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Instantaneous current in coil

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    In the inductive circuit shown in Figure 2, the switch S1 is closed 0.75s after switch S2 is closed. Calculate the instantaneous current in the coil 1.5s after switch S1 is closed.
    upload_2016-3-27_13-10-3.png

    2. Relevant equations
    T=L/R
    I=V/R

    3. The attempt at a solution
    When S2 is closed:
    T=L/R
    =2.4/7.2
    =0.333
    =333ms

    IMAX=V/R
    =24/7.2
    =3.33A

    At time 0.75s S1 is closed, the current is:
    i1=IMAX(1-e^-t/T)
    =24/3.6(1-e^-750/333)
    =5.97A

    When S1 is closed current flows through the 3.6ohm resistor and decays:
    Time constant is:
    T=L/R
    =2.4/3.6
    =0.666
    =666ms

    Current after 1.5s after S1 is closed:
    i2=i1e^-t/T
    =5.97e^-(1.5x10^-3/666)
    =5.97A

    This cannot be correct?
     
  2. jcsd
  3. Mar 27, 2016 #2

    gneill

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    Staff: Mentor

    What's the total resistance in the series circuit when only S2 is closed?
     
  4. Mar 27, 2016 #3
    7.2ohms+3.6ohms = 10.8ohms

    Paris_Tuileries_Garden_Facepalm_statue.jpg
     
  5. Mar 27, 2016 #4
    I get an answer of 6.44A for i1 and i2
     
  6. Mar 27, 2016 #5

    gneill

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    Staff: Mentor

    That doesn't look right to me. Can you show some details?

    What was the current just before S1 closed?
    After S1 closes, what final current is the circuit heading for?
     
  7. Mar 27, 2016 #6
    When S2 is closed:
    T=L/R
    =2.4/7.2+3.6
    =0.222
    =222ms

    IMAX=V/R
    =24/7.2+3.6
    =2.22A

    At time 0.75s S1 is closed, the current is:
    i1=IMAX(1-e^-t/T)
    =24/3.6(1-e^-750/222)
    =6.44A

    When S1 is closed current flows through the 3.6ohm resistor and decays:
    Time constant is:
    T=L/R
    =2.4/3.6
    =0.666
    =666ms

    Current after 1.5s after S1 is closed:
    i2=i1e^-t/T
    =6.44e^-(1.5x10^-3/666)
    =6.44A
     
  8. Mar 27, 2016 #7

    gneill

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    Staff: Mentor

    Why didn't you use the IMAX value that you found? You've dropped the 7.2 Ω resistor again.

    Edit: Sorry, I just realized that you were writing an expression for I(t) for the second circuit and that you're implying a new IMAX. Two problems though, you used the same time constant as for the previous circuit configuration (it should change when the total resistance changes), and you didn't take into account the initial current that the circuit starts with at the instant S1 closed.
    Why does it decay? What is the final current that the circuit is heading for with a 24 V supply and 3.6 Ω? Is this current smaller or greater than the initial current (at the instant S1 closes)?
     
    Last edited: Mar 27, 2016
  9. Mar 27, 2016 #8
    Final current is I=V/R 24/3.6=6.66A the 7.2ohm resistor is bypassed when S1 is closed? The current is greater because when S1 is open we pass through both the 7..2ohm and 3.6ohm resistors.

    Which step looks wrong to you? Am I even headed in the right direction with my attempt?
     
  10. Mar 27, 2016 #9

    gneill

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    Staff: Mentor

    You haven't determined the current in the circuit for the instant *before* S1 closes. That current is the result of the the initial configuration (with both resistors) running for 0.75 seconds. That current is also the initial current that the new configuration starts with when S1 closes. You need to take that initial current into account.
     
  11. Mar 27, 2016 #10
    I=V/R
    I=24/10.8
    =2.22A

    AT 0.75S:
    i1=24/10.8(1-e^-(750/222)
    = 2.15A
     
  12. Mar 27, 2016 #11

    gneill

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    Staff: Mentor

    Yup. You might want to keep a couple more digits as this is an intermediate step value that will be used for further calculations. Only round final results for presentation.

    Now, how are you going to use this as the initial current for the next step (S1 closes)?
     
  13. Mar 27, 2016 #12
    Thank you for your help thus far :)

    Okay how does this look:
    When S2 is closed:
    T=L/R
    =2.4/10.8
    =0.222
    =222ms

    IMAX=V/R
    =24/10.8
    =2.22A

    At time 0.75s S1 is closed, the current is:
    i1=IMAX(1-e^-t/T)
    =24/10.8(1-e^-750/222)
    =2.15A

    When S1 is closed current flows through the 3.6ohm resistor:
    Time constant is:
    T=L/R
    =2.4/3.6
    =0.666
    =666ms

    Current after 1.5s after S1 is closed:
    i2=i1e^-t/T
    =2.15e^-(1500/666) Note: 1.5 seconds = 1500ms right?
    =0.23A
     
  14. Mar 27, 2016 #13

    cnh1995

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    Homework Helper

    Right. Now from this value, the current increases exponentially to 6.66A when S1 is closed. Write an equation which gives you i=2.15A for t=0 and i=6.66A for t=∞.(assuming S1 is closed at t=0)
     
    Last edited: Mar 27, 2016
  15. Mar 27, 2016 #14

    gneill

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    Staff: Mentor

    The current will start at ~ 2.15A, but where will it be headed? What would be the final current value after a long time has passed?
     
  16. Mar 27, 2016 #15

    cnh1995

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    Homework Helper

    How will the current decay? Check your equation for i2.
    Assume switch S1 is closed at t=0 and initial current as 2.15A.
     
  17. Mar 27, 2016 #16
    Final current is I=V/R 24/10.8=2.22A?
    So current increases..
    I'm confused?
     
  18. Mar 27, 2016 #17

    gneill

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    Staff: Mentor

    No, for this part switch S1 is closed. What is the steady state current for this configuration?
     
  19. Mar 27, 2016 #18

    cnh1995

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    Homework Helper

    You've got it right already..
     
  20. Mar 27, 2016 #19
    Sorry that is my mistake I should have written I=V/R 24/3.6=6.66A

    Okay so initial current is 2.15A and final current is 6.66A

    How do I find instantaneous current in the coil 1.5s after S1 closes?
     
    Last edited: Mar 27, 2016
  21. Mar 27, 2016 #20

    gneill

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    Staff: Mentor

    So you know that the current will start at 2.15 A and head towards 6.67 A, following an exponential curve. How might you write an expression for this?
     
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