In the RL circuit in the figure, E = 12 V, R = 6 Ω and L = 3 H, and no current flows initially. Which of the following is closest to the voltage vL across the coil one time constant τ after the switch is closed at t = 0?

Answer: A. About 4.4 V

Method 1 (differential equation): solving L di/dt + Ri = E with i(0) = 0 gives i(t) = (E/R)(1 − e^(−t/τ)) with τ = L/R = 0.5 s, and vL = L di/dt = E e^(−t/τ). At t = τ, vL = 12 × e^(−1) ≈ 4.4 V. Method 2 (Laplace transform): I(s) = E/(s(R + sL)), so VL(s) = sL I(s) = E/(s + R/L), whose inverse transform is vL(t) = E e^(−Rt/L). Method 3 (initial and final values): the coil current cannot change suddenly, so at t = 0+ the coil acts like an open circuit with vL = E = 12 V; after a long time it acts like a short with vL = 0, and in between vL changes exponentially with the time constant τ.

• A. Correct. vL(τ) = 12 × e^(−1) ≈ 4.4 V.
• B. About 7.6 V is the resistor voltage at t = τ, vR = 12(1 − e^(−1)).
• C. 12 V is vL just after the switch closes, at t = 0+.