202211250902 Solution to 2018-DSE-PHY-IA-26

A mobile phone battery labelled 2800\,\mathrm{mA\,h} capacity is fully charged initially. What percentage of capacity remains after it has delivered a current of 200\,\mathrm{mA} for 3 hours?


Background. (C-rate)

C-rate (in \mathrm{h^{-1}}) is defined the charging or discharging current (in \mathrm{A}) divided by the charge capacity (in \mathrm{A\,h}).

Background. (Capacity's)

Capacity C rated in ampere hours:

\begin{aligned} 1\,\mathrm{A\,h} & = 1\,\mathrm{A}\times 1\,\mathrm{h} \\ & = 1\cdot\mathrm{C\,s^{-1}}\times 3600\cdot\mathrm{s} \\ & = 3600\,\mathrm{C} \\ \end{aligned}

is equivalent to a measure of charge as in coulometry; whereas in watt hours

\begin{aligned} 1\,\mathrm{W\,h} & = 1\,\mathrm{W}\times 1\,\mathrm{h} \\ & = 1\cdot\mathrm{J\,s^{-1}}\times 3600\cdot\mathrm{s} \\ & = 3600\,\mathrm{J} \\ \end{aligned}

a measure of energy as in voltammetry.

Background. (state of charge/depth of discharge)

\textrm{SoC}+\textrm{DoD}=1

Background. (full/nominal/cutoff voltage)

\displaystyle{V_\textrm{nominal}=\frac{V_\textrm{full}-V_\textrm{cutoff}}{2}}


Limited in my knowledge, may I write

\begin{aligned} \textrm{SoC} & = \frac{2800-200\times 3}{2800} \times 100\% \\ & = 78.6\%\quad \textrm{(3 s.f.)} \\ \end{aligned}