Capacitor in series

Thread Starter

tatoearashiga

Joined May 23, 2012
4
Check attached.

So, all elements are series connected

The circuit (initially uncharged , V(0) =0 ) is consisted of a current source Is, series connected with C1 = 6uF and C2 =3uF. Vo=Vof3uF

From the graph I got,

For 0<t<1 sec, I = 90t mA.
For 1<t<2 sec, I= 180-90t mA

Then, transforming it to voltage

For 0<t<1 sec,
Vo(t)=(1/C) integral [90t^{2}]dt+ 0

My problem is what value of capacitor to plug in the formula.
 

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Last edited:

shteii01

Joined Feb 19, 2010
4,644
Umm... How did integral of t^2 became t^2?
For that matter. Where did the integral of t^2 come from?
You said that at the interval 0< t <1 you have 90t, not 90t^2. So. Where did the t^2 come from?

Integral of t with respect to t is ½*t^2. So you should have had (30 kV/2)*t^2=15t^2 kV.
 
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