Colby's Asgn

Find the Thevenin equivalent impedance seen looking into the terminals of the following circuit:

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To find the Thevenin equivalent impedance I will need to take the Laplace transform of each branch of the circuit and add them in series.

The Laplace transform of a resistor is ${ℒ}\{resistor\}=R$

The Laplace transform of an inductor is ${ℒ}\{inductor\}=sL$

The Laplace transform of a capacitor is ${ℒ}\{\text{capacitor}\}=\frac{1}{sC}$

Taking the Laplace transform of the original circuit gives the following.

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The Thevenin equivalent impedance of this circuit is given by the equation

$Z=\frac{1}{\frac{1}{R1+\frac{1}{sC}}+\frac{1}{R2+sL}}\mathrm{\Omega }$

Substituting the values for R1, R2, C and L into the equation gives

$Z=\frac{1}{\frac{1}{50+\frac{1}{.002s}}+\frac{1}{100+.5s}}\mathrm{\Omega }=\frac{1}{\frac{s}{50(s+10)}+.5(s+200)}\mathrm{\Omega }=\frac{2(s+10)}{s^2+210.04s+2000}\mathrm{\Omega }$