Laplace transforms:Series RLC circuit: Difference between revisions
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==Laplace Transform Example: Series RLC Circuit== |
==Laplace Transform Example: Series RLC Circuit== |
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===Problem=== |
===Problem=== |
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Given a series RLC circuit with <math>R=10 |
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>. |
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===Solution=== |
===Solution=== |
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Substituting numbers, we get |
Substituting numbers, we get |
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<math> |
<math>10cos(20t)=10i+0.1\dfrac{di}{dt}+10^{5}\int{i dt}</math> |
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<math>\Rightarrow cos(20t)= |
<math>\Rightarrow cos(20t)=i+0.01\dfrac{di}{dt}+10000\int{idt}</math> |
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Now, we take the Laplace Transform and get |
Now, we take the Laplace Transform and get |
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<math>\dfrac{s}{s^2+20^2}= |
<math>\dfrac{s}{s^2+20^2}=I+0.01[sI-i(0)]+10000\dfrac{I}{s}</math> |
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Using the fact that <math>i(0)=0A</math>, we get |
Using the fact that <math>i(0)=0A</math>, we get |
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<math>\dfrac{s}{s^2+400}= |
<math>\dfrac{s}{s^2+400}=I+0.01sI+10000\dfrac{I}{s}</math> |
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<math>\Rightarrow \dfrac{s^2}{s^2+400}= |
<math>\Rightarrow \dfrac{s^2}{s^2+400}=sI+0.01s^2I+10000I</math> |
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<math>\Rightarrow \dfrac{s^2}{s^2+400}=(0. |
<math>\Rightarrow \dfrac{s^2}{s^2+400}=(0.01s^2+s+10000)I</math> |
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<math>\Rightarrow I(s)=\dfrac{s^2}{(s^2+400)(0. |
<math>\Rightarrow I(s)=\dfrac{s^2}{(s^2+400)(0.01s^2+s+10000)}</math> |
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Using partial fraction decomposition, we find that |
Using partial fraction decomposition, we find that |
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<math>I(s)=\dfrac{1. |
<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> |
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<math>\Rightarrow I(s)=\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> |
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<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> |
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<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> |
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<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> |
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Finally, we take the inverse Laplace transform to obtain |
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<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> |
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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.