Laplace transforms for transient analysis

  • #31
agata78 said:
I used this website to educate myself a bit more then i remembered from school www.mash.dept.shef.ac.uk/Resources/web-partialfractions.pdf

But does it matter which way i used to calculate A and B.

i(t)= 20e 1 + (-20) -2t

But what next?

Your transformation of the Laplace terms does not look right! You have found the that in the Laplace domain:
$$I(s) = \frac{20}{s} - \frac{20}{s + 2}$$
Find the forms of those two terms in your Laplace Transform tables and convert them to time domain terms. The result should look familiar!
 
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  • #32
the only way it could be is:

20-20e(-2t)
 
  • #33
agata78 said:
the only way it could be is:

20-20e(-2t)

Right. Technically each term is multiplied by a unit step function to indicate the the voltage is "switched on" at time t = 0. Mathematically it means that this solution applies only for t ≥ 0.

So, writing it in proper form for a function:

[STRIKE]I(t)[/STRIKE]Vc(t) = 20-20e(-2t) = 20(1 - e(-2t))
[STRIKE]
Now you can move on to finding Vc(t). Use the same methods.[/STRIKE]

EDIT: Sorry about that. I misremembered where we were at in the problem sequence! I think you're done with this one!
 
Last edited:
  • #34
Yes i was actually trying to ask you why would i have to calculate it again.

Hurrayyyyy!

Thank you so much for help all the way! One to go!
 

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