I have to simplify the following circuit by using equivalent resistance. I know it probably should be very simple to do it, but for some reason, I can’t tell if the two resistors in the “internal” mesh are in series or in parallel with the other resistor R, in the external mesh. circuit


I was able to simplify the circuit only to this point, by using the fact that:

  • There is a short circuit between node a and node b
  • The two resistors in the internal mesh are equivalent to one resistor of resistance 2R, since they are in series

enter image description here

My question is: are these two in series or in parallel? I would say in parallel, but the fact that there is just one node confuses me, I think.

  • 2
    \$\begingroup\$ Do you see the direct short between a and b? \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2023 at 9:53
  • \$\begingroup\$ @Andyaka Isn’t it what I did in my guess? Or maybe I didn’t understand what you mean \$\endgroup\$
    – selenio34
    Commented Mar 21, 2023 at 9:57
  • 2
    \$\begingroup\$ Are you sure that you have reproduced the circuit accurately? It make little sense to label nodes "a" and "b" which are obviously directly shorted, (it is all the same node.) Just seems odd even for a trick question. \$\endgroup\$
    – Nedd
    Commented Mar 21, 2023 at 10:23
  • 1
    \$\begingroup\$ I mean that the equivalent resistance will be zero irrespective of any other components that appear to be connected between a and b. @selenio34 \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2023 at 10:24
  • 1
    \$\begingroup\$ Since there is a wire between points A and B, creating a short circuit or Zero Ohms, there is no need to do any further calculation - whatever else may be connected between A and B will have no effect. \$\endgroup\$ Commented Mar 21, 2023 at 15:41

3 Answers 3


You did well. You can imagine that the dot in parallel with 2R is a wire. In other words they are shorted.

However, normally, an equivalent resistance is usually taken between 2 nodes (sometimes, one of them is ground). In your network, after getting rid of the 2R, I still see a loop of wire in parallel with R. In that case, I'd say the total R of the network is 0.

  • \$\begingroup\$ Ah okay, now I see it, thanks for the explanation. So in this case, “unifying” the two nodes a-b that are shorted doesn’t really help me with solving the exercise, instead it confused me, visually speaking \$\endgroup\$
    – selenio34
    Commented Mar 21, 2023 at 10:08
  • \$\begingroup\$ @selenio34 looks that way. The impedance is normally taken across two nodes. If you delete that loop, then you're just left with a shorted resistor. \$\endgroup\$
    – Designalog
    Commented Mar 21, 2023 at 11:55

I'll try to follow a straightforward approach, directly from the initial schematic.

Between a and b there are three paths in parallel:

  • the wire connection at the top,
  • the two R resistors in the "internal mesh",
  • the one R resistor in the "external mesh".

(If you have difficulty seeing that fact, it helps to "pull" at the two points, a to the left and b to the right, treat the wires as rubber bands, and watch how the elements re-arrange.)

The resistances on these three paths are:

  • 0
  • 2R
  • R

Any parallel circuit where one path has zero resistance gives zero resistance, so the result is zero.


I have to simplify the following circuit by using equivalent resistance.

If the only things you're interested in happen "between a and b", then the question and the suggested approach are misleading at best, and nonsense at worst.

Nodes in circuit analysis are abstract things. The connected lines don't mark wires - they mark nodes. So, if letters "a" and "b" are on the same connected line, they denote the same node, and by definition a node has no (zero) resistance to itself.

The entire line with letters "a" and "b" on it is just one node. The letters are a distraction.

Whatever is connected to this node doesn't matter at all. There could be no circuit at all, or an arbitrarily complex circuit. From the point of view of circuit analysis - and that's what you're doing here - there's one node of interest only.

Trying to apply some reasoning that would treat "a" and "b" nodes as being separate may be an interesting thought exercise but is pointless until they actually become separate, not just two names for one node!

So, to apply any sort of equivalent resistance analysis, there must be two nodes to have resistance between. How can we do that? We must split the "ab" node into two nodes "a" and "b", and add a "shunt" resistance between them. Like this:


simulate this circuit – Schematic created using CircuitLab

Now we can actually use equivalent resistance, because there are two nodes!

The circuit reduces to \$R_s\$ we just added, in parallel with "the rest of the circuit", i.e. \$\frac{2}{3}R\$. The resistance between "a" and "b" $$ R_{ab} = \left(\frac{2}{3}R \right)||R_s = \frac{(2/3)R\cdot R_s}{(2/3)R + R_s} = \frac{R\cdot R_s}{R + (3/2)R_s}.$$ We can now ask "what happens as \$R_s\$ goes to zero?" $$ \lim_{R_s\to 0} \frac{R\cdot R_s}{R + (3/2)R_s} = \frac{R\cdot 0}{R+0} = 1\cdot0 = 0.$$

You can see that the impedance of the entire circuit beyond \$R_s\$ doesn't matter - it can be whatever, can be complex, imaginary, even nonlinear (as long as it's finite).

So, we don't even need to figure out what the equivalent impedance of the rest of the circuit is. We could just call it \$R_x\$, and see if we need to know it. We don't: the math is the same, the result is independent of \$R_x\$: $$ \lim_{R_s\to0}R_{ab}=\lim_{R_s\to0} R_x||R_s = \lim_{R_s\to0} \frac{R_x\cdot R_s}{R_x+R_s} = \frac{R_x\cdot0}{R_x+0} = 0 \text{ iff } R_x\ne0. $$

The simple substitution limit-taking method doesn't apply when \$R_x=0\$. De l'Hospital's rule to the rescue: $$ \lim_{R_s\to0}R_{ab} = \lim_{R_s\to0} \frac{R_x\cdot R_s}{R_x+R_s} \overset{\rm DH}{=} \lim_{R_s\to0} \frac{\frac{d}{dR_s} (0\cdot R_s)} {\frac{d}{dR_s} (0+R_s)} = \lim_{R_s\to0} \frac{0}{1} = 0.$$


simulate this circuit


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