In the given circuits a and b , switches S 1 and S 2 are closed at t = 0 and kept close for a long time. The…

In C-R series circuit,
$\mathit{\text{q}}={\mathit{\text{q}}}_0 \left(1-{\text{e}}^{\frac{-t}{\text{τ}}}\right)$
$⟹\mathit{\text{q}}=\mathit{\text{CE}} \left(1-{\text{e}}^{\frac{-\mathit{\text{t}}}{\mathit{\text{RC}}}}\right)$
$\therefore \mathit{\text{I}}=\frac{\text{d}\mathit{\text{q}}}{\text{d}\mathit{\text{t}}}$
$=\frac{\mathit{\text{CE}}}{\mathit{\text{RC}}}\left(+{\text{e}}^{\frac{-\mathit{\text{t}}}{\mathit{\text{RC}}}}\right)$
$\mathit{\text{I}}=\frac{\mathit{\text{E}}}{\mathit{\text{R}}} {\text{e}}^{\frac{-\mathit{\text{t}}}{\mathit{\text{RC}}}}$
$⟹$ the urrent decays exponentially. In $L-R$ series circuit,
$\mathit{\text{I}}={\mathit{\text{I}}}_0(1-{\text{e}}^{\frac{-\mathit{\text{t}}}{\mathit{\text{τ}}}})$
$\Rightarrow {\mathit{\text{I}}}_0=\frac{\mathit{\text{E}}}{\mathit{\text{R}}}$
$\text{⇒}\mathit{\text{I}}=\frac{\mathit{\text{E}}}{\mathit{\text{R}}} \left(1-{\text{e}}^{\frac{-\mathit{\text{Rt}}}{\mathit{\text{L}}}}\right)$
$⟹$ current grows exponentially.

$\therefore$ For $C-R$ circuit,



For $L-R$ circuit,