Physics - Current Electricity Question with Solution | TestHub

Given that
Current $I=I_0+\beta t$, $I_0=20 \mathrm{A}$ and $\beta =3 \mathrm{A} {\mathrm{s}}^{-1}$

The instantaneous current is given by

$I=20+3t ...\left(1\right)$

Integrating equation (1), we have

$$\begin{gathered} \frac{dq}{dt}=20+3t \\ \Rightarrow {\int }_0^{q}dq={\int }_0^{20}(20+3t)dt \\ \Rightarrow q={\int }_0^{20}20dt+{\int }_0^{20}3tdt \\ \Rightarrow q={\left[20t+\frac{3t^2}{2}\right]}_0^{20} \\ =1000 \mathrm{C} \end{gathered}$$