# Capacitor in series

Discussion in 'Homework Help' started by tatoearashiga, Apr 8, 2014.

1. ### tatoearashiga Thread Starter New Member

May 23, 2012
4
0
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.

• ###### 2capacitors.png
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Last edited: Apr 8, 2014
2. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,529
718
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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3. ### tatoearashiga Thread Starter New Member

May 23, 2012
4
0
Thanks, I got it. I did an error in integrating.