Laplace transforms:Series RLC circuit

Laplace Transform Example: Series RLC Circuit

Problem

Given a series RLC circuit with Failed to parse (syntax error): {\displaystyle R=10 Ω} , $L=0.08H$, and $C=10^{-5}F$, having power source $v(t)=50cos(20t)$, 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

$50cos(20t)=10i+0.08\frac{di}{dt}+10^5\int idt$

$\Rightarrow cos(20t)=0.2i+0.0016\frac{di}{dt}+20000\int idt$

Now, we take the Laplace Transform and get

$\frac{s}{s^2+20^2}=0.2I+0.08[sI-i(0)]+20000\frac{I}{s}$

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

$\frac{s}{s^2+400}=0.2I+0.08sI+20000\frac{I}{s}$

$\Rightarrow \frac{s^2}{s^2+400}=0.2sI+0.08s^{2}I+20000I$

$\Rightarrow \frac{s^2}{s^2+400}=(0.08s^2+0.2s+20000)I$

$\Rightarrow I(s)=\frac{s^2}{(s^2+400)(0.08s^2+0.2s+20000)}$

Using partial fraction decomposition, we find that

$I(s)=\frac{1.0016-1.605*10^{-8}s}{0.08s^2+0.2s+20000}+\frac{2.046*10^{-7}s-.02003}{s^2+400}$

<math>\Rightarrow I(s)=\dfrac{6.24*10^7-s}{4.98*10^6s^2+1.25*10^7s+1.25*10^{12}}+\dfrac{