# Determine resistance

1. Nov 24, 2005

### phyznut

Given the following circuit

How do you determine the total resistance?
The book tells me the answer is R2 || (R1 + R3). How would I analyze the circuit to get that answer? I really cant see how R2 is in parallel with the rest. To me it looks like R2 is in parallel with a short.

Last edited: Nov 24, 2005
2. Nov 24, 2005

### mezarashi

Um, where are the nodes in which you are taking your equivalent resistance? Across the R2 resistor?

3. Nov 24, 2005

### phyznut

yes. I'm taking it across R2. Why did you ask that question? Does that affect something?

4. Nov 24, 2005

### Gokul43201

Staff Emeritus
Phyznut: Make sure you post the complete question, EXACTLY as it appears in your homework/text. As it is, there's a few things wrong with the problem.

Edit : Okay, that answers the main difficulty.

5. Nov 24, 2005

### mezarashi

Well if you take it between the R3 resistor, it'd be a totally different story no?

In anycase, to help visualize your case, do this.
You see the piece of wire on the left separating the R1 and R3? Continually shrink it until the two nodes of R1 and R3 touch.

6. Nov 24, 2005

### Gokul43201

Staff Emeritus
1. Keep in mind that the actual shape formed by the wires means absolutely nothing.

2. Whenever, there's a length of resistanceless wire (like the entire stretch between R1 and R3), you can reduce this to as tiny a length as you want without affecting anything (the resistance between those two points is still zero).

3. So, from 1 and 2 above, you find that you can redraw the circuit with r2 and r3 essentially right next to each other. If it makes things easier, you can replace the upper, left and lower segments of the rectangle with a semicircular arc, and place R2, R3 right beside each other, anywhere along this arc.

Does that make it more clear then ?

Edit : mez had roughly the same idea (I was writing before his post was up).

7. Nov 24, 2005

### phyznut

So when I shrink it, it looks like the two of them are in series. And R2 being parallel to them, yes?

IF I were taking it across the R3 resistor, shrinking the wire between R1 and R2 shows that they are in series, and R3 is in series also?

8. Nov 24, 2005

### phyznut

Lets say that it didnt. Is there another method of breaking the problem down?

9. Nov 24, 2005

### Gokul43201

Staff Emeritus
This is perfectly correct. But if you want an alternative approach, here's the more correct way to think about things :

1. First identify the two points across which you are to find the effective resistance. Without naming a pair of points, it makes no sense to ask for an effective resistance.

2. Imagine you are now going to drive a current from one point to the other. Follow the path of the current as it splits up or recombines along the various paths and branches that it must take.

3. Whenever, the current is made to split up, the two - or more - different branches (defined as the two - or more - complete paths that each of the two - or more - portions of the current take until they recombine to give you back the same original current) are said to be in parallel. The effective resistance in the two - or more - branches are in parallel with each other.

4. If any path has more than one resistor in it, but there is no branching (or splitting up of currents), these resistors are said to be in series. What this means is that the current traveling through the first first of these resistors would travel through all the rest of them.

5. Using 3 and 4 you can identify series and parallel paths and appropriately use the math to calculate the effective resistance.

Last edited: Nov 24, 2005
10. Nov 24, 2005

### phyznut

Umm..so if the current splits at point A, then recombnines at point B, all the resisotrs between A and B are parallel then?
But in that case (taking my original problem) Here is the circuit I get after shrinking.

Even if I didnt shrink it, the current looks like it only took one path. Am I missing something?

11. Nov 24, 2005

### mezarashi

You are misunderstanding the concept of node...

Your circuit simplification is correct. Let me ask you one more question then. Where does your voltage source go? Add the voltage source.

12. Nov 24, 2005

### phyznut

Since I'm finding resistance, I took the voltage source out. (Thevenin)
Misunderstanding? Please correct me in that case.

13. Nov 24, 2005

### mezarashi

The concept of equivalent resistance is "the equivalent resistance that the voltage or current source sees". What does it see? Where is it?

You say that "the current looks like it only took one path". Where is the current coming from? Prove it me it is only taking one path. Without input/output nodes for the voltage/current source to go in and exit, it doesn't work. Hint: You told me earlier that the input/output nodes are across R2.

14. Nov 24, 2005

### phyznut

It could very well come from a ammeter couldnt it? But thats not the case with this. Ok then, so the nodes are across R2. Current leaves from one end and enetrs at the opposite end. But for it to get there, it must first travel through the loop. I'm looking at the loop and I can see only one path. I dont see where the current splits up and rejoins.

Sorry for being annoying, but I fail to see the same thing(s) you see.

15. Nov 24, 2005

### mezarashi

I'll give it one more try without an aid of a diagram.

Okay, so you agree that there will be current entering from the top of resistor R2. How exactly does this current get to the bottom side of R2 (where the exit is)?

16. Nov 24, 2005

### phyznut

Well it must first go across resistors 1 and 3 before reaching its destination. But in doing so, the current takes the following path (in red):

I traced only one path. Are you guys counting it as a different path when the current "turns" at each of the four cornes of the rectangle?

This is my concept of more that one path (I added an extra loop)

Last edited: Nov 24, 2005
17. Nov 24, 2005

### mezarashi

The current wants to go to the bottom of R2. The current can split when it is at the top of R2. It sees a path going through R2 and also a path going through R1+R3. Hmmm, can think of what would satisfy your condition of resistors being in parallel then?

18. Nov 24, 2005

### phyznut

I think I'm beginning to understand a little now. DId I illustrate it correctly. If we call the point to the right of R1 A, and the bottom of R2 B. The top of R2 C. The current could flow from C through A to B. It could also go from C directly to B right?

Since the current through R1 R3 would be the same then they are in series. Since R2 has a different current, it must be in parallel, yes?

Last edited: Nov 24, 2005
19. Nov 24, 2005

### mezarashi

Yes, you are starting to pick up on the node concept. If you consider the point above R2 to be node A and the point below R2 to be node B. Looking from A-B, you will see the equivalent resistance R2//(R1+R3). When you 'look' into a circuit for an equivalent resistance, there must be two points (in and out).

Suppose we combine R1+R3 into one resistor, R4. Then you have R2 and R4. Does this convince you definitely that they are in parallel (R2//R4). If not, I refer to my old question, what situation would?

20. Nov 24, 2005

### phyznut

Here is how I would analyze it. Since R4 would have a different resistance, be it higer or lower than that of R2. They must obviously be in parallel because both of them would have different amounts of current through them.

If R4 = R2, then they would have the same current. But the current could arrive at its destination in two ways. Either going through R4 to reach its destination or by going directly across R2. SO I guess the answer to your question is the amount of paths the current has to take?