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Use positive current in the charging direction and passive voltage drops across $R,L,C$. The five ideal elements obeyThe battery maintains its emf independently of current. An ideal open switch has $I=0$ with its voltage unconstrained; an ideal closed switch has $v_{\mathrm{switch}}=0$ with its current unconstrained. These statements are the switch's piecewise current-voltage relation. Inductor current continuity and capacitor-voltage continuity hold when no impulse voltage or current is imposed.

The source diagram has $2R$ in series with two parallel branches: one containing $R$ then $L$, and the other containing $2R$. Let their common voltage be $u(t)$, measured from the upper node to the lower return. Kirchhoff's current law and the branch laws giveUsing $I_2=u/(2R)$ in the source equation yields $u=V/2-R{I}_1$. Substitution into the coil branch givesThe switch was open for a long time, so $I_1(0^{-})=0$. Inductor current continuity gives $I_1(0^{+})=0$ for a finite applied voltage.

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 giveThus $I^{\prime }(0^{+})=(V_0-R{I}_1)/L\simeq 115.0\mathrm{A}{\mathrm{s}}^{-1}$, not zero. Differentiating the loop equation gives the homogeneous current equationThe 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 areThe full current satisfying both initial conditions isIts amplitude envelope decays as $e^{-\alpha \tau }$. ThereforeThis is measured after reopening; on the clock starting with the first closure it is $t\simeq 0.02098\mathrm{s}$. An instantaneous current zero is not an amplitude decay measurement.