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
    104
    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)

    Is my answer correct?:confused:

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

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


    Thank you in advance for your help :D
     
  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
     
    abhaymv likes this.
  3. abhaymv

    Thread Starter Active Member

    Aug 6, 2011
    104
    4
    Thank you!that explains it.got to work more on these type of problems.
     
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