In the series RLC circuit in the figure, E = 6 V, R = 4 Ω, L = 1 H and C = 1/3 F, and all initial values are 0. Which of the following is the current i(t) [A] after the switch is closed at t = 0?

Answer: B. 3e^(−t) − 3e^(−3t)

Method 1 (Laplace transform and partial fractions): in the s-domain, I(s) = (E/s)/(R + sL + 1/(sC)) = 6/(s² + 4s + 3) = 6/((s + 1)(s + 3)). Splitting into partial fractions as I(s) = A/(s + 1) + B/(s + 3) gives A = 6/(−1 + 3) = 3 and B = 6/(−3 + 1) = −3, so i(t) = 3e^(−t) − 3e^(−3t) A. Method 2 (differential equation): the roots −1 and −3 of the characteristic equation s² + 4s + 3 = 0 give i = Ae^(−t) + Be^(−3t); i(0+) = 0 (continuity of the coil current) and di/dt(0+) = E/L = 6 A/s give A + B = 0 and −A − 3B = 6, so A = 3 and B = −3. The current peaks at about 1.15 A at t = (ln 3)/2 ≈ 0.55 s and returns to 0 as the capacitor charges.

• A. At t = 0 this gives i = 6 A, which contradicts the coil current starting from 0.
• B. Correct. i(t) = 3e^(−t) − 3e^(−3t), with i(0) = 0.
• C. This forgets to divide by (s + 3) − (s + 1) = 2 when finding the partial-fraction coefficients.