Laplace transforms:Series RLC circuit: Difference between revisions
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(New page: ==Laplace Transform Example: Series RLC Circuit== ===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 sour...) |
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<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{1.0016-1.605*10^{-8}s}{0.08s^2+0.2s+20000}+\dfrac{2.046*10^{-7}s-.02003}{s^2+400}</math> |
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<math>\Rightarrow I(s)=\dfrac{6.24*10^7-s}{4.98*10^6s^2+1.25*10^7s+1.25*10^{12}}+\dfrac{ |
Revision as of 20:26, 19 October 2009
Laplace Transform Example: Series RLC Circuit
Problem
Given a series RLC circuit with Failed to parse (syntax error): {\displaystyle R=10 Ω} , , 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
<math>\Rightarrow I(s)=\dfrac{6.24*10^7-s}{4.98*10^6s^2+1.25*10^7s+1.25*10^{12}}+\dfrac{