# Equivalent resistance problem

1. Sep 14, 2008

### elcotufa

1. The problem statement, all variables and given/known data

Edit Circuit looks messed up, if you can quote me you can see the text of the circuit
This is how the circuit looks

*--------------------*
| / \
| R2 R3
R1 / \
| *------- *--------a
| \ /
| R4 R5
| \ /
*-------------------*-------------b

R1=4ohms
R2=20ohms
R3=30ohms
R4=60ohms
R5=40ohms

The answer is supposed to be Rab=9.6 ohms, but I have tried everything and cannot get anything close

Any input is greatly appreciated

2. Sep 14, 2008

### LowlyPion

What have you tried?

What are the equivalent resistances for R23 and R45?

3. Sep 14, 2008

### elcotufa

R23 and R45 are both in parallel right?

R23= (20)(30)/(50) R45(60*40)/100

I tried them in series too, what I want to understand is how to deduce the pathways for the current in circuits like this one. Diagonals seem to mess me up

Edit do I have to do a wye delta conversion?

4. Sep 14, 2008

### elcotufa

nm finally got it

The R23 is in parallel and then in series with the R1

R45 are in parallel and also in parallel with R123

Problem is it was by trial and error, if I get another circuit with diagonals I would probably miss it in a test

5. Sep 14, 2008

### elcotufa

if I had a circuit like this
Code (Text):

a------------*
/   \
R2    R3
/       \
*-------*
\      /
R4   R5
\  /
b----------*

Would R23 be in parallel, R45 be in parallel, and R23 in parallel with R45 ? or would R23 be in series with R45?

6. Sep 14, 2008

### LowlyPion

No need for it to be trial and error. The more you simplify it the easier it becomes. Just calculate equivalent resistances and work toward the answer of what's asked. I think you will be able to do it now even on a test. That's the beauty of homework.

7. Sep 14, 2008

### elcotufa

What is the approach for that other problem?

And also for this one

Sorry, but I really want to master these diagonal circuits :)
Code (Text):

|------------*
|          /   \
|         R2    R3
|        /       \
|       *--a   b-*
|        \       /
|        R4    R5
|         \   /
|----------*

Approach
R23 in series R25 in series and R45 in series are the 3 routes from a to b

I don't know how I would combine them together

INput greatly appreciated

8. Sep 14, 2008

### LowlyPion

This one is different. The short from R2 R3 to R4 R5 could as easily be drawn down the middle which reveals it to be simply R2 R4 in parallel and then in series with R3 R5 in parallel. Then calculation is simple.

9. Sep 14, 2008

### elcotufa

Thanks alot man, still not so obvious to me

10. Sep 15, 2008

### elcotufa

For this one the TA said that R2 R4 are in series and R3 R5 are in series, and R24 in parallel with R35

She explained it as R2 R4 having the same current, same as R3 R5 and drew it like this
Code (Text):

_________________
|        |       |
R2       |       R3
|___a    |       |
|        |    b__|
|        |       |
R4       |       R5
|________|_______|

It made more sense to me this way, but the book doesn't give a right answer for the different values of resistors

Input appreciated

11. Sep 15, 2008

### LowlyPion

That was how I suggested you draw it before. In that configuration, the current flowing from a to b will only be equal in R2 and R4 if their resistances are equal. For the topology you've drawn you need to calculate R2||R4 and then add to R3||R5.

Using the original values in the earlier problem,
R2=20, R4=60, R3=30, R5=40

R2||R4 looks like 15 and R3||R5 looks like 17.14 with the total equivalent as 32.14.

12. Sep 15, 2008

### elcotufa

Ahhh cool
Now I get it

TA probably thought all resistances were equal and confused me more