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:

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

Substituting numbers, we get

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

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

Now, we take the Laplace Transform and get

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

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

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

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

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

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

Using partial fraction decomposition, we find that

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