# Effective resistance between A and B

## Homework Statement

Calculate the effective resistance betwen A and B!

## The Attempt at a Solution

1/ER=(1/2R)+(1/5R)+(1/R)
ER=10R/17

#### Attachments

• Diagram.JPG
3.6 KB · Views: 695

cepheid
Staff Emeritus
Gold Member
There are a couple of shorts in that diagram. So...what's the path of *least* resistance between A and B?

Are you familar with transformations with resistor/impedance networks? This looks to be be one of those.

I recommend drawing the circuit another way.

There are a couple of shorts in that diagram. So...what's the path of *least* resistance between A and B?
The path with least resistance would be "R"!

Are you familar with transformations with resistor/impedance networks? This looks to be be one of those.

I recommend drawing the circuit another way.

No, I am not familiar with that. Yes, I can resolve the diagram into 3 parallel resistances of R, 2R and 5R. And then I calculated the effective resistance. Is my answer correct?

I was thinking star-delta transformations, but I've just moved them around again and the circuit is simpler than I though ... doh!

Where do you get 5R from?

I was thinking star-delta transformations, but I've just moved them around again and the circuit is simpler than I though ... doh!

Where do you get 5R from?

I got the 5R from the path if the current flows straight from A to B without using any of the short circuits. Am i right?

Try moving the left resistor above the middle resistor, and the right resistor at the bottom. Keep the wires connected as they are, and see what the circuit looks like.

cepheid
Staff Emeritus
Gold Member
Although I didn't follow those exact directions, I did redraw the circuit using my own methods and arrived at the result that I think you were hinting at, Delta, which is that the three resistors are in parallel. The only thing that confuses me is that if you draw them this way, the topology really seems to downplay the fact that it is possible for current to travel from A to B through all three resistors, a fact that I usually don't consider when looking at resistors in parallel, whereas the topology in the OP's diagram really emphasizes this point. That's where he got 5R from.

EDIT: Nevermind. I realize what the mistake is. If there is a current, then the electric potential at A is higher than at B (or vice versa), which means the path through all three resistors is not valid, because the current would be flowing from a lower potential to a higher one if it went through the middle resistor after having gone through the leftmost one. This is much much clearer in the redrawn topology, in which you can actually see that the current would have to "double back on itself" by going down one branch from A to B and then back up another from B to A, in order to take this 5R path. Therefore, the 5R path is not valid, and there are only three parallel routes (one through each resistor).

i cant understand what you all mean to say. Will you please be so good to draw the simplified diagram in paint for me? Please.

cepheid
Staff Emeritus
Gold Member
i cant understand what you all mean to say. Will you please be so good to draw the simplified diagram in paint for me? Please.

Have you tried drawing it out yourself? It's definitely worth learning how to do this. The good part is that it's simple. Start at node A, and trace out all possible paths from node A to node B. For example, you can go through the leftmost resistor having 2R resistance, and that will take you from A to B. That's one path. What other paths are there?

Have you tried drawing it out yourself? It's definitely worth learning how to do this. The good part is that it's simple. Start at node A, and trace out all possible paths from node A to node B. For example, you can go through the leftmost resistor having 2R resistance, and that will take you from A to B. That's one path. What other paths are there?

yes, i tried it. from A to B:
1. through 2R and then through short circuit to B
2. through short circuit and the via R to B
3. via all three resistances without using any short circuits

cepheid
Staff Emeritus
Gold Member
yes, i tried it. from A to B:
1. through 2R and then through short circuit to B
2. through short circuit and the via R to B

Good! Can you see a third path that passes through ONLY the middle resistor? 3. via all three resistances without using any short circuits

I'm afraid this path is not valid. Here is a very important point in circuit analysis: if two points are connected directly by ideal wires (i.e. they are shorted together), then those two points are electrically connected and can be considered to be the same point. What I mean by this is that the two points and the entire wire connecting them are all at the same potential (there is no voltage between them). As a result, from the point of view of circuit analysis, they can be considered a single "node". I have modified your diagram to illustrate this:

http://img74.imageshack.us/my.php?image=resistorsif6.jpg

Everything circled in red is at the same potential as point B, and therefore is, in a sense, electrical point B. Likewise for everything circled in blue being A electrically. As a result, can you see that if current passes through the leftmost 2R resistor, it has already travelled from A to B in the sense of having lost potential equivalent to the voltage between A and B?

Now for the point all of this was building up to: if current goes through the left resistor, and then the middle one, then it will have gone from A, to B, and then back to A again! Clearly this won't happen in practice: either A or B will be at a higher potential if current is to flow. As a result, the path passing through all three resistors is not valid, for it goes from A to B to A to B. This is what I was trying to explain in my edit to post #8, but redrawing the circuit makes this point really clear.

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cepheid
Staff Emeritus
Gold Member
Since you've clearly been racking your brains for some time now, and have put a fair amount of thought into the quesiton, I thought I'd help you out. This post will show you how you can redraw the circuit, step by step, just by considering all possible paths from A to B. So, current can either pass from A, through the left 2R resistor, and continue to B along the wire, or it can pass through the other wire, and then through the rightmost R resistor to B. That allows us to draw the following so far:

http://img76.imageshack.us/my.php?image=resistorsqt7.png

So, that takes care of the leftmost and rightmost resistors. What about the middle one? We can add it to our diagram by carefully noting where it starts and ends. From the argument in my previous post, it starts at B and ends at A.

Another way of looking at it: Once you've gone through the left resistor, there are two choices: 1. go straight to B through the wire. 2. Go back to A through the middle resistor (which could never actually happen). So clearly we need to add a junction on the far side of the left resistor. This junction is the starting point for the middle resistor, like so:

http://img64.imageshack.us/my.php?image=resistors2dl8.png

However, it's pretty clear that the diagram above is equivalent to the one below (and I figured out how to do thumbnails):

http://img64.imageshack.us/img64/6791/resistors3if9.th.png [Broken]

Redrawing the circuit in this way makes two things very clear that were hard to see in your original picture:

1. There is a current path from A to B that passes ONLY through the middle resistor.

2. The current path through all three resistors is NOT VALID, because it would entail current going from A to B, then back to A, then back to B.

Last edited by a moderator:
Good! Can you see a third path that passes through ONLY the middle resistor? I'm afraid this path is not valid. Here is a very important point in circuit analysis: if two points are connected directly by ideal wires (i.e. they are shorted together), then those two points are electrically connected and can be considered to be the same point. What I mean by this is that the two points and the entire wire connecting them are all at the same potential (there is no voltage between them). As a result, from the point of view of circuit analysis, they can be considered a single "node". I have modified your diagram to illustrate this:

http://img74.imageshack.us/my.php?image=resistorsif6.jpg

Everything circled in red is at the same potential as point B, and therefore is, in a sense, electrical point B. Likewise for everything circled in blue being A electrically. As a result, can you see that if current passes through the leftmost 2R resistor, it has already travelled from A to B in the sense of having lost potential equivalent to the voltage between A and B?

Now for the point all of this was building up to: if current goes through the left resistor, and then the middle one, then it will have gone from A, to B, and then back to A again! Clearly this won't happen in practice: either A or B will be at a higher potential if current is to flow. As a result, the path passing through all three resistors is not valid, for it goes from A to B to A to B. This is what I was trying to explain in my edit to post #8, but redrawing the circuit makes this point really clear.

Since you've clearly been racking your brains for some time now, and have put a fair amount of thought into the quesiton, I thought I'd help you out. This post will show you how you can redraw the circuit, step by step, just by considering all possible paths from A to B. So, current can either pass from A, through the left 2R resistor, and continue to B along the wire, or it can pass through the other wire, and then through the rightmost R resistor to B. That allows us to draw the following so far:

http://img76.imageshack.us/my.php?image=resistorsqt7.png

So, that takes care of the leftmost and rightmost resistors. What about the middle one? We can add it to our diagram by carefully noting where it starts and ends. From the argument in my previous post, it starts at B and ends at A.

Another way of looking at it: Once you've gone through the left resistor, there are two choices: 1. go straight to B through the wire. 2. Go back to A through the middle resistor (which could never actually happen). So clearly we need to add a junction on the far side of the left resistor. This junction is the starting point for the middle resistor, like so:

http://img64.imageshack.us/my.php?image=resistors2dl8.png

However, it's pretty clear that the diagram above is equivalent to the one below (and I figured out how to do thumbnails):

http://img64.imageshack.us/img64/6791/resistors3if9.th.png [Broken]

Redrawing the circuit in this way makes two things very clear that were hard to see in your original picture:

1. There is a current path from A to B that passes ONLY through the middle resistor.

2. The current path through all three resistors is NOT VALID, because it would entail current going from A to B, then back to A, then back to B.

Thanks a lot sir for giving your precious time to that. I had been trying to solve this for over a week now. Thanks once again for the help. Physics really sems interesting!!!!!!!!!

Last edited by a moderator: