Calculating Equivalent Resistance in a Hexagonal Prism Circuit

[zorro]

Homework Statement

All the wires on the front and the back hexagonal face (of the Skelton hexagonal prism) have resistance R. All the wires along the lines joining the vertices of two hexagons have resistance 2R. The equivalent resistance between P and Q as shown in the figure is

The Attempt at a Solution

To me the wires joining the two faces of hexagon are redundant as the current through them is 0 (balanced wheatstone bridges). So the resistance between P and Q is 3R/2. But this is not the correct answer.

 

[SammyS]

Homework Statement

All the wires on the front and the back hexagonal face (of the Skelton hexagonal prism) have resistance R. All the wires along the lines joining the vertices of two hexagons have resistance 2R. The equivalent resistance between P and Q as shown in the figure is

The Attempt at a Solution

To me the wires joining the two faces of hexagon are redundant as the current through them is 0 (balanced Wheatstone bridges). So the resistance between P and Q is 3R/2. But this is not the correct answer.

This is not correct. If you have a potential difference between P & Q, current will flow to and from the back face.

Draw the equivalent circuit.

There is quite a bit of symmetry in this circuit. That usually aids in analyzing these kinds of circuits.

It will help to label all of the junctions in some systematic way.

Also, are you familiar with Y - Δ transform methods for circuits?

I get (73/36)R . - but it's too late at night for me to be too confident with that.

 

[zorro]

This is not correct. If you have a potential difference between P & Q, current will flow to and from the back face.

But the resistances are proportional (R/R = R/R, 2R is redundant) - which implies a balanced wheatstone bridge.

Also, are you familiar with Y - Δ transform methods for circuits?

Yeah, I am familiar with it but we are not supposed to use it.

I get (73/36)R . - but it's too late at night for me to be too confident with that.

Its wrong.

 

[Metaleer]

Ah, interesting problem. :)

I find that by connecting a 1 A current source between P and Q, solving the circuit using a systematic method such as nodal analysis, and then remembering Thévenin's theorem (that the equivalent resistance the current source "sees" can be derived from V = IR, V being the voltage drop in the current source and I being its current, 1 A), you can solve for the equivalent resistance R.

 

[zorro]

Can you explain why there is no balanced wheatstone bridge?

 

[cupid.callin]

I have to say ... its a different question ... i was bored of those cubes ! LOL

This is my attempt:

just extended branches a little :tongue:

I just simply used symmetry !

And i have numbered the currents .. and green branches have no current !

So net resistance becomes ... oops .. can't write answers ... or i'll have another Infractions ...

oops ... but you can figure it out anyway ...

please some expert tell me if I'm right

anyway here's my answer :
(Point on it to see the answer ... but please do it yourself first before checking my answer ...)

Spoiler

... 21R/20 ...

 

[cupid.callin]

Also, are you familiar with Y - Δ transform methods for circuits?

Where Can we apply Y - Δ transformation ... where will it be useful

If you are talking about all the Y junctions then won't it be too long ?

 

[SammyS]

I have to say ... its a different question ... i was bored of those cubes ! LOL

This is my attempt:
...

Spoiler

... 21R/20 ...

I got 23R/20 - after some sleep.

I'll go through it again after a while.

@ A-Q:

Yes, in some sense the front face (hexagon) is like a balanced Wheatstone bridge -- but more complicated. That's how I & (I presume cupid.c) used symmetry to simplify the circuit. That in no way means that no current passes through the 2R wires.

If you change the problem by replacing the hexagons with squares, then the front square (on which P & Q lie), IS indeed like a balanced Wheatstone bridge. With this configuration you will definitely have current passing through the 2R wires going to the rear square. I may try posting a sketch of an equivalent circuit for this configuration.

 

[SammyS]

Where Can we apply Y - Δ transformation ... where will it be useful

If you are talking about all the Y junctions then won't it be too long ?

After simplifying the circuit using symmetry, I did have a delta at each end, joined by two wires, each of R. I changed each delta (They were identical) to a Y which gives a pretty simple series/parallel circuit.

Added id Edit:

Looking at your circuit diagram, I see that you don't have arrows on the
.green wires. There is no reason that current won't pass through them. They are the reason that I used the Y-Delta transform.

 

[zorro]

look at my diagram:

The ratios R/R and R/R are equal. They form a balanced wheatstone bridge, which means that there is no current through the 2R resistance. We can apply the same to all the corresponding branches and remove them one by one. I still don't understand where is my mistake.

 

[SammyS]

You have 6 wires labeled with R & 2 labeled with 2R. Which are you taking the ratio of, and where do you connect the supply voltage?

 

[zorro]

This might be more clear. Supply voltage is connected across PQ.

 

[cupid.callin]

I got 23R/20

You might be correct ... i thought all the wires have resistance R

I'll be back in minutes with my answer

 

[cupid.callin]

Ok so my answer is 33R/28

is it correct Abdul ?

 

[zorro]

Ok so my answer is 33R/28

is it correct Abdul ?

Sadly, I don't have any access to the correct answer now. So I have no idea :((

 

[cupid.callin]

When you get to know the answer .. do post it here ...

 

[SammyS]

This might be more clear. Supply voltage is connected across PQ.

It looks like your Wheatstone bridge has the 2R wire where you would normally place the galvanometer. The problem with this is that the legs of the bridge (2 pairs of R wires) are not connected directly together. The legs are connected by 2R wires (unlabeled here). These extra wires put the bridge WAY out of balance.Yes, for your double hexagon circuit, the supply voltage is across P & Q -- that was very clear. There are lots of other resistors & connections between P & Q and what you want to use as a Wheatstone bridge.

 

[SammyS]

I have labeled the junctions on the Original Image, and merged a copy of cupid's image.

By symmetry, A & E are at the same potential, so they can be treated as if the are connected by 0Ω wires, i.e. they form a node. The same is true for the pairs, B & F, C & G, and D & H.

Notice, that b/c A & B are not at the same potential, current will flow from A to D to C to B, as well as directly from A to B.

 

[cupid.callin]

I have labeled the junctions on the Original Image, and merged a copy of cupid's image.

By symmetry, A & E are at the same potential, so they can be treated as if the are connected by 0Ω wires, i.e. they form a node. The same is true for the pairs, B & F, C & G, and D & H.

Notice, that b/c A & B are not at the same potential, current will flow from A to D to C to B, as well as directly from A to B.

But if you imagine A&E to be connected by 0Ω wire the current can flow from A to E or vice versa.

currents don't flow b/w two same potential connected points if there is some resistance connected in between.

* I'm not talking about this case but in general taking this assumption will not be correct

 

[zorro]

It looks like your Wheatstone bridge has the 2R wire where you would normally place the galvanometer. The problem with this is that the legs of the bridge (2 pairs of R wires) are not connected directly together. The legs are connected by 2R wires (unlabeled here). These extra wires put the bridge WAY out of balance.

Ok, this is how I proceed to solve the problem-

If you push the hexagon surface containing P and Q to form the outer hexagon in 2-D, you get the following figure-

I connected same potential points with a red line. In the 2nd figure, the 5R/2 resistance is redundant as it forms the 'galvanometer' of the balanced wheat stone bridge. Simplifying the circuit further, I got 17R/12 as the equivalent resistance. Why different?

 

[SammyS]

But if you imagine A&E to be connected by 0Ω wire the current can flow from A to E or vice versa.

currents don't flow b/w two same potential connected points if there is some resistance connected in between.

* I'm not talking about this case but in general taking this assumption will not be correct

Points A & E are at the same potential because of symmetry. If they are at the same potential, then connecting them with a perfect conductor will NOT change how current flows, but it will make the problem much easier to solve. This is one of the main features of solving a circuit problem by use of symmetry.

 

[SammyS]

Ok, this is how I proceed to solve the problem-

If you push the hexagon surface containing P and Q to form the outer hexagon in 2-D, you get the following figure-

I connected same potential points with a red line. In the 2nd figure, the 5R/2 resistance is redundant as it forms the 'galvanometer' of the balanced wheat stone bridge. Simplifying the circuit further, I got 17R/12 as the equivalent resistance. Why different?

Where did the 5R/2 wires go ? (2nd to 3rd sketches)

 

[zorro]

I wrote the reason below the image-

In the 2nd figure, the 5R/2 resistance is redundant as it forms the 'galvanometer' of the balanced wheat stone bridge

 

[cupid.callin]

Points A & E are at the same potential because of symmetry. If they are at the same potential, then connecting them with a perfect conductor will NOT change how current flows, but it will make the problem much easier to solve. This is one of the main features of solving a circuit problem by use of symmetry.

But then won't the current flow through least resistant path (conductor in this case)

This is how short-circuiting is done

 

[SammyS]

But then won't the current flow through least resistant path (conductor in this case)

This is how short-circuiting is done

Cupid is correct here!

That's a balanced Wheatstone bridge connected "sideways". You can't eliminate the 5R/2 by this method.

 

[gneill]

The nested hexagon diagram elucidates the symmetries in the circuit that can be exploited. If you take that diagram and draw a line through PQ, the portion of the circuit above the line is completely symmetrical with the portion below. That means that all "mirror image" currents and voltages are equal. This allows us to commit the following act of violence upon the circuit.

If you "fold" the circuit along line PQ so that all the corresponding points of the upper half join in parallel with their lower half, then no new currents should flow (because all corresponding points have the same potential). The net result is that we can dispense with half the circuit provided that we divide all the "folded" resistances in half (note that components lying along the fold line are not "folded", but remain the same). The resulting simpler circuit is quite amenable to further simplification.

 

[cupid.callin]

The nested hexagon diagram elucidates the symmetries in the circuit that can be exploited. If you take that diagram and draw a line through PQ, the portion of the circuit above the line is completely symmetrical with the portion below. That means that all "mirror image" currents and voltages are equal. This allows us to commit the following act of violence upon the circuit.

If you "fold" the circuit along line PQ so that all the corresponding points of the upper half join in parallel with their lower half, then no new currents should flow (because all corresponding points have the same potential). The net result is that we can dispense with half the circuit provided that we divide all the "folded" resistances in half (note that components lying along the fold line are not "folded", but remain the same). The resulting simpler circuit is quite amenable to further simplification.

What answer do you get by this method

 

[zorro]

That's a balanced Wheatstone bridge connected "sideways". You can't eliminate the 5R/2 by this method.

Does a sideways connection matter? Now I have to make a list of precautions for using Wheatstone bridge principle.

If you "fold" the circuit along line PQ so that all the corresponding points of the upper half join in parallel with their lower half,

Care to show your idea through a diagram?

 

[cupid.callin]

Care to show your idea through a diagram?

He's also using symmetry we all have been talking about ... but just in more step by step method ...

my teacher also taught me this method ... but i like the traditional way

 

[gneill]

What answer do you get by this method

The correct one

I'm not sure it's the right time yet to give away the answer.

Care to show your idea through a diagram?

Sure. I've only labeled some of the resistances, the rest should be obvious by color code and symmetry.

 

[cupid.callin]

The correct one
I'm not sure it's the right time yet to give away the answer.

I did the same you did ...

So your answer is 33R/28 ?

 

[gneill]

I did the same you did ...

So your answer is 33R/28 ?

Nope. A bit smaller.

 

[cupid.callin]

smaller than 1.7R ?

Here's how i solved ... Bold black lines are the resistances

 

[gneill]

What happened to the resistances ab, cd, a'b', c'd' ?

 

[cupid.callin]

there won't be any current through them right? so we can just remove them. Is that wrong ?