Engineering RLC circuit solved with Laplace transformation

[MaxR2018]

Hi, i need some help here. Can you help me?

Here is the problem.

Exercise statement: The switch have been closed for a long time y is opened at t=0. Using Laplace's transtormation calculate V0(t) for t ≥ 0

This is what i made to solve it:

1) I know while the switch is closed, the current trough the circuit is i=12v/200, so i=60mA.

2) When the switch is opened at t=0, i use the voltages law of kirchoff:

12 = vc + vr + vl

12= (1/c)*(integrate of i dt from 0 to t) + vc(0) + iR + L(di/dt)

i know that vc(0) = 0

so : 12= (1/c)*(integrate of i dt from 0 to t) + iR + L(di/dt)

Then i used the laplace transformation:

12/s = I(s)/Sc + RI(S) + LSI(S) - LI(0)

And i know that LI(0)=60mA

so:

12/s = I(s)/Sc + RI(S) + LSI(S) - 60mA

Finally i calculate I(S) and then i obtain i(t) with the antitransformation of Laplace.

Then, with i(t) i calculate vt knowing that:

VL=L*(di/dt), but i obtain a diferent solution.

I obtain that V0(t) = -300000*e^(-5000t) + 12e^(-5000t)

What I am doing wrong??

I thing I am having a mistake with some signs.

Pd: Sorry for my bad english.

Thanks for reading me!

 

[Chestermiller]

What do you get for the Laplace Transform of the current I(s)?

 

[Chestermiller]

For I(s), I get:

$$I(s)=\frac{300}{(s+5000)^2}+\frac{0.06}{(s+5000)}$$
and, for v(s), I get $$v(s)=\frac{30000}{(s+5000)^2}+\frac{12}{(s+5000)}$$

 

[MaxR2018]

For I(s), I get:

$$I(s)=\frac{300}{(s+5000)^2}+\frac{0.06}{(s+5000)}$$
and, for v(s), I get $$v(s)=\frac{30000}{(s+5000)^2}+\frac{12}{(s+5000)}$$

One time you have I(s), you find V(s) with V(s) = LSI(S) - Li(0) ??

 

[Chestermiller]

One time you have I(s), you find V(s) with V(s) = LSI(S) - Li(0) ??

No. $$v(s)=RI(s)+LsI(s)-Li(0)$$You left out RI(s)

 

[MaxR2018]

Ah, ok thanks! i thought the exercise was asking me for VL, but it really asking for the the voltage in R and L together! Thank you so much!