Answer:
Approximately [tex]208[/tex] ampere-hours.
Explanation:
In this question, the amount of energy stored in the battery is given (in kilowatt-hours, [tex]{\rm kWh}[/tex].) To find the capacity of the battery in ampere-hours, divide the amount of energy stored in the battery by the rated voltage of the battery. Note that [tex]1\; {\rm W} = (1\; {\rm V})\, (1\; {\rm A})[/tex].
[tex]\begin{aligned}\frac{7.70\; {\rm kW\, h}}{37.00\; {\rm V}} &= \frac{7.70 \times 10^{3}\; {\rm W\, h}}{37.00\; {\rm V}} \approx 208\; {\rm A\, h}\end{aligned}[/tex].
In other words, the capacity of this battery would be approximately [tex]208[/tex] ampere-hours.
To convert the 37.00 V flashlight battery's rating of 7.70 kWh to ampere-hours, we need to use the relationship Energy (Wh) = Voltage (V) * Capacity (Ah). The battery's rating is 208.11 ampere-hours (Ah).
To find the battery's rating in ampere-hours (Ah), we first need to convert the energy rating from kilowatt-hours (kWh) to watt-hours (Wh). Since 1 kWh = 1000 Wh, the battery's energy rating is:
7.70 kWh * 1000 = 7700 Wh
Next, we use the relationship between energy (in watt-hours), voltage (in volts), and capacity (in ampere-hours). This relationship is given by:
Rearranging the formula to solve for capacity (Ah):
Substituting the given values:
Capacity (Ah) = 7700 Wh / 37.00 V = 208.11 Ah
Therefore, the battery's rating is 208.11 ampere-hours (Ah).