A current through a wire depends on time

(4)

given : $i=\alpha_{0} t+\beta t^{2}$

$\alpha=20 \mathrm{~A} / \mathrm{s}$ and $\beta=8 \mathrm{As}^{-2}$

$\mathrm{t}=15 \mathrm{sec}$

we know that, $i=\frac{d q}{d t} \Rightarrow \int_{0}^{t} i d t=\int_{0}^{Q} d q$\

$\Rightarrow \int_{0}^{15}\left(\alpha_{0} t+\beta t^{2}\right) d t=\int_{0}^{Q} d q$

$\Rightarrow Q=\left[\frac{\alpha_{0} t^{2}}{2}+\frac{\beta t^{3}}{3}\right]_{0}^{15}$

$\Rightarrow Q=\frac{20 \times 15 \times 15}{2}+\frac{8 \times 15 \times 15 \times 15}{3}-0$

$\Rightarrow Q=11250 \mathrm{c}$