Question

For the R-L circuit shown in the figure, the input voltage $V_i\left(t\right)=u\left(t\right)$. The current $i\left(t\right)$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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

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

$i\left(t\right)=\frac{1}{2}(1-e^{-2t})$

$Att=0,i\left(t\right)=0$

$t=\infty ,i\left(t\right)=0.5A$

$t=\frac{1}{2},i\left(t\right)=0.31A$

Graph (c) satisfies all conditions.