Battery capacity application: A 6 V battery supplies a 300 Ω load. If the battery is rated at 40 Ah, for approximately how many hours can it supply the load (ideal conditions)?

Introduction / Context:
Battery capacity, expressed in ampere-hours (Ah), indicates how long a battery can supply a given current. Translating load resistance and source voltage into current allows estimating runtime under idealized conditions (ignoring Peukert effects and cutoff limits).

Given Data / Assumptions:

• Battery voltage V ≈ 6 V (assumed constant).
• Load resistance R = 300 Ω.
• Battery capacity = 40 Ah.
• Ideal conditions: no internal resistance limits or voltage sag considered.

Concept / Approach:
First compute the current drawn by the load using Ohm's law. Then divide the battery's ampere-hour rating by the load current to estimate hours of operation: time (h) = Ah / A.

Step-by-Step Solution:

Verification / Alternative check:
Energy viewpoint: Average power P = V * I = 6 * 0.02 = 0.12 W; over 2,000 h, energy ≈ 240 Wh, consistent with 6 V * 40 Ah = 240 Wh nominal capacity.

Why Other Options Are Wrong:

• 1 h, 10 h, 200 h: Far below the correct runtime; each implies much larger load current than actually drawn by 300 Ω at 6 V.

Common Pitfalls:

• Forgetting to convert milliamperes to amperes when using Ah.
• Confusing energy (Wh) with capacity (Ah).

Final Answer:
2,000 h