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RL circuit problem

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    The attachment shows the circuit in question. Pardon the crude drawing. The top circuit is the circuit drawn in the text. The middle circuit is how the circuit is supposed to behave for t < 0 (switched closed). The bottom circuit is how the circuit is supposed to behave for t > 0 (switch open). The current source in the top circuit is constant.


    The switch is closed for a very long time and then at t = 0 it opens.
    Derive an expression for i(t) for t > 0.

    2. Relevant equations
    V=iR
    τ=L/R
    i(t)=i(0)e^-(t/τ)

    3. The attempt at a solution
    Looking at the top circuit:
    I handled the case for t<0 to find the Initial current through the inductor (is the initial current in the equation the initial current through the inductor or is it the current flowing through the resistor?) by closing the switch for a longtime and then did a source transformation transforming the current source with a resistor in parallel to a voltage source with a resistor in series.
    vsc=(15A)(24Ω)=360V


    Now I'm at the middle circuit:
    I understand that direct current shorts an inductor.
    This in turn shorts out the 5ohm resistor ( Why? I know that current takes the path of least resistance but...)

    So I can combine the 12Ω and 8Ω resistor and add it to the 24Ω resistor.
    Req=(12||8)+24=36.5Ω

    Therefore Ieq=360/36.5=9.86A

    Then I used current division across the 12Ω and 8Ω resistors in parallel.
    Iο=((8)/(12+8))(9.86A)=3.94A

    After finding the initial current I open the switch and thus arrive at the bottom circuit:
    From here I find RL and then find τ which should be my final answer.

    The answer is 5e^(-2t) A. Could someone please help? Thank you.
     

    Attached Files:

    Last edited: Apr 21, 2015
  2. jcsd
  3. Apr 21, 2015 #2
    The issue is with how you're drawing the middle circuit.

    The 5 Ohm resistor is still there. The thing is, there's no voltage difference across it, because the end of the other two resistors are both connected to ground (so the voltage on the left is 0 and the voltage on the right is 0). So you can conclude that no current passes through the 5 Ohm resistor. So now, you know you only have to find the current through the 12 Ohm resistor to find the current through the inductor (which is short circuited since the switch has been closed for a long time).

    I found it easier to leave the current source in and use current division twice, but to each his own.
     
  4. Apr 21, 2015 #3
    Thank you axmls.
     
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