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Homework Help: RLC Circuit w/ switch

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Provided in the picture below.


    2. Relevant equations



    3. The attempt at a solution
    So what I've done so far is use KVL to obtain Vc(t) +Vr(t) + Vo(t) = 1/2Vin(t).
    Could someone tell me if I started this problem correctly? Would I just take the derivative of the equation next?
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2010 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. I think the input Vi is meant to have the value of 1/2V, but it's hard to tell for sure from the drawing.

    And yes, writing the KVL around the loop is a good approach, but you need to use the differential equations that relate I and V for the capacitor and inductor. That's where you end up with one differential and one integral. And yes, you then differentiate that equation to get a 2nd order DE, which you then solve and apply the initial conditions to.
     
  4. Apr 14, 2010 #3
    Thanks for the reply! And yes I think you're right, it is probably 1/2 V. So then the differential equation would look like: Vc(t) + VL(t) + Vo(t) = 1/2 .
    Vo(t) = Ri(t)
    VL(t) = Ldi(t)/dt
    Vc(t) = 1/C integral from -infinity to t of i(a)da
    So if I plug these into the equation and take the derivative, would I get:
    1/C*i(t) + Ld^2i(t)/dt^t + Rdi(t)/dt = 0? So the 1/2 would just be irrelevant?

    So for these C, L, and R variables, would I just substitute these for the numbers given in the problem? Say C = 1/2, L= 1/4, and R = 1?


    PS: Is there a program/website I can use to make these equations look nicer? I'm sure what I'm typing must be hard to read lol.
     
  5. Apr 15, 2010 #4
    So I was hoping if someone could still help me with this equation?
    From the above equation: 1/C*i(t) + Ld^2i(t)/dt^t + Rdi(t)/dt = 0?
    Since i(t) = 1/R * Vo(t), I would substitue i(t) and the values of C, L and R and get:
    1/4*d^2Vo(t)/dt + dVo(t)/dt + 2Vo(t) = 0
    So I assume Vo(t) = Ae^st and get the roots: s = -2+-2i
    Can anyone tell me what to do next or if I'm doing this correctly? Thanks
     
  6. Apr 16, 2010 #5

    berkeman

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    Staff: Mentor

    The response will be a damped sinusoid, so your exponential term should have both sigma and j*omege in it. Something like:

    [tex]V_o(t) = A e^{B(\sigma + j\omega)}[/tex]

    Use that, differentiate, substitute back, and apply initial conditions to solve for the constants.
     
  7. Apr 16, 2010 #6
    Thanks so much! Now the trouble I'm having is knowing what are the initial conditions. I can't seem to understand it based on the circuit given. Do you think you could help explain it to me?
     
  8. Apr 16, 2010 #7

    berkeman

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    Staff: Mentor

    Glad that helped. BTW, the coefficient B may just be 1, but I'm not sure. You should be able to tell as you apply the ICs.

    At time t=0-, the switch is open, so there is zero current, and zero voltage across the elements to the right of the switch. At t=0, the switch is closed, so all of a sudden you have the supply voltage across the series RLC combination. That will let the current start to build, and the voltage division between those 3 components will start to change with time...
     
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