The charge following through a resistance $$R$$ varies with time $$t$$ as $$Q = at - b{t^2},$$   where $$a$$ and $$b$$ are positive constants. The total heat produced in $$R$$ is

Solution :
Given, charge $$Q = at - b{t^2}\,.......\left( {\text{i}} \right)$$
$$\because $$ We know that current, $$I = \frac{{d\theta }}{{dt}}$$
So, eq. (i) can be written as
$$I = \frac{d}{{dt}}\left( {at - b{t^2}} \right) \Rightarrow I = a - 2bt\,......\left( {{\text{ii}}} \right)$$
For maximum value of $$t,$$ till the current exist is given by
$$\eqalign{ & \Rightarrow a - 2bt = 0 \cr & \therefore t = \frac{a}{{2b}}\,......\left( {{\text{iii}}} \right) \cr} $$
$$\because $$ The total heat produced $$\left( H \right)$$ can be given as
$$\eqalign{ & H = \int_0^t {{I^2}Rdt} \cr & = \int_0^{\frac{a}{{2b}}} {{{\left( {a - 2bt} \right)}^2}R.dt\,\left\{ {\because t = \frac{a}{{2b}}} \right\}} \cr & = \int_0^{\frac{a}{{2b}}} {\left( {{a^2} + 4{b^2}{t^2} - 4abt} \right)Rdt} \cr & H = \left[ {{a^2}t + 4{b^2}\frac{{{t^3}}}{3} - \frac{{4ab{t^2}}}{2}} \right]_0^{\frac{a}{{2b}}}R \cr} $$
Solving above equation, we get
$$ \Rightarrow H = \frac{{{a^3}R}}{{6b}}$$