 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013/8/iv/solution

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 gives

V_{o} (t) - V_{o} (0) = - \frac{1}{C _{f}} \int_{0}^{t} I_{i} (t^{'}) d t^{'}, \frac{Δ V _{o}}{Q _{in}} = - \frac{1}{C _{f}} .

If the calorimeter delivers

Q_{in} = κ E

for deposited energy

E

, then

Δ V_{o} = - κ 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.

 New to topics? Read the docs here!