a
$q=\int \mathrm{idt}=$ area of $\mathrm{i} \mathrm{v} / \mathrm{s}$ $t$ graph
$=\frac{1}{2}\left(I_{0}\right)(\Delta t)$
$I_{0}=\frac{2 q}{\Delta t}$
also, $\mathrm{i}=-\left(\frac{\mathrm{I}_{0}}{\Delta \mathrm{t}}\right) \mathrm{t}+\mathrm{I}_{0}$
heat loss $=\int i^{2} \mathrm{Rdt}=\frac{4}{3} \frac{\mathrm{q}^{2} \mathrm{R}}{\Delta \mathrm{t}}$
