In the circuit in the figure, E = 12 V, R1 = 2 Ω and R2 = 4 Ω, and the capacitor and the coil initially store no energy. Which pair correctly gives the current i(0+) just after the switch is closed at t = 0 and the capacitor voltage vC after a long time?

Answer: C. i(0+) = 6 A, vC = 8 V

Method 1 (continuity and the DC steady state): the capacitor voltage and the coil current cannot change suddenly, so at t = 0+, vC = 0 (the capacitor acts like a short) and iL = 0 (the coil acts like an open circuit). All of the source current flows into the capacitor, so i(0+) = E/R1 = 12/2 = 6 A. After a long time the capacitor acts like an open circuit and the coil like a short, so i = 12/(2 + 4) = 2 A and vC is the voltage across R2, 2 × 4 = 8 V. Method 2 (Laplace transform with the initial- and final-value theorems): with the impedance seen from the source Z(s) = R1 + (sL + R2 in parallel with 1/(sC)), I(s) = E/(s Z(s)). As s → ∞, Z → R1, so i(0+) = E/R1; as s → 0, Z → R1 + R2, so i(∞) = E/(R1 + R2), the same results.

• A. 12 V assumes the coil still blocks current after a long time, so the full source voltage appears across the capacitor. Just after switching, the capacitor voltage is 0, so the capacitor acts like a short and current flows.
• B. 2 A is the current after a long time, not just after switching.
• C. Correct. i(0+) = 6 A and vC(∞) = 8 V.