Find the equivalent resistance between AB

[arutor]

Homework Statement

Find the equivalent resistance between AB.

Homework Equations

The Attempt at a Solution

I could not find any Wheatstone bridge or series parallel combination. Should I use star delta transformation ? If yes, how exactly should I use it ?

 

[berkeman]

Homework Statement

https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/77/77761-73b9806372fec877fb82679f18f05f16.jpg Find the equivalent resistance between AB.

Homework Equations

The Attempt at a Solution

I could not find any Wheatstone bridge or series parallel combination. Should I use star delta transformation ? If yes, how exactly should I use it ?

Yoiks! There might be a simpler way, but I'm not seeing it right off the bat.

Instead, I'd just put a 1V voltage source between points A & B, and write all the KCL equations for the nodes in the circuit. It will be a lot of equations, but just solve them simultaneously, and then solve for the overall current flowing in the 1V source. That will get you to the equivalent resistance.

BTW, are all the resistors of the same value?

 

[SammyS]

There is some symmetry to this circuit. In the configuration that it's present, you can see a little of that symmetry.

I suggest labeling the resistors (or the nodes if you prefer).

Redraw (and redraw and redraw...) emphasizing triangles formed by the resistors, until you get a very symmetric looking array.

Make maximum use of the symmetry

In the end, I found it handy to split that center resistor into two resistors, R/2, in series.

You can perform many Y-Δ transformations combining a few resistors each time. (After drawing many Δs I finally saw that the original circuit has lots of symmetry.)

 

[arutor]

Yoiks! There might be a simpler way, but I'm not seeing it right off the bat.

Instead, I'd just put a 1V voltage source between points A & B, and write all the KCL equations for the nodes in the circuit. It will be a lot of equations, but just solve them simultaneously, and then solve for the overall current flowing in the 1V source. That will get you to the equivalent resistance.

BTW, are all the resistors of the same value?

Yes, the resistors are of same value.

 

[arutor]

There is some symmetry to this circuit. In the configuration that it's present, you can see a little of that symmetry.

I suggest labeling the resistors (or the nodes if you prefer).

Redraw (and redraw and redraw...) emphasizing triangles formed by the resistors, until you get a very symmetric looking array.

Make maximum use of the symmetry

In the end, I found it handy to split that center resistor into two resistors, R/2, in series.

You can perform many Y-Δ transformations combining a few resistors each time. (After drawing many Δs I finally saw that the original circuit has lots of symmetry.)

How to use symmetry to simplify circuit ?

 

[SammyS]

How to use symmetry to simplify circuit ?

I made some suggestions. Have you tried them?

 

[NascentOxygen]

I could not find any Wheatstone bridge or series parallel combination. Should I use star delta transformation ? If yes, how exactly should I use it ?

Solving by symmetry is a valuable technique to master. You may not be taught it, and just have to pick it up by practice. There are probably many ways to solve your particular problem, I'll start you off with the one I used.

You can see symmetry here by noting how a rotation of 180 degrees around the red central dot causes each element and each wire to superimpose on one identical to itself in an identical location, as though each has a twin. This symmetry means that each element carries an identical current and voltage as its twin.

To help you I have identified one set of twins by colouring them green. For convenience, I'll assume V_B is at zero volts.

Suppose there is a voltage drop of V₁ across each green resistor, then symmetry let's us label the voltage at two nodes, V₁ and V_A-V₁. I also did the same for nodes around another set of twins I coloured purple.

That's the hard part finished with! You will notice how we now have every node labelled with its voltage.

Now all that's needed are two equations in two unknowns to eliminate V₁ and V₂. See whether you can apply Kirchoffs Current Law to do this, and finally obtain the circuit's resistance. (= V_A / current).

Does your textbook give the answer...so that I can check my working?

 

[cnh1995]

Does your textbook give the answer...so that I can check my working?

I ran a simulation and got R_eq=0.714R

 

[NascentOxygen]

I ran a simulation and got R_eq=0.714R

[emoji106] That's my answer, too. Thanks.

 

[SammyS]

I ran a simulation and got R_eq=0.714RView attachment 94935

Using symmetry, lots of symmetry, I get (5/7)R .

 

[cnh1995]

Using symmetry, lots of symmetry, I get (5/7)R .

That's same as 0.714R. Cool! I'll try working it out by symmetry.

 

[SammyS]

That's same as 0.714R. Cool! I'll try working it out by symmetry.

(So you are pursuing this thread. Good.)Here is a start.

Resistors are numbered. The six nodes are circled and marked A through F, using the original A & B .

I suggest placing nodes A, E, F, and B roughly spaced equally in a line. Place nodes C & D on opposite sides of that line on its perpendicular bisector.
Connect the nodes with the resistors.

       D

A   E     F    B

       C

 

[SammyS]

Here is the circuit element drawn showing more of the symmetry.

Resistors and nodes as in the previous post.

 

[NascentOxygen]

Using a bit of symmetry, and a Y-∆ transformation followed by a ∆-Y transformation, I again find resistance to be 5/7 Ω. (Though I wouldn't expect anyone to memorize these transforms for unbalanced networks.)

 

[SammyS]

Using a bit of symmetry, and a Y-∆ transformation followed by a ∆-Y transformation, I again find resistance to be 5/7 Ω. (Though I wouldn't expect anyone to memorize these transforms for unbalanced networks.)

From Post #3 ("that center resistor" being the #6 resistor.):

In the end, I found it handy to split that center resistor into two resistors, R/2, in series.

Call the node between these two resistors Node G.

Then nodes C, D, & G can be collapsed into one node. The whole mess can be dealt with using only series/parallel analysis.

 

[NascentOxygen]

Call the node between these two resistors Node G.

Then nodes C, D, & G can be collapsed into one node. The whole mess can be dealt with using only series/parallel analysis.

Now I can see how you did it. Yes, that would be the easiest. It's always good to have two methods, so you can confirm the answer to the first.