= Solution
Keep the source diagram's topology: the arm from the left supply node to $X$ contains $R_0$ and $C_0$ in series; the arm from $X$ to the right supply node is $C_1$; the arm from the left node to $Y$ is $R_1$; and the arm from $Y$ to the right node contains $R$ and $C$ in parallel. Their <complex impedances> are
$$
Z_0=R_0+\frac1{i\omega C_0},\qquad Z_1=\frac1{i\omega C_1},\qquad Z_p=\left(\frac1R+i\omega C\right)^{-1}.
$$
For this <capacitor bridge with series loss and a parallel standard arm>, the detector carries no <electric current> at balance. Taking the right supply node as zero <electric potential> and the left one as the source <voltage> $V_s$, the two <voltage dividers> give $V_X=V_sZ_1/(Z_0+Z_1)$ and $V_Y=V_sZ_p/(R_1+Z_p)$. Equality of these complex <voltages> means equality at every time, rather than merely equality of their magnitudes. Cross multiplication yields $R_1Z_1=Z_0Z_p$. Thus
$$
R_0+\frac1{i\omega C_0}=R_1\frac{Z_1}{Z_p}=\frac{R_1C}{C_1}+\frac{R_1}{i\omega RC_1}.
$$
Equating real and imaginary parts gives
$$
\boxed{R_0=\frac{R_1C}{C_1},\qquad C_0=\frac{RC_1}{R_1}.}
$$
The two adjustable standards determine the two components of the unknown <complex impedance>. Treating the lower-right arm as a series combination would give a different and incorrect balance equation for the source circuit. The result also gives the dimensionless <dielectric loss tangent> $\omega C_0R_0=\omega RC$ for the series-loss model.
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