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Heat generated in coil

  1. Feb 12, 2015 #1
    1. The problem statement, all variables and given/known data

    What amount of heat will be generated in a coil of resistance R due to a charge q passing through it if the current in the coil decreases down to zero uniformly during a time interval ##\Delta t##
    2. Relevant equations

    ##dH=i^2Rdt##
    ##i=\frac{dq}{dt}##

    3. The attempt at a solution
    ##dH=i^2Rdt##
    ##dH={\frac{dq}{dt}}^2Rdt##
    ##dH=\frac{{dq}^2R}{dt}##
    How do I solve this multi variable differential equation?
     
  2. jcsd
  3. Feb 12, 2015 #2

    haruspex

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    Your are told the current decreases uniformly with time. Write that as an equation.
     
  4. Feb 13, 2015 #3
    ##i=-at##
    ##di=-adt##
    ##dH=iR^2dt##
    ##dH=-aR^2tdt##
    ##H=-aR^2\frac{t^2}{2}##
     
  5. Feb 13, 2015 #4

    BvU

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    I see an unknown a appearing, whereas the known (given) q has disappeared.
    And the math leaves to be desired:
    • ##i = -at## can't be right (try ## t=0## and ## t = \Delta t##)
    • Any reason for ##i^2R## changing into ##iR^2## ?
    With ##t## in the answer I suppose you mean ##\Delta t## ?
    Any chance to present the answer in terms of the givens ?
     
  6. Feb 13, 2015 #5
    Big typing mistake sorry
    ##di/dt = -a##
    This shows that i decreases with time, just like dN/dt=-N for nuclear decay.
    Hence you will again get I=-at.
     
  7. Feb 13, 2015 #6

    haruspex

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    You're forgetting the constant of integration. The current starts positive and finishes at zero. And you need to relate a to q.
    And it is not like nuclear decay which follows a negative exponential with time, not linear.
     
  8. Feb 13, 2015 #7
    ##i = i_0 - at##
    ##dH= (i_0 - at)^2Rdt##
    ##H=i_0^2Rt - aRt^2/2##
     
  9. Feb 13, 2015 #8

    BvU

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    Still this annoying a ...

    And a strange result for the integral...
     
  10. Feb 13, 2015 #9
    ##i=i_0-bt##
    Please help me solve this
     
  11. Feb 13, 2015 #10
    You can't exclude a because the equation will become dimensionally incorrect
     
  12. Feb 13, 2015 #11

    BvU

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    Now it fits for t=0. Next: make it fit for ## t = \Delta t##.
    And (to keep in the back of your mind): ##q = \int_0^{\Delta t}\; i \;dt ## to eliminate ##I_0##

    [edit] from zero to Delta t
     
  13. Feb 13, 2015 #12
    ##i=\frac{q}{\Delta t} - at##
     
  14. Feb 13, 2015 #13
    No that doesn't seem right.
     
  15. Feb 13, 2015 #14
    ##i=i_0 - \frac{i_0t}{\Delta t}##
     
  16. Feb 13, 2015 #15

    BvU

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    ##i=i_0-at## not good ? try ##i=i_0-bt## or ##i=i_0-ct## :)

    Reminded me of this joke where you see Einstein in front of a blackboard with ## E=ma^2 ## and ## E=mb^2 ## stricken through, and then ## E=mc^2 ## ! (with an exclamation mark). Sorry.

    What I am pushing towards: ##i=i_0-at## is correct, but not good enough. t = 0 works fine, now fill in ##\Delta t## when you know ##i(\Delta t) = 0 ## you get a.
    Bit now you have ##i_0## as unknown. However, ... (see post #11)

    [edit] crossed post #14. Bingo !
     
  17. Feb 13, 2015 #16
    Ok...here comes my updated equation
    ##i=\frac{q}{\Delta t} - \frac{q}{(\Delta t)^2}t##
     
  18. Feb 13, 2015 #17

    BvU

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    Almost :)
     
  19. Feb 13, 2015 #18
    Is it correct?
    Now can I use i2Rt?
     
  20. Feb 13, 2015 #19

    BvU

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    Almost means: not there yet: $$q \; {?\atop =}\ \int_0^{\Delta t}\; i \;dt = {q\over \Delta t} \int_0^{\Delta t} \;(1-{t\over \Delta t}) \; dt = {q} \int_0^{\Delta t} \;(1-{t\over \Delta t}) \; d({t\over \Delta t}) = q\int_0^1 (1-u) du = {q\over 2}$$so there is a 2 missing. Obvious if you sketch i(t), a triangle.
     
  21. Feb 13, 2015 #20
    At t=0,you get ##q=i\Delta t##.
    Here you get Q=q/2
    What is the difference between the two 'q's?
    (I got the answer after using post 19. Thank you very much)
     
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