Question

In the given figure assume that the switch was in position 1 for a long time and thrown to position 2 at $t=0$. $I_1\left(s\right)$ and $I_2\left(s\right)$ are the Laplace transform of $i_1\left(t\right)$ and $i_2\left(t\right)$
respectively. The equation for the loop currents $I_1\left(s\right)$ and $I_2\left(s\right)$ for the circuit after switch is brought form position 1 to position 2 at$t=0$ is

• A. $\left[\begin{array}{cc}5+\frac{1}{s}+2s& -2s\\ -2s& 2+\frac{1}{s}+2s\end{array}\right]\left[\begin{array}{c}{I}_1\left(s\right)\\ I_2\left(s\right)\end{array}\right]=\left[\begin{array}{c}0\\ \frac{-10}{s}\end{array}\right]$
  
• B.
  $\left[\begin{array}{cc}5+\frac{1}{s}+2s& -2s\\ -2s& 2+\frac{1}{s}+2s\end{array}\right]\left[\begin{array}{c}{I}_1\left(s\right)\\ I_2\left(s\right)\end{array}\right]=\left[\begin{array}{c}\frac{10}{s}\\ 0\end{array}\right]$
• C. $\left[\begin{array}{cc}5+\frac{1}{s}+2s& -2s\\ -2s& 2+\frac{1}{2s}+2s\end{array}\right]\left[\begin{array}{c}{I}_1\left(s\right)\\ I_2\left(s\right)\end{array}\right]=\left[\begin{array}{c}\frac{-10}{s}\\ 0\end{array}\right]$
  
• D. $\left[\begin{array}{cc}5+\frac{1}{s}+2s& -2s\\ -2s& 2+\frac{1}{s}+2s\end{array}\right]\left[\begin{array}{c}{I}_1\left(s\right)\\ I_2\left(s\right)\end{array}\right]=\left[\begin{array}{c}0\\ \frac{-10}{s}\end{array}\right]$
  

Answer 1

The correct option is C $\left[\begin{array}{cc}5+\frac{1}{s}+2s& -2s\\ -2s& 2+\frac{1}{2s}+2s\end{array}\right]\left[\begin{array}{c}{I}_1\left(s\right)\\ I_2\left(s\right)\end{array}\right]=\left[\begin{array}{c}\frac{-10}{s}\\ 0\end{array}\right]$


When switch is in poistion -2,


Applying KVL in loop -1, we get
$I_1\left(s\right)5+\frac{10}{s}+I_1\left(s\right)\frac{1}{s}+\left[I_1\right(s)-I_2(s\left)\right]2s=0$
$I_1\left(s\right)[5+\frac{1}{s}+2s]-I_2\left(s\right).2s=-\frac{10}{s}$
Applying KVL in loop-2 we get
$\left[I_2\right(s)-I_1(s\left)\right]2s+I_2\left(s\right).2+I_2\left(s\right).\frac{1}{2s}=0$
$I_1\left(s\right)(-2s)+I_2\left(s\right)[2+2s+\frac{1}{2s}]=0$
$\left[\begin{array}{cc}5+\frac{1}{s}+2s& -2s\\ -2s& 2+\frac{1}{2s}+2s\end{array}\right]\left[\begin{array}{c}{I}_1\left(s\right)\\ I_2\left(s\right)\end{array}\right]=\left[\begin{array}{c}\frac{-10}{s}\\ 0\end{array}\right]$