The charge flowing through a resistance $R$ varies with time according to $Q = at -bt^2.$ The total heat produced in $R$ is : (assume that direction of current not reversed)
Diffcult
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$Q=a t-b t^{2}$
$\mathrm{I}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{a}-2 \mathrm{bt}$
$I=0$ for $t=t_{0}=(a / 2 b)$
Current flows from $t=0$ to $t=t_{0}$
Heat produced $ = \int_0^{{t_0}} {{{\rm{I}}^2}} {\rm{R\,dt}}$
$ = \int_0^{{t_0}} {{{(a - 2bt)}^2}} R\,dt$
$\int_0^{{t_0}} {\left( {{a^2}R - 4abRt + 4{b^2}R{t^2}} \right)} dt$
Solving above equation and putting $t_{0}=(a / 2 b)$
we get Heat produced $=\left(\frac{a^{3} R}{6 b}\right)$
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