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V(t) on a Capacitor from i(t)

  1. Feb 25, 2016 #1
    1. The problem statement, all variables and given/known data

    zGhum8Q.png

    imgur link: http://i.imgur.com/zGhum8Q.png

    The answer given in the back of the text is

    hAprWPV.png

    imgur link: http://i.imgur.com/hAprWPV.png

    2. Relevant equations

    [tex]v(t) = \frac{1}{C_{eq}}\int i_s dt[/tex]

    3. The attempt at a solution

    I get the last piece of the piece-wise v(t) to be [itex]0.75t^2 - 7.5t + 7.5[/itex], and I can't figure out where 23.25 could come from.

    If we just look at the integral of i(t) over the range 0 to 5s as being area under its curve, then when we get to the next piece of the function, we need to make the constant of integration of that next piece equal to where the last piece brought us to, yes?

    So, the first piece takes us from 0 to 1.5 kV, no worries, 1.5 becomes the constant of integration when integrating the second piece of i(t), and then we move for 2 seconds through the second piece, and by that time we should be at 7.5 kV, because that second piece of the i(t) function should add 6 kV to where we are when we integrate it (ie, the area under that portion of the curve), hence once we're in the third piece of the i(t) and integrating that piece, shouldn't we be starting at 7.5 kV, why would we be starting at 23.25 kV?
     
  2. jcsd
  3. Feb 25, 2016 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Text color change mine.

    Yes, in the third piece we are starting at 7.5 kV. Just keep in mind that that piece starts when t = 3.0 sec (not t = 0.0 sec).

    If it helps, evaluate [itex] [0.75 t^2 - 7.5 t +23.25] \ \mathrm{kV} [/itex] when t = 3.0 sec. :wink:
     
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