Obtaining Anderson's Bridge Balance Equation
By definition of bridge's balance, when the bridge is balanced the voltage \$V_\mathrm{D}\$ across the detector D is zero.

If you write the Kirchhoff's voltage law to the loop comprising branches 1, 2, C and D, you get

$$I_1(r_1+R_1 + \mathrm{j}\omega L_1)+V_\mathrm{D}-rI_C -R_2I_2=0$$

Since \$V_\mathrm{D}=0\$, you obtain equation (15.24).

Note, from the schematic, that the current \$I_C\$ is not only the current crossing the capacitor, but it's also the current which crosses the resistance \$r\$, because at balance the current which crosses the detector is zero.