Inductance Bridge

The resistance wire can be represented by two partial resistors $R_x$ and $R - R_x$ with $R_x = \frac{x}{A} \rho = \frac{x}{l} R$ If you also replace the speaker with the resistor $R_S$ , the result is the following circuit:

The loudspeaker is mute when there is no current flowing through it, and voltages $U_R$ and $U_L$ are the same on both sides of the circuit. Therefore, the currents through both resistors on the left are equal to $I_R$ . The same applies also to both inductances with a corresponding current $I_L$ $\begin{aligned} I_R &= \frac{U_0 - U_R}{R_x} e^{i \omega t} = \frac{U_R}{R - R_x} e^{i \omega t} \\ I_L &= \frac{U_0 - U_R}{i \omega L_x} e^{i \omega t} = \frac{U_L}{i \omega L} e^{i \omega t} \end{aligned}$ Here we used the complex representation of current, voltage and impedances. If we convert both equations according to the voltages, we get $\begin{aligned} U_R &= \frac{R - R_x}{R} U_0, \\ U_L &= \frac{L}{L_x + L} U_0 \\ \stackrel{U_R = U_L}{\Rightarrow} \quad L_x &= \frac{R}{R - R_x} L - L = \frac{R_x}{R - R_x} L = \frac{x}{l-x} L \end{aligned}$ Therefore, $L_x = \frac{10 \,\text{cm}}{50 \,\text{cm} - 10\,\text{cm}} \cdot 500 \,\mathrm{\mu H} = 125 \, \mathrm{\mu H}$