Laplace transforms:Series RLC circuit: Difference between revisions

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==Laplace Transform Example: Series RLC Circuit==
==Laplace Transform Example: Series RLC Circuit==
===Problem===
===Problem===
Given a series RLC circuit with <math>R=10 Ω</math>, <math>L=0.08 H</math>, and <math>C=10^{-5} F</math>, having power source <math>v(t)=50cos(20t)</math>, find an expression for <math>i(t)</math> if <math>i(0)=0 A</math> and <math>v_c(0)=0 V</math>.
Given a series RLC circuit with <math>R=10 Ohms</math>, <math>L=0.1 H</math>, and <math>C=10^{-5} F</math>, having power source <math>v(t)=10cos(20t)</math>, find an expression for <math>i(t)</math> if <math>i(0)=0 A</math> and <math>v_c(0)=0 V</math>.


===Solution===
===Solution===
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Substituting numbers, we get
Substituting numbers, we get


<math>50cos(20t)=10i+0.08\dfrac{di}{dt}+10^{5}\int{i dt}</math>
<math>10cos(20t)=10i+0.1\dfrac{di}{dt}+10^{5}\int{i dt}</math>


<math>\Rightarrow cos(20t)=0.2i+0.0016\dfrac{di}{dt}+20000\int{idt}</math>
<math>\Rightarrow cos(20t)=i+0.01\dfrac{di}{dt}+10000\int{idt}</math>


Now, we take the Laplace Transform and get
Now, we take the Laplace Transform and get


<math>\dfrac{s}{s^2+20^2}=0.2I+0.08[sI-i(0)]+20000\dfrac{I}{s}</math>
<math>\dfrac{s}{s^2+20^2}=I+0.01[sI-i(0)]+10000\dfrac{I}{s}</math>


Using the fact that <math>i(0)=0A</math>, we get
Using the fact that <math>i(0)=0A</math>, we get


<math>\dfrac{s}{s^2+400}=0.2I+0.08sI+20000\dfrac{I}{s}</math>
<math>\dfrac{s}{s^2+400}=I+0.01sI+10000\dfrac{I}{s}</math>


<math>\Rightarrow \dfrac{s^2}{s^2+400}=0.2sI+0.08s^2I+20000I</math>
<math>\Rightarrow \dfrac{s^2}{s^2+400}=sI+0.01s^2I+10000I</math>


<math>\Rightarrow \dfrac{s^2}{s^2+400}=(0.08s^2+0.2s+20000)I</math>
<math>\Rightarrow \dfrac{s^2}{s^2+400}=(0.01s^2+s+10000)I</math>


<math>\Rightarrow I(s)=\dfrac{s^2}{(s^2+400)(0.08s^2+0.2s+20000)}</math>
<math>\Rightarrow I(s)=\dfrac{s^2}{(s^2+400)(0.01s^2+s+10000)}</math>


Using partial fraction decomposition, we find that
Using partial fraction decomposition, we find that


<math>I(s)=\dfrac{1.0016-1.605*10^{-8}s}{0.08s^2+0.2s+20000}+\dfrac{2.046*10^{-7}s-.02003}{s^2+400}</math>
<math>I(s)=\dfrac{1.0004-4.003*10^{-8}s}{0.01s^2+s+10000}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}</math>


<math>\Rightarrow I(s)=\dfrac{6.24*10^7-s}{4.98*10^6s^2+1.25*10^7s+1.25*10^{12}}+\dfrac{
<math>\Rightarrow I(s)=\dfrac{100.04-4.003*10^{-6}s}{s^2+100s+1000000}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}</math>

<math>\Rightarrow I(s)=\dfrac{100.04-4.003*10^{-6}s}{(s+50)^2+997500}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}</math>

<math>\Rightarrow I(s)=\dfrac{100.038}{(s+50)^2+(50\sqrt{399})^2}-\dfrac{4.003*10^{-6}s+.002}{(s+50)^2+(50\sqrt{399})^2}+\dfrac{4.003*10^{-6}s}{s^2+20^2}-\dfrac{0.04002}{s^2+20^2}</math>

<math> \Rightarrow I(s)=\dfrac{10.038}{50\sqrt{399}}\dfrac{50\sqrt{399}}{(s+50)^2+(50\sqrt{399})^2}-4.003*10^{-6}\dfrac{s+50}{(s+50)^2+(50\sqrt{399})^2}+4.003*10^{-6}\dfrac{s}{s^2+20^2}-\dfrac{0.04002}{20}\dfrac{20}{s^2+20^2}</math>

Finally, we take the inverse Laplace transform to obtain

<math> i(t)=0.01e^{-50t}sin(998.8t)-(4.003*10^{-6})e^{-50t}cos(998.8t)+(4.003*10^{-6})cos(20t)-0.002sin(20t) \,</math>

which is our answer.

Revision as of 21:11, 19 October 2009

Laplace Transform Example: Series RLC Circuit

Problem

Given a series RLC circuit with , , and , having power source , find an expression for if and .

Solution

We begin with the general formula for voltage drops around the circuit:

Substituting numbers, we get

Now, we take the Laplace Transform and get

Using the fact that , we get

Using partial fraction decomposition, we find that

Finally, we take the inverse Laplace transform to obtain

which is our answer.