 Solution

For an initially uncharged capacitor and zero initial current, Kirchhoff's voltage law is $L\dot{I}+RI+Q/C=V_0$, with $I=\dot{Q}$. LetDifferentiating the circuit equation at $t>0$ gives $\ddot{I}+2\alpha \dot{I}+{\omega }_n^{2}I=0$, while the original equation fixes $I(0)=0$ and $\dot{I}(0)=V_0/L$. The underdamped response isFor critical damping, $\alpha ={\omega }_n$, its limit is $I=(V_0/L)t{e}^{-\alpha t}$. For overdamping, put $\beta =\sqrt{{\alpha }^2-{\omega }_n^2}$ and obtain $I=V_0{e}^{-\alpha t}\sinh (\beta t)/(L\beta )$. These forms cover every positive $R,L,C$.

The source does not explicitly specify the capacitor's initial charge. If its initial voltage is $V_{C0}$, replace $V_0$ in the numerators by $V_0-V_{C0}$; an open switch alone does not discharge a capacitor. The boxed expression uses the usual uncharged initial condition.