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

Replace the feedback resistor by a capacitor $C_f$, retaining the input resistor $R_i$. The virtual ground makes the input charge flow onto the feedback capacitor rather than into the ideal amplifier. Integrating its current givesIf the calorimeter delivers $Q_{\mathrm{in}}=\kappa \mathcal{E}$ for deposited energy $\mathcal{E}$, then $\Delta V_o=-\kappa \mathcal{E}/C_f$. A reset establishes the initial output; a practical large resistor across $C_f$ limits long-time drift, but is omitted from the ideal integration interval.