Laplace transforms:Series RLC circuit

Laplace Transform Example: Series RLC Circuit

Problem

Given a series RLC circuit with $R=10Ohms$, $L=0.1H$, and $C=10^{-5}F$, having power source $v(t)=10cos(20t)$, find an expression for $i(t)$ if $i(0)=0A$ and $v_c(0)=0V$.

Solution

We begin with the general formula for voltage drops around the circuit:

$v(t)=Ri+L\frac{di}{dt}+\frac{1}{C}\int idt$

Substituting numbers, we get

$10cos(20t)=10i+0.1\frac{di}{dt}+10^5\int idt$

$\Rightarrow cos(20t)=i+0.01\frac{di}{dt}+10000\int idt$

Now, we take the Laplace Transform and get

$\frac{s}{s^2+20^2}=I+0.01[sI-i(0)]+10000\frac{I}{s}$

Using the fact that $i(0)=0A$, we get

$\frac{s}{s^2+400}=I+0.01sI+10000\frac{I}{s}$

$\Rightarrow \frac{s^2}{s^2+400}=sI+0.01s^{2}I+10000I$

$\Rightarrow \frac{s^2}{s^2+400}=(0.01s^2+s+10000)I$

$\Rightarrow I(s)=\frac{s^2}{(s^2+400)(0.01s^2+s+10000)}$

Using partial fraction decomposition, we find that

$I(s)=\frac{1.0004-4.003*10^{-8}s}{0.01s^2+s+10000}+\frac{4.003*10^{-6}s-0.04002}{s^2+400}$

$\Rightarrow I(s)=\frac{100.04-4.003*10^{-6}s}{s^2+100s+1000000}+\frac{4.003*10^{-6}s-0.04002}{s^2+400}$

$\Rightarrow I(s)=\frac{100.04-4.003*10^{-6}s}{(s+50{)}^2+997500}+\frac{4.003*10^{-6}s-0.04002}{s^2+400}$

$\Rightarrow I(s)=\frac{100.038}{(s+50{)}^2+(50\sqrt{399}{)}^2}-\frac{4.003*10^{-6}s+.002}{(s+50{)}^2+(50\sqrt{399}{)}^2}+\frac{4.003*10^{-6}s}{s^2+20^2}-\frac{0.04002}{s^2+20^2}$

$\Rightarrow I(s)=\frac{10.038}{50\sqrt{399}}\frac{50\sqrt{399}}{(s+50{)}^2+(50\sqrt{399}{)}^2}-4.003*10^{-6}\frac{s+50}{(s+50{)}^2+(50\sqrt{399}{)}^2}+4.003*10^{-6}\frac{s}{s^2+20^2}-\frac{0.04002}{20}\frac{20}{s^2+20^2}$

Finally, we take the inverse Laplace transform to obtain

$i(t)=0.01e^{-50t}sin(998.8t)-(4.003*10^{-6})e^{-50t}cos(998.8t)+(4.003*10^{-6})cos(20t)-0.002sin(20t)$

which is our answer.

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Written by Nathan Reeves Checked by