# Circuit Analysis: Using LaPlace Transform/ Understanding

Discussion in 'Homework Help' started by kevlawliet, Jun 11, 2017.

1. ### kevlawliet Thread Starter New Member

Jun 11, 2017
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I am trying to understand why I'm not getting the same answer of using a certain method for solving circuits. For the following example, the switch opens at t = 0, I am to find the capacitor voltage at t>0.

The first method I tried was to write the differential equation for the capacitor voltage and find the LaPlace transform of it :
$frac{1}{2}frac{d^2v}{dt}+frac{3}{2}frac{dv}{dt}+v=\0$ => $V(s)=\frac{12s+28}{3(s+1)(s+2)$ => $v(t)=frac{16}{3}e^(-t)-frac{4}{3}e^(-2t)$

However, representing it in the s-domain leads me to another equation:
$V(s)=frac{-8s-24}{3s(s+1)(s+2)$ => $v(t)=[frac{16}{3}e^(-t)-frac{4}{3}e^(-2t)-4]u(t)$

Moderators note : removed the plain tags to show the latex

Last edited by a moderator: Jun 11, 2017
2. ### kevlawliet Thread Starter New Member

Jun 11, 2017
3
0
Using differential equation:
$\frac{1}{2}\frac{d^2v}{dt}+\frac{3}{2}\frac{dv}{dt}+v=0$ =>$V(s)=\frac{12s+28}{3(s+1)(s+2)$ => $v(t)=[\frac{16}{3}e^-^t-\frac{4}{3}e^-^2^t]u(t)$
Using s-domain representation:
$V(s)=\frac{-8s-24}{3s(s+1)(s+2)$ => $v(t)=[\frac{16}{3}e^-^t-\frac{4}{3}e^-^2^t-4]u(t)$
As you can see similar answers, but the second one includes -4V. I don' t know why...

Last edited: Jun 11, 2017
3. ### MrAl AAC Fanatic!

Jun 17, 2014
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Last edited: Jun 11, 2017
4. ### MrAl AAC Fanatic!

Jun 17, 2014
5,117
1,109

Hello,

The voltage they are asking for is the voltage across the ORIGINAL cap in the original circuit, so for the second circuit that means the voltage across BOTH the cap and the initial current generator for the cap. This means your second drawing is not correct because you must show the cap voltage as being across both of those elements. If you try to solve for the voltage across the cap alone in the second circuit you'll get a -4 in the result because of that IC generator.

Try that and see if you get the right result. I dont see how you could get this wrong now though

BTW a more direct IC generator for the inductor is a stepped current source.

Last edited: Jun 11, 2017
5. ### kevlawliet Thread Starter New Member

Jun 11, 2017
3
0
Thank you so much, it helps clear up some misunderstandings for me.