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[Intro Circuits] Simple problem, need help finding initial

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Two independent sources are given. They are active at different times.

    Find the conditions at t = 0- DCSS (aka t < 0 DCSS) and t = ∞ DCSS for IL and Vc

    U9nbJ1X.jpg

    2. Relevant equations

    At DCSS, replace inductor with short circuit
    At DCSS, replace capacitor with open circuit

    3. The attempt at a solution

    So I need to find the initial and final conditions.

    Let's start with initial conditions at t = 0- DCSS. The [2 + 5u(t)]V becomes 2 V. The 4u(t) I am not 100% sure about, but I think it becomes a short circuit. Then I replace the inductor and capacitor with a short and open circuit respectively because it is in DCSS.

    Is this correct? The switches at 2 + 5u(t) and 4u(t) are giving me a bit of a problem understanding and I'm not sure if I'm doing it right.

    As for the final conditions at t = ∞, I have no clue where to start.
     
  2. jcsd
  3. Dec 6, 2015 #2

    gneill

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    Staff: Mentor

    I think you're almost there. Before the "t = 0" event where the 5u(t) and 4u(t) sources switch on, the leftmost voltage source will have been producing 2 V since time began, and the rightmost source will have been producing 0 V (your short circuit) for all that time. So you have to find the steady state conditions for (let's call them) V1 = 2 V and V2 = 0 V.

    Do the open capacitor / short inductor analysis and find the voltages and currents.

    When t = 0 arrives V1 suddenly becomes 2 V + 5 V = 7 V. V2 becomes 4 V. They stay that way from then on, so to the steady state analysis again with these new sources in place in order to find the conditions at t → ∞.
     
  4. Dec 6, 2015 #3
    Ok thanks, so for the t = 0- DCSS, the 2+5u(t) becomes 2v and the 4u(t) becomes a short.

    Now for the t = ∞, the 2+5u(t) becomes 7V and the 4u(t) becomes 4V? I was thinking that the 2+5u(t) becomes 2V and the 4u(t) becomes 4V
     
  5. Dec 6, 2015 #4

    gneill

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    Staff: Mentor

    u(t) is a unit step function. It's zero for all times before t=0, then 1 for all times t > 0.

    So once the u(t)'s have turned on, they stay on.
     
  6. Dec 6, 2015 #5
    Ahhh I see.

    So just to confirm... At both t = 0 and t = ∞, the left side is 7V and the right side is 4V.

    That makes a lot more sense than what I was thinking.
     
  7. Dec 6, 2015 #6

    gneill

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    Staff: Mentor

    Yup. To be specific, for all time up to and including t = 0- the left side is 2 V and the right is 0 V. At t = 0+ the left side becomes 7 V and the right side 4 V and they stay that way indefinitely.
     
  8. Dec 6, 2015 #7
    ok I think I got the final answers. Mind checking if I did them right. I was a bit confused on the open circuit at VC, so I wanted to make sure it's correct.

    At DCSS, replace inductor with short circuit
    At DCSS, replace capacitor with open circuit


    For t = 0- DCSS, I got IL = 0A and VC = 2V

    For t = ∞ DCSS, I got IL = -4A and VC = 11V
     
  9. Dec 6, 2015 #8

    gneill

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    Staff: Mentor

    Looks good except for your Vc at t = ∞. The top end of the resistor should be at 4 V, the left end of the cap at 7V. I don't see an 11 V difference there...
     
  10. Dec 6, 2015 #9
    Hmm, I must have done a sign error or something, this is the work I did for it

    iitbqhi.jpg

    I first found IL which was -4 A. Then I did KVL on the left side which I found to be -VC + 7V - V1 = 0. I subsituted V1 to be (-4 A)*(1 ohm) and solved for the rest. Ended up with 11V.
     
  11. Dec 6, 2015 #10

    donpacino

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    Gold Member

    -VC + 7V - V1 = 0.

    that's your problem. take another look at that equation
     
  12. Dec 6, 2015 #11

    gneill

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    Staff: Mentor

    Look at the circuit. The capacitor is an open so the only current path is for the 4V supply through the resistor. As you say, IL is -4 A, so it's flowing down through the resistor leaving +4 V at its top end. That's the right side of the cap. The left side of the cap is fixed at +7V.
     
  13. Dec 6, 2015 #12
    -VC + 7V + V1 = 0.

    VC = 3 V
     
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