 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022/8/b/i/solution

At zero angular frequency, the lower right capacitor is an open circuit, while the upper right electrical impedance tends to

R_{x}

. With the detector removed, no electric current flows down that branch, so

V_{R} = V_{0}

and

V_{L} = V_{0} /2

. Therefore

V_{oc} (0) = \frac{V _{0}}{2}, V_{A} (0) = 0 for an ideal ammeter .

For the connected ideal ammeter, the resulting electric current from right to left is

V_{0} / (R + 2 R_{x})

, despite its zero terminal voltage. This follows from

I_{A} = V_{oc} / Z_{th}

with

Z_{th} = R /2 + R_{x}

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