Laplace transforms: Simple Electrical Network

Problem Statement

Using the formulas

$E(t)=L\frac{d{i}_R}{dt}+R{i}_C$
$RC\frac{d{i}_R}{dt}+i_R-i_C=0$

Solve the system when V0 = 50 V, L = 0.5 h, R = 80 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

$0.5\frac{d{i}_1}{dt}+80i_2=50$

$80(10^{-4})\frac{d{i}_2}{dt}+i_2-i_1=0$

Applying the Laplace transform to each equation gives

$\frac{1}{2}(s{ℒ}\left\{i_1\right\}-i_1(0))+80{ℒ}\left\{i_2\right\}=50$

$0.008(s{ℒ}{i}_2-i_2(0))+{ℒ}\left\{i_2\right\}-{ℒ}\left\{i_1\right\}=0$

$\Rightarrow \frac{1}{2}s{I}_1(s)+80I_2(s)=\frac{50}{s}$

$-125I_1(s)+[s+125]I_2(s)=0$

Solving for $I_2(s)$

$I_2(s)=\frac{12500}{s(s^2+125s+20000)}$

We find the partial decomposition

Let $I_2(s)=\frac{12500}{s(s^2+125s+20000)}=\frac{A}{s}+\frac{Bs+C}{s^2+125s+20000}$

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

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

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

$I_1(s)=\frac{100s+12500}{s(s^2+125s+20000)}=\frac{5}{8s}+\frac{-5s+175}{s^2+125s+20000}$