# Effective Resistance between 2 points

Given that the resistance of each resistor is 1.0 ohm, find the effective resistance between P and Q.
I do not know how to simplify the circuit.
A little help on how to redraw the circuit such that the Reff can be calculated is appreciated.

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Chestermiller
Mentor
You need to consider the symmetry.

You need to consider the symmetry.
May I know a little bit more about that? I always have difficulties whenever I encounter questions that need me to redraw the circuit. I have been stuck in this question for 1 day and I still do not get it. @[email protected]

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cnh1995
Homework Helper
Gold Member
May I know a little bit more about that? I always have difficulties whenever I encounter questions that need me to redraw the circuit. I have been stuck in this question for 1 day and I still do not get it. @[email protected]
In this question, you can see that the circuit is symmetric about the line joining P and Q. You can replace the two horizontal 1 ohm resistors with parallel combinations of two 2 ohm resistors and then cut the circuit along the line PQ such that each part contains a 2 ohm resistor. This way, you'll have two exactly identical circuits which are connected in parallel in the original circuit. You can analyze any one circuit and get the result.

Abysmaltan
NascentOxygen
Staff Emeritus
Hi Abysmaltan.

It's always wise to look at the foot of this page to see what earlier threads identified as similar sounding might be of assistance. On rare occasions you might even discover your very problem has been solved!

The technique of symmetry can be used here. Imagine P=+10V and Q=-10V and try to identify nodes in the circuit where the potential might be 0V.

To confirm this, now imagine the voltage sources to be reversed so that P=-10V and Q=+10V and see whether the nodes you think should be 0V are those same nodes you identified in the step above.

Once you have identified points of equal potential, you can simplify the drawing by showing those points connected together using a thick piece of wire to form a single node because you have shown they are all already at the same potential. (Connecting up points of identical potentials changes nothing because with the ends of each added wire being at the same potential then no current is going to flow through that added wire, anyway.)

You then determine the resistance of this much-simplified circuit.

To gain a good understanding of this requires that you work through practice exercises such as this homework problem.

Chestermiller, Abysmaltan and CWatters
Hi Abysmaltan.

It's always wise to look at the foot of this page to see what earlier threads identified as similar sounding might be of assistance. On rare occasions you might even discover your very problem has been solved!

The technique of symmetry can be used here. Imagine P=+10V and Q=-10V and try to identify nodes in the circuit where the potential might be 0V.

To confirm this, now imagine the voltage sources to be reversed so that P=-10V and Q=+10V and see whether the nodes you think should be 0V are those same nodes you identified in the step above.

Once you have identified points of equal potential, you can simplify the drawing by showing those points connected together using a thick piece of wire to form a single node because you have shown they are all already at the same potential. (Connecting up points of identical potentials changes nothing because with the ends of each added wire being at the same potential then no current is going to flow through that added wire, anyway.)

You then determine the resistance of this much-simplified circuit.

To gain a good understanding of this requires that you work through practice exercises.
However, sadly, I still do not quite understand the idea of using symmetry to solve complex circuits.
Regarding the first part of the explanation (imagine P=+10V and Q=-10V then...), I do not know how to identify the nodes where the potential might be 0V. (I am still a beginner in circuits T___T)
Also, you mentioned working through exercises. Unfortunately, I cannot find any similar question in my book, and I am not sure where to find on the Internet. Do you know of any websites that will allow me to understand more about solving circuit using symmetry and that have practice questions.

cnh1995
Homework Helper
Gold Member
Can you do series-parallel reductions? Have you studied Wheatstone's bridge?

Abysmaltan
Can you do series-parallel reductions? Have you studied Wheatstone's bridge?
I can do series-parallel reduction, but have not studied Wheatstone's Bridge.

cnh1995
Homework Helper
Gold Member
I can do series-parallel reduction,
So you can do this..
In this question, you can see that the circuit is symmetric about the line joining P and Q. You can replace the two horizontal 1 ohm resistors with parallel combinations of two 2 ohm resistors and then cut the circuit along the line PQ such that each part contains a 2 ohm resistor. This way, you'll have two exactly identical circuits which are connected in parallel in the original circuit. You can analyze any one circuit and get the result.

So you can do this..
I will still get 2 unbalanced brigdes (I just read it up), right?

cnh1995
Homework Helper
Gold Member
I will still get 2 unbalanced brigdes (I just read it up), right?
You'll get two balanced bridges.

You'll get two balanced bridges.
Oh, I see. So I just need to calculate accordingly, then.
I am pretty sure there are other methods of doing it, like what @NascentOxygen suggested (Just that I do not understand). Thanks for your responses, though!

cnh1995
Homework Helper
Gold Member
Oh, I see. So I just need to calculate accordingly, then.
Yeah. Only 3 steps are required...Good luck!

NascentOxygen and Abysmaltan
In this question, you can see that the circuit is symmetric about the line joining P and Q. You can replace the two horizontal 1 ohm resistors with parallel combinations of two 2 ohm resistors and then cut the circuit along the line PQ such that each part contains a 2 ohm resistor. This way, you'll have two exactly identical circuits which are connected in parallel in the original circuit. You can analyze any one circuit and get the result.
How do I know that the two exactly identical circuits are connected in parallel in the original circuit? Is it linked to the concept of symmetry?
Sorry, haha, I am an inquisitive but obtuse high-schooler, who is struggling to survive for physics. :(

cnh1995
Homework Helper
Gold Member
How do I know that the two exactly identical circuits are connected in parallel in the original circuit? Is it linked to the concept of symmetry?
You can see the two configurations. They'll be exactly identical. The circuit is symmetrical about the PQ axis.

You can see the two configurations. They'll be exactly identical. The circuit is symmetrical about the PQ axis.
So the idea is that when a circuit is symmetrical about an axis, the two identical circuits, separated by line of symmetry, are in parallel?

cnh1995
Homework Helper
Gold Member
How do I know that the two exactly identical circuits are connected in parallel in the original circuit? Is it linked to the concept of symmetry?
Sorry, haha, I am an inquisitive but obtuse high-schooler, who is struggling to survive for physics. :(
When you separate the two circuits along the line PQ, their ends are still connected between P and Q.

Abysmaltan
When you separate the two circuits along the line PQ, their ends are still connected between P and Q.
OOOHHHH, understood!!! Thank you!

cnh1995
Homework Helper
Gold Member
So the idea is that when a circuit is symmetrical about an axis, the two identical circuits, separated by line of symmetry, are in parallel?
Not always. Here, the line joining the points between which the effective resistance is asked is the useful line of symmetry. If the line of symmetry coincides with the line joining the two points between which the effective resistance is asked, then the two circuits are in parallel. The circuit is symmetrical about vertical axis too. But that will not give two parallel circuits between P and Q.

Abysmaltan
Not always. Here, the line joining the points between which the effective resistance is asked is the useful line of symmetry. If the line of symmetry coincides with the line joining the two points between which the effective resistance is asked, then the two circuits are in parallel. The circuit is symmetrical about vertical axis too. But that will not give two parallel circuits between P and Q.
OK. Understood. Thanks a lot! :)

NascentOxygen
Staff Emeritus
However, sadly, I still do not quite understand the idea of using symmetry to solve complex circuits.
Regarding the first part of the explanation (imagine P=+10V and Q=-10V then...), I do not know how to identify the nodes where the potential might be 0V.
For the approach I outlined, all nodes along the vertical axis of symmetry here will be at 0V when P and Q have equal but opposite polarity voltages. Points of symmetry are those which are electrically "equally distant" from P as from Q, so their potential is midway between -10V and +10V.

Abysmaltan
Chestermiller
Mentor
For the approach I outlined, all nodes along the vertical axis of symmetry here will be at 0V when P and Q have equal but opposite polarity voltages. Points of symmetry are those which are electrically "equally distant" from P as from Q, so their potential is midway between -10V and +10V.
And that means that there is no current flow through those two resistors right in the middle, because the voltage difference across each of them is zero. This means that, if they were removed from the circuit, it would have no effect on the equivalent resistance.

Abysmaltan
Given that the resistance of each resistor is 1.0 ohm, find the effective resistance between P and Q.
I do not know how to simplify the circuit.
A little help on how to redraw the circuit such that the Reff can be calculated is appreciated.

You just need to recall ohm law..............!

Chestermiller
Mentor
You just need to recall ohm law..............!
Please tell us specifically how you personally would apply ohm'(s) law to obtain a rapid and simple solution to this problem.

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For the approach I outlined, all nodes along the vertical axis of symmetry here will be at 0V when P and Q have equal but opposite polarity voltages. Points of symmetry are those which are electrically "equally distant" from P as from Q, so their potential is midway between -10V and +10V.
Understood! Thank you for your explanation!