Laplace transforms: Simple Electrical Network

Problem Statement

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 = 0.5 h, R = 60 Ω, C = 10^{-4} f, and the currents are initially zero. Substituting numbers into the equations, we have

$0.5{\frac {di_{1}}{dt}}+80i_{2}=50$

$80(10^{-4}){\frac {di_{2}}{dt}}+i_{2}-i_{1}=0$

Applying the Laplace transform to each equation gives

${\frac {1}{2}}(s{\mathcal {L}}\left\{i_{1}\right\}-i_{1}(0))+80{\mathcal {L}}\left\{i_{2}\right\}=50$

$0.008(s{\mathcal {L}}{i_{2}}-i_{2}(0))+{\mathcal {L}}\left\{i_{2}\right\}-{\mathcal {L}}\left\{i_{1}\right\}=0$

$\Rightarrow {\frac {1}{2}}sI_{1}(s)+80I_{2}(s)={\frac {50}{s}}$

$-125I_{1}(s)+[s+125]I_{2}(s)=0$

Solving for $I_{2}(s)$

$I_{2}(s)={\frac {12500}{s(s^{2}+125s+20000)}}$

We find the partial decomposition

Let $I_{2}(s)={\frac {12500}{s(s^{2}+125s+20000)}}={\frac {A}{s}}+{\frac {Bs+C}{s^{2}+125s+20000}}$

$\Rightarrow 12500=A(s^{2}+125s+20000)+(Bs+C)s$

$\Rightarrow 12500=As^{2}+125As+20000A+Bs^{2}+Cs$

Comparing the coefficients we get

$A={\frac {5}{8}},B=-5,C=-625$

Thus $I_{2}(s)={\frac {5}{8s}}-{\frac {5s+625}{s^{2}+125s+20000}}$

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

$I_{1}(s)={\frac {100s+12500}{s(s^{2}+125s+20000)}}={\frac {5}{8s}}+{\frac {-5s+175}{s^{2}+125s+20000}}$