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Superposition with AC voltage source and DC current source

  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Using the superposition principle, determine the current i(t.
    oT07MYo.png

    2. Relevant equations

    Zc = 1/jwC
    Zl = jwL
    V = I*R
    I = V/Z

    3. The attempt at a solution
    First, I converted inductor/capacitor to impedance:
    L = 1.5H -> jwL = j * 10 * 1.5 = 15j
    C = 10mf = 1/(jwC) = 1/(j*10*10*10^-3) = -0.1j

    Then, I transformed the 3A DC current source and 10 ohm resistor into a DC voltage source:
    IIz1hcB.png

    I then shorted the 30V DC source to find i(t) with only Vs(t) = 10cos(10t)V
    After shorting the 30V DC source, C2 and R3 are in parallel. Their impedance is:

    (0.1j * 10)/(10-0.1j) = -j/(10-0.1j)
    After rationalizing: (0.1 - j)/100.01

    Now that, L2 and R4 are in series, total impedance:
    15j + 5 + (0.1 - j)/100.01 = (500.15-1499.15j)/100.01 = 15.8 < -71.55°.
    I = V/Z = (10<0°)/(15.88<-71.55° ) = 10/15.88<71.55°A

    Now I short the AC source:
    D9em2vN.png


    The DC source had a frequency of 0 rad/s, so the impedance of C3 is infinity (1/0) and L3 is 0 (j*0*L).
    The inductor is shorted and the capacitor is opened. There is a voltage source of 30V and a combined resistance of 15ohms. V=IR -> I = 30/15 = 2A<0°
    [(500.15/100.01 + 2) + (1499.15j/100.01)]
    Is 16.54<64.96° A the correct answer?
     
    Last edited: Oct 10, 2016
  2. jcsd
  3. Oct 10, 2016 #2

    gneill

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    Staff: Mentor

    Check your capacitor impedance.
     
  4. Oct 10, 2016 #3
    Is it -10j instead of -0.1j?
     
  5. Oct 10, 2016 #4

    gneill

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    It is indeed!
     
  6. Oct 10, 2016 #5
    My new current for part 1 is 10/14.14 <-45°
    Is the final answer (2.5 + 0.5j) A?
     
  7. Oct 10, 2016 #6

    gneill

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    That looks good!
     
  8. Oct 10, 2016 #7

    gneill

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    Oops, hang on. I think your signs may be incorrect. The diagram defines the current direction as flowing out of the AC source. So the DC contribution to that must be negative.
     
  9. Oct 10, 2016 #8

    gneill

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    Also, if the AC current angle is -45°, its imaginary part should be negative...
     
  10. Oct 10, 2016 #9
    It was negative on my paper, I don't know why I typed positive.
    The final answer should be:
    (-1.5 - 0.5j)A, correct?
     
  11. Oct 10, 2016 #10

    gneill

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    Yes, that looks good.
     
  12. Oct 10, 2016 #11
    Thank you so much for your help!
     
  13. Oct 13, 2016 #12
    the final answer is -1.5 - 0.5j amps
    If w = 10, would the answer in time domain be:
    [tex]\sqrt{2.5} * cos(10t + 18.435°)A[/tex]
    ?
     
  14. Oct 13, 2016 #13

    gneill

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    The DC and AC components must be kept separate. There will be a constant DC current with an AC component superimposed on it, or if you like, the AC signal will be offset by a DC bias.
     
  15. Oct 13, 2016 #14
    How would I write it out? I need the answer in the time domain.
     
  16. Oct 13, 2016 #15

    gneill

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    A DC current is just a fixed value in the time domain; a number. So your result should have the form:

    ##I(t) = I_{dc} + I_{ac}##

    where ##I_{ac}## is your cosine expression (the part that varies with time).
     
  17. Oct 13, 2016 #16
    So:

    I(t) = sqrt(2)/2 * cos(10t - 45°) - 2
     
  18. Oct 13, 2016 #17

    gneill

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    Staff: Mentor

    Yes. That looks better!
     
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