Node analysis of RLC circuit using Laplace transform

Discussion in 'Homework Help' started by abhaymv, Aug 7, 2011.

1. abhaymv Thread Starter Active Member

Aug 6, 2011
105
4
Hello everyone,

I have been having some problems with the circuit attached.We were asked to find V1 and V2 at the nodes.The circuit opens at t=0 and disconnects from the Voltage source.The source voltage is 1 Volt.The initial values are:
Vc(0+)=1 V
iL(0+)=1 A
L=0.5 H
C=1 F
G=1 mho(Or Resistance,R = 1 ohm)

I have to solve this by using laplace transforms, mathematically.I think I found part of the solution,but I am unsure.

Equation 1:

(1/L)*∫(V1(t)-V2(t))dt+C(d/dt)(V1(t))=0

Corresponding Laplace equation:
(2+s)*V1(s)-2*V2(s)=s-1-----------------(1)

Equation 2:

(1/L)*∫(V2(t)-V1(t))dt+V2*G=0
(Conductance,G is taklen)

Laplace:

-2*V1(s)+(2+s)*V2(s)=-1----------------(2)

Solving (1) and (2):

2*(1)+(2)*(2+s):

we get:

V2(s)=(s-4)/(s^2+4*s)

or V2(t)=-1+2*exp(-4t)

Similarly,I got:
V1(s)=(s^2+s-2)/(s^2+4s)

This doesn't seem right..does it?Have I gone wrong somewhere?

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2. Zazoo Member

Jul 27, 2011
114
43
I have the same node equations, but when I ran through the Lapalace transforms I arrived at the following:

Equation 1:
(2 + s^2)V1(s) - 2V2(s) = s - 1

Equation 2:
-2V1(s) + (2 + s)V2(s) = 1

Assuming I didn't err somewhere,
In eq. 1 I think you forgot to multiply the sV1(s) term by s when clearing the fractions after the transform.
In eq. 2 it looks like you reversed the direction of the initial current through the inductor when performing the integral transform.

Mike

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3. abhaymv Thread Starter Active Member

Aug 6, 2011
105
4
Thank you!that explains it.got to work more on these type of problems.