The circuit for $\text{Q. 33-34}$ are given in the figure. For both are the questions, assume that the switch $S$ is in position $1$ for a long time and thrown to position $2$ at $t=0$.
$I_{1}(s)$ and $I_{2}(s)$ are the Laplace transforms of $i_{1}(t)$ and $i_{2}(t)$ respectively. The equations for the loop currents $I_{1}(s)$ and $I_{2}(s)$ for the circuit shown in the figure after the switch is brought from position $1$ to position $2$ at $t=0$, are
- $\left[\begin{array}{ll}\mathrm{R}+\mathrm{Ls}+\frac{1}{\mathrm{Cs}} & -\mathrm{Ls} \\ -\mathrm{Ls} & \mathrm{R}+\frac{1}{\mathrm{Cs}}\end{array}\right]\left[\begin{array}{l}\mathrm{I}_{1}(s) \\ \mathrm{I}_{2}(s)\end{array}\right]=\left[\begin{array}{c}\frac{\mathrm{V}}{\mathrm{s}} \\ 0\end{array}\right]$
- $\left[\begin{array}{ll}\mathrm{R}+\mathrm{Ls}+\frac{1}{\mathrm{Cs}} & -\mathrm{Ls} \\ -\mathrm{Ls} & \mathrm{R}+\frac{1}{\mathrm{Cs}}\end{array}\right]\left[\begin{array}{l}\mathrm{I}_{1}(s) \\ \mathrm{I}_{2}(s)\end{array}\right]=\left[\begin{array}{c}-\frac{\mathrm{V}}{\mathrm{s}} \\ 0\end{array}\right]$
- $\left[\begin{array}{ll}\mathrm{R}+\mathrm{Ls}+\frac{1}{\mathrm{Cs}} & -\mathrm{Ls} \\ -\mathrm{Ls} & \mathrm{R \; Ls} +\frac{1}{\mathrm{Cs}}\end{array}\right]\left[\begin{array}{l}\mathbf{l}_{1}(s) \\ \mathbf{l}_{2}(s)\end{array}\right]=\left[\begin{array}{c}-\frac{\mathrm{V}}{\mathrm{s}} \\ 0\end{array}\right]$
- $\left[\begin{array}{ll}\text{R + Ls +}\frac{1}{\text{Cs}} & \text{– Ls} \\ – \text{Ls} & \text{R + Ls +}\frac{1}{\text{Cs}}\end{array}\right]\left[\begin{array}{l}\text{I}_{1}(s) \\ \text{I}_{2}(s)\end{array}\right]=\left[\begin{array}{l}\frac{\text{V}}{\mathrm{s}} \\ 0\end{array}\right]$