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Maximum power dissipation for resistors

  1. Jul 13, 2011 #1
    there are 3 resistors of, different values. 470o,22o and 100o. but each one is 500W
    I want to connect them in series/parallel in order to dissipate the maximum power 500W.

    I have connected 4,22o in series and 2, 100o in series, 1, 470 in parallel. but the dissipation is not maximum.
    the supply is 415

    i would like to know if its possible to get max power dissipation
     
  2. jcsd
  3. Jul 13, 2011 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    You only have one each of 22, 100 and 470 Ohm resistors? And you want to maximize the power dissipation with no single component exceeding 500W, or you want the total dissipation of the sum to equal 500W?
     
  4. Jul 13, 2011 #3
    i have each resistor in a large quantity. here is the circuit
     

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    Last edited: Jul 13, 2011
  5. Jul 13, 2011 #4
    want to maximize with no single component exceeding 500W. each is rated for 500W
     
  6. Jul 13, 2011 #5

    berkeman

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    So what *exactly* are you trying to do? Do you want the total power dissipated to be 500W?
     
  7. Jul 13, 2011 #6
    No, total power will be the sum of the all the dissipated power for each resitor
     
  8. Jul 13, 2011 #7

    berkeman

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    The maximum power dissipated will depend on how many resistors you have. If you have an unlimited number of resistors, you can dissipate infinite power.

    Please post the *exact* text of the question.
     
  9. Jul 13, 2011 #8
    ok sorry for the confusion

    What i have in the attached circuit.

    I will have maximum power for each as followed based on P=I^2*R:

    22ohm: 489W (that would be multiplied by 4 since 4, 22ohm are used)
    100ohm: 430W (that would be multiplied by 2)
    470ohm: 366 (that would be single)

    Now if the circuit is connected in such a way that each resistor is connected to dissipated at maximum rating 500W. the total would be the sum of all

    As long as each is dissipating close to 500W. the sum of each would be the total
     
  10. Jul 13, 2011 #9
    Did you get my question this time

    lets make it simple. i donot care whats the total output power going to be when all the resistors are dissipating 500W each.

    for the time i just want each to dissipate 500W (or close).

    With the current circuit its way to less than 500w for resistors 100, 470
     
  11. Jul 13, 2011 #10

    berkeman

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    No, sorry.
    Try this -- what voltage do you need across each of those 3 resistor values to get to 500W? Does that help you start to figure out how you could combine them to always have close to that voltage across them?
     
  12. Jul 13, 2011 #11
    using P=V^2/R,the voltages across each would be

    22ohm: 104.9V
    100ohm:223.6V
    470ohm: 484V

    any idea how to combine resistors to get these voltages across it
     
  13. Jul 13, 2011 #12

    berkeman

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    Staff: Mentor

    That's your task, right?
     
  14. Jul 13, 2011 #13
    thats why i am here after not been able to figure out

    I dont see of combinations since P=I^2*R= V^2/R

    which would be the same as i was trying to achieve current.
     
  15. Jul 13, 2011 #14

    berkeman

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    Can you please show us the original problem statement, word for word? If it's in a language other than English, show us the original and then please give us a word-for-word translation.

    The problem makes no sense to me as stated.
     
  16. Jul 13, 2011 #15
    if i were to combine then in order to have required voltages across them, the I^2 current is either goin to burn the resistor or not provide sufficient power.
     
  17. Jul 13, 2011 #16
    Problem: Construct a 3-ph resistor load bank provided the following resistors:

    22ohm:500W (Quantity : 15)
    470ohm:500W (Quantity:15)
    100ohm: 500W (Quantity: 15)

    Each power resistor should dissipate nearly 500W (full wattage) connected in series and/or parallel. The supply line is 415ac. The delta-wye combination can also be mixed in order to connect the resistors.
     
  18. Jul 13, 2011 #17

    berkeman

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    Whew. That's a very different problem from the one you've been trying to describe so far. This is a 3-phase voltage supply, and you have a finite number of the resistors. Check.

    Okay, so what do the voltages look like in a 3-phase "415Vrms" system? What is a Delta-Wye Combination?
     
  19. Jul 14, 2011 #18
    please see the attached print

    I dont know if this information should have make difference in understanding the situation.
    Since even with the 3phase balanced load the current flowing through the resistors are same. This is just delta configuration.

    it does not seem to help anyway combining wye connection in that, since Phase voltage would be 240V

    Do u understand the taks this time
     

    Attached Files:

  20. Jul 14, 2011 #19

    gneill

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    What is the phase-to-neutral voltage for a 415V 3 phase system? What is the phase-to-phase voltage? Are they the same or different?
     
  21. Jul 14, 2011 #20
    As you know

    in delta configuration

    Phase voltage= Line voltage= So Phase-Phase is 415

    If i were to connect it in Wye
    Thenk phase to neutral votlage would be 240V
     
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