 Solution

ID: 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 ω t

V_{o} (t) = V_{o} (0) - \frac{V _{0}}{ω R _{i} C _{f}} sin ω t .

V_{0} / (ω R_{i} C_{f})

and leads the cosine input by

9 0^{\circ}

; the constant is set by the initial capacitor charge. This assumes the output remains within the amplifier's linear range.

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