Effective Resistance between 2 points

[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 R_eff can be calculated is appreciated.

 

[Chestermiller]

You need to consider the symmetry.

 

[Abysmaltan]

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. @.@

 

[cnh1995]

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.

 

[NascentOxygen]

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.

 

[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.

Thank you for your reply!
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]

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]

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 anyone circuit and get the result.

 

[Abysmaltan]

So you can do this..

I will still get 2 unbalanced brigdes (I just read it up), right?

 

[cnh1995]

I will still get 2 unbalanced brigdes (I just read it up), right?

You'll get two balanced bridges.

 

[Abysmaltan]

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]

Oh, I see. So I just need to calculate accordingly, then.

Yeah. Only 3 steps are required...Good luck!

 

[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 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. :(

 

[cnh1995]

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.

 

[Abysmaltan]

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]

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. :(

If you are asking about the parallel connection part:
When you separate the two circuits along the line PQ, their ends are still connected between P and Q.

 

[Abysmaltan]

If you are asking about the parallel connection part:
When you separate the two circuits along the line PQ, their ends are still connected between P and Q.

OOOHHHH, understood! Thank you!

 

[cnh1995]

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]

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.

 

[Chestermiller]

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.

 

[Edmundcarlov]

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 R_eff can be calculated is appreciated.

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

 

[Chestermiller]

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.

 

[Abysmaltan]

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!

 

[Abysmaltan]

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!

 

[NascentOxygen]

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.