Physics - Current Electricity Question with Solution | TestHub

Given that
Current $I=I_0+\beta t$, $I_0=20\text{A}$ and $\beta =3\text{A}{\text{s}}^{-1}$The instantaneous current is given by

$I=20+3t...\left(1\right)$Integrating equation (1), we have

$$\begin{gathered} \frac{dq}{dt}=20+3t \\ \Rightarrow \in t_0^{q}dq=\in t_0^{20}(20+3t)dt \\ \Rightarrow q=\in t_0^{20}20dt+\in t_0^{20}3tdt \\ \Rightarrow q={\left[20t+\frac{3t^2}{2}\right]}_0^{20} \\ =1000\mathrm{C} \end{gathered}$$