Question:
A current through a wire depends on time as $\mathrm{i}=\alpha_{0} \mathrm{t}+\beta \mathrm{t}^{2}$ where $\alpha_{0}=20 \mathrm{~A} / \mathrm{s}$ and $\beta=8 \mathrm{As}^{-2}$. Find the charge crossed through a section of the wire in $15 \mathrm{~s}$.
Correct Option: , 2
Solution:
$\mathrm{i}=20 \mathrm{t}+8 \mathrm{t}^{2}$
$\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}} \Rightarrow \int \mathrm{dq}=\int \mathrm{idt}$
$\Rightarrow \mathrm{q}=\int_{0}^{15}\left(20 \mathrm{t}+8 \mathrm{t}^{2}\right) \mathrm{dt}$
$q=\left(\frac{20 t^{2}}{2}+\frac{8 t^{3}}{3}\right)_{0}^{15}$
$\mathrm{q}=10 \times(15)^{2}+\frac{8(15)^{3}}{3}$
$q=2250+9000$
$q=11250 \mathrm{C}$