 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2021/5/a/solution

R_{a}

R_{f}

, followed by an op amp integrator with input resistor

R_{i}

C_{i}

. Ground both non-inverting inputs. Each ideal operational amplifier has infinite input impedance, so its inputs draw no electric current; negative feedback makes the inverting input a virtual ground.

Kirchhoff's current law at the first summing junction gives

V_{1} = - (R_{f} / R_{a}) V_{in}

. At the second summing junction it gives

V_{1} / R_{i} + C_{i} d V_{out} / d t = 0

. Consequently the two inversions cancel:

V_{out} (t) = V_{out} (t_{0}) + \frac{R _{f}}{R _{a} R _{i} C _{i}} \int_{t_{0}}^{t} V_{in} (t^{'}) d t^{'} .

The initial output voltage records the initial capacitor charge. A reset switch across

C_{i}

sets it to zero. The formula holds while both stages remain within their supply rails; a persistent input voltage otherwise produces amplifier saturation.

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