Question

The circuit shown in the figure has initial current $i_L\left(0^{-}\right)=1A$ through the inductor and an initial voltage $v_c\left(0^{-}\right)=-1V$ across the capacitor. For input v(t) = u(t), the Laplace transform of the current i(t) for $t\ge 0$ is

• A. $\frac{s}{s^2+s+1}$
• B. $\frac{s+2}{s^2+s+1}$
• C. $\frac{s-2}{s^2+s-1}$
• D. $\frac{s-2}{s^2+s+1}$

Answer 1

The correct option is B

$\frac{s+2}{s^2+s+1}$

KVL,
$v\left(t\right)=R.i\left(t\right)+\frac{Ldi\left(t\right)}{dt}+\frac{1}{C}\underset{0}{\overset{\infty }{\int }}i\left(t\right)dt$

Taking Laplace transform on both sides,

$V\left(s\right)=RI\left(s\right)+LsI\left(s\right)-LI\left(0^{+}\right)+\frac{I\left(s\right)}{sC}+\frac{V_c\left(0^{+}\right)}{s}$

$\Rightarrow \frac{1}{s}=I\left(s\right)+sI\left(s\right)-1+\frac{I\left(s\right)}{s}-\frac{1}{s}$

$\frac{2}{s}+1=\frac{I\left(s\right)}{s}[s^2+s+1]$

$I\left(s\right)=\frac{s+2}{s^2+s+1}$