Worked example: C1 = 2 μF and C2 = 4 μF are in series across 12 V; then the same two are reconnected in parallel across 12 V. Compare total charge drawn. Step 1 (series): 1/Ceq = 1/2 + 1/4 = 3/4 μF⁻¹, so Ceq = 4/3 ≈ 1.33 μF. Q_total = Ceq·V = 16 μC — this same 16 μC sits on each capacitor. Voltages: V1 = Q/C1 = 8 V, V2 = Q/C2 = 4 V (sum = 12 V ✓). Step 2 (parallel): Ceq = 2 + 4 = 6 μF. Q_total = 6 × 12 = 72 μC (24 μC on C1, 48 μC on C2). Step 3: The parallel combination draws 4.5× more charge — parallel multiplies storage, series divides it but raises voltage tolerance (here each capacitor sees only part of 12 V).