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I stumbled across this circuit here and I don't really know how to approach calculate the resistance measured between A and B here:

enter image description here

I had a go at it but I assumed that I could "uncross" the 2R resistances but I am not sure if I am allowed to do so and I did not get the answer from the textbook.

Any ideas?

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  • \$\begingroup\$ Is the correct answer 1.3333 ohms? \$\endgroup\$
    – Andy aka
    Mar 12, 2023 at 12:59
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    \$\begingroup\$ @Andyaka of couse not, the value of R is not given. Thus the answer can never be like "xxx ohms" \$\endgroup\$ Mar 13, 2023 at 14:31
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    \$\begingroup\$ I meant to say 1.333 R \$\endgroup\$
    – Andy aka
    Mar 13, 2023 at 16:36

3 Answers 3

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Circuit transformation

STEP 0: Original circuit folded

schematic

simulate this circuit – Schematic created using CircuitLab

STEP 1: Original circuit with potentiometers

Since the circuit is symmetrical, if we look inside the two 2k resistors (for example, by replacing them with potentiometers with K = 0.5 or replacing them with networks of 1k resistors in series), we will see that their middle points have the same potentials. In the simulation below, by the help of Live DC simulation, hover the mouse over the pot wipers to see that the midpoint voltages are equal.

schematic

simulate this circuit

STEP 2: Original circuit unfolded

So we can join the midpoints.

schematic

simulate this circuit

STEP 3: Unfolded circuit split

Then we see that we can swap the right 1k (halves of the) resistors and assemble new 2k resistors.

schematic

simulate this circuit

STEP 4: Split circuit simplified...

Or (a simpler idea) we can just swap the right ends of the resistors because they are also at the same potential. Now it is easy: We find the equivalent resistance of two resistors in parallel...

schematic

simulate this circuit

STEP 5: ...more...

... then find the equivalent resistance of three resistors in series...

schematic

simulate this circuit

STEP 6: ...and more

... and of two resistors in parallel...

schematic

simulate this circuit

STEP 7: Еquivalent total resistance found

...thus obtaining the total equivalent resistance.

schematic

simulate this circuit

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  • \$\begingroup\$ @greybeard, I found another scenario but the previous one was more beautiful:-) \$\endgroup\$ Mar 12, 2023 at 16:17
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    \$\begingroup\$ Can you shed a little light on the Step 0 to Step 1 transformation? \$\endgroup\$
    – Dancrumb
    Mar 12, 2023 at 21:22
  • \$\begingroup\$ Replace each 2k resistor with 2 1k resistors in series. Note that the voltage at the junction of both pairs of these new resistors is identical, as they have the same resistance to either end of the circuit. Connect the two identical voltages together - no current will flow. \$\endgroup\$
    – KFW
    Mar 13, 2023 at 2:07
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    \$\begingroup\$ definitely nicer than "set up matrix and invert :--)" \$\endgroup\$
    – Sascha
    Mar 13, 2023 at 21:25
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    \$\begingroup\$ the key thing is to figure out step 2 and step 3, which thanks to symmetrical, that suggests it can not work for asymmetric circuit, so I want to know how to solve asymmetric circuit look like crossed network \$\endgroup\$
    – http8086
    Mar 14, 2023 at 11:02
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Here's a hint....

Because of loading symmetries, you can convert the original diagram into the one below with the red lines on: -

enter image description here

And that makes it easier to solve.

In fact, you can make the resistors with values 2R any value you want and you'll still get this symmetry.

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This answer leverages the circuit symmetry like the others, but considers the common mode and differential mode circuits to arrive at the answer. This technique is often used in both analog and RF circuit analysis.

Generalizing the network in question as a resistance R, we can calculate the value of R by dividing the voltage across R by the current through it: \$R = V/I\$

schematic

simulate this circuit – Schematic created using CircuitLab

Because the network is linear (i.e. allows super-position), we can decompose the voltage \$V\$ and the current \$I\$ from one source into common mode (CM) and differential mode (DM) voltages and currents. The two figures below show the common mode (top) and differential mode (bottom) circuits. You can see that if you add the voltage sources from the CM and DM circuits, you will get the original circuit, i.e., \$V= V_{DM} + V_{CM} \$ and \$I = I_{DM} + I_{CM}\$.

schematic

simulate this circuit

schematic

simulate this circuit

We can see that \$I_{CM} = 0\$, since the voltage across R is zero.

Due to symmetry, the voltage in the middle of the DM circuit is zero. We can split the DM circuit into two half circuits as shown below, and get that \$R = 2\cdot (R/2)\$. We will see that finding \$R/2\$ is easy!

schematic

simulate this circuit

Below is a schematic of the original circuit, with some of the resistors broken up into two series resistors to help see the line of symmetry in the middle.

schematic

simulate this circuit

We can then create the half circuit shown below, and easily find the resistance of the half circuit.

schematic

simulate this circuit

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  • \$\begingroup\$ Interesting.... \$\endgroup\$ Mar 16, 2023 at 19:02

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