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 = 0.5 h, R = 60 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

$0.5{\frac {di_{1}}{dt}}+60i_{2}=50$

$60(10^{-4}){\frac {di_{2}}{dt}}+i_{2}-i_{1}=0$

Applying the Laplace transform to each equation gives

${\frac {1}{2}}(s{\mathcal {L}}\left\{i_{1}\right\}-i_{1}(0))+60{\mathcal {L}}\left\{i_{2}\right\}=50$

$0.006(s{\mathcal {L}}{i_{2}}-i_{2}(0))+{\mathcal {L}}\left\{i_{2}\right\}-{\mathcal {L}}\left\{i_{1}\right\}=0$

$\Rightarrow {\frac {1}{2}}s{\mathcal {L}}\left\{i_{1}\right\}+60{\mathcal {L}}\left\{i_{2}\right\}={\frac {50}{s}}$

$-200{\mathcal {L}}\left\{i_{1}\right\}+(s+200){\mathcal {L}}\left\{i_{2}\right\}=0$

Solving for ${\mathcal {L}}\left\{i_{1}\right\}$ gives