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RL Circuit, need help

  1. Jun 13, 2015 #1
    1. The problem statement, all variables and given/known data

    Using Kirchhoff's current law, determine:

    1) The nodal equations
    2) The expression and solution for the inductor voltage
    3) The expression and solution for the inductor current
    4) Determine the initial and final inductor current

    Circuit diagram:

    WP_20150613_004.jpg


    2. Relevant equations


    3. The attempt at a solution

    I'm really stuck with this question. The best I could try and do was question one which I don't think is correct but here it is

    1) (9-V1)/12 = (V1-V2)/Zl + V1/6

    V1 is the voltage across the 6 ohm resistor and V2 is the voltage across the inductor
     
  2. jcsd
  3. Jun 13, 2015 #2

    mfb

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    Do you expect those voltages to be different?
    Probably just a typo in the V1 definition.

    Independent of the definition of V1 and V2, this term looks wrong.
     
  4. Jun 13, 2015 #3

    gneill

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    Since you've got a reactive component in the circuit (an inductor), you should realize that you are going to end up with a differential equation. Impedance won't help unless you move to the Laplace Domain, which I'm guessing you haven't covered yet.

    What do you know about the relationship between the voltage and current for an inductor? Can you write an expressions for the current as a function of the voltage and vice versa?
     
  5. Jun 13, 2015 #4
    So if the voltages are the same since they are in parallel, then from the equation

    (9-V1)/12 = (V1-V2)/Zl +V1/6

    since V1-V2 = 0

    then, solving for V1 I get 3V which is the voltage across the 6 ohm resistor and the inductor ?
     
  6. Jun 13, 2015 #5

    mfb

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    If V1 and V2 are as you described, then the equation you set up for current does not make sense.
     
  7. Jun 13, 2015 #6

    phinds

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    You should answer the question gneill asked in post #3. Until you get that right, there is no point in proceeding with your effort.
     
  8. Jun 13, 2015 #7
    Ok, I know that V = L*di/dt , so

    I(t) = 1/L integral V2 dt

    so, 9-V2/12 = V2/6 + 1/L integral V2 dt ?, so I just have to make an expression for V2 in terms of t ?
     
  9. Jun 13, 2015 #8

    mfb

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    Integrating over the voltage is not so helpful. Better look at voltages and derivatives of the current.
     
  10. Jun 13, 2015 #9

    gneill

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    That's the idea; you write the node equation incorporating the ##I(t) = \frac{1}{L} \int V_2 dt## for the inductor branch's current. That gives you an integro-differential equation. Collect the ##V_2## terms together.

    Hint: you will want to differentiate the whole thing to clear the integral and leave the equation in purely differential form.
     
  11. Jun 14, 2015 #10
    Okay thanks, I found the voltage expression to be 3*e^-800t

    Also the general expression for the voltage of an inductor in an RL circuit is Vo*e^-(Rp/L)t, so to do this without integration you could have found Vo by removing the inductor from the circuit because it acts as an open at t=0 and measure the voltage across the 6 ohm resistor which you can get using the voltage divider rule. Rp would just be the 12 and 6 ohm resistors and the inductance is 5mh

    Anyway I think they wanted you to derive the expression which is why they asked for the nodal equations
     
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