# Circuit analysis - Laplace

1. Apr 3, 2008

### ineedmunchies

1. The problem statement, all variables and given/known data

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

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

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