the current won't be. If you're solving the voltage across \$R_2\$ and \$R_3\$ then combining them won't alter your results. The proof is rather conceptual but if you want to verify it, you can apply Kirchhoff's Current Law. You'll find that the current across the resistors are different but when you apply Ohm's Law, the voltage will be the same. \$\endgroup\$ – KingDuken Nov 18 '19 at 15:35
  • \$\begingroup\$ I gave that picture only for example, I want a proof that it is true in general. \$\endgroup\$ – asv Nov 18 '19 at 15:40
  • 1
    \$\begingroup\$ Is KCL not sufficient proof? If you want "rigorous proof" then set up an experiment. \$\endgroup\$ – KingDuken Nov 18 '19 at 15:43
  • \$\begingroup\$ I made an update \$\endgroup\$ – asv Nov 18 '19 at 15:56
  • 2
    \$\begingroup\$ HEY WAIT A MINUTE... This question is starting to look like your previous question. Why are you asking the same thing? Was the answer that you accepted not clear? Don't accept answers unless you fully understand what they're talking about. If it didn't make sense to you, then you should ask the person who wrote that answer. If my assumption is wrong, then can you tell us why this question is different from your previous question? \$\endgroup\$ – KingDuken Nov 18 '19 at 16:49
0
\$\begingroup\$

Well comments are enough to answer your query already but still I'll put it in the words you want using your network A, Network B terminology.

By your definition of equivalency you're just stating that, loosely speaking equivalent resistance of A (say Ra) and B(say Rb) is same i. e Ra=Rb(then & then only your definition would be satisfied) (technically by your definition of equivalency you're saying that input impedance of both networks are equal).

Now since Ra=Rb and resistance of your network C, Say Rc is fixed, applying voltage divider rule, of course and always,

V12= Rb/(Rb+Rc) = Ra/(Ra+Rc) = V1=V2.

I hope you understand how you're stating what's quite obvious.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ But I want to prove it in general, the picture is only an example. I want to prova a theorem that I can apply. \$\endgroup\$ – asv Nov 19 '19 at 12:03
  • \$\begingroup\$ I want to prove that if I have two equivalent circuits A, B respect to a pair of terminals and I plug them to another circuit C so we have C+A and C+B, plug using the terminal in which A and B are equivalent, then the voltage on A is equals on B in both the circuits on the terminals. \$\endgroup\$ – asv Nov 19 '19 at 12:04
  • \$\begingroup\$ oh God that's what I wrote above, by Ra and Rb I meant input resistances of any (well any linear passive) network A and B. Network A and B can be anything as far as the elements within are linear and passive and what I wrote above would still hold true. It is perfectly general for any linear passive network, by any means have you studied 2 port parameters for network analysis? \$\endgroup\$ – Deep Nov 19 '19 at 17:37
  • \$\begingroup\$ If you think I still don't understand your point, make a clear comment stating what in above proof you find that is not general? Why do you think it is not general case? Any counter-examples? \$\endgroup\$ – Deep Nov 19 '19 at 17:40
  • \$\begingroup\$ It is general not by magic but the by very way we have defined term "equivalent resistances", the rest is just Ohm's law. \$\endgroup\$ – Deep Nov 19 '19 at 17:42
0
\$\begingroup\$

I don't know what a "rigorous proof" would be, but consider that you combined R2 and R3 because they are parallel. By definition, elements in parallel have the same voltage across them, and as you observe the equivalent resistor must also have the same voltage across it. So, if you find the voltage across R23 then you have also found the voltage across R2 and R3. This is because of how the terms parallel and equivalent are defined in this context.

Based on OP's edits: When you combine elements in parallel the voltage across them is the same by definition, and is the same across the equivalent resistor that replaces them, but the current through them may be different. When you combine elements in series the voltage across them may be different but the current through them is the same by definition, and is the same through the equivalent resistor that replaces them.

This is inherent in the definition of parallel and series. I don't know how they would be proved more rigorously.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ I gave that picture only for example, I want a proof that it is true in general. \$\endgroup\$ – asv Nov 18 '19 at 15:41
  • 1
    \$\begingroup\$ Well all we have is your example, and my explanation can be generalized to any two elements in parallel. You haven't told us formally what you mean by "it". What exactly, precisely do you want to prove? \$\endgroup\$ – Elliot Alderson Nov 18 '19 at 15:48
  • \$\begingroup\$ I made an update \$\endgroup\$ – asv Nov 18 '19 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.