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Find the resistance and inductance ensuing a short-circuit

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    To measure the inductance and resistance of a real inductor, a physicist first connects the inductor to a 3V battery. At these conditions, the final, steady, current is 24 A. Then the physicist suddenly short-circuits the inductor with a thick (resistance-less) wire place across its terminals. The current then decreases from 24 A to 12 A in 0.22 s. Find the resistance and the inductance.


    2. Relevant equations
    ε = 3V
    I1 = 24A
    I2 = 12A
    t = 0.22s

    L = (εL X dt) / DI
    R = L / time constant


    3. The attempt at a solution

    For L: (3v X .22s) / 12A = 0.0275H

    For R: 0.0275H / 0.22s = 0.125Ω

    Would that be right? It seems like what I did is so simple for a physics assignment, I get suspicious of it..
     
  2. jcsd
  3. Nov 12, 2012 #2

    gneill

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    Staff: Mentor

    Find the resistance first; You've got the steady-state current and the potential drop.

    For the inductance, note that the current will follow an exponential function from its initial value, finally decaying down to zero as the energy is bled off by the resistance. What's the equation for the current with respect to time?
     
  4. Nov 12, 2012 #3
    Should I assume it's an RL circuit then? Because if I do then I would get a completely different set of equation...

    For R, would (assuming it's an RL circuit) the equation be:

    I = (ε/R) (I being the initial current at t = 0) ==> R = ε/I = 0.125Ω.. which I just noticed is what I previously found. Does that make sense?
     
  5. Nov 12, 2012 #4

    gneill

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    Well, it's clearly an RL circuit, since it's an inductance in series with a resistance.
    Yes, that's correct.
     
  6. Nov 12, 2012 #5
    Ok, then the next step would be to find the time constant to get from 24A to 12A, here's how I would do it:

    I = ε/R X (1 - e-t/time constant, isolate for time constant:
    IR / ε = 1 - e-t/time constant
    1 - (IR / ε) = e-t/time constant
    ln (1 - (IR / ε) = -t / time constant
    time constant = -t / ln (1 - (IR / ε)

    Once I have the time constant, I would then use:

    Time constant = L/R;

    L = Time constant X R

    I would you 12A as my I since we have the time value for how long it took to reach 24 to 12A

    Does that work as well for part B?
     
  7. Nov 12, 2012 #6

    gneill

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    The current starts with a maximum value (24A) and then decays to zero over time. What's the shape of the resulting curve? What's the shape of the curve that you've specified in the above equation?

    <snip>
    That would work, yes.
     
  8. Nov 12, 2012 #7
    Ok well the current decaying over time, I'm assuming it's non-linear, otherwise it would have been specified, so it's an inverse non-linear graph (I don't know it that is the right term, it's been a while since I studied graphs)

    The formula that I have posted, wouldn't also yield a non-linear result?

    Oh wait I think I see what you mean... The formula, should it simply be: I = ε/R X (e-t/time constant)
     
  9. Nov 12, 2012 #8

    gneill

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    Bingo! :smile:
     
  10. Nov 12, 2012 #9
    Awesome thanks, I just want to make sure I understand correctly though.

    On one end, we have the current that decay from 24A to 0A with a sudden drop from 24A to 12A, thus giving an inversely proportional graph

    On the other end, according to my book (I don't have the reflex to picture the graphs from a formula) the second formula I gave you gives out that graph.

    Makes sense. But I don't understand how that formula gives that graph, do you mind giving me a bit of insight?
     
  11. Nov 12, 2012 #10

    gneill

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    Okay, consider the formula:

    ##y = e^{-x}##

    What value does y take on when x = 0, and when x → +∞ ?
     
  12. Nov 12, 2012 #11
    Ahh put it like that, it makes sense.. Awesome, thanks!
     
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