Effective resistance between two points problem

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Homework Help Overview

The discussion revolves around calculating the effective resistance between two points in a circuit involving multiple resistors arranged in a symmetric configuration. Participants explore the implications of symmetry in circuit analysis and the validity of removing certain resistors based on their positions in the circuit.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss using symmetry arguments to simplify the circuit analysis, questioning the removal of resistors based on current flow and voltage distribution. There are inquiries about the voltage at specific points in the circuit and the implications of connecting nodes together.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some suggest alternative approaches, such as the Delta-Star transformation, while others express uncertainty about the validity of symmetry arguments. There is a recognition of the complexity involved in determining the effective resistance and the need for careful consideration of the circuit's configuration.

Contextual Notes

Participants note the presence of two Wheatstone bridges in the circuit and discuss whether certain resistors can be ignored during analysis. There is an acknowledgment of the challenges posed by the circuit's symmetry and the need for precise values of current to fully understand the voltage distribution.

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Homework Statement


resistance1.jpg


Homework Equations

The Attempt at a Solution



I will call the topmost node T , middle node M , bottommost left node L and bottommost right node Q .

I tried to use symmetry arguments . If I connect a battery across AB , i1 flows through branch AT , i2 through AM , i3 through AL . From symmetry , i1 flows through branch TB , i2 through MB , i3 through QB .

The resistors between T And M , L and M , Q and M can be removed .

Now we have three parallel branches 2R||2R||3R between A and B . Their effective resistance is 3R/4 .

But this is not an option . Could someone help me with the problem .
 

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Jahnavi said:

Homework Statement


View attachment 223607

Homework Equations

The Attempt at a Solution



I will call the topmost node T , middle node M , bottommost left node L and bottommost right node Q .

I tried to use symmetry arguments . If I connect a battery across AB , i1 flows through branch AT , i2 through AM , i3 through AL . From symmetry , i1 flows through branch TB , i2 through MB , i3 through QB .

The resistors between T And M , L and M , Q and M can be removed .

Now we have three parallel branches 2R||2R||3R between A and B . Their effective resistance is 3R/4 .

But this is not an option . Could someone help me with the problem .
Can you really delete the resistors LM and QM? What is the voltage halfway between L and Q? At M? At T? Could you connect those points together without affecting the current flow?
 
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Jahnavi said:
L and M , Q and M can be removed .
What is your argument for this?
 
tnich said:
Can you really delete the resistors LM and QM? What is the voltage halfway between L and Q? At M? At T? Could you connect those points together without affecting the current flow?

I understand my reasoning is wrong . Could you please explain why the symmetry argument in the OP is wrong . I am not good with these types of problems .
 
Orodruin said:
What is your argument for this?

I think by symmetry whatever current flows in branch AL same current flows in branch QB . Similarly whatever current flows in branch AM flows in branch MB .
 
Jahnavi said:
I understand my reasoning is wrong . Could you please explain why the symmetry argument in the OP is wrong . I am not good with these types of problems .
If the voltage at L is not the same as the voltage at M, then current must flow between those two points, so the resistance on that branch will contribute to the total resistance.
 
tnich said:
If the voltage at L is not the same as the voltage at M, then current must flow between those two points, so the resistance on that branch will contribute to the total resistance.
Try the approach I suggested in #2. It simplifies the problem into simple series and parallel circuits.
 
tnich said:
What is the voltage halfway between L and Q?
There is a resistor between L and Q . What do you mean by voltage halfway between L and Q ?

tnich said:
Try the approach I suggested in #2. It simplifies the problem into simple series and parallel circuits.

Sorry . I don't understand your approach . Please elaborate .
 
Jahnavi said:
There is a resistor between L and Q . What do you mean by voltage halfway between L and Q ?
Sorry . I don't understand your approach . Please elaborate .
Replace the resistor between L and Q with two resistors in series, each with resistance R/2. What is the voltage at the point between the two resistors?
 
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  • #10
tnich said:
Replace the resistor between L and Q with two resistors in series, each with resistance R/2. What is the voltage at the point between the two resistors?

Half of the voltage between L and Q
 
  • #11
Jahnavi said:
Half of the volatge between L and Q
You said you wanted to connect a battery across A and B. Let the voltage at A be 0 and the voltage at B be ##V_B##. What is the voltage at T, M and halfway between L and Q?
 
  • #12
Jahnavi said:
There is a resistor between L and Q . What do you mean by voltage halfway between L and Q ?
Sorry . I don't understand your approach . Please elaborate .
I converted the delta formed by nodes LMQ into star, and I am getting one of the options.
You could try that later after completing the analysis using tnich's method.
 
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  • #13
tnich said:
What is the voltage at T, M and halfway between L and Q?

This cannot be determined until precise values of current flowing through the resistors are known .

Could you briefly explain the strategy you are applying .
 
  • #14
Jahnavi said:
This cannot be determined until precise values of current flowing through the resistors are known .
Yes it can. Look at the symmetry of the circuit.
 
  • #15
cnh1995 said:
I converted the delta formed by nodes LMT into star, and I am getting one of the options.

OK . Is Delta Star method the first thing that struck in your mind on seeing this question OR is it the more easier approach ?
 
  • #16
tnich said:
Yes it can. Look at the symmetry of the circuit.
See post #2 for a brief explanation of the strategy I am applying.
 
  • #17
Jahnavi said:
Is Delta Star method the first thing that struck in your mind on seeing this question OR is it the more easier approach .
For me, it is easier as I am not so good at using symmetry arguments. And yes, seeing FIVE deltas at once, that's the first thing that struck my mind.
 
  • #18
Jahnavi said:
OK . Is Delta Star method the first thing that struck in your mind on seeing this question OR is it the more easier approach ?
BTW, I edited that post a bit. The delta I converted is LMQ and not LMT.
 
  • #19
cnh1995 said:
For me, it is easier as I am not so good at using symmetry arguments. And yes, seeing FIVE deltas at once, that's the first thing that struck my mind.

OK .

Do you understand what tnich is trying to convey to me ? I am looking quite dumb at this moment following him/her :smile:
 
  • #20
cnh1995 said:
BTW, I edited that post a bit. The delta I converted is LMQ and not LMT.

Would converting one Delta LMQ solve our purpose ? Don't we need to convert another Delta AMT or BMT ?
 
  • #21
Jahnavi said:
OK .

Do you understand what tnich is trying to convey to me ? I am looking quite dumb at this moment following him/her :smile:
The circuit is symmetric - the left side is the mirror image of the right side. So if the voltage at A is 0 and the voltage at B is ##V_B## and you draw a line down the middle of the circuit, what does the voltage have to be along that line?
 
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  • #22
tnich said:
The circuit is symmetric - the left side is the mirror image of the right side. So if the voltage at A is 0 and the voltage at B is ##V_B## and you draw a line down the middle of the circuit, what does the voltage have to be along that line?
Look at it another way. If you knew the resistance of the left side of the circuit was x, what would the resistance of the right side of the circuit be?
 
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  • #23
Jahnavi said:
Don't we need to convert another Delta AMT or BMT ?
No.
Hint: Look for balanced wheatstone bridge.
 
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  • #24
tnich said:
The circuit is symmetric - the left side is the mirror image of the right side. So if the voltage at A is 0 and the voltage at B is ##V_B## and you draw a line down the middle of the circuit, what does the voltage have to be along that line?

Is it 8R/11 i.e option a) ?
 
  • #25
Jahnavi said:
Is it 8R/11 i.e option a) ?
Yes.
 
  • #26
cnh1995 said:
No.
Hint: Look for balanced wheatstone bridge.

There are two Wheatstone bridges . Can we ignore the bottom one while analysing the top one or vica versa ?
 
  • #27
Jahnavi said:
There are two Wheatstone bridges . Can we ignore the bottom one while analysing the top one or vica versa ?
You can consider them simultaneously. The two bridge-resistors R and R/3 will be redundant.
 
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  • #28
tnich said:
The circuit is symmetric - the left side is the mirror image of the right side. So if the voltage at A is 0 and the voltage at B is ##V_B## and you draw a line down the middle of the circuit, what does the voltage have to be along that line?

Thank you very much @tnich .

I think I understand why I was wrong in removing resistors between LM and QM .

Even though I have got the right answer , I am still doubtful about why we could remove resistor between T and M .

The thing which is troubling me is that - The middle line we have drawn is coinciding with the resistor between T and M .
 
  • #29
Jahnavi said:
Thank you very much @tnich .

I think I understand why I was wrong in removing resistors between LM and QM .

Even though I have got the right answer , I am still doubtful about why we could remove resistor between T and M .

The thing which is troubling me is that - The middle line we have drawn is coinciding with the resistor between T and M .
Since you have solved the problem (and I assume that you used the delta star transformation to do that), I will stop hinting around and explain what I meant. Substitute two R/2 resistors in series for the resistor between L and Q. Call the point between those two resistors K. If we draw a line through T, M and K, the part of the circuit on the right is the mirror image of the part on the left. That means that the voltage drop from B to T, M or K must be the same as the voltage drop from T, M or K to A. So if the voltage at A is 0 and voltage at B is ##V_B##, then the voltage at T, M and K must be ##V_B/2##. Since T, M and K are all at the same voltage, you can combine them into one point without changing the current flow through any of the resistors. When you do that you have a circuit that you can solve using just the series and parallel rules. Also, now the resistor between T and M has both ends connected to the same point (because T and M are the same point), so no current flows through it and it does not contribute anything to the effective resistance between A and B.
 
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  • #30
This is exactly how I got the answer first time in post#24 :smile:

tnich said:
and I assume that you used the delta star transformation to do that)

No . I had actually employed the strategy as suggested by you when I wrote the answer in post#24 . After getting the right answer , then I tried applying Star Delta approach .

Your approach is quite admirable !

Please help me understand one thing i.e how could the middle line drawn ( to exploit symmetry ) coincide with the resistor between T and M .

I am still unsure how we could remove resistor between T and M .

Please reflect a little more on this symmetry aspect .
 

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