# Determine Capacitor's Voltage graph by its Current graph...

Status
Not open for further replies.

#### BlackOps

##### Full Member level 5
I am very confused with very simple thing...

i have developed simple problem, for example on the pic u see the graph of the Current which was applied to the capacitor... now i must know how will the Voltage graph look like...

Capacitor = 3µF

so:

v(t) = 1/C * (∫t0-t i(t')dt') + v0

the graph of I(t) on the interval from 0mS to 3mS is x^2, is parabola.

i can say that the V-t graph on interval from 3mS to 6mS will look like on picture... it will be straignt line. because: V(t) = 1/C * area of the rectangle ===> V(t) = (1/3 * 10^-6) * 3 * 10^-3 * 9 * 10^-3 = 9V...

but i am confused about how will look the graph on V-t from 0mS to 3mS interval...

and please derive the step by step integral equation for computing the expression from 0 to 3 ms

thanks, sometimes simple things make me confused i will be very gratefull if u sketch correct graph and upload it here...

Re: Determine Capacitor's Voltage graph by its Current graph

ok i figured it out correctly i think...here is the graph, correct me if it is wrong. thanks

Re: Determine Capacitor's Voltage graph by its Current graph

hello, I think it can be
sence i= C*du/dt
=> U=1/C*∫i*dt + const
on period 0< t < 3 ms
i=t²
We are easy to see
u=1/C*∫t²dt (t=0..k / k<3ms )
u1= k³/(3*C)
on on period 3 ms < t <6ms
do example with i= 9ma
u2=1/C*∫9dt + Const2
u2=(9/C) *t + Const2

when t=3ms u1=u2 you can see const2 = ?
so that grapphich is

Points: 2