Topic: Network Theory
In the network shown below, switch is moved from position a to b at $t=0$. The current $i_{L}(t)$ for $t>0$ is given as
- A. $(4-6t)e^{4t}A$
- B. $(3-6t)e^{-4t}A$
- C. $(3-9t)e^{-5t}A$
- D. $(3-8t)e^{-5t}A$
Answer: C. $(3-9t)e^{-5t}A$
Explanation: For $t<0$, the inductor is shorted by position a. $i_{L}(0^-) = 3\text{ A}$.
For $t>0$, taking the Laplace transform of the circuit:
$(\frac{4}{s} - I_L)6 + 6 = \frac{12}{s} + I_L(2s + 14 + \frac{50}{s})$
$\frac{12}{s} + 6 = I_L(2s + 20 + \frac{50}{s})$
$I_L = \frac{6s+12}{2s^2+20s+50} = \frac{3s+6}{s²+10s+25} = \frac{3s+6}{(s+5)^2} = \frac{3(s+5) - 9}{(s+5)^2} = \frac{3}{s+5} - \frac{9}{(s+5)^2}$.
Taking the inverse Laplace transform gives $i_L(t) = 3e^{-5t} - 9te^{-5t} = (3-9t)e^{-5t}\text{ A}$.