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Math Inductance help

  1. Jul 14, 2006 #1
    I'm having some problems with inductances and I want to know on i) what is the value of L at its terminals for t<0...
    I did an equivalent circuit of the original one when t=0 and this circuit consists of the inductor and R2+R3 in series...
    I considered t(seconds)<0 and the L acts like a short Circuit
    thats gives me the inical current of L which is -10ma.

    Then I used the following formula : L *di/dt + Ri <=> di/i=-R/L *i * dt

    integer of i(t) i(t0) dx/x = -R/L integer (t) (t0) 1dy

    ln( i(t)/i(0) ) = -R/Lt
    i(0+)=i(0-)=I=Io
    i(t)= Io * e^-((R/L)t)

    http://i75.photobucket.com/albums/i281/esmeco/inductance.jpg

    Is this okay?Thanks in advance for the help!
     
  2. jcsd
  3. Jul 14, 2006 #2

    berkeman

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    Staff: Mentor

    This is a little confusing, but here goes. You are correct that for a DC situation (and ignoring the winding resistance of the inductor), that the inductor looks like a short. So before t=0 (if I understand the figure correctly), R1 and R2 are the parallel-connected loads driven by I1. So I through L is not the full 10mA before t=0. What is it instead?

    And it looks like at t=0, U3 opens and U1 closes. What was the initial voltage across R2 just before t=0 (from the current solved for above)? When the switch closes, the current through the inductor will not instantaneously change, so the current through R2 will do likewise. What does that mean that the voltage across R2 has to become, in order for the loop voltage to equal zero around the right-hand loop?

    Once you have these initial voltages, then you just have the RL decay of the current over time, with the resulting decay of the voltages. Your exponential decay equation (the last one) looks okay, as long as you figure out the starting current Io correctly, and figure out what the voltages do as a result. Does that help?
     
  4. Jul 14, 2006 #3
    Although I've understood the 1st part,I'm not quite understanding the part where you ask "What was the initial voltage across R2 just before t=0 (from the current solved for above)?"....Could you explain it again?
     
  5. Jul 14, 2006 #4

    berkeman

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    Staff: Mentor

    If you understood the first part, then you know that the initial conditions at t=0- are I=5mA through the inductor and R2. That gives you 5V across R2 and 0V across the inductor. As the 5mA inductor current is diverted for t>0 through R3, that gives the initial -5V across R3 so that the sum of the loop voltages is zero. Then as the inductor current decays exponentially, the voltages across R2 and R3 do as well. BTW, what did the voltage across R1 do right at t=0+?
     
  6. Jul 14, 2006 #5
    Sorry,I made a mistake...The exercise I have doubts in is ii) where we want to know the current through L for t>=0....

    In exercise i) where we want to know the current through the inductance for t<0 I did an equation for the node when the switch is closed on the left sideand open on the right side...

    Eq.: -10 + v1/1k + v1/1k =0

    I'm not sure if this's okay,so any help is appreciated...
     
  7. Jul 14, 2006 #6

    berkeman

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    Staff: Mentor

    Yeah, that's how you get 5mA for the initial current.
     
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