The PDF gives $V=H(t)$, the <Heaviside step function>; the TeX incorrectly changes this to $V=I(t)$. A unit voltage step has derivative the <Dirac delta function>, so the current is the causal <Green function> for $L\,d^2/dt^2+R\,d/dt+1/C$.

The current must be continuous at zero: a jump would create a delta derivative through $LI''$, absent on the right-hand side. Integrating across zero therefore gives $L[I']+R[I]=1$, so $I(0^+)=0$ and $I'(0^+)=1/L$. Put
$$
r_\pm=\frac{-R\pm\sqrt{R^2-4L/C}}{2L}.
$$
These distinct negative <characteristic roots> give the <overdamped RLC response>
$$
\boxed{I(t)=\frac{e^{r_+t}-e^{r_-t}}{L(r_+-r_-)}
=\frac{e^{r_+t}-e^{r_-t}}{\sqrt{R^2-4L/C}},\qquad t\geq0.}
$$
Together with $I(t)=0$ for $t<0$, this solves the step response. The eventual current is zero: the <capacitor> charges and blocks steady current in the <series RLC circuit>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 5C / b / Solution

{wiki}

A capacitor stores opposite electric charges on two conductors separated by an insulator. Its capacitance is $C=Q/V$.


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