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Circuit analysis - Laplace

  • Engineering
  • Thread starter ineedmunchies
  • Start date
  • #1
1. Homework Statement

The question is as shown in the first picture. Question1.jpg

2. Homework Equations
3. The Attempt at a Solution
It asks that initial condition current generators are used, what I believe models the circuit for t>0 is shown in the second picture with the switch open, and a current generator added. (I am not sure if this is correct.)

I then used KCL to write equations at node 1 and node 2.

Node 1:
[tex]\frac{5}{s}[/tex] - [tex]\frac{2}{s}[/tex]=[tex]\frac{V_{1}}{1}[/tex]+[tex]\frac{V_{1}-V_{2}}{2s}[/tex]


Node 2:
[tex]\frac{2}{s}[/tex]+[tex]\frac{2}{s}[/tex]=[tex]\frac{V_{2}}{1}[/tex]-([tex]\frac{V_{1}-V_{2}}{2s}[/tex])


These can then be rearraged to give

[tex]\frac{3}{s}[/tex]=[tex]V_{1}[/tex](1+[tex]\frac{1}{2s}[/tex])-[tex]V_{2}[/tex]([tex]\frac{1}{2s}[/tex])

and

[tex]\frac{4}{s}[/tex]=[tex]V_{1}[/tex]([tex]\frac{-1}{2s}[/tex])+[tex]V_{2}[/tex](1+[tex]\frac{1}{2s}[/tex])

Which I then put into a matrix and solved for [tex]V_{1}[/tex] and [tex]V_{2}[/tex]

Giving
[tex]V_{1}[/tex] = [tex]\frac{3}{s}[/tex]-[tex]\frac{2}{1+\frac{1}{2s}}[/tex]
which can be simplified to
[tex]V_{1}[/tex] = [tex]\frac{3}{s}[/tex]-[tex]\frac{4}{s+2}[/tex]

and
[tex]V_{2}[/tex] = [tex]\frac{\frac{-3}{2}}{1+\frac{1}{2s}}[/tex]+[tex]\frac{4}{s}[/tex]
Which can be simplified to
[tex]V_{2}[/tex] = [tex]\frac{4}{s}[/tex]-[tex]\frac{3}{s+2}[/tex]

Then convert these back to the time domain to give:
[tex]V_{1}[/tex](t)=3-4[tex]e^{-2t}[/tex]
and [tex]V_{2}[/tex](t)=4-3[tex]e^{-2t}[/tex]

Can anyone tell if there is a mistake here or not?
I don't feel confident this is the correct answer. I think the 5 ohm resistor and 5A current should effect the circuit somehow but do not know how to encorporate it into my equations.
 

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