Question 4.6.2: Solve the system in (5) under the conditions E(t) = 60 V, L ......
Solve the system in (5)
\begin{aligned}L \frac{d i_1}{d t}+R i_2 & =E(t) \\R C \frac{d i_2}{d t}+i_2-i_1 & =0 .\end{aligned} (5)
under the conditions E(t)=60 \mathrm{~V}, L=1 \mathrm{~h}, R=50 \Omega, C=10^{-4} \mathrm{f}, and the currents i_{1} and i_{2} are initially zero.
We must solve
\begin{array}{r} \frac{d i_{1}}{d t}+50 i_{2}=60 \\ 50\left(10^{-4}\right) \frac{d i_{2}}{d t}+i_{2}-i_{1}=0, \end{array}
subject to i_{1}(0)=0, i_{2}(0)=0.
Applying the Laplace transform to each equation of the system and simplifying gives
\begin{aligned} s I_{1}(s)+\quad 50 I_{2}(s) & =\frac{60}{s} \\ -200 I_{1}(s)+(s+200) I_{2}(s) & =0 \end{aligned}
where I_{1}(s)=\mathscr{L}\left\{i_{1}(t)\right\} and I_{2}(s)=\mathscr{L}\left\{i_{2}(t)\right\}. Solving the system for I_{1} and I_{2} and decomposing the results into partial fractions gives
\begin{aligned} & I_{1}(s)=\frac{60 s+12,000}{s(s+100)^{2}}=\frac{6 / 5}{s}-\frac{6 / 5}{s+100}-\frac{60}{(s+100)^{2}} \\ & I_{2}(s)=\frac{12,000}{s(s+100)^{2}}=\frac{6 / 5}{s}-\frac{6 / 5}{s+100}-\frac{120}{(s+100)^{2}}. \end{aligned}
Taking the inverse Laplace transform, we find the currents to be
i_{1}(t)=\frac{6}{5}-\frac{6}{5} e^{-100 t}-60 t e^{-100 t}
i_{2}(t)=\frac{6}{5}-\frac{6}{5} e^{-100 t}-120 t e^{-100 t}.