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Resistor / Watts calculation in parallel

Discussion in 'Electronic Basics' started by Andrew Holme, Mar 3, 2005.

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  1. Andrew Holme

    Andrew Holme Guest

    No. This is 1 Watt 100k
     
  2. mike

    mike Guest

    Yes,
    If you know the voltage or current, you can calculate the wattage
    in each resistor for combinations of different values.

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  3. Under the assumption that the rating for the individual
    resistors remains unchanged when in "parallel", then
    your written statement is not false.

    Something to keep in mind, however, is that the rated
    dissipation holds for certain standard conditions that
    may not pertain when you parallel the parts. As an
    example, significant heat flows out of a lead-mounted
    resistor to the circuit board, keeping the resistor's
    temperature lower than if that path did not exist. If
    you connect two resistors in parallel and mount them
    in the hole pair intended for a single resistor, the heat
    flow into the circuit board will not double and there
    will be a greater temperature rise for the part bodies.
    And if N paralleled resistors are bunched together,
    they will not be cooled by air as well as they would
    be if mounted in the standard fashion.
    A stronger statement, assuming you equalize the
    dissipation in N resistors, (and retaining the "no
    change in rating" assumption), is that the rated
    dissipation is multiplied by N.
    I would hope so.

    [Cut some resistor diagrams.]
     
  4. Of the 3 posts answering your question I have seen,
    (by "mike", "Andrew Holme" and "Larry Brasfield"),
    the answers are all correct given the ambiguity of
    your question.
    I would concur, at this point.
    Yes, (under the "unchanged rating" assumption).
    That's true, but not does not contradict Andrew's correction.
    The second drawing does not state "could be
    replaced by ...". It seems (to me) to state an
    equivalence. The rating of the 4 resistor glom is
    closer to 1W than 1/2 W, under the "unchanged
    rating" assumption. Under the more realistic
    "changed mounting changes rating" assumption,
    the 2nd drawing 1/2W rating could be closer
    to correct and the 1st drawing deemed wrong.
     
  5. Chretien

    Chretien Guest

    Im just trying to confirm my understanding of Parallel circuits and Watts
    calculations when using resistors.

    This is not for any project. But sometimes projects call for 1/2 watt
    resistors. It means I have to run to the store since I don't keep any of
    them on hand. This would be a way around this.

    Are my simple calculations right?

    Basically you can double the wattage if you put two or more resistors in
    parallel. This will of course decrease the R Value.



    All Resistors are
    100k 1/4 Watt

    ___
    o-|___|-o
    | |1/2 Watt 50k
    -----| |---
    | ___ |
    o-|___|-o




    ___ ___
    o-|___|--|___|-o
    | |
    | ___ ___ |1/2 Watt 100k
    o----o-|___|--|___|-o------o

    All Resistors are
    100k 1/4 Watt
    (created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)
     
  6. John Fields

    John Fields Guest

    ---
    The total resistance of two resistors in series is the sum of the
    resistances.

    The total resistance of two resistances in parallel is:

    R1 R2
    Rt = ---------
    R1 + R2

    In your first example you have:


    1E5R * 1E5R 10E9R
    Rt = ---------------- = ------- = 5E4R = 50K ohms
    1E5R + 1E5R 2E5R

    and since you have two resistors, each of which can dissipate 0.25W,
    the pair can dissipate 0.5W, just like two 100 watt light bulbs
    dissipate 200 watts when they're both on.


    If we label your second example:


    ___ ___
    o-|___|--|___|-o
    | R1 R2 |
    | ___ ___ |1/2 Watt 100k
    o----o-|___|--|___|-o------o
    R3 R4


    Then R1 and R2 in series will be


    Rt = R1 + R2 = 200k ohms


    and so will R3 and R4 be, so you now have, in effect:


    ___
    o-|___|-o
    | 200k |
    | ___ |
    o----o-|___|-o---o
    200k

    which, since they're in parallel further reduces to:


    ___
    o----o-|___|-o---o
    100k


    and since you've got four 1/4 watt resistors in there, they can
    dissipate 4 * 1.4 watt = 1 watt.
     
  7. Chretien

    Chretien Guest

    Well I have two posters with two different answers.

    Perhaps I have labeled my drawings poorly and the confusion starts there.

    Each of my expamples consists of 100k 1/4 watt Resistors.
    1. The first schematic in this configuration could be replaced with a 1/2
    watt 50k Resistor.

    2. The second drawing could be replaced with a 1/2 Watt 100k Resistor.

    Are you saying that I'm wrong on both drawings, or the second drawing and
    why?

    Thanks.
     
  8. Chretien

    Chretien Guest

    Thank you Larry.

    Yes I understand that you can't keep adding resistors and not take into
    consideration the ramifications, eg heat and draw on the rest of the
    circuit. Thanks for the tip.

    If anyone cares I found this usage of resistors to increase wattage on a
    document

    http://www.sarts.org.sg/lesson2.pdf

    Search for the word Even Hotter where it says.


    EVEN HOTTER

    If an even higher power rating is required then several resistors may be
    connected in parallel. Ten 5 Watt resistors connected in parallel would give
    a total power rating of 50 Watts. The resulting resistance will, of course,
    be a tenth of the individual resistors.



    It was from there that I thought mmm I wonder if I can do the same when I
    get stuck.



    Regards

    Novice
     
  9. Rich Grise

    Rich Grise Guest

    Because 4 * 1/4 = 1.

    Cheers!
    Rich
     
  10. news

    news Guest

    I see. I neglected to take into consideration the series resistors. I
    suppose then that two 1/4 watt 100k resistors equals a 200k 1/2 watt.

    thanks.
     
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