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RLC Series Circuit

  • Engineering
  • Thread starter danilorj
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  • #1
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Guys,
Let's suppose we have RLC series circuit feeded by tension source initially with no energy stored. When we apply an unitary step of tension at t=0 at the terminals of the circuit, we observe a current of i(t)=(125/24) * exp(-700t)* Sin(2400t) mA that goes around the circuit for t>0s.
I want to find the values of R, L, C.
I was trying to match the laplace transformation of the current above to the laplace transformation of the circuit for the current but no success.
 

Answers and Replies

  • #2
berkeman
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Guys,
Let's suppose we have RLC series circuit feeded by tension source initially with no energy stored. When we apply an unitary step of tension at t=0 at the terminals of the circuit, we observe a current of i(t)=(125/24) * exp(-700t)* Sin(2400t) mA that goes around the circuit for t>0s.
I want to find the values of R, L, C.
I was trying to match the laplace transformation of the current above to the laplace transformation of the circuit for the current but no success.
Can you just write the differential equation for the current and voltage of the circuit, solve the DE and apply your initial conditions? That should get you the answer as well. Or are you required to use transforms?
 
  • #3
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Let me explain it better
The Laplace transformation of the given current is I(s)=12500/[(s+700)^2+2400^2]
And if I apply the Kirchhoff Law Tension around the RLC series circuit I get
I(s)=1/[s*(LC*s^2+RC*s+1)]. But I don't know what to do with this. They seem to be different, one has degree 3 and the other has degree 2 in the denominator.
 
  • #4
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And I think the best way of doing this is using the Laplace transformation, cause solving the differential equation would take a very hard work, and I don't think it's gonna work as well.
 
  • #5
vela
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Let me explain it better
The Laplace transformation of the given current is I(s)=12500/[(s+700)^2+2400^2]
And if I apply the Kirchhoff Law Tension around the RLC series circuit I get
I(s)=1/[s*(LC*s^2+RC*s+1)].
Show us how you got this. I don't think it's correct, which is why it's not matching up with the I(s) you were given.
But I don't know what to do with this. They seem to be different, one has degree 3 and the other has degree 2 in the denominator.
 

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