# Circuit analysis problem - Laplace

Discussion in 'Homework Help' started by ineedmunchies, Apr 3, 2008.

1. ### ineedmunchies Thread Starter New Member

Apr 3, 2008
1
0
The question is as shown in the first picture. Question1.jpg

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:
$\frac{5}{s}$ - $\frac{2}{s}$=$\frac{V_{1}}{1}$+$\frac{V_{1}-V_{2}}{2s}$

Node 2:
$\frac{2}{s}$+$\frac{2}{s}$=$\frac{V_{2}}{1}$-($\frac{V_{1}-V_{2}}{2s}$)

These can then be rearraged to give

$\frac{3}{s}$=$V_{1}$(1+$\frac{1}{2s}$)-$V_{2}$($\frac{1}{2s}$)

and

$\frac{4}{s}$=$V_{1}$($\frac{-1}{2s}$)+$V_{2}$(1+$\frac{1}{2s}$)

Which I then put into a matrix and solved for $V_{1}$ and $V_{2}$

Giving
$V_{1}$ = $\frac{3}{s}$ - $\frac{2}{1+\frac{1}{2s}}$
which can be simplified to
$V_{1}$ = $\frac{3}{s}$ - $\frac{4}{s+2}$

and
$V_{2}$ = ($\frac{\frac{-3}{2}}{1+\frac{1}{2s}}$)+$\frac{4}{s}$

(*Note the 4/s is not part of the denominator, I couldn't get the brackets to work properly.)

Which can be simplified to
$V_{2}$ = $\frac{4}{s}$-$\frac{3}{s+2}$

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

and $V_{2}$(t) = 4-3$e^{-2t}$

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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2. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
I think your algebra is off in your solutions for V1 and V2. Starting with your original equations for nodes 1 and 2, I get:
V1 = (3s + 7/2)/(s(s + 1))
V2 = (4s + 7/2)/(s(s + 1))

I've checked, and these values satisfy your original equations, so I'm pretty confident about them.
To do the inverse Laplace transforms, you'll need to break each of these apart into a sum of two rational expressions. The denominators will be s and s + 1, not s and s + 2 that you show.
Mark

3. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
In case the bit about breaking up the expressions for V1 and V2 is not clear enough, you'll need to do a partial fraction decomposition for each of them.

For V1, you need to solve for A and B in this equation:
(3s + 7/2)/(s(s+1)) = A/s + B/(s + 1)

When you do the inverse Laplace transform, you'll end up with v1(t) = A + B*e^(-t)

Do the same thing to find v2(t).