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Second-order linear voltage equation

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data
    0156ee252979510.jpg
    vs = 300 (t>0)
    vs = 150 (t<0)
    v(0+)≠0
    i(0+)≠0
    R=50ohm
    L=10H
    C=1/100F

    Determine v(t) when t>0

    2. Relevant equations

    3. The attempt at a solution
    I wrote node voltage equation
    (1/R)*v+(1/L)∫(v-vs)dt+C.(dv/dt)=0

    First I do is get derivation to two sides of the equation to escape the integral, after that, replace all known datas, I got the equation
    v''+2v+10v=10vs
    So we have A(s)=s2+2s+10=0 ⇔ s=-1±3i
    v(t)=vn+vp
    which vn = 150V (cause we have t>0)
    → v(t)=150 + e[itex]^{-t}[/itex].(Acos3t+Bsin3t)

    I am stucking on the way to find A and B constants, all I can do is below
    v(0+)=150+A
    v(0+)'=-A+3B
     
  2. jcsd
  3. May 6, 2013 #2

    rude man

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    EDIT: determine the initial conditions v (0+) and dv/dt (0+) and then solve your differential equation with those two initial conditions.

    You should be able to determine v(0+) and v'(0+) by inspecting the diagram. v(0+) is fairly obvious (voltage cannot change instantaneously across a capacitor) but v'(0+) requires some thought.

    Hint: i = C dv/dt. What is initial current thru the capacitor?

    Second hint: your equation to solve is
    v'' + 2v' + 10 = 10Vs = 3000. The 150V for t < 0 does not come into the equation directly as you have shown.

    Your equation for v is actually almost correct except for the "150". But you need to work on the initial conditions. When you have those you can determine A and B.
     
    Last edited: May 6, 2013
  4. May 6, 2013 #3
    So my correct equation for v(t) should be v(t)= 300 + e[itex]^{-t}[/itex].(Asin3t+Bcos3t), right ?
    First hint : i(0+)=C.d(v0+)/dt ?
    Second hint : I think the equation should be v'' + 2v' + 10v = 3000, but what is the reason to solve it ?
     
  5. May 6, 2013 #4

    rude man

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    right
    can you use that to determine v'(0+)? Think about what the currents are doing going thru L, R and C just after the switch is thrown at t = 0+.
    Well, by now you already have solved part of it. I just meant that's the full equation you need to solve. And you haven't until you figured out A and B.

    Now you need to finish up by solving for A and B given the initial conditions on v and v'.
     
  6. May 6, 2013 #5
    I am stucking at this point. Have no idea for it .
     
  7. May 7, 2013 #6

    rude man

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    1. Have you come up with v(0+) yet?

    2. What is the current thru the inductor just before switching the source from 150V to 300V, i.e. t = 0-?
    What is the current thru R at that time?
    What is the current into C? (Obviously zero since the source had been = 150V for a long time).

    OK, so now we switch the source voltage from 150V to 300V and it's t = t(0+). What now is the current thru L and R? And C?
    Then, iC = C dv/dt. Doesn't that tell you what v'(0+) = dv/dt at t = t(0+) must be?
     
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