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cal. short circuit power, in a series circuit.

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

  1. ljcox

    ljcox Well-Known Member

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    I've decided that I don't want to leave the issue hanging, so I'll make one more attempt to explain why I have a problem with your method.

    There are 4 equal resistors in series dissipating a total of 20 W.

    You divide 20 by 3. Why? Dividing by 3 does not fit the information given.

    The 20W is associated with 4 resistors, not 3. So there is no reason to divide by 3 at this point.

    It would make sense to divide by 4 since it gives the dissipation per resistor.

    Then you add 20. Why? 20W is the total power dissipated by the 4 resistors.

    This procedure makes no sense.
     
  2. north

    north New Member

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    good because I'd like to clear up this issue as well.

    There are 4 equal resistors in series dissipating a total of 20 W.

    why? because of the short, remember, in one resistor. which means that the power, 20W is now dissipated over 3 resistors rather than 4.

    I just automaticly divided 20 by 3 it made sense to me.

    at what point do you mean? after or before the short?

    sure at first. I don't think that I'm following you now. for why worry about a a resistor that no longer is involved in the circuit?


    I didn't add 20, I added the result of 20/3 TO 20 = 26.67W

    but the power is no longer dissipated by 4 resistors the power is now dissipated by 3. is it not?
     
  3. Roff

    Roff Well-Known Member

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    Ok, here's my take. To me, the logic goes as follows:

    We know that the voltage is a constant. Let's call it V. :D
    Lets call the power dissipated by 4 resistors P4, and the power dissipated by 3 resistors P3.

    P4=V^2/(4*R)=20
    P3=V^2/(3*R)
    therefore,
    P3/P4=4/3
    P3=P4*(4/3)

    Ok, now let's look at North's solution:

    P3=P4/3 + P4
    simplifying,
    P3=P4*(4/3)
    Same answer (which I think we already knew). :D

    I probably just repeated whan Len (and maybe several others) said.

    I have the same problem as Len. I don't understand how you would come up with North's solution unless you knew the answer and juggled numbers until you got there.
     
  4. dave

    Dave New Member

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  5. north

    north New Member

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    okay I see what your saying but

    you could come up with the answer the way I did it by thinking that because there is a short , there is an increase in current going into the remaining 3 resistors , therefore there is a increase in power dissipated by the remaining 3 resistors. this what I originally failed to consider. so when I divided 20/3 and got 6.67W I should have relised that this the total increase of power by the resistors. had I done so , relised that 6.67W was a total increase of power, it would have made sense to add this to the original 20W.

    inotherwords I was thinking more in terms of the physical dynamics involved than I was mathematically.

    and quite frankly I didn't know how to approach this scenario mathematically otherwise.

    for instance it didn't make sense to me to say multiply 20W*4. why would I do this? because dynamically this isn't happening in the circuit. do you see my point?

    but what is interesting is this , if you take the 6.67W increase and divide this number by 3( each remaining resistor) you get 2.2233W add this to the original 5W when there was 4 resistors and you get 7.2W but the answer is 8.89W which when you divide this answer by 3 you get an increase of 2.96W:confused: why?
     
    Last edited: Mar 3, 2007
  6. ljcox

    ljcox Well-Known Member

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    Ron's method is a variant on mine. We both express the powers as functions of V^2/R. and then eliminate the V^2/R.

    Let's go back to what I posted originally.
    V and R are unknown. But we can find V^2/R.

    For the 4 res case, V^2/4R = 20, so V^/R = 80.

    Now substitute this is the second equation, ie. P = V^2/3R.

    So P = 80/3, then divide by 3 again to obtain the per resistor power.

    P = 80/3/3 = 80/9 = 8.89W.

    It is simply a matter of maths. Once you set up the formulae, you use the techniques learnt in maths to solve it.
     
  7. ljcox

    ljcox Well-Known Member

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    This is irrelevant, it is simply part of solving the maths.
    This is because your basic reasoning is flawed.

    As I wrote earlier, you are just juggling figures. You will not succeed in electronics if you can't apply mathematical reasoning.

    This problem is very simple. If you applied your reasoning to a more difficult problem you would fail.
     
  8. north

    north New Member

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    your right , if I take the diffence between 8.9W and 6.67W=2.22W which is what I get when I divide 6.67 by 3.

    and interstingly enough is that if I mulitply 6.67*4=26.68W , then if I divide this answer by 3 , I get 8.89W. interesting.

    although I'm not quite sure why , yet , I should still include the forth resistor into the equation?

    thanks by the way for hanging in there with me:D as frustrated as you might get.
     
  9. Roff

    Roff Well-Known Member

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    Here's another way to look at it:
    You have a constant voltage source. The load goes from 4*R to 3*R. By what ratio does the current from the voltage source increase? Answer that, then remember that power is V*I.
     
  10. north

    north New Member

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    I think I have understood why the 4th resistor must be included.

    what I did was 4/3= 1.33333W. but what I also did was take 0.33333 and *20 and got = 6.666W.

    so what 0.33333 represents is that for every 1 watt that was originally there(20W) , there is an increase per watt of 0.33333 of the total original wattage because the 4th resistor is no longer considered. and this makes sense since the current in the circuit has increased.

    so now if we take the answer of 4/3=1.33333*20=26.6666 and then 26.6666/3=8.8888W=8.89W

    is my thinking sound here?
     
  11. north

    north New Member

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    1.3:1 , I think

    what do you think of my thinking on my last post? does it make sense?
     
  12. Roff

    Roff Well-Known Member

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    Yep, that makes sense to me, except 4/3 does not equal 1.33333 Watts. 4/3 is just the ratio of the number of resistors you started with to the number you ended with. It has no units.
     
  13. north

    north New Member

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    finally I get it:D although with a small flaw ( ratio )

    thanks Ron and Ijcox for making me think and your both your time.

    as always appreciated

    north
     
  14. ljcox

    ljcox Well-Known Member

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    As I wrote earlier, the 4th resistor must be included since that is part of the initial data, ie. 20W is being dissipated in the 4 resistors. This is your starting point.

    You did not need to do the 0.33333 part.

    As Ron suggested, the current has increased by a factor of 4/3.

    Since P = V * I and V is constant, then P has increased by the same proportion. So P3 = 4/3 * 20. Hence for one resistor the power is

    4/3 * 20/3 = 80/9 = 8.89W.
     
    Last edited: Mar 3, 2007
  15. ljcox

    ljcox Well-Known Member

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    You're welcome.
     
  16. north

    north New Member

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    thanks again.

    I just did not get , at the time , the significance of the 4th resistor, even though it was shorted.

    I just have to get use to, at times , thinking in terms of ratio or porportion. right now its not easy for me to do. it will take time I just hope not to long.
     
  17. ljcox

    ljcox Well-Known Member

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    You still don't understand the process.

    Initially the 4th resistor is NOT shorted. The resistors are assumed to be equal and are dissipating a total of 20W. These facts lead to the first equation (see my maths or do it yourself).

    THEN you short one resistor and write the second equation.

    Then you eliminate the unknowns and find the power dissipated by the 3 resistors. Finally divide by 3 to obtain the per resistor dissipation.
    There is more to it than ratios and proportions.

    I did not simply accept Ron's (no offence to Ron) statement that the currents are in a 4/3 ratio. Firstly, I wrote down the equations and proved that this is true.

    I suggest that you do the same.
     
  18. Roff

    Roff Well-Known Member

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    Len I didn't just make an unsubstantiated statement. I wrote:
    No offense taken.:)
     
  19. ljcox

    ljcox Well-Known Member

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    Ron,
    Sorry, I wrote my post from memory, I should have re-read your post, ie. the one I had in mind.

    I just re-read it and you did not actually say the currents were in a 4/3 ratio, you asked North to determine what the ratio is.

    That's what I did, I did the maths to show that the currents are in a 4/3 ratio.
     

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