= 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
$$
\boxed{V_{\rm oc}(0)=\frac{V_0}{2},\qquad V_A(0)=0\ \text{for an ideal ammeter}.}
$$
For the connected ideal <ammeter>, the resulting <electric current> from right to left is $V_0/(R+2R_x)$, despite its zero terminal <voltage>. This follows from $I_A=V_{\rm oc}/Z_{\rm th}$ with $Z_{\rm th}=R/2+R_x$.
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