What is Partial fraction decomposition?

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.

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.