= Solution
Choose positive <electric current> in the initial discharge direction, since the source diagram does not label the initial plate polarity or a current arrow. Let $q$ be the charge remaining on the initially positive plate, so $I=-\dot q$. <Kirchhoff's voltage law> for the <series RLC circuit> gives
$$
L\ddot q+R\dot q+\frac qC=0,\qquad q(0)=CV_0=150\,\mu\mathrm C,\quad \dot q(0)=0.
$$
The natural <angular frequency>, <damping> coefficient and <quality factor> are
$$
\omega_0=\frac1{\sqrt{LC}}=816.50\,\mathrm{rad\,s^{-1}},\qquad
\alpha=\frac R{2L}=33.333\,\mathrm{s^{-1}},\qquad
\boxed{Q=\frac{\omega_0L}{R}=12.25.}
$$
Since $\alpha<\omega_0$ (equivalently $Q>1/2$), this is an <underdamped RLC response>. With $\omega_d=\sqrt{\omega_0^2-\alpha^2}=815.82\,\mathrm{rad\,s^{-1}}$, satisfying the initial conditions gives
$$
q(t)=CV_0e^{-\alpha t}\left(\cos\omega_dt+\frac\alpha{\omega_d}\sin\omega_dt\right),
$$
and differentiating gives
$$
\boxed{I(t)=\frac{V_0}{L\omega_d}e^{-\alpha t}\sin\omega_dt
=(0.12258\,\mathrm A)e^{-t/(30.0\,\mathrm{ms})}\sin\left[(815.82\,\mathrm{rad\,s^{-1}})t\right].}
$$
The <electric current> starts at zero with $\dot I(0)=V_0/L=100\,\mathrm{A\,s^{-1}}$. Its <wave period> is $T_d=2\pi/\omega_d=7.702\,\mathrm{ms}$, its zeros occur every $3.851\,\mathrm{ms}$, and its positive and negative <wave envelopes> decay with time constant $30.0\,\mathrm{ms}$. The first maximum obeys $\tan\omega_dt=\omega_d/\alpha$ and is $0.1151\,\mathrm A$ at $1.875\,\mathrm{ms}$. The <electric current> tends to zero as the initial <capacitor> <energy> is dissipated in the <resistor>.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-2-a3-current.png]
{title=Original calculated <electric current> and decay <wave envelopes> for the <underdamped RLC response>}
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