In the RC circuit in the figure, E = 10 V, R = 10 kΩ and C = 100 µF, and the capacitor is initially uncharged. Which of the following is closest to the capacitor voltage vC 2 seconds after the switch is closed at t = 0?

Answer: B. About 8.6 V

Method 1 (differential equation): the circuit equation is RC dvC/dt + vC = E, and its solution with vC(0) = 0 is vC(t) = E(1 − e^(−t/RC)). Since τ = RC = 10×10³ × 100×10⁻⁶ = 1 s, vC(2) = 10(1 − e^(−2)) ≈ 10 × 0.865 ≈ 8.6 V. Method 2 (Laplace transform): in the s-domain, VC(s) = (E/s) × (1/(sC))/(R + 1/(sC)) = E/(s(1 + sRC)), which splits into partial fractions as E(1/s − 1/(s + 1/RC)) and transforms back to the same formula. Method 3 (time-constant rule of thumb): 63% at 1τ, and at 2τ a further 63% of the remaining 37% is added, reaching about 86%, so about 8.6 V can be estimated.

• A. About 6.3 V is the value at t = τ = 1 s.
• B. Correct. vC(2 s) = 10(1 − e^(−2)) ≈ 8.6 V.
• C. About 9.5 V is the value at t = 3τ = 3 s.