= Solution
For an initially uncharged capacitor and zero initial current, <Kirchhoff's voltage law> is $L\dot I+RI+Q/C=V_0$, with $I=\dot Q$. Let
$$
\alpha=\frac{R}{2L},\qquad\omega_n=\frac1{\sqrt{LC}},\qquad\omega_d=\sqrt{\omega_n^2-\alpha^2}.
$$
Differentiating the circuit equation at $t>0$ gives $\ddot I+2\alpha\dot I+\omega_n^2I=0$, while the original equation fixes $I(0)=0$ and $\dot I(0)=V_0/L$. The underdamped response is
$$
\boxed{I(t)=\frac{V_0}{L\omega_d}e^{-\alpha t}\sin\omega_dt.}
$$
For 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 $I=V_0e^{-\alpha t}\sinh(\beta t)/(L\beta)$. 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.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-resit-rlc-step.png]
{title=Current after a voltage step in a series RLC circuit}
{description=Underdamped, critically damped, and overdamped responses for an initially uncharged capacitor. Time and current use the natural LC scales.}
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