Answer
The reserve capacity rating (2)
Work Step by Step
(1) cranking amps number$=905A$
$E_{1}=P\Delta t=IV\Delta t$
$=(905\mathrm{A})(7.2\mathrm{V})(30.0\mathrm{s})$
$=2.0\times 10^{5}\mathrm{J}$
(2) reserve capacity$=155\displaystyle \min$
$E_{2}=P\Delta t=IV\Delta t$
$=(25\displaystyle \mathrm{A})(10.5\mathrm{V})(155\min\times\frac{60\mathrm{s}}{1\min})$
$=2.4\times 10^{6}\mathrm{J}$
The reserve capacity rating represents a greater amount of energy delivered by the battery.
