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Equivalent Power In series and parallel combination

  1. Aug 14, 2015 #1
    How can we derive the formula for finding equivalent power in a series and parallel combination of 'n' resistors (with fixed resistance)?
  2. jcsd
  3. Aug 14, 2015 #2
    Use that formula for the particular combination , in which a particular variable is common to all .
    Example - Use P = ∑i2Rn . What is the advantage of using this ?

    Thus , what would you use for parallel ?

    Hope this helps .
  4. Aug 14, 2015 #3
    n=no. of resistors.
    R=value of resistor.
    Last edited: Aug 14, 2015
  5. Aug 14, 2015 #4
    I think i should put forward the question as under:

    we have 2 identical resistors R(1) and R(2),
    P(1) = Power of R(1)
    P(2) = Power of R(2)
    P(s) = Equivalent power of resistors R(1) and R(2) when connected in series.

    P(s) = 1/P(1) + 1/P(2) HOW?

    and if

    P(p) = Equivalent power of resistors R(1) and R(2) when connected in parallel.


    P(p) = P(1) + P(2) HOW?
  6. Aug 14, 2015 #5


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    Welcome to the PF.

    Can you give more details about that equation that you wrote?
  7. Aug 15, 2015 #6
  8. Aug 15, 2015 #7


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    The question still makes no sense. Are any things being held constant? Like the source voltage or the source current? You need to show the circuit diagrams involved, along with the source voltages in each configuration. Then show your math based on the circuit diagrams.
  9. Aug 16, 2015 #8


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    Energy conservation principle. Where else would the input power come from or where would any surplus power go?
  10. Aug 16, 2015 #9
    Sir, i myself have not done any math. I just seek to find a short way to calculate equivalent power in a circuit where devices of equal power rating are connected.
    The source VOLTAGE is kept constant as you said.

    for eg-

    When three bulbs of 60W-200V rating are connected in series to a 200V supply, the power drawn by them will be : (Ans= 20W)

    Now, i am asking for the proof of the formula that i have highlighted.
    Have provided the circuit diagram also this time :)

    Attached Files:

  11. Aug 16, 2015 #10
    Well the answer is very simple;
    If bulbs have 60W at 200V this means that the bulbs resistance is equal to:
    P = V^2/R ---> R = V^2/P = 200V^2/60W = 667Ω and because we have three light bulbs connected in series the total resistance is equal to 3*667 = 2kΩ so the total power is equal to P = 200V^2/2kΩ = 20W This means that the book give you the correct equation.
    But you should remember that this formula is only true for a device with a constant resistance. But this equation is not true for the light bulb. Becomes the light bulb resistance is not constant but will change with the apply voltage. See the graph of resistance of a 100W/230V light bulb in function of a supply voltages.


    The resistance for 230V is equal 530Ω so current is equal 0.434A
    But for 120V the current will not be equal to I = 120V/530Ω = 0.226A
    As we can read form the graph the current will be equal I = 120V/ 380Ω = 0.315A
  12. Aug 16, 2015 #11


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    Even this answer is too simple, I think. You cannot tell, without measurement, how an individual bulb will behave. The surface temperature will surely depend upon the actual filament and envelope design and not just on its power rating. I really don't like the idea of putting filament bulbs as 'examples' of Resistors - presumably in an attempt to make the topic more approachable so the original question is 'questionable'. Science teaching is full of dodgy examples, thought up by people who don't have a lot of practical experience or detailed knowledge of the example items they include in their question.

    Proof of the formulae in the red squares is easy. Start with the formula for parallel resistors. The Total Power will be V2/Rp so multiply both sides of the parallel resistor equation by V2 and that produces what you want. Likewise, write down the formula for series resistors and remember Power is I2R. The answer comes oput by multi[plying both sides of the resistance formula by I2.
  13. Aug 16, 2015 #12
    Yes perfect. Now i've got the answer and understood it completely.

    Thanks to all for guiding.
  14. Aug 16, 2015 #13


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    Thanks for asking the 1/P question. To be honest, I'd never thought of it in that way before. :smile:
  15. Aug 16, 2015 #14
    no its not

    Why insist on making things more complicated than they are?

    The purpose of the question was to deal with a simplified 'ideal' system for a complete begineer to start to understand; not a real world system.

    When the students here are asking about, for example, projectile motion, nobody is going over the top with air resistance, wind, the dimples and imperfections in the projectile; they solve simple SUVAT equations. How far do you go? You cannot model everything, its impossible.

    There really is a tendency to make things more complicated for no gain.
  16. Aug 16, 2015 #15


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    It's halfway between a simple resistor and a real device. If you want to help someone get to the real world then your graph should be coupled with a caveat or they could take it as gospel. If you had added "one example of a 100W light bulb" I wouldn't have picked it up.
    There are a whole lot of possible circuit problems that include non-Ohmic devices and they are good problem-solving fodder. Resistor/diode networks Resistor/Thermistor combinations etc etc. They all result in a nice equation to solve.
    I'm all for starting simple but with the simplification being made clear.
  17. Jul 22, 2016 #16
    from where did you get this picture

    which book it is?
    give the name of the author
  18. Jul 23, 2016 #17


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    There isn't a short way to get a reliable answer. Can you not accept that? The "short way" is only available for Resistors that do not change their value over their operating range. If you haven't done any Maths then I would advise you to avoid Electronic Engineering questions.
  19. May 13, 2017 #18
    since in a series circuit having e resistances r1, r2, r3, R = r1+r2+r3
    then I²R = i1²*r1+i2²r2+i3²r3, (since I = i1 = i2 = i3)
    then P = p1+p2+p3
    but how actually it be 1/P = 1/p1+1/p2+1/p3
  20. May 13, 2017 #19


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    I had a problem with the implications of that equation. Power must be conserved so there are only certain values for p1, p2, p3 that can satisfy both that equation and the equation P = p1+p2+p3
    Another way of looking at is is that there is only a limited set of a,b,c for which the mean is the same as the harmonic mean - so that equation is just a bit of Maths jiggery pokery and one shouldn't struggle too much with interpreting it in physical terms.
    It's good to get us thinking though.
  21. May 13, 2017 #20

    jim hardy

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    Beginners should start simple and derive general equations for themselves only after they've enough experience to do so. The old "What before Why" teaching method.

    Resistors in series have same current through them.
    So total power = I2 X R total = I2 X (R1 +R2 + R3) = P1 + P2 + P3

    Resistors in parallel have not the same current through but the same voltage across them.
    So power is E2/Requivalent = E2/(1/(1/R1 + 1/R2 + 1/R3)) = E2 X (1/R1 + 1/R2 + 1/R3) = P1 + P2 + P3

    Can anyone really look at this picture and say total power is other than 180 watts?


    I'd say the grad student who wrote that exercise is a nincompoop and so is the author of the book who didn't check up on him..
    Mathematical tap-dancing doesn't impress me at all. Well, maybe Cyd Charisse could ....
    Last edited: May 13, 2017
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