A 10 μF capacitor in series with a 100 kΩ resistor is connected to a 12 V battery at t = 0. Find (a) the time constant, (b) the initial current, (c) the charge after one time constant, (d) roughly when it is fully charged. Step 1: τ = RC = (100×10³)(10×10⁻⁶) = 1.0 s. Step 2: Initial current I₀ = V₀/R = 12/(100×10³) = 1.2×10⁻⁴ A = 0.12 mA (capacitor initially uncharged, acts as short). Step 3: Maximum charge Q₀ = CV₀ = 10×10⁻⁶ × 12 = 1.2×10⁻⁴ C = 120 μC. After t = τ: Q = Q₀(1 − e⁻¹) = 120 × 0.632 ≈ 75.8 μC (63% charged); current has fallen to 37%: 0.044 mA. Step 4: After 5τ = 5 s, charge is (1 − e⁻⁵) ≈ 99.3% of maximum — for practical purposes fully charged, and current is essentially zero. If R were 10 kΩ instead, τ = 0.1 s and charging would finish in 0.5 s — smaller resistance, faster transient.