1. Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
    Dismiss Notice

cal. short circuit power, in a series circuit.

Discussion in 'Mathematics and Physics' started by north, Feb 17, 2007.

  1. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    this is the problem, there are 4 resistors in a series circuit, and the total power is 20W. one resistor burns out( which should mean that there should zero current or voltage running through the circuit, so I think they mean a short circuit) anyway, what is the power dissipated by the remaining resistors?

    now the answer is suppose to be 8.9W. but the only way I could get this answer is by dividing 20W by 3( which is 6.67W per resistor) and then adding the 6.67W to 20W and then dividing by 3 again. hence 8.9W dissipated per resistor.( rounded up)

    this is not making much sense to me. is the answer wrong or I am I just missing some little detail here.

    any help would be appreciated.

    north
     
  2. Hero999

    Hero999 Banned

    Joined:
    Apr 6, 2006
    Messages:
    14,902
    Likes:
    79
    Location:
    England
    Is this college work, we don't mind helping you but we won't doo all the work for you.

    It depends on which resistor burns out and the resistor values.

    For example if four 5:eek:hm: reisitors are connected to a 10V supply as two series pairs in parallel then the power dissipation will be 20W but if any single resistor burns out the power dissipation will be 10W.
     
  3. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    it is a college course I'm taking. its a home based course though. so I'm teaching myself.( so the question is computer generated)

    anyway the only information I'm given is 4 resistors ,in a series circuit, total power is 20W and one burns out( I think it should be short circuit). so I've assumed that each resistor dissipates 5W. I have no choice but to assume this since no other info. is given.

    maybe the computer question is not valid in the first place? it seems to make no sense. a glich perhaps on the computers side?
     
    Last edited: Feb 17, 2007
  4. dave

    Dave New Member

    Joined:
    Jan 12, 1997
    Messages:
    -
    Likes:
    0


     
  5. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA

    nobody has any idea(s) thought(s) on my problem?

    surely someone does:)
     
  6. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,233
    Likes:
    645
    Location:
    Ex Yorks' Hants UK
    hi north,

    The way I read the question is:

    Four resistors of equal value,connected in series, have a total dissapation of 20Watts.

    One resistor goes short circuit, so whats the dissapation per resistor?

    Working back from the known answer, should give you a clue.

    8.9W * 3 = 26.7W total.

    So (4/3) * 20 = 26.7W

    Hope this helps, leave the rest to you.

    EricG
     
  7. RODALCO

    RODALCO Well-Known Member

    Joined:
    May 21, 2006
    Messages:
    1,234
    Likes:
    43
    Location:
    Akld, New Zealand
    There are no R values given or no voltage.

    The way i work it out as a 20 watt only question.

    20 watts over 4 R's, if one burns out it usually goes open cct, no current flow no power in a series cct dissipated.

    Say one R short circuited then 20W x 4 R's / 3 working R's = 26.6 watts over 3 R's.

    The 20 watt dissipated over 4 R's now has to be dissipated over 3 R's. hence an increase in current and wattage by a factor 4 / 3.

    26.6 / 3 = 8.86 watts per resisitor dissipated as heat
     
    Last edited: Feb 18, 2007
  8. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

    Joined:
    Nov 17, 2003
    Messages:
    39,323
    Likes:
    653
    Location:
    Derbyshire, UK
    There appears to be two possible problems here, firstly the question as posted isn't very clear (this may, or may not, be the posters fault), secondly the question may originally be very badly written, this is all too common in courses and exams!.

    In the second case you not only have to work out the answer, you have to work out the question as well - and hope you answer the question that they are expecting!.
     
  9. Roff

    Roff Well-Known Member

    Joined:
    May 16, 2003
    Messages:
    7,757
    Likes:
    89
    Location:
    Idaho, USA
    When resistors "burn out", they rarely become shorted. They may become shorted from other causes, but not from "burning out". I would assume the resistor is open. This also makes the value of the other resistors irrelevant.
    My guess is that this isn't an arithmetic problem, it's a check on your understanding of series circuits.
     
  10. ljcox

    ljcox Well-Known Member

    Joined:
    Dec 25, 2003
    Messages:
    3,226
    Likes:
    28
    Location:
    Melbourne Australia
    It's a simple case of maths logic.

    Assume that the burnt out resistor is S/C not O/C.

    Initial power P = V^2/4R = 20 Wwhere V is the supply voltage and R is the value of each res. Therefore, V^2/R = 80

    Now, with only 3 R, P = V^2/3R, but we know that V^2/R = 80

    Thus P = 80/3 = 26.7 W, and so power per resistor = 26.7/3 = 8.89 W
     
  11. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA

    but I did not know the answer originally( it was a practice exam question).

    the question was this;

    " The total power in a series circuit with 4 resistors of equal value is 20 W. if one of the resistors becomes short circuited , what is the total power dissipated by each of the remaining resistors? " I did make sure of the wording this time.

    thanks for your reply:) always appreiciated

    north
     
  12. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    in a way this what my tuter did ( it was the weekend and I could not ask him this question and I wanted to take my section exam sunday night) but on monday I did phone and ask him how he would do this situation.

    he used the equation P(T)=E^2/R

    80W=E^2

    and hence both your answers agree. boy there are a few ways of thinking towards the right answer:)

    thanks though your response is appreciated.
     
    Last edited: Feb 23, 2007
  13. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    I did sort out the problem... eventually , thankgoodness!!

    short circuit was the only logical situation. I did work out the answer though , my way.

    I divided 20W/3= 6.67W per resistor added it to the original W of 20( because it turned out to be a short-circuit, hence an increase in current, therefore an increase in the power dissapated by each resistor) = 26.67W and again / 3 = 8.89W.

    thanks for your response it is appreicated!!
     
  14. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    exactly what my tutor did ( I phoned him on monday to see how he would approach the problam).

    thanks for your response. it is , as well , appreciated.


    by the way, and this goes out to all of you that have responsed to my question , is there any vaild reason why my approach is not sound? if not why?

    to reiterate> I took 20W/3R = 6.67W , then added my answer to the original , total P of 20W , = 26.67W , which is again is /3 = 8.89W per R dissipated. I know I got the right answer but thats not my point , could I use this method again and again?
     
  15. ljcox

    ljcox Well-Known Member

    Joined:
    Dec 25, 2003
    Messages:
    3,226
    Likes:
    28
    Location:
    Melbourne Australia
    The problem with your method is that you don't know how you got the right answer and neither do I.

    It may be just a coincidence that your method leads to the right answer in this case.

    Whereas, the method that your tutor and I used is analytical, ie. you obtain the answer by logic and deduction.

    In your first post, you appeared to know that the answer is 8.9W.

    In real life, we don't normally know the answer to a problem otherwise it would not be a problem.

    So you can't do what you appear to have done, ie. juggle the figures until you get the answer.

    In real life, we have to deduce the answer.
     
    Last edited: Feb 24, 2007
  16. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    I think I do know, now at least anyway. it makes sense when you think about it. at least I think so.

    you have originally 20W and 4 resistors. there is a short. which implies a decrease in resistance, increase in current which also implies an increase in power. which means that the overall power , or total power has increased.



    but have I not used deduction if not originally mathematical then by the physical dynamics of the situation?

    for from now on when I come accross this situation there is no reason for me to think , that by my method , one should not come up with the right answer every time.
     
  17. ljcox

    ljcox Well-Known Member

    Joined:
    Dec 25, 2003
    Messages:
    3,226
    Likes:
    28
    Location:
    Melbourne Australia
    I don't understand tour reasoning. You appear to have just juggled the figures until you got the answer.

    Could you have done it if you did not know the answer?
     
  18. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    at first this what I did, I had to , I'm teaching myself ( although I do have access to tutors , during the week , but not on weekends , and I wanted to take the exam on this module on sunday night , therefore I had to figure it out , by juggling the numbers) but I also tried to reason it out.

    since there is a short , and therefore one less resistor , which therefore leads to an increase in current and therefore power which each of the remaining resistors had to dissipate.

    so you had to ADD additional power to the original power quantity.

    which both methods do. its just that the method I chose is faster.

    let me ask you a question that came up last night.

    how would you approach this question;

    you have 2 resistors, R1 has a resistance of 650ohms, R2 ? (is unknown)

    the total resistance is 250ohms. its a parallel circuit. what equation(s) would you use to solve this problem? this how I did it ;

    1/250-1/650=1/R2, then to get R2 alone I divided the answer on the right side by 1 to get rid of the 1 over R2.( which changes nothing on the right). I got the right answer. do you find a problem with my above method that I used here?

    one of my tutors , who returned my call from last night , used the " product over sum rule " equation.

    ( I couldn't ask my tutors last night because a snowstorm shut down the college. so I had to figure it out on my own.)

    anybody else of course can suggest their approach as well!! and please do.

    yes I could've and did to a point I just didn't go far enough and added the answer of 20/3 to the original power quantity and divide by 3 again. but being a student and learning I didn't have the confidence in what I was doing.
     
    Last edited: Mar 2, 2007
  19. ljcox

    ljcox Well-Known Member

    Joined:
    Dec 25, 2003
    Messages:
    3,226
    Likes:
    28
    Location:
    Melbourne Australia
    My problem with your method is that I can't see any rationale behind it.

    If you want to be able to solve any given problem, then you need an analytical approach.

    Your method for the R1 R2 problem is correct.

    Do you know what the " product over sum rule " is?

    If not, see if you can derive it. Hint, it is used to calculate the resultant resistance of 2 resistors in parallel.

    It is generally easier than adding the reciprocols. And it can be varied for this case.
     
  20. north

    north New Member

    Joined:
    Jan 12, 2007
    Messages:
    25
    Likes:
    0
    Location:
    CANADA
    I,ve explained my rational. what specifically do you not understand , in my rational? you keep saying this without giving any specifics.

    sure you do. but not just from a mathematical point of view. but also from a physical dynamic point of view. they ty together.

    I know this. this " product over sum rule " is in the module I just completed.

    explain further
     
  21. ljcox

    ljcox Well-Known Member

    Joined:
    Dec 25, 2003
    Messages:
    3,226
    Likes:
    28
    Location:
    Melbourne Australia
    I think we have to agree to disagree about your method.

    As for the last point, you tell me how to use the " product over sum rule " for the R1 R2 problem. I think you'll learn more by doing it yourself.

    If you can't then I'll write it out for you.
     

Share This Page