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Inductor time calculations

  1. Jun 23, 2014 #1
    I was so happy with the assistance provide here by NascentOxygen! Great contributor!
    I have another question:

    1. The problem statement, all variables and given/known data
    A 20V battery is connected to an inductor with 1000μH that has 1000 turns with a cross section area of 1cm^2 is connected for 0.3 Sec during where 20J energy are lost over that period.

    Calculate how much flux has accumulated after 0.3 Sec?



    2. Relevant equations

    This is where I need help. What equation do I start using? I understand it will be an integration with respect to time, but I am not sure where to start

    3. The attempt at a solution
     
  2. jcsd
  3. Jun 23, 2014 #2

    NascentOxygen

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    You are needing an equation that relates current to time, for a circuit comprising the elements relevant to the problem. :wink:
     
  4. Jun 23, 2014 #3
    Yes, i figured that. But not sure where to start looking for that.
    If possible could you post the equation i need to start using.

    so far I understand the 1J = 1A per second. So 20J = 20A per second.
    I'm having difficulty associating the Joules in there.
     
  5. Jun 23, 2014 #4

    NascentOxygen

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    you first need to represent the situation as an electrical circuit, an equivalent circuit. How will you draw it using all ideal electrical elements?
     
  6. Jun 23, 2014 #5
    Drawing it would be a DC supply connected to an inductor via a switch. I've seen many diagrams. I've read there is a small amount of resistance, but I think I should ignore that.

    I've seen the formula I = Vb/R * (1-e^-tR/L)
     
  7. Jun 24, 2014 #6

    NascentOxygen

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    Sometimes it is appropriate to make approximations, or overlook non-idealities, yes. Which element in your equivalent circuit is going to account for the 20 J mentioned?

    Something like that could be useful.
     
  8. Jun 24, 2014 #7
    The inductor has a work measured in Joules = 1/2Li^2 but I just can't get my head around the time component.
    The formula says that the amount of work capacity is capable of inductor, but doesn't allow to incorporate the current in time.
     
  9. Jun 24, 2014 #8

    NascentOxygen

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    Whatever the current happens to be, that determines the stored energy at that moment.

    The inductor stores energy, to release it later. The details about the 20 J uses the word "lost".
     
  10. Jun 24, 2014 #9
    Thankyou for the reply, but I'm still lost myself. What do you mean by the very last sentence?

    I understand the concept after reading up on inductors that the inductor stores energy where current rises steadily, but due the electromagnetic motive force the inductor is resisting that with an opposing force where increasing current increase in flux. When the supply is disconnected the magnetic field dissappears, discharging and causing energy in the opposite direction.
    I've dug up so many formulas but I'm going around in circles with this trying to put it in terms of formula that defines time..
     
    Last edited: Jun 24, 2014
  11. Jun 24, 2014 #10

    NascentOxygen

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    Inductance does tend to impede any change in its current, as you describe.

    Inductance stores energy, later releasing it all back into the circuit. Inductance does not cause a loss of energy from the circuit.
     
  12. Jun 24, 2014 #11
    Is it lost as heat then???
     
  13. Jun 24, 2014 #12

    NascentOxygen

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    Yes.
     
  14. Jun 24, 2014 #13
    I think I have to go back and read more...
    I'm reading a book here... energy is lost to the magnetic field...
     
  15. Jun 24, 2014 #14
    So would that mean the 20J information is not needed for the overall part of calculations since it is converted to heat?
     
  16. Jun 24, 2014 #15

    NascentOxygen

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    Loss due to straying magnetic field lines is typically infinitesimally tiny here.
     
  17. Jun 24, 2014 #16

    NascentOxygen

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    If electrical energy is continually being lost from the circuit, then the inductor will be storing less energy than it otherwise would be. So you can't ignore this energy loss. You'll have to account for it.
     
  18. Jun 25, 2014 #17

    NascentOxygen

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    You account for losses by including the inductor's resistance. You cannot pretend it is zero.
     
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