Question

A current through a wire depends on time as $i={\alpha }_{o}t+\beta t^2$ where ${\alpha }_o=20\text{A/s}$ and $\beta =8{\text{As}}^{-2}$. Find the charge crossed through a section of the wire in $15\text{s}$.

• A. $2100\text{C}$
• B. $260\text{C}$
• C. $2250\text{C}$
• D. $11250\text{C}$

Answer 1

The correct option is D $11250\text{C}$

Given :
$i={\alpha }_{o}t+\beta t^2$
${\alpha }_o=20\text{A/s}$
$\beta =8{\text{As}}^{-2}$
$t=15\text{s}$

We know that,
$i=\frac{dq}{dt}$

$\Rightarrow {\int }_0^{t}idt={\int }_0^{Q}dq$

$\Rightarrow {\int }_0^{15}({\alpha }_{o}t+\beta t^2)dt={\int }_0^{Q}dq$

$\Rightarrow Q={[\frac{{\alpha }_o{t}^2}{2}+\frac{\beta t^3}{3}]}_0^{15}$

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

​​​​​​​$\Rightarrow$ $Q=11250\text{C}$