$(ii)$ $\mathrm{t} \uparrow \mathrm{I} \downarrow$
$(iii)$ For $\mathrm{I}=0 ; \mathrm{a}-2 \mathrm{bt}=0 ; \mathrm{t}=\mathrm{a} / 2 \mathrm{b}$
$(iv)$ $\frac{\mathrm{dI}}{\mathrm{dt}}=-2 \mathrm{b}$
Hence option $(4)$ is incorrect.



Reason : The current flows towards the point of the higher potential, as it does in such a circuit from the negative to the positive terminal.


