Worked example: Three identical cells (E = 1.5 V, r = 0.2 Ω each) drive a 4 Ω bulb, first all in series, then all in parallel. Compare current and terminal voltage. Step 1 (series): Eeq = 4.5 V, req = 0.6 Ω. I = 4.5/(4 + 0.6) = 0.978 A ≈ 0.98 A. Terminal V = Eeq − I·req = 4.5 − 0.59 = 3.91 V. Bulb power = I²R ≈ 3.83 W — bright. Step 2 (parallel): Eeq = 1.5 V, req = 0.2/3 ≈ 0.067 Ω. I = 1.5/(4.067) = 0.369 A. Terminal V = 1.5 − 0.025 = 1.475 V. Power ≈ 0.54 W — dim, but the set lasts ~3× longer and can push 3× the current into a heavy load. Step 3: Moral — series for voltage, parallel for current capacity and life.