Past exam of the physics 2 paper of the University of Oxford / 2022 / 8 / b / iii / Solution

As $\omega \to \infty$, both capacitors approach short circuits. Thus $Z\to 0$, $W\to R_v>0$, and the unloaded right junction tends to the source's top-terminal voltage. HenceThe connected ideal ammeter carries limiting electric current $V_0/R$ from right to left because $Z_{\mathrm{th}}\to R/2$. Reversing the output polarity reverses both nonzero open-circuit answers, without changing their magnitudes.