Topic: Network Theory
A unit step voltage $2u(t-\theta)$ is applied in a series RC circuit with $R=2~\Omega$, $C=1\text{F}$. Assuming zero initial conditions, find $i(t)$
- A. $i(t)=u(t)e^{-0.5t}\text{ A}$
- B. $i(t)=u(t-2\theta)e^{-0.5(t-2\theta)}\text{ A}$
- C. $i(t)=u(t-\theta)e^{-0.5(t-\theta)}\text{ A}$
- D. $i(t)=u(t-\theta)e^{-(t-\theta)}\text{ A}$
Answer: C. $i(t)=u(t-\theta)e^{-0.5(t-\theta)}\text{ A}$
Explanation: Given $v_{s}(t)=2u(t-\theta)$, its Laplace transform is $V_{s}(s)=\frac{2e^{-\theta s}}{s}$.
$I(s)=\frac{V_{s}(s)}{R+\frac{1}{Cs}} = \frac{\frac{2e^{-\theta s}}{s}}{2+\frac{1}{s}} = \frac{2e^{-\theta s}}{2s+1} = \frac{e^{-\theta s}}{s+0.5}$.
Taking the inverse Laplace transform, $i(t)=u(t-\theta)e^{-0.5(t-\theta)}\text{ A}$.