# Effective Resistance between 2 points

• Abysmaltan
In summary, The effective resistance between P and Q can be found by using the technique of symmetry. By replacing the two horizontal 1 ohm resistors with parallel combinations of two 2 ohm resistors and cutting the circuit along the line PQ, two identical circuits are formed. Since these circuits are connected in parallel in the original circuit, the effective resistance can be calculated by analyzing either one of the identical circuits. It is important to practice exercises such as this in order to fully understand the concept of using symmetry to solve complex circuits.

#### 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.

#### Attachments

• hellooooo.jpg
25.6 KB · Views: 452
Last edited:
You need to consider the symmetry.

Chestermiller said:
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. @.@

Last edited:
Abysmaltan said:
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. @.@
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 anyone circuit and get the result.

Abysmaltan
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
NascentOxygen said:
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.

Can you do series-parallel reductions? Have you studied Wheatstone's bridge?

Abysmaltan
cnh1995 said:
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.

Abysmaltan said:
I can do series-parallel reduction,
So you can do this..
cnh1995 said:
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 anyone circuit and get the result.

cnh1995 said:
So you can do this..
I will still get 2 unbalanced brigdes (I just read it up), right?

Abysmaltan said:
I will still get 2 unbalanced brigdes (I just read it up), right?
You'll get two balanced bridges.

cnh1995 said:
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!

Abysmaltan said:
Oh, I see. So I just need to calculate accordingly, then.
Yeah. Only 3 steps are required...Good luck!

NascentOxygen and Abysmaltan
cnh1995 said:
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 anyone 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. :(

Abysmaltan said:
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.

cnh1995 said:
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?

Abysmaltan said:
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
cnh1995 said:
When you separate the two circuits along the line PQ, their ends are still connected between P and Q.
OOOHHHH, understood! Thank you!

Abysmaltan said:
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
cnh1995 said:
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! :)

Abysmaltan said:
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
NascentOxygen said:
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
Abysmaltan said:
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.....!

Edmundcarlov said:
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.

Last edited:
NascentOxygen said:
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!

Chestermiller said:
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.
OK. Understood. Thank you for your explanation!

Chestermiller said:
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.
That's right. But I didn't want to muddy the water by saying nodes could be joined, or elements could be removed, so I restricted my description to just one technique. Either way, I think there won't be much difference in the amount of work involved.

Happily, OP has now had both explained.

## 1. What is the definition of effective resistance between two points?

The effective resistance between two points is a measure of the total opposition to electric current flow between those two points in a circuit. It takes into account the resistance of all components in the circuit, including resistors, wires, and any other components.

## 2. How is effective resistance calculated?

The effective resistance between two points can be calculated using Ohm's law, which states that resistance is equal to the voltage difference between two points divided by the current flowing between those points. It can also be calculated by adding the individual resistances of all components in series or using the parallel resistance formula for components in parallel.

## 3. How does the length of a wire affect effective resistance?

The length of a wire can affect effective resistance by increasing the total resistance in a circuit. This is because longer wires have more resistance than shorter wires due to the increased distance that electric current must travel. Therefore, the longer the wire, the higher the effective resistance between two points.

## 4. What factors can affect effective resistance?

There are several factors that can affect effective resistance, including the type of material the wire is made of, the thickness or gauge of the wire, and the temperature of the wire. These factors can all impact the overall resistance of a wire and, in turn, affect the effective resistance between two points in a circuit.

## 5. How can effective resistance be reduced?

Effective resistance can be reduced by using thicker wires with lower resistance, minimizing the length of wires in a circuit, and using components with lower resistance values. Additionally, circuits can be designed with components in parallel to reduce the overall resistance between two points. Lowering the temperature of the wire can also decrease its resistance and, therefore, the effective resistance between two points.