# Is equivalent resistance always lower if we add a resistor to a passive electronic circuit?

How to prove that equivalent resistance of any passive network is always lower (or equal) if we add a resistor between arbitrary two nodes?

Note that this is not necessarily a parallel circuit, 2 nodes that we connect with a resistor are not the same 2 nodes between we want to measure equivalent resistance but completely arbitrary 2 nodes in passive network.

So imagine a circuit with 4 access points: A, B, C and D. We want to measure equivalent resistance between A and B. How can you prove that Rab will be less if we add a resistor between C and D?

I tried to search the web but didn't have any success.

To clarify, I am interested in equivalent DC resistance in an arbitrary network made of resistors only. How can we prove that the resistance Rab is not higher if we connect nodes C and D with any resistor?

tl;dr: 2 port network. Prove that connecting a load will reduce input resistance. In terms of z-parameters we need a prof that $Z_{12} \cdot Z_{21} > 0$.

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Sounds like homework. Explain what you've done so far and exactly where you're stuck. It's hard to create a good answer to an open ended question. – Trygve Laugstøl Dec 13 '12 at 7:34
I don't think that you read the question correctly, it would be too hard for a homework, it is a theoretical question for any circuit (that can have a million nodes, branches and resistors inside). – Serge Dec 13 '12 at 9:56
You say "any passive network". Do you mean resistors only? What about capacitors or inductors? What about nonlinear passive elements? Are you talking about DC resistance only, or complex impedance in general? – Dave Tweed Dec 13 '12 at 12:18
DC only, so resistors only, sorry for not clarifying that earlier. Thanks for commenting. – Serge Dec 13 '12 at 12:36

To clarify, I am interested in equivalent DC resistance in an arbitrary network made of resistors only. How can we prove that the resistance Rab is not higher if we connect nodes C and D with any resistor?

I believe it is the case that to increase Rab, the added resistor must be in series with any of the other resistors thereby increasing the resistance of that branch.

But, this would create a new node in the circuit.

Since your problem requires that the resistor be placed across two existing nodes, this added resistance is in parallel with the equivalent resistance between those nodes thereby decreasing the resistance of that branch.

To see that the Rab must decrease, consider terminals A & B to be port 1 and terminals C & D to be port 2 of a two-port network.

Looking into port 1, the equivalent resistance is, in terms of the Z parameters:

$R_{ab} = z_{11} - \dfrac{z_{12}z_{21}}{z_{22}+R_L}$

where $R_L$ is the resistance of the resistor to be connected across port 2 (here the impedances are all real and positive since this two-port is a network of resistors.)

Without the added resistor, $R_{ab} = z_{11}$ since $R_L = \infty$

For $0 \leq R_L < \infty$ , $R_{ab} < z_{11}$

Actually this is not a complete prof as we don't know that z12 and z21 are >0. How can we derive that? We actually just need a prof that z21*z12 is greater or equal zero.

I quote from your problem statement: To clarify, I am interested in equivalent DC resistance in an arbitrary network made of resistors only.

Thus, we do know that all the impedance parameters, for a network of resistors only, are real and positive.

Even if all elements are resistors z12 can be real and negative! For example just change the direction of I2 and you will have new Z12 = - old Z12.

The following defines the Z parameters.

$\begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} z_{11} & z_{12} \\ z_{21} & z_{22} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix}$

If you'll stop to think about this a bit, you should see that the Z parameters are real and positive for a resistor network.

For a worked example, see this.

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Nope, that resistor (CD) is not necessarily in parallel with the Eq resistance Rab. – Serge Dec 13 '12 at 13:08
@serge, I didn't write that it was in parallel with Rab; you've misunderstood what I wrote. – Alfred Centauri Dec 13 '12 at 13:12
@serge, try thinking about it this way; if you reduce the value of any resistor in your network, is it possible that Rab increases? – Alfred Centauri Dec 13 '12 at 13:18
Yep, I can see from impedance relations (when we connect a load). Thanks. – Serge Dec 13 '12 at 14:01
I see the addition but it is not nonsense as you said. Even if all elements are resistors z12 can be real and negative! For example just change the direction of I2 and you will have new Z12 = - old Z12. – Serge Dec 14 '12 at 2:47

My informal answer wouldn't probably satisfy a teacher, but I'd simply say: Let's see each connection between nodes as some kind of a path for a current to flow in the circuit. If there is no connection, resistance is infinite and current cannot flow. If we connect some nodes, we create a path between them to let the current flow. If we add more connections, we create more or wider paths to let more current flow there. Althoug it is called "resistor" and it creates "resistance", it actually creates a path for current. The biggest resistor is no resistor (no connection between nodes) at all.

Also, resistors are similar to roads for cars. Low resistance = wide road, high resistance = narrow road. If you build a new road between any two cities, you always increase traffic throughput. (Conversely, if you would build a new road in series with an existing road, you would then lower the traffic throughput.)

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I don't need to satisfy a teacher because there are no teachers and uni days are long gone. Of course I understood intuitively why this is true but I was just wondering if there is a formal prof for that. Anyways I have my answer and formal prof is a derivation of Z-parameters when we have a load connected. – Serge Dec 13 '12 at 22:43

To complete Alfred's proof, we have this, from Wikipedia:

A network is said to be reciprocal if the voltage appearing at port 2 due to a current applied at port 1 is the same as the voltage appearing at port 1 when the same current is applied to port 2. ... In general, a network will be reciprocal if it consists entirely of linear passive components (that is, resistors, capacitors and inductors). In general, it will not be reciprocal if it contains active components such as generators.

So, because your problem calls for all passive components in the network, you have a reciprocal network. I'll assume there's an unstated assumption that the elements are also linear, or we are working in a linearized small-signal approximation, because we're talking about Z-parameters.

Therefore Z12 = Z21 because that's what reciprocal means.

Therefore Z12 and Z21 have the same sign and Z12Z21 >= 0.

Edit Before I remove this answer since I essentially re-used it in the follow-up question, here's a trivial example of a 2-port network, made entirely of resistors, with negative Z21:

Here Z21 is -100 Ohms.

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Thanks for your feedback but do you know a prof that a network will be reciprocal if it consists entirely of linear passive components? I tried to prove that but it doesn't look easy at all. And because I am new in the community can you tell me if I can see why is my question marked down? Is it that electronic community here is not that friendly? And why everyone thinks that this is a homework, I don't remember that I was getting this types of homework when I was in uni as this doesn't look easy at all. It looks to me that it requires some matrix derivations and theorems. – Serge Dec 14 '12 at 1:14
This claims to "proove" it, but don't ask me to explain it without staring at it for an hour or so: en.wikipedia.org/wiki/… – The Photon Dec 14 '12 at 1:17
As for why the downvote, I can't say. Somebody might have dv'ed because of the combination of it's a fairly tricky problem and your English is not perfect making it slightly more difficult to understand what you are asking (and somebody not being willing to take the effort to figure out what you're asking). I could certainly see this being asked as a homework in a university network theory class, so there's that aspect as well. – The Photon Dec 14 '12 at 1:21
Given the "proof" goes back to Maxwell's equations and fundamental electromagnetics, you might also try this question at physics.stackexchange.com – The Photon Dec 14 '12 at 1:23
His problem not only calls for passive components, it calls for the DC resistance of a network of resistors only. While the reciprocal nature of the transimpedances of a passive network is important and generally relevant, it's not needed for the case of a network of resistors only. – Alfred Centauri Dec 14 '12 at 2:06

This looks like homework, so you should at least derive the equations by yourself.

1. In any case, for a mathematical proof, I would start by drawing all the possible resistance combinations inside the 4-node black box: Rab, Rac, Rad, Rbc, Rbd, Rcd.

2. Now take a pair of nodes, say A-B, and calculate the equivalent resistance. This will be a big and ugly equation.

3. Now, calculate the derivative of this equation with respect to Rcd. What this tells us is the exactly how will Rab change if we change Rcd.

4. By checking the monotonicity of this derivetive, you know EXACTLY whether a negative change of Rcd will result in a negative change in Rab.

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This is not a homework, why do you think it is? You cannot derive equation because I am interested in any passive circuit that can have a million branches, resistors and nodes inside. So how to prove that for any passive circuit Rab is not bigger if we connect nodes C and D? – Serge Dec 13 '12 at 9:51
I think this is a homework because it blatantly looks like one. I've done a few of these. Usually people don't care much for mathematical proofs otherwise. Anyway, if your circuit has N=100000 nodes instead of 4, it's not much different at all. You'll just have more combinations of node-to-node resistances, silly. That's how you prove it mathematically. If you want a gut feeling, intuitive understanding, then try imagining this: HOW could the resistance increase after putting a positive resistance in parallel? Give me ONE example of a linear circuit that will exhibit this behaviour. – Jonny B Good Dec 13 '12 at 19:33
No, I don't want a gut feeling or intuitive understanding because I had that already, that is why I formed the question in that way. And what example do you need from me? Like circuit with resistor of 1 Ohm and then you add another resistor of 1 Ohm. You totally misunderstood the question if you think that it is as easy to prove this for a network with a million of nodes and a simple circuit. For a simple circuit you can just calculate everything but for an arbitrary circuit you don't have a final formula. – Serge Dec 13 '12 at 20:57
I didn't say that original circuit have only 4 nodes so I don't know why you said "instead of 4"? That is why I think that you misunderstood the question. This has nothing to do with a parallel circuit. Anyways I got my answer, prof is actually a derivation of formulas for z parameters when you add a load to the other and. I never ever had a homework like this, I was trying to solve another problem with infinite grid and was looking for lower bound which I can get by connecting some nodes. – Serge Dec 13 '12 at 21:07
If you add a resistor in series, you create a new node. Either you cannot understand your own problem statement, or your problem statement is incorrect. – Jonny B Good Dec 13 '12 at 21:27

I believe there is a relatively simple proof of this assertion. Apply a voltage, E, to terminals A and B. Measure the current, I, being drawn from this voltage. Then, by definition, the resistance between A and B is E/I. Now consider terminals C and D. Find the Thevenin equivalent voltage and resistance at these terminals with voltage E connected to terminals A and B. Denote them as Eth and Rth. If there is no added resistor at terminals C and D, then no current flows out of these terminals since the equivalent circuit consists of Eth in series with Rth and terminals C and D are open circuited. If a resistor, Rcd, is connected to terminals C and D, then the equivalent circuit now consists of Eth in series with Rth and Rcd. Then a current, Icd, will flow into resistor Rcd and equal to Eth/(Rth + Rcd). This current is in addition to the current, I, that flowed into terminals A and B without Rcd connected. So now the resistance between terminals A and B is equal to E/(I + Icd) which is less than E/I. Therefore the resistance between terminals A and B will be reduced if a resistor is added to terminals C and D. Terminals C and D could be placed anywhere within the network so the proof holds for any resistor connected to any 2 terminals of the network.

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"This current is in addition to the current, I, that flowed into terminals A and B without Rcd connected." Why? – Serge Dec 14 '12 at 6:35
@Serge The existing network will continue to draw a current I because nothing in that network has changed. Adding the resistor to terminals C and D has created a new current path that causes additional current to flow. That's why the resistance looking into terminals A and B decreases. – Barry Dec 15 '12 at 1:49

I have another way to prove it. We need to prove that every passive network with 2 ports has equivalent Pi or T configuration. After that it is easy to prove that placing a resistor on the output would reduce the input resistance.

By reducing number of nodes one by one we can reduce our network to Pi or T configuration. We can do that by y-delta transform or star-mesh transform.

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