Graphing the voltage through a capacitor

Discussion in 'Homework Help' started by de1337ed, Mar 2, 2012.

  1. de1337ed

    Thread Starter New Member

    Mar 6, 2011
    23
    0
    Hello, I'm given a current source i(t) = 5sin(10t)*u(t). Where u(t) is the heaviside unit step function (0 at t<0, 1 at t>0).
    I'm given that the capacitance is 2F and that the current source is directly attached to the capacitor.

    I need to graph the current i(t) (this is trivial, and I already did this), and I also need to graph the Voltage Vc.

    I know that i(t) = C*\frac{dVc}{dt}.

    So, I split up the problem into two parts, I said that if
    t<0, it is:
    \int 0dt = K,
    so Vc(t<0) = \frac{K}{2}
    (because capacitance is 2F)

    At t>0:
    \int 5sin(10t)dt = \frac{-cos(10t)}{2} +b
    So Vc = \frac{-cos(10t)}{4} + \frac{b}{2}

    Am I on the right track? I have no idea how to graph this because there are so many constants (like K and b) that I can't seem to get rid of. Thank you
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You simply need to know what you have already found ....

    v(t)=-\frac{1}{4}cos(10t)+const

    at t=0 v(0)=0

    so

    v(0)=0=-\frac{1}{4}+const

    so this will give you the constant as 0.25

    Therefore you simply need to plot

    v(t)=0.25(1-cos(10t))
     
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