Calculating Parallel Resistance

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

  1. Dal1980

    Thread Starter New Member

    Aug 30, 2015
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    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.


    ontheboards.jpg
     
  2. bertus

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  3. DerStrom8

    Well-Known Member

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    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]
     
    gisdude likes this.
  4. Dal1980

    Thread Starter New Member

    Aug 30, 2015
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    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

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    Mar 24, 2008
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    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
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    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)

    :confused::confused::confused:
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    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
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    :confused:

    Thanks for the reply.

    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? :eek:
     
  9. Wendy

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    You caught me mid edit and mid brain fart.

    Only series is additive:

    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

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    Mar 24, 2008
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    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
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    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
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    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

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    Mar 24, 2008
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    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
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    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
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    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 :D) 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

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    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
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    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

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    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

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    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

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    Mar 24, 2008
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    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.
     
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