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Circuit and Resistance

  1. May 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Why do series circuits tend to have more error than parallel circuits. It is a lab question.


    2. Relevant equations
    None


    3. The attempt at a solution

    I believe that since series has more resistance than parallel, then more current travels through the circuit, dissipating more electrons to heat. Also, since electrons have to travel through both resistors, there is a greater probability of error.
     
  2. jcsd
  3. May 9, 2007 #2

    chroot

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    Have you copied the problem exactly as it appears in your lab manual? I have a hard time believing this was actually in your lab manual. It makes absolutely no sense at all; nor does your answer.

    - Warren
     
  4. May 10, 2007 #3

    andrevdh

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    The person who set the question might have ment that series circuits are more prone to defects than parallel circuits. If one component goes in a series circuit the whole circuit goes defective, which is not the case with parallel circuits.
     
  5. May 11, 2007 #4
    Yes, it is copied correctly. What I am trying to say is that if there is a margin of error for each of the resistors, then a series circuit adds the error while a parrallel circuit averages the error. Just a thought; based on my data, I thought it was a reasonable concept.

    Younglearner
     
  6. May 11, 2007 #5

    chroot

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    That is indeed a reasonable concept! I just didn't understand what you meant by "error."

    Your conclusion is correct.

    - Warren
     
  7. May 11, 2007 #6

    berkeman

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    Just calculate the error for two 5% resistors in series versus in parallel, and compare the largest variations....

    1.05 + 1.05 = 2.10, 0.95 + 0.95 = 1.90 (the series sum is still 5% error)

    1/ (1/1.05 + 1/1.05) = ? 1/ (1/0.95 + 1/0.95) = ?

    What is the resulting percentage error for the parallel combination? Hmmm, I get an interesting answer (not what I expected)....
     
    Last edited: May 11, 2007
  8. May 11, 2007 #7

    berkeman

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    And just to follow-up.... If there were a difference for resistors, then it would be the other way for capacitors, right?
     
  9. May 11, 2007 #8

    berkeman

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    Interesting.... When I combine a 1% resistor and a 5% resistor, I finally get a small difference between series and parallel circuits, but still not much.

    I wonder if it's a distribution thing.... If you assume a Gaussian distribution of error for the two components, do you get a different distribution for the series versus parallel case?

    Except, remember that the distribution of error for real resistors is not generally Gaussian. Quiz Question -- why not? :biggrin:
     
  10. May 11, 2007 #9

    chroot

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    The endpoints of the range are not a useful way to analyze this problem, berkeman. Instead, you do indeed need to look at resistor's error as a random process, and look at the distribution of the series and parellel combinations -- particularly, the variances of the distributions.

    Give me a bit and I'll write up a proof.

    - Warren
     
  11. May 11, 2007 #10

    chroot

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    Okay, berke buddy, you asked for it!

    In English: Let's assume we have two resistors, which we'll wire either in series or in parallel. The value of the first resistor is random (since it has some random error), and we'll call its value X. Let's say its nominal value is one ohm, and the amount of error is normally distributed around one ohm. The value of the second resistor is also random, but it's independent of the first. We'll call its value Y.

    In math speak: Assume two random variables, X and Y, which are indpendent, distributed randomly with Gaussian distribution, with mean one and variances [itex]\sigma _X ^2[/itex] and [itex]\sigma _Y ^2[/itex], respectively.

    [itex]
    \begin{gathered}
    X \sim \left( {1,\sigma _X ^2 } \right) \hfill \\
    Y \sim \left( {1,\sigma _Y ^2 } \right) \hfill \\
    \end{gathered}
    [/itex]

    For the series case:

    The sum of the two resistances in series is Z = X + Y. Since X and Y are independent and Gaussian, the sum Z is Gaussian, with mean two and variance equal to the sum of the variances of X and Y.

    In math speak:

    [itex]
    \begin{gathered}
    Z = X + Y \hfill \\
    Z \sim \left( {2,\sigma _X ^2 + \sigma _Y ^2 } \right) \hfill \\
    Var\left( Z \right) = \sigma _X ^2 + \sigma _Y ^2 \hfill \\
    \end{gathered}
    [/itex]

    If the variance of X is, say, 0.05, and the variance of Y is also 0.05, then the variance of the two resistors in series is 0.05 + 0.05 = 0.10. Keep this number in your head!

    For the parallel case:

    [itex]
    Z = \frac{1}
    {{\frac{1}
    {X} + \frac{1}
    {Y}}} = \frac{{XY}}
    {{X + Y}}
    [/itex]

    We want to find the variance of Z:

    [itex]
    Var\left( Z \right) = E\left( {Z^2 } \right) - E\left( Z \right)^2
    [/itex]

    so we need to find the second moment and expected value of Z.
    [itex]
    \begin{gathered}
    E\left( Z \right) = \frac{{E\left( {XY} \right)}}
    {{E\left( {X + Y} \right)}} \hfill \\
    = \frac{{E\left( X \right)E\left( Y \right)}}
    {{E\left( X \right) + E\left( Y \right)}} \hfill \\
    = \frac{1}
    {2} \hfill \\
    \end{gathered}
    [/itex]

    The expected value of Z is 1/2, which makes perfectly good sense. If you put two resistors in parallel, each with resistance one ohm, you expect to get a combined resistance of 1/2 ohm.

    The second moment of Z takes a bit more care to calculate:

    [itex]
    \begin{gathered}
    E\left( {Z^2 } \right) = \frac{{E\left( {X^2 Y^2 } \right)}}
    {{E\left( {X^2 + 2XY + Y^2 } \right)}} \hfill \\
    = \frac{{E\left( {X^2 } \right)E\left( {Y^2 } \right)}}
    {{E\left( {X^2 } \right) + 2E\left( X \right)E\left( Y \right) + E\left( {Y^2 } \right)}} \hfill \\
    = \frac{{\left( {1 + \sigma _X ^2 } \right)\left( {1 + \sigma _Y ^2 } \right)}}
    {{\left( {1 + \sigma _X ^2 } \right) + 2 + \left( {1 + \sigma _Y ^2 } \right)}} \hfill \\
    = \frac{{\left( {1 + \sigma _X ^2 } \right)\left( {1 + \sigma _Y ^2 } \right)}}
    {{4 + \sigma _X ^2 + \sigma _Y ^2 }} \hfill \\
    \end{gathered}
    [/itex]

    Now we can find the variance of Z:

    [itex]
    \begin{gathered}
    Var\left( Z \right) = E\left( {Z^2 } \right) - E\left( Z \right)^2 \hfill \\
    = \frac{{\left( {1 + \sigma _X ^2 } \right)\left( {1 + \sigma _Y ^2 } \right)}}
    {{4 + \sigma _X ^2 + \sigma _Y ^2 }} - \frac{1}
    {4} \hfill \\
    \end{gathered}
    [/itex]

    Now, if we assume the variances of X and Y are each 0.05, like we assumed above, the variance of Z, the parallel combination of the two, is only 0.019. The variance of the parallel combination of resistors is about 80% smaller than the variance of the series combination of resistors.

    Thus, you're right -- the parallel combination is less sensitive to variations in each resistor's resistance than the series combination.

    (This is graduate-level statistical inference, so I don't expect Younglearner to be able to follow all of it -- but the conclusions, I hope, are of some value!)

    - Warren
     
    Last edited: May 11, 2007
  12. May 11, 2007 #11

    berkeman

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    Great stuff chroot, thanks. So the answer to the OP's question is that yes, for resistors the parallel combination is less sensitive to variations, but for capacitors, the series combination is less sensitive. Interesting.
     
  13. May 11, 2007 #12

    berkeman

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    But nobody answered my Quiz Question yet :grumpy: :smile:

     
  14. May 11, 2007 #13
    Because the error in resistance changes as a function of power, i.e. temperature.

    Also, Chroot's way is the way I would have done it. Error analysis!
     
  15. May 11, 2007 #14

    berkeman

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    Nope, that's not it. Hint -- remember that you can buy 1% resistors and pay extra for 0.1% resistors in the same form factor...
     
  16. May 12, 2007 #15
    Must be a material thing of some sort. I seem to remember that standard resistors are made out of carbon, whereas the fancy ones must use a different method. Are you trying to give us a quiz question that doesn't involve integration?
     
  17. May 12, 2007 #16
    thanks everyone for participating! I ended up putting my original answer in my lab and it turned out to be correct. I was not able to follow all of chroot's proof, but the concepts make sense and I am glad to know that my logic was correct (though I could not prove it)
     
  18. May 12, 2007 #17
    I am thinking about your quiz question, but I do not completely understand what you are asking. could you rephrase it a bit. I am sorry that I don't get it yet. (Younglearner=not skilled yet, but working at it)
     
  19. May 14, 2007 #18

    berkeman

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    No worries. Just remember that 0.1% resistors are obtained by sorting/binning 1% resistors. So the distribution of 1% resistors will be sort of Gaussian, but with a hole in the middle where the most accurate resistors were. You would also see this sort of effect in ICs that are speed graded -- the fastest ICs are marked with a faster speed grade.
     
  20. Jan 10, 2011 #19
    :confused:

    I stumbled upon this (old) discussion and because I found the results very interesting I also did some calculations on the matter. I am just wondering, aren't you neglecting signing?
    If two resistor have a positive deviation, the parallel circuit will give more weight to the lowest one so indeed the parallel circuit will give best results. But if both resistors have a negative deviation also the parallel circuit will give more weight to the lowest one, this time resulting in a win for the serial circuit. If one deviation is positive and one is negative, roughly the biggest deviation wins, so biggest deviation negative you need a serial circuit for best performance, biggest deviation positive, go for the parallel circuit.
    I say roughly because while the errors zero out at the exact opposites with a serial circuit in parallel this zero error point shifts; however, this gain on the one side will be the loss on the other, so statistic wise won't matter.

    devR1 devR2 error%serial error%par win?
    1 -1 0.000 0.010 serial
    1 3 2.000 1.990 parallel
    -1 -3 2.000 2.010 serial
    1 -2 0.500 0.523 serial
    -1 2 0.500 0.478 parallel
    -1 -1 1.000 1.000 draw
    -1 0 0.500 0.503 serial
    1 0 0.500 0.498 parallel

    Remco
     
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