= Solution
The source's variable inductor and variable resistor are in \b[parallel]. Let their combined <complex impedance> be
$$
Z_V=\frac{i\omega L_Vr_V}{r_V+i\omega L_V}.
$$
Use the left and right source terminals as the drive nodes. The upper divider has arms $Z_V$ and $i\omega L_1$; the lower divider has arms $R_1$ and $R+i\omega L$. At bridge balance, the detector draws no <electric current> and the divider midpoint <electric potential differences> agree, so
$$
\frac{Z_V}{i\omega L_1}=\frac{R_1}{R+i\omega L},\qquad Z_V(R+i\omega L)=i\omega L_1R_1.
$$
Substitution and cancellation at a nonzero drive <frequency> give
$$
L_Vr_V(R+i\omega L)=L_1R_1(r_V+i\omega L_V).
$$
Equating real and imaginary parts yields
$$
\boxed{R=\frac{R_1L_1}{L_V},\qquad L=\frac{R_1L_1}{r_V}}.
$$
The first expression has units of resistance and the second units of <inductance>. Both are independent of the drive amplitude and, for ideal frequency-independent components, of the chosen nonzero drive <frequency>. At zero <frequency> the ideal inductors short their arms and the cancellation would not be valid as a measurement of $L$.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-inductive-bridge-balance.png]
{title=Bridge topology with a parallel variable arm and a series lossy-inductor arm}
Two applications are precision comparison of an unknown <electrical resistance> with calibrated standards in a <Wheatstone bridge>, and detecting small resistance changes in a <strain gauge> or resistance <thermometer>. Alternating-current bridges additionally compare <capacitance>, <inductance> and dielectric losses; this particular bridge determines the two parameters of a non-ideal <inductor> from a null measurement.
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