= Solution
Put $t_1=0.005\,\mathrm s$, $\tau=t-t_1$, and $I_1=2(1-e^{-1.875})$. When the switch reopens, the circuit again contains the battery and all three series elements. <Inductor current continuity> and <capacitor voltage continuity> give
$$
I(0^+)=I_1,\qquad V_C(0^+)=0,\qquad L I'+RI+V_C=V_0,\qquad C V_C'=I.
$$
Thus $I'(0^+)=(V_0-RI_1)/L\simeq115.0\,\mathrm{A\,s^{-1}}$, not zero. Differentiating the loop equation gives the homogeneous current equation
$$
I''+\frac RL I'+\frac1{LC}I=0.
$$
The battery is still connected; its constant voltage disappears on differentiating, and the eventual capacitor voltage is $V_0$.
With $\alpha=R/(2L)=187.5\,\mathrm{s^{-1}}$ and $\omega_0=1/\sqrt{LC}$, the inequality $\alpha<\omega_0$ makes the circuit underdamped. The damped <angular frequency> and ordinary <frequency> are
$$
\boxed{\omega_d=\sqrt{\frac1{LC}-\frac{R^2}{4L^2}}\simeq1.7677\times10^4\,\mathrm{rad\,s^{-1}},\qquad f_d=\frac{\omega_d}{2\pi}\simeq2.813\,\mathrm{kHz}.}
$$
The full current satisfying both initial conditions is
$$
I(\tau)=e^{-\alpha\tau}\left[I_1\cos(\omega_d\tau)+\frac{I'(0^+)+\alpha I_1}{\omega_d}\sin(\omega_d\tau)\right].
$$
Its amplitude envelope decays as $e^{-\alpha\tau}$. Therefore
$$
\boxed{\tau_{5\%}=\frac{\ln20}{\alpha}=\frac{2L}{R}\ln20\simeq0.01598\,\mathrm s.}
$$
This is measured after reopening; on the clock starting with the first closure it is $t\simeq0.02098\,\mathrm s$. An instantaneous current zero is not an amplitude decay measurement.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-resit-rlc-switching-response.png]
{title=Capacitor bypass and reopening: RL current growth, the first oscillations, and the decaying RLC amplitude envelope}
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