Equivalent resistance between two points on a circuit

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

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The circuit is symmetric about AC and BD , but I am not sure how to use this to simplify the circuit .

Could someone help me in solving this circuit using symmetry considerations .

 

[haruspex]

not sure how to use this to simplify the circuit .

Consider splitting node C into two nodes.

 

[Jahnavi]

Consider splitting node C into two nodes.

Could you please elaborate .

 

[haruspex]

Could you please elaborate .

Of the current which flows from O to C, how much then flows to B? Can you replace C by two nodes, one connected to O and B, the other to O and D, in such a way that the currents to B and D do not change?

By the way, all of the given options are wrong.

 

[cnh1995]

Could you please elaborate .

You can convert the star BC-OC-DC into delta and then apply symmetry argument to the resulting circuit. It will be easier IMO.
(And yes, as haruspex said, all the options are incorrect.)

 

[Jahnavi]

all the options are incorrect

Are you getting 7/15 Ohms ?

 

[haruspex]

You can convert the star BC-OC-DC into delta and then apply symmetry argument to the resulting circuit. It will be easier IMO.
(And yes, as haruspex said, all the options are incorrect.)

Yes, I was deliberately avoiding the need to know the star/delta rules.

Edit: anyway, my way is quite easy. Split OC into OC' and OC", 2Ω each. OC'B is 3Ω. Combining with OB gives 3/4Ω. Tacking BA onto that gives 7/4Ω. Similarly on the right.
This leaves us with three parallel paths, having resistances 7/4, 1, 7/4.

 

[cnh1995]

Are you getting 7/15 Ohms ?

Yes.

 

[Jahnavi]

You can convert the star BC-OC-DC

Are you sure to convert BC-OC-DC ? Will this be sufficient ?

 

[cnh1995]

Are you sure to convert BC-OC-DC ? Will this be sufficient ?

Yes. Then apply symmetry argument for further simplification.

 

[Jahnavi]

Then apply symmetry argument

These symmetry arguments are killing me

 

[cnh1995]

These symmetry arguments are killing me

Well, this will be easier than using symmetry from the beginning (at least for me).
BTW, how did you calculate 7/15?

 

[Jahnavi]

BTW, how did you calculate 7/15?

Symmetry .

I am just shooting in the dark .

 

[Jahnavi]

Edit: anyway, my way is quite easy. Split OC into OC' and OC", 2Ω each. OC'B is 3Ω. Combining with OB gives 3/4Ω. Tacking BA onto that gives 7/4Ω. Similarly on the right.
This leaves us with three parallel paths, having resistances 7/4, 1, 7/4.

Very impressive .

Thanks !

 

[cnh1995]

I am just shooting in the dark

Looks like you nailed it!

 

[Jahnavi]

@cnh1995 please give me some time . I will try your approach and see if I can come up with the same answer .

 

[BvU]

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[edit] a appears to have a fixed size . Only place you need symmetry is from pic 2 to pic 3 (no current in that little piece of wire because both ends are at same potential)

 

[BvU]

By the way, all of the given options are wrong

Maybe teacher overlooked the currents in the upper half...

My post 17 was inspired by

The problem is solved . But I am still not very confident about exploiting symmetries and arguing which points are equipotential. If you have some insight in that problem or would like to share your reasoning as to how you would tackle that problem , please do so .

 

[Jahnavi]

@BvU ,

In the first picture in post#17 , from left to right , the first middle node is O , second middle node is B and third is D . Right ?

 

[BvU]

No. It is: B -- O -- D
O is in the very center in pic 1, 2, 3. And I pull it to the top in pic 4.



B and D stay in place all the time. C is split in C and C' (both at same potential -- because of symmetry) in pic 3 . In pic 4 C is two-thirds down the left 3 $\Omega$, C' two thirds down the right 3 $\Omega$

Like haru, I shy away from star and delta as much as I can (almost always).

Oh, and the exercise would be a lot easier if it asked for the resistance between C and A. And in that dcase, the right answer IS among the choices given... maybe teacher had a bad day ?

 

[TSny]

Just to show a somewhat different (not better) way. By symmetry it should be clear that points B and D will be at the same potential.

So you can fold the network about line AC so that triangle ABC is folded over on top of triangle ADB and let junctions B and D become the same junction. Then resistors BC and DC are in parallel, resistors BO and DO are in parallel, and resistors BA and DA are in parallel. So, the circuit reduces to

which is easy to reduce to an equivalent resistance between A and O.

 

[Jahnavi]

Just to show a somewhat different (not better) way. By symmetry it should be clear that points B and D will be at the same potential.
View attachment 223844

So you can fold the network about line AC so that triangle ABC is folded over on top of triangle ADB and let junctions B and D become the same junction. Then resistors BC and DC are in parallel, resistors BO and DO are in parallel, and resistors BA and DA are in parallel. So, the circuit reduces to

View attachment 223845

which is easy to reduce to an equivalent resistance.

I don't know what to say . It is as if you have read my mind

This is exactly what I have been thinking all the time and in fact this is how I got the answer in post#6 .

By symmetry it should be clear that points B and D will be at the same potential.

Precisely what I would like to understand . The most obvious thing to you is not clear to me .How did you conclude that B and D are equipotential ? Is there a systematic way to identify which points are equipotential while using symmetry arguments ?

 

[TSny]

I don't know what to say . It is as if you have read my mind

This is exactly what I have been thinking all the time and in fact this is how I got the answer in post#6 .

Good!

How did you conclude that B and D are equipotential ? Is there a systematic way to identify which points are equipotential while using symmetry arguments ?

In finding the equivalent resistance between A and O, I imagine using a source of emf to apply a potential difference $V$ between A and O. Current $I$ from the source enters the network at A and returns to the source at O. The equivalent resistance is the ratio $V/I$. From symmetry, the amount of current from A to B will be the same as the current from A to D. So, the potential drop from A to B will be the same as the potential drop from A to D. Thus, B and D can be assumed to be at the same potential when finding the equivalent resistance between A and O.

You would not necessarily be able to treat B and D as being at the same potential if you are finding the equivalent resistance between other points of the network.

 

[Jahnavi]

From symmetry, the amount of current from A to B will be the same as the current from A to D.

What type of symmetry ?

Is it because if we draw a line along AC , the circuit is divided in two parts with left part being the mirror image of right part and analogous nodes at the same potential ?

And if we interchange the left and right parts circuit remains same ?

 

[TSny]

What type of symmetry ?

Is it because if we draw a line along AC , the circuit is divided in two parts with left part being the mirror image of right part ?

And if we interchange the left and right parts circuit remains same ?

Yes. Flipping the network over about the line AC doesn't change the network.

 

[Jahnavi]

OK .

Possibly one last question

You would not necessarily be able to treat B and D as being at the same potential if you are finding the equivalent resistance between other points of the network.

How would you find which points are equipotential if equivalent resistance is to be found between , say A and D ?

 

[TSny]

How would you find which points are equipotential if equivalent resistance is to be found between , say A and D ?

If you imagine applying a potential difference between A and D, then symmetry can be used to see which branches of the network would have the same current. However, symmetry does not imply in this case that any of the junctions would be at the same potential. Nevertheless, the reduction in the number of independent currents due to symmetry allows the problem to be easily solved using junction and loop equations.

 

[Jahnavi]

However, symmetry does not imply in this case that any of the junctions would be at the same potential.

Thanks

I was having this misconception .

Do you get 8/15 Ohms as equivalent resistance between A and D ?

 

[TSny]

Do you get 8/15 Ohms as equivalent resistance between A and D ?

Yes, that's what I get also.

 

[haruspex]

Just to show a somewhat different (not better) way. By symmetry it should be clear that points B and D will be at the same potential.
View attachment 223844

So you can fold the network about line AC so that triangle ABC is folded over on top of triangle ADB and let junctions B and D become the same junction. Then resistors BC and DC are in parallel, resistors BO and DO are in parallel, and resistors BA and DA are in parallel. So, the circuit reduces to

View attachment 223845

which is easy to reduce to an equivalent resistance between A and O.

Yes, that is essentially the same as in post #7, but definitely a neater way to express it

 

[Jahnavi]

Yes, that's what I get also.

Thank you very much TSny .

Thank you everyone for contributing in this thread .

 

[Jahnavi]

@TSny ,

I came across a similar problem (hope I am not wrong ). Please see the attached picture . I do not have the answer key . Could you please confirm whether you get 22/35 Ω as the answer .

Thank you

 

[TSny]

@TSny ,

I came across a similar problem (hope I am not wrong ). Please see the attached picture . I do not have the answer key . Could you please confirm whether you get 22/35 Ω as the answer .

Thank you

Yes, I also get 22/35 Ω.