- #1
ineedmunchies
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Homework Statement
The question is as shown in the first picture. Question1.jpg
Homework Equations
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.