Question

For a series RLC circuit excited by an impulse voltage of magnitude 1 and having R = 4Ω, L = 1H, C = \(\frac{1}{3}\) F, the value of the current I(t) is ___________

(a) \(\frac{3}{2} e^{3t} – \frac{1}{2} e^{-t}\)

(b) \(\frac{3}{2} e^{3t} – \frac{1}{2} e^t\)

(c) \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^{-t}\)

(d) \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^t\)

I have been asked this question in an interview for job.

The origin of the question is Characterization and Nature of Systems topic in portion Laplace Transform and System Design of Signals and Systems

Answer 1

The correct choice is (c) \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^{-t}\)

To explain I would say: Z(s) = R + Ls + \(\frac{1}{Cs}\)

= 4 + s + \(\frac{3}{s} = \frac{(s+3)(s+1)}{s}\)

Now, v (t) = δ(t)

∴ V(s) = 1

∴ I(s) = \(\frac{V(s)}{Z(s)} = \frac{1}{Z(s)}\)

∴ I(s) = \(\frac{s}{(s+3)(s+1)}\)

Taking inverse, we get, I (t) = \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^{-t}\).