Laplace transforms: Simple Electrical Network

Using the formulas

 $E(t)=L{\dfrac {di_{R}}{dt}}+Ri_{C}$ 
 $RC{\dfrac {di_{R}}{dt}}+i_{R}-i_{C}=0$

Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10^-4 f, and the currents are initially zero.

Solution

Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10^-4 f, and the currents are initially zero.

$4{\frac {di_{1}}{dt}}+20i_{2}=50$

$20(10^{-4}){\frac {di_{2}}{dt}}+i_{2}-i_{1}=0$

Applying the Laplace transform to each equation gives

$4(s{\mathcal {L}}\left\{i_{1}\right\}-i_{1}(0))+20{\mathcal {L}}\left\{i_{2}\right\}=50$

$\Rightarrow 4sI_{1}(s)+20I_{2}(s)={\frac {50}{s}}$

$0.005(s{\mathcal {L}}{i_{2}}-i_{2}(0))+{\mathcal {L}}\left\{i_{2}\right\}-{\mathcal {L}}\left\{i_{1}\right\}=0$

$\Rightarrow -500I_{1}(s)+[s+500]I_{2}(s)=0$

Solving for $I_{2}(s)$

$I_{2}(s)={\frac {6250}{s(s^{2}+500s+2500)}}$

We find the partial decomposition

Let $I_{2}(s)={\frac {6250}{s(s^{2}+500s+2500)}}={\frac {A}{s}}+{\frac {Bs+C}{s^{2}+500s+2500}}$

$\Rightarrow 6250=A(s^{2}+500s+2500)+(Bs+C)s$

$\Rightarrow 62500=As^{2}+500As+2500A+Bs^{2}+Cs$

Comparing the coefficients we get

$A={\frac {5}{2}},B=-5,C=-1250$

Thus $I_{2}(s)={\frac {5}{2s}}-{\frac {5s+1250}{s^{2}+500s+2500}}$

Now we do the same for $I_{1}$ where we solve the function in terms of $I_{1}$ and decomposing the partial fraction resulting in

$I_{1}(s)={\frac {25s+12500}{s(s^{2}+500s+2500)}}={\frac {5}{s}}-{\frac {5s+2475}{s^{2}+500s+2500}}$

In order to make it nicer on us we need to complete the square as follows

$s^{2}+500s+2500=0$

$\Rightarrow s^{2}+500s=-2500$

$\Rightarrow s^{2}+500s+[{\frac {500}{2}}]^{2}=-2500+({\frac {500}{2}})^{2}$

$\Rightarrow s^{2}+500s+62500=6000$

$\Rightarrow (s^{2}+250)^{2}-(20{\sqrt {15}})^{2}=0</math)wedothisbydivinding<math>({\frac {b}{2}})^{2}$

Taking the Inverse Laplace transform yields

${\mathcal {L}}^{-1}\left\{I_{1}(s)\right\}={\frac {5}{8}}+{\frac {39{\sqrt {103}}}{824}}sin*({\frac {25}{2}}{\sqrt {103}}*t)$

${\mathcal {L}}^{-1}\left\{I_{2}(s)\right\}=$