RLC circuit solved with Laplace transformation

  • #1
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Hi, i need some help here. Can you help me?:sorry:

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

culonzo.png



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 im doing wrong??

I thing im having a mistake with some signs.

Pd: Sorry for my bad english.

Thanks for reading me!!
 

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Answers and Replies

  • #3
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For I(s), I get:

$$I(s)=\frac{300}{(s+5000)^2}+\frac{0.06}{(s+5000)}$$
and, for v(s), I get $$v(s)=\frac{30000}{(s+5000)^2}+\frac{12}{(s+5000)}$$
 
  • #4
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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) ??
 
  • #6
5
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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!!
 

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