Past exam of the physics 2 paper of the University of Oxford / 2013 / 8 / v / Solution

For the modified circuit, $I_i=V_i/R_i$ and ${\dot{V}}_o=-V_i/(R_i{C}_f)$. With $V_i=V_0\cos \omega t$,The sinusoidal component has amplitude $V_0/(\omega R_i{C}_f)$ and leads the cosine input by $90^{\circ }$; the constant is set by the initial capacitor charge. This assumes the output remains within the amplifier's linear range.