In the RC circuit in the figure, E = 10 V, R = 2 kΩ and C = 1 µF, and the capacitor is initially charged to 4 V with its + side in the figure at the higher potential. Which of the following is closest to the current i one time constant τ after the switch is closed at t = 0?

Answer: A. About 1.1 mA

Method 1 (initial and final values): the capacitor voltage cannot change suddenly, so at t = 0+ it is still vC = 4 V and i(0+) = (10 − 4)/2 kΩ = 3 mA. After a long time i = 0, and in between the current changes exponentially with the time constant τ = RC = 2 ms, so i(t) = 3e^(−t/τ) mA and i(τ) = 3 × e^(−1) ≈ 1.1 mA. Method 2 (Laplace transform): representing the capacitor as 1/(sC) in series with a voltage source V0/s gives I(s) = ((E − V0)/s)/(R + 1/(sC)) = ((E − V0)/R)/(s + 1/(RC)), whose inverse transform is the same i(t). Method 3 (differential equation): solving RC dvC/dt + vC = E with vC(0) = 4 V gives vC = 10 − 6e^(−t/τ), and i = C dvC/dt gives the same formula.

• A. Correct. i(τ) = 3 mA × e^(−1) ≈ 1.1 mA.
• B. About 1.8 mA ignores the initial voltage: (10 V/2 kΩ) × e^(−1).
• C. 3.0 mA is the current just after the switch closes, at t = 0+.