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>

<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 19:26, 19 October 2009

Laplace Transform Example: Series RLC Circuit

Problem

Given a series RLC circuit with Failed to parse (SVG with PNG fallback (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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{