# Calculating Parallel Resistance

Discussion in 'General Electronics Chat' started by Dal1980, Aug 30, 2015.

1. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3
I'm working through the book "Starting Electronics 2nd Edition" by Keith Brindley.

Apart from the fact that I'm sitting with pages falling out of the book all around me (really bad binding) the math has me a little confused.

The formula for calculating 2 resistors in parallel is simple enough:

(R1 x R2) / (R1 + R2)

The book then tells me that this will not work for calculating resistance for more than 2 resistors in parallel?

So (R1 x R2 x R3) / (R1 + R2 + R3) won't work?
Neither should (R1 x R2 x R3 x R4 x R5) / (R1 + R2 + R3 + R4 + R5) right?

But the long calculation that he says I need to use looks like the same thing...

Step 1: R1 x R2 = x
Step 2: R1 + R2 = y
Step 3: x / y = Resistance

Book quote "If there are only two resistors in parallel, you don't have to calculate it the way we've just done here - there is a simpler way, given by the expression: (R1 x R2) / (R1 + R2). But if there are three or more resistors in parallel you have to use the long method, I'm afraid."

It's hard to show exactly the way the book has them so please find attached a scan of the book pages below.

Apr 5, 2008
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3. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
That is correct. Those last two formulas you posted will not work.

Interesting they would use that as an example, that's not correct for more than two resistors. You need to use the 1/Rtot = 1/R1 + 1/R2 + 1/R3 +.... method for more than two resistors.[/QUOTE]

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4. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3
Thanks guys, I guess my question has turned to: How can I tell if a book is good or not if I don't have the knowledge to check if what it is telling me is correct. The book got good reviews when I bought it.

5. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
The 1st 2 formula are correct, all else follows from that. I keep a note pad next to my computer so I can do quick algebra manipulation when needed. I don't participate in most of the math section because I am a math brick, to borrow a phrase from another member. Algebra I know though.

6. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3
lol thanks.

Just so I understand the math behind calculating resistance....

So

4 resistors in parallel

R1 = 220
R2 = 220
R3 = 220
R4 = 220

R1 x R2 x R3 x R4 = 2342560000
R1 + R2 + R3 + R4 = 880
2342560000 / 880 = 2662000
2662000 / 1000 = 262.000

1 / 262.000 = 0.0038Ω

or

1 / R1 = 0.00454545454
1 / R2 = 0.00454545454
1 / R3 = 0.00454545454
1 / R4 = 0.00454545454
Sum = 0.01818181818

R1 + R2 + R3 + R4 = 2342560000

2342560000 / 0.01818181818 = 128840800013
128840800013 / 1000 = 128840800.013

1 / 128840800.013 = 7.76Ω (2 sig fig)

So which is the right calculation/answer?

a) 0.0038Ω
b) 7.76Ω
c) You've annihilated the very core of maths

(out of interest, if I divide 1 by 0.00076 on the calculator I get 1315.7894 which is not really 1304 as the book suggests)

7. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
4 resistors in parallel

R1 = 220
R2 = 220
R3 = 220
R4 = 220

1/Rt = 1/R1 + 1/R2 + 1/R3 + 1/R4 = 1/220 + 1/220 + 1/220 + 1/220 = 4/220

Flipping the inverses on both sides...

Rt = 220/4 = 55 ohms

Shortcuts are OK, but be sure they go where you want them to go.

Calculators with parenthesis are also your friend.

Rt = 1 / ((1/R1) + (1/R2) + (1/R3) + (1/R4))

The base formula:

1/Rt =1/R1 + 1/R2

is usually the best starting point.

Last edited: Aug 30, 2015
8. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3

I take it Rt is a representation of Resistor Total?

Rt = 880 (220 + 220 + 220 + 220)

I'm so confused by all of this I'm not even sure how you calculated 200?

1 / 200 = 0.005... eh? Did I miss something?

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
You caught me mid edit and mid brain fart.

Rt = R1 + R2

Parallel is inverse:

1/Rt = 1/R1 + 1/R2

Rt is indeed total resistance.

Like I said, a piece of paper to lay out the equation is always helpful for me.

10. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Try another one.

R1 = 240 0hms
R2 = 330 ohms

What is the series and parallel Rt for these pair?

11. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3
LOL ok, is this correct then

R1 = 220
R2 = 220
R3 = 220
R4 = 220

(I'll just used 4 sig fig)

1 / 220 = .0045
.0045 x 4 = .0181

1 / .0181 = 55.24Ω (or basically 55Ω)

Lets me just try your example
R1 = 200
R2 = 200
R3 = 200
R4 = 200

1 / 200 = .005
.005 x 4 = .02

1 / .02 = 50Ω (Yay!)

So...

R1 = 220
R2 = 50
R3 = 120
R4 = 450

(4 sig fig)
1 / 220 = .0045
1 / 50 = .0200
1 / 120 = .0083
1 / 450 = .0022

Sum = .215

1 / .215 = 4.65Ω

Do I have this right?

12. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3

Series

R1 + R2 = Rov
240 + 330 = 570Ω

Parallel

1 / (1/R1 + 1/R2)
1 / 240 = .0041
1 / 330 = .0030
.0041 + .0030 = .0071
1 / .0071 = 140.84Ω

Is that right?

13. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Rt = 1 / ((1/220) + (1/50)+(1/120)+(1/450)) = 1/(.00455+.02+.00833+.00222) = 1/.0351 = 28.49 ohms

1/(1/240+1/330)=138.95

I'm using the windows calculator.

The second answer is error creeping in, you are using 2 digits, not 3, rounding.

Rounding the intermediate answer is .00720.

Inverted that is 13.88.

Last edited: Aug 30, 2015
14. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
For 3 resistors you can use:

$\frac{R_1R_2R_3}{(R_1+R_2)R_3+R_1R_2}$

The question is can you remember the form of the expression in a pinch without having to derive it from first principles?

For 4 resistors you can use:

$\frac{R_1R_2R_3R_4}{((R_1+R_2)R_3+R_1R_2)R_4+R_1R_2R_3}$

Things just get hairier as the number of resistors goes up. I think there is a pattern however.

Last edited: Aug 30, 2015
15. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3
Sorry - must have missed something on my calculator - redone and got 0.035 as my total and not .215

1 / .035 = 28.57Ω

They now put me in the same ball park as your answers.

Depending upon the significant figures used (i.e. the number of decimal places to keep) and the calculator used (I'm using a standard physical desktop calculator made by Texet from Asda ) shows slightly different results. Would these discrepancies be acceptable in the real world or would that be the difference between blowing something up and having a working circuit?

16. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
If you want 3 digits accuracy you probably ought to use 4 digit entering your values. With inversion very small changes add up fast.

Of course, they taught slide rules in my high school, so I am used to 3 digits.

17. ### Dal1980 Thread Starter New Member

Aug 30, 2015
19
3
That may be the final wedge I needed to become unhinged lol

Trying to remember all this from school as I've not really needed to use formulae since. Is that like saying:

R1 x R2 x R3 at the top (z)
Divided by
(R1 + R2) = x
(R1 x R2) = y
a = x + y + R3

z / a?

18. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Like I said, stick with the basics. We are not all good with math.

In the real world +/- 5% is usually acceptable in electronics.

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19. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
It may give you some peace to know that in a career spanning half a century, I've never had to deal with more than two resistors in parallel, except as an academic exercise.

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20. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
That is funny, because it is a common problem for me.

1. It allows me to derive resistors I do not have in my box.

2. As a troubleshooting tool, it allows me to predict with fair accuracy what resistance I should see in any leg.

I don't use it every day, or even every month, but it is a useful tool.