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
790
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

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