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Power of resistors in parallel vs series circuit

  1. May 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Two Ohm's resistors in parallel consume power of 76 W and 24 W.
    What power will each of them consume if transfer them to the serial circuit with the same voltage of source.

    Sorry for my bad english translations I m not use to writing questions in English.. So what confuses me is that on this forum similar question was asked and it stated that Power in serial circuit is 1/P=1/P1+1/P2. (https://www.physicsforums.com/threa...series-and-parallel-combination.827649/page-2 ) Should I use that formula or not? I was thinking that both in parallel and serial power is just P=P1+P1, and power in parallel is therefore 100W, but power of serial is less because total serial resistance is greater than individual resistance of 1 resistor, while in parallel total resistance is less than individual resistance of 1 resistor, so total power would be greater in parallel circuit than in serial circuit?

    2. Relevant equations

    3. The attempt at a solution

    In parallel 1/R=1/R1+1/R2 and power is P=V^2/R
    IN series R=R+R2 but power here is less since we have the same R1 and R2 but greater R(total) in serial makes it that P(total) in serial is less thatn P(total) in parallel???
    Does this make sense or should it be like the previously asked question here where P in series is 1/P=1/P1+1/P2 ?
     
    Last edited by a moderator: May 29, 2017
  2. jcsd
  3. May 29, 2017 #2

    BvU

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    Hello crostud, :welcome:

    It would be better if you understood the characteristics of parallel and series circuits. The proper formulas to use follow from that understanding.

    If the resistors are in parallel, the voltage over each of them is the same.

    You already have ## P=V^2/R ## They give the power ratio, so $${V^2/R_1 \over 76 } = {V^2/R_2 \over 24 }\ . $$Now what is asked is the power for the serial circuit, with resistance ##R_1 + R_2##. The expression for that power is ##V^2\over R_1+R_2##. From now on it is math: express e.g. ##R_2## in terms of ##R_1## and calculate the power.
     
  4. May 29, 2017 #3
    Thank you very much but I got the relation between R2 and R1 and I still can't calculate P1 and P2 in serial (power of each resistant like it's asked from me). I don't know where to go from after I got the relation between R1 and R2 since I don't know whats the V, I only know that it's the same in parallel and serial.
     
  5. May 29, 2017 #4
    You know the power with V across each resistor.
    With the resistors in series and their ratios known, you can calculate the voltage across each in terms of V e.g. V-x,x.
    You can then calculate the power with V-x or x across each resistor compared to V.
    You can't calculate and don't need the actual values of V or R1 or R2 or the current.
     
  6. May 29, 2017 #5
    Sorry, still quite don't understand. I always get to a dead end :(
     
  7. May 29, 2017 #6

    BvU

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    Show your work. Just shouting I dunno doesn't help you further.
    You have ##V^2/R_1##; what doe you have for ##R_2## in terms of ##R_1## ? Why is it so difficult to calculate ##V^2 \over R_1+R_2## ?
     
  8. May 29, 2017 #7
    I would find it difficult to do this; as it is not the question asked I won't try.


    This is correct.
    From #2
    $${V^2/R_1 \over 76} = {V^2/R_2 \over 24 }\ . $$
    You can work out from this the numeric ratio ##{R_1 \over R_2}##.
    WRONG!!! or English error... the voltage across each resistor in series (not serial) is less than V.

    Hint: a related simpler problem:
    If you have (say) two resistors in series, one twice as big as the other, connected across any nonzero voltage (say X), can you say anything about the ratio of the voltages across those resistors and express the voltages in terms of X? What must be the sum of those two voltages?


    I expect you to also get a similar expression for the main problem; you may then see how to solve it.

    Your original attempt was a good start; this is as far as I can presently go without (I hope) breaching forum rules.
     
  9. May 30, 2017 #8

    BvU

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    Once you have the sum of powers, it's easy to find the separate ones.
    You don't have to: it's crostud's exercise.
     
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