Colby's Asgn

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

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 ${\mathcal {L}}\{resistor\}=R$

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

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

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

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

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

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

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