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Finding current in RC circuit

  1. May 8, 2014 #1

    dwn

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

    Image

    3. The attempt at a solution

    I know how to solve for a and c.

    a: 300V / (300Ω) = 1 A
    c: 120V / (200Ω) = 0.6 A

    what I don't understand is how they solved for B (0.6A), considering the current is 3A provided by the capacitor.

    I initially tried to use a current divider, but that didn't give me the correct result.
    Since the capacitor acts as a voltage source, I considered trying to find the voltage of the capacitor, v = Ve(-t/RC), but the value of C is not given.
     

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    Last edited: May 8, 2014
  2. jcsd
  3. May 8, 2014 #2

    donpacino

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    Gold Member

    I cannot view the image. Try re-uploading it
     
  4. May 8, 2014 #3

    dwn

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    Hopefully it will work now.
     
  5. May 8, 2014 #4

    donpacino

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    Do you know what the 'u' means in the equations?

    u is the identifier for unit step function.

    if I have an equation X=u(t), that mean X=0 when t<0, and 1 when t>=0

    so if iC=3u(-t). iC=3u(-t) when t<=0, and iC=0 when t>0
     
  6. May 8, 2014 #5

    dwn

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    Yes, I was aware of that. That's how I solved the for a and c. For part b we are to solve for -0.5 second, which means that neither voltage sources are active, and the only source comes from the capacitor (3 amps). Correct?
     
  7. May 8, 2014 #6

    donpacino

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    so by 'a' you mean you're solving for i1 at t=-1.5?
    if that is the case, then your method of solving the problem is incorrect, and you happened to get the correct answer.

    at t=-1.5:
    Va=0
    Vb=0
    iC=3
    you use current division to find i1
    i1=iC*100/(100+200)=1A

    same with C:
    at t=1.5
    vA=300
    vB=-120
    iC=0

    write a kvl
    vA-100*i1+vB-200*i1=0
    i1=0.6 A

    at t=-0.5:
    vA=300u(t-1) since t-1=-1.5, vA=0
    vB=120(t+1) since t+1=0.5, vB=-120
    iC=3u(-t) since -t=0.5, iC=3

    .....
     
  8. May 8, 2014 #7

    dwn

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    Oh wow, I think I was jumping to conclusions and just got lucky. I see the mistake now, thank you very much! Thought I was doing right thing since the answer just happened to work out that way. Wish the textbook wouldn't do that, set's one off in the wrong direction! haha
     
  9. May 8, 2014 #8

    donpacino

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    you're welcome. Mistakes will always happen. The important thing is to rectify them!
     
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