Current $I=I_0+\beta t$
$ I_0=20 \mathrm{~A} $
$ \beta=3 \mathrm{~A} / \mathrm{s}$
$ \mathrm{I}=20+3 \mathrm{t} $
$ \frac{\mathrm{dq}}{\mathrm{dt}}=20+3 \mathrm{t} $
$ \int_0^q \mathrm{dq}=\int_0^{20}(20+3 \mathrm{t}) \mathrm{dt} $
$ \mathrm{q}=\int_0^{20} 20 \mathrm{dt}+\int_0^{20} 3 \mathrm{tdt} $
$ \mathrm{q}=\left[20 \mathrm{t}+\frac{3 \mathrm{t}^2}{2}\right]_0^{20}=1000\ \mathrm{C}$

where $i$ is the current in the potentiometer

($A$) The voltmeter displays $-5 \mathrm{~V}$ as soon as the key is pressed, znd displays $+5 \mathrm{~V}$ after a long time
($B$) The voltmeter will display $0 \mathrm{~V}$ at time $t=\ln 2$ seconds
($C$) The current in the ammeter becomes $1 / e$ of the initial value after $1$ second
($D$) The current in the ammeter becomes zero after a long time
