Past exam of the physics 2 paper of the University of Oxford / 2015 / 8 / c

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 areFor 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 $V_X=V_s{Z}_1/(Z_0+Z_1)$ and $V_Y=V_s{Z}_p/(R_1+Z_p)$. Equality of these complex voltages means equality at every time, rather than merely equality of their magnitudes. Cross multiplication yields $R_1{Z}_1=Z_0{Z}_p$. ThusEquating real and imaginary parts givesThe 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 $\omega C_0{R}_0=\omega RC$ for the series-loss model.