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.

 

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

 

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).

 

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It could be something as this...

 

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.

Hello there,

It does not look like either of your expressions came out right.
To check, do a DC analysis at t=0 and check your time equations at t=0, or use your Laplace solutions using the initial value theorem. The DC analysis should match the time solution at t=0 or else you did something wrong.
Hint: At t=0 the current through the inductor in the second drawing is zero, although the initial current generator is still in effect and the first solution at t=0 is with the switch closed. The second solution at t=0 is the very same but with the switch open. You can then use the switch open time solution to finish the complete solution.

You have to get the first two parts right first anyway or else you wont know the correct starting voltages for both v1(t) and v2(t) at t=0 and that is part of the solution.

So the steps to the solution are as follows...
1. Find the solutions at t=0 with the switch closed, note you only need solutions at t=0 for this step.
2. Find the solutions at t=0 with the switch open.
3. Find the time solutions for times greater than t=0 with the switch open.
4. Combine all your solutions starting with the one from step 1.
Note you may combine step 3 with step 2 finding the time solution and then setting t=0 to get step 2, but it would be worthwhile to do the DC solution for step 2 also to check your time solution result.
Also note that you can check your solutions at t going to infinity also.
Keep in mind that in your second circuit drawing the inductor is open at t=0, and a short circuit at t going to infinity. This allows you to use DC analysis to check your results at t=0 and t at infinity.