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

 

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

\(v(t)=0.25(1-cos(10t))\)