# Calculating Rv

1. Jun 19, 2007

### SD43

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

I need to calculate the replacement resistance Rv (Ohm) of the following
R1 = 12 ohm
R2 = 8 ohm
R3 = 6 ohm
R4 = 4 ohm
R5 = 9 ohm
R6 = 12 ohm
R7 = 4 ohm

I've been trying to include a picture that I've drawn in paint but apparently a couple of lines exceeds the 100Kb line. So I'm afraid I'm going to have picture the drawing in your mind.
R1 and R2 are right next to each other so it's alright if you put them together so 12 + 8 makes 20 Ohm
Under R1 and R2 there's R3
Directly under R3 there's R4.
Next to R4 on the right side there's R5
Under R5 there's R6
Between R5 and R6 there's R7 which isn't parallel to them.
I'm sorry for the confusion, but there's just no other way.

My teacher said the answer should be 9,7 ohm with lots of variables behind the 7

2. Relevant equations

1/Rv = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5 +1/R6

3. The attempt at a solution

1/20 + 1/6 + 1/4 = 7/ 15
15/7 = 2,14

1/9 + 1/12 = 7/36
36/7 = 5,14
5,14 + 2,14 = 7,28
1/4 = 0,25
0,25 + 7,28 doesn't make 9,7

If you just use the equation 1/Rv = 1/R1 + 1/R2 ... you should get 41/45 which again doesn't make 9,77.

So after 2 days of being puzzled I decided the internet was a good place to find the right way to get to 9,7

2. Jun 19, 2007

### Staff: Mentor

Welcome to the PF. We really do need to see a diagram in order to help you on this one. Once you've drawn it in Paint, save it as a PDF, and that will reduce the size enough to post. If you don't have an easy way to save-as a PDF, you can download PrimoPDF for free.

EDIT -- Oops, I think I meant save it as a JPG file. PDF might compress it too, but JPG is the traditional compressed picture file format.

Last edited: Jun 19, 2007
3. Jun 20, 2007

### SD43

Thank you, I never knew I could save images as JPG.

Anyway I attached the file and it's oddly enough only 9,7 Kb.

#### Attached Files:

• ###### sdsdd.JPG
File size:
7 KB
Views:
42
4. Jun 20, 2007

### Staff: Mentor

Ah, that helps a lot.

You are correct in your combination of R1-R4, but then you try to combine R5 (9 Ohms) as if it were in parallel with R6, but R9 is in series with the running total resistance that you have for R1-R4, and you need to do that series addition before you have a clean parallel combination with R6. Just be careful about the combinations, and try again.

EDIT -- BTW, I also get 9.7777777... as the answer.

Last edited: Jun 20, 2007
5. Jun 20, 2007

### SD43

I'm sorry, but I keep getting 9,8 as the answer without any variables behind it. So what I do is 12 + 8 = 20
1/20 + 1/6 + 1/4 + 9 = 9,46666667
9,46666667 + 1/12 + 1/4 = 9,8

Though it's close it's still completely false. What am I doing wrong?

6. Jun 20, 2007

### nrqed

Watch out. The equivalent resistance of the three resistors in parallel will be

$(\frac{1}{20} + \frac{1}{6} + \frac{1}{4})^{-1}$

You must take the inverse before going on to add it to the 9 ohms!
No. The result of combining R1, R2,R3 , R4 and R5 will be in parallel with R6. You must calculate that before adding the result to R7 (it seems as if you wanted to treat R7 as being in parallel. It's not)

7. Jun 20, 2007

### SD43

I did what you said
1/20 + 1/4 + 1/6 ^ -1 = 2,14
2,14 + 9 = 11,14
1/11,14 + 1/12 ^ -1 = 5,77
5,77 + 4 = 9,77!

Wow thanks! So do any of you know sites where I can practice these sort of questions because I'm getting a final test for physics about this the law of Archimedes, pressure and significant numbers.

8. Jun 20, 2007

### Staff: Mentor

This website is a little unfinished, but look under the DC circuits section for some more info on resistors, etc.

http://hypertextbook.com/physics/electricity/