Capacitor in series

Discussion in 'Homework Help' started by tatoearashiga, Apr 8, 2014.

  1. tatoearashiga

    Thread Starter New Member

    May 23, 2012
    4
    0
    Check attached.

    So, all elements are series connected

    The circuit (initially uncharged , V(0) =0 ) is consisted of a current source Is, series connected with C1 = 6uF and C2 =3uF. Vo=Vof3uF

    From the graph I got,

    For 0<t<1 sec, I = 90t mA.
    For 1<t<2 sec, I= 180-90t mA

    Then, transforming it to voltage

    For 0<t<1 sec,
    Vo(t)=(1/C) integral [90t^{2}]dt+ 0

    My problem is what value of capacitor to plug in the formula.
     
    Last edited: Apr 8, 2014
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,398
    497
    Umm... How did integral of t^2 became t^2?
    For that matter. Where did the integral of t^2 come from?
    You said that at the interval 0< t <1 you have 90t, not 90t^2. So. Where did the t^2 come from?

    Integral of t with respect to t is ½*t^2. So you should have had (30 kV/2)*t^2=15t^2 kV.
     
    tatoearashiga likes this.
  3. tatoearashiga

    Thread Starter New Member

    May 23, 2012
    4
    0
    Thanks, I got it. I did an error in integrating.
     
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