# Finding the voltage across capacitor in s-domain circuit (solution provided)

Discussion in 'Homework Help' started by trougnouf, Nov 24, 2014.

1. ### trougnouf Thread Starter New Member

Nov 24, 2014
2
0
Hello all,
I am stuck on the following problem: http://wstaw.org/m/2014/11/24/plasma-desktopGXk814.png
The result is circled in blue but I don't understand how it got there.

The way I would solve this problem would be to apply kvl on both sides of the circuit, with I_L going right on the left circuit, I_B going right on the right circuit, and I_A going down on the right circuit. (I_B being the current going across the capacitor)
I get the following equations:
1 - s*I_L + 1 - I_L - I_A = 0
(-1/s)(I_B) - 1/s + I_A = 0
I_L = I_A + I_B

At this point I solve the equations and get I_B = (s-2) / (s^2 + 2s + 2)
Now I'm not sure whether V_c(s) is just the voltage across the capacitor (I_B * 1/s) or if I should also add the voltage source created during the s-domain transformation (1/s), but either way my answer isn't right so I am doing this wrong.

2. ### trougnouf Thread Starter New Member

Nov 24, 2014
2
0
My question has been answered elsewhere:

"Look at the circuit right above it. You have a node at Vc. Use KCL. Looking to the left you have Vc-1/s+2 Looking down you have Vc/1 and looking to the right you have (Vc-1/s)/(1/s). The equations come from that node at Vc in that circuit."