Topic: Network Theory
In the network shown below, switch is opened at $t=0$ after long time. The current $i_L(t)$ for $t>0$ is given as
- A. $e^{-2t}(2\cos t+4\sin t)\text{ A}$
- B. $e^{-2t}(3\sin t-4\cos t)\text{ A}$
- C. $e^{-2t}(-4\sin t+2\cos t)\text{ A}$
- D. $e^{-2t}(2\sin t-4\cos t)\text{ A}$
Answer: D. $e^{-2t}(2\sin t-4\cos t)\text{ A}$
Explanation: For $t<0$, $i_L = -4 \text{ A}$, $V_c = 8 \text{ V}$. For $t>0$, using the equivalent circuit in Laplace domain and superposition theorem:
$I_L(s) = \frac{-4(s+2)}{s²+4s+5} + \frac{2}{s²+4s+5} = \frac{-4(s+2)}{(s+2)^2+1} + \frac{2}{(s+2)^2+1}$
Taking inverse Laplace transform:
$I_L(t) = -4e^{-2t}\cos t + 2e^{-2t}\sin t = e^{-2t}(2\sin t - 4\cos t) \text{ A}$.