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Find current through inductor that parallels a resistor

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data

    [Broken]


    The switch in the circuit has been closed for a long time before it is opened at t = 0.


    Find:

    a. IL(t) for t > 0

    b. i0(t) for t > 0

    c. V0(t) for t > 0


    2. Relevant equations

    equivalent resistance, equivalent current, equivalent voltage

    voltage division, current division,

    V through inductor = L*(dI/dt)

    Thevenin / Norton procedures

    3. The attempt at a solution


    So at the beginning I changed the above to this and simplified the resistors to the right of the inductor to 10Ω with 10 parallel 40 which is 8 and then added to 2 in series to get 10:

    [Broken]


    Then I changed the current source to the Thevenin equivalent voltage to get this:


    [Broken]


    But since the switched was closed for a long time and the current through the inductor wasn't changing, the voltage through the inductor is 0.

    But what I DO NOT know now...


    Does this mean that there's ALSO NO voltage through the 10Ω resistor which now parallels it?


    This would mean that there's only voltage across the 0.1Ω resistor which is 2V across. Therefore the current across the 0.1Ω would be

    2/0.1 = 20A

    So would all the current go across the inductor and not the 10Ω?


    If so, then the current through the inductor at the beginning with the switch closed for a long time would be 20A at t=0 .

    So then with the switch open the equivalent resistance in the loop left over is 10Ω and so the total V is

    10Ω * I +L(dI/dt) = 0;

    10Ω*I = -2H*(dI/dt)

    (-10Ω/2H)dt = dI/I

    integrating both sides gets:

    -5t + c = ln I

    then if t = 0 then I = 20A. Substituting this initial condition gets C = 3 so

    -5t + 3 = ln I; so


    e(-5t + 3) = e(ln I)


    So ILt = 20e(-5t)A. And this is the total current for the rest of the loop in the circuit with the switch opened too.

    Using current division between the 10Ω and 40Ω gets

    I = 4e(-5t)A through the 40Ω

    Since i0 is going in the opposition direction from the diagram it would be

    i0t = -4e(-5t)A

    then V = IR so

    V0t = 4e(-5t)A * 40Ω

    = 80e(-5t)V





    Did I do all of this right or did I go wrong somewhere? Thank you.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 2, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    You've got the right results for IL(t) and io(t). Check your math for Vo(t); what's -4 x 40?
     
  4. May 2, 2013 #3
    Thanks again gneill.

    Rushed the end of my post there though so V0t is actually

    -160e(-5t)V

    So the drop is negative because the current goes in the direction from - to + across V0,

    and would be positive if it was + to - instead, right?
     
  5. May 2, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Yup. Looks good.
     
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