A battery stores 36.129 Wh. How long will it run a load of 8 W, 10 W, 15 W or 25 W?

Step-by-step solution

1. 1

   Get the units to do the work. A watt-hour is one watt sustained for one hour, so energy divided by power leaves a time:

   $\text{runtime (h)}=\frac{\text{energy (Wh)}}{\text{power (W)}}$

   Dimensionally $\dfrac{\text{W}\cdot\text{h}}{\text{W}}=\text{h}$ ✓. This is why a Wh rating is more directly useful than a mAh rating, which needs the voltage before it can be converted.

2. 2

   Divide for a light load. At $8$ W and at $10$ W:

   $\frac{36.129}{8}=4.5161\ \text{h},\qquad\frac{36.129}{10}=3.6129\ \text{h}$

   So roughly $3.6$ to $4.5$ hours — about $4\tfrac12$ hours at the lightest draw. Dividing by $10$ is just a decimal shift, a handy sanity anchor for the other figures.

3. 3

   Divide for a moderate and a heavy load.

   $\frac{36.129}{15}=2.4086\ \text{h},\qquad\frac{36.129}{25}=1.4452\ \text{h}$

   That is about $2.4$ hours and about $1.4$ hours. Note the inverse relationship: tripling the load from $8$ W to $25$ W cuts the runtime to roughly a third, not by a third.

4. 4

   Convert to hours and minutes. $0.4086$ h $\times60\approx24.5$ min, so $2.4086$ h is about $2$ h $25$ min; similarly $1.4452$ h is about $1$ h $27$ min and $4.5161$ h is about $4$ h $31$ min. Multiplying the fractional part by $60$ is the only conversion needed.

5. 5

   State the caveats that make these upper bounds. The arithmetic assumes the full rated energy is usable at a constant draw. In practice the reserve cut-off, conversion losses and capacity fade with age all reduce the figure, so treat these as optimistic ceilings — real-world runtime typically lands $10$–$20\%$ lower.