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
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