= Solution
Let the source's top terminal have <voltage> <phasor> $V_0$ and its bottom terminal zero. Denote the left and right junctions by $L$ and $R$. The two left <resistors> are equal, so at balance $V_L=V_0/2$. Zero <electric current> through the <ammeter> requires $V_R=V_L$, and therefore equal upper and lower right-arm <electrical impedances>:
$$
Z=\frac{R_x}{1+i\omega R_xC_x}=W,\qquad
W=R_v+\frac1{i\omega C}=R_v-\frac{i}{\omega C}.
$$
Write $x=\omega R_xC_x$. Separating <real parts> and <imaginary parts> gives
$$
R_v=\frac{R_x}{1+x^2},\qquad
\frac1{\omega C}=\frac{\omega R_x^2C_x}{1+x^2}.
$$
The second equation implies $1+x^2=\omega^2R_x^2CC_x$. Subtract $x^2=\omega^2R_x^2C_x^2$ to obtain
$$
\boxed{\omega_0=\frac1{R_x\sqrt{C_x(C-C_x)}},\qquad
\frac{C_x}{C}=1-\frac{R_v}{R_x}.}
$$
A finite balance with positive components requires $0<C_x<C$, and then $0<R_v<R_x$. This <equal-ratio RC bridge> uses equality of arm <electrical impedances>, independent of the conjugate-matching convention in part (a).
\b[Detector convention.] The stated negligible-<electrical impedance> <ammeter> is an ideal <short circuit>, whose actual terminal <voltage> is zero at every <angular frequency>. The nontrivial bridge output usually intended is the \i[open-circuit] junction difference, obtained with the detector removed or replaced by a high-<input impedance> <voltmeter>. Define its polarity by
$$
V_{\rm oc}=V_R-V_L
=V_0\left(\frac{W}{Z+W}-\frac12\right)
=\frac{V_0(W-Z)}{2(W+Z)}.
$$
For a detector of finite <electrical impedance> $Z_A$, its terminal <voltage> is instead
$$
V_A=V_{\rm oc}\frac{Z_A}{Z_A+Z_{\rm th}},\qquad
Z_{\rm th}=\frac R2+\frac{ZW}{Z+W}.
$$
Here $Z_{\rm th}$ is the <Thévenin equivalent> <electrical impedance>, found by shorting the ideal source. This exhibits why the ideal-<ammeter> limit $Z_A\to0$ has zero terminal <voltage> even when the bridge is unbalanced.
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