= Solution
\b[Complex impedance.] In sinusoidal steady state write physical voltage and <electric current> as real parts of $\widetilde V e^{i\omega t}$ and $\widetilde I e^{i\omega t}$. The <electrical impedance> $Z=\widetilde V/\widetilde I$ is a ratio of <phasors>; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in <ohms>.
For an ideal <capacitor>, $q=Cv$ and $i=dq/dt=C\,dv/dt$. Consequently $\widetilde I=i\omega C\widetilde V$, giving
$$
\boxed{Z_C=\frac1{i\omega C}=-\frac i{\omega C}.}
$$
With this time convention the voltage lags the <electric current> by $\pi/2$.
\b[Bridge output.] The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible <electric current>, the two <voltage dividers> give
$$
V_A=V_i\frac{Z_2}{Z_1+Z_2},\qquad
V_B=V_i\frac{R_4}{R_3+R_4}.
$$
Their difference is therefore
$$
\boxed{\frac{V_{AB}}{V_i}=\frac{Z_2}{Z_1+Z_2}-\frac{R_4}{R_3+R_4}.}
$$
The reactive-arm <electrical impedances> are
$$
Z_1=\left(\frac1{R_1}+i\omega C_1\right)^{-1},\qquad
Z_2=R_2+\frac1{i\omega C_2}.
$$
\b[Balance.] Setting $V_{AB}=0$ and cross-multiplying gives $R_3Z_2=R_4Z_1$. For this <Wien bridge>,
$$
\frac{Z_2}{Z_1}
=\left(R_2+\frac1{i\omega C_2}\right)\left(\frac1{R_1}+i\omega C_1\right)
=\frac{R_2}{R_1}+\frac{C_1}{C_2}
+i\left(\omega R_2C_1-\frac1{\omega R_1C_2}\right).
$$
The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts gives
$$
\boxed{\omega=\frac1{\sqrt{R_1R_2C_1C_2}},\qquad
\frac{C_1}{C_2}=\frac{R_4}{R_3}-\frac{R_2}{R_1}.}
$$
Positive <capacitances> require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to
$$
\boxed{\omega=\frac1{RC},\qquad R_4=2R_3.}
$$
\b[Band-pass output and phase.] Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,
$$
\frac{V_{AC}}{V_i}=\frac{Z_2}{Z_1+Z_2}-1
=-\frac{Z_1}{Z_1+Z_2}.
$$
Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the <Wien bridge band-pass response> is
$$
\boxed{H(\omega)=\frac{V_{AC}}{V_i}
=-\frac1{3+i(u-u^{-1})}
=-\frac{iu}{1-u^2+3iu}.}
$$
At finite positive <frequency>, its imaginary part vanishes exactly when $u-u^{-1}=0$. Thus
$$
\boxed{\omega_r=\frac1{RC},\qquad H(\omega_r)=-\frac13.}
$$
The magnitude $|H|=[9+(u-u^{-1})^2]^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a <band-pass filter>.
A convenient continuous, unwrapped <wave phase> is
$$
\boxed{\phi(\omega)=-\pi-\arctan\!\left(\frac{u-u^{-1}}3\right).}
$$
It decreases from $-\pi/2$ as $\omega\to0^+$, passes through $-\pi$ at $\omega_r$, and tends to $-3\pi/2$ as $\omega\to\infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same <phasor>. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018-ia-physics-q11-bridge-response.png]
{title=Band-pass magnitude and continuous phase for the printed output polarity VA minus VC}
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