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DC Steady State (RLC circuit)

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data

    In the Figure Q1 (c), the switch S was closed for very long time before it is opened at time t = 0. Find the value of vC(t) and iL(t) at t = 0+ c L i.e. immediately after the switch was opened.

    2ewzh1i.jpg


    2. Relevant equations

    V = IR

    3. The attempt at a solution

    From DC steady state, we know that C acts as an open circuit while L acts as a short wire, hence, we will have:

    Current in circuit = 20 / (10+10) = 1A
    vC(t) is thus 20 - (10*1) = 10V.

    However, for iL(t), I am a bit confused, because since this is a short wire, will the 2A current affect the current in this wire? This is because my answer is 1A, however, when I simulate this circuit in LTSpice, it gave me an answer of 3A, which is a bit puzzling. The direction of current would be different, and thus, even if the 2A current source affects the current in the inductor (which is a short), it should be 1-2 = -1A right?

    Thanks for any assistance rendered! :D
     
  2. jcsd
  3. Nov 9, 2014 #2

    Simon Bridge

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    Try replacing the 2A source and 50Omh resistor with their Thevinin equivalent - see if that clears it up for you.
     
  4. Nov 9, 2014 #3
    otngo7.jpg

    I think I know why now. I must add 2A to the 1A already present from the 20V source. Thanks! So that means, unless the right of the inductor are purely resistors, I cannot just ignore the right branches even though the inductor is shorting, right?
     
  5. Nov 9, 2014 #4

    gneill

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    Staff: Mentor

    In general you can't ignore any sources that might contribute to the voltage or current you're interested in. When in doubt, try looking at the circuit via superposition (consider the contributions of each source individually).
     
  6. Nov 9, 2014 #5
    I see. I will keep that in mind. Thank you! :D
     
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