I'm trying to learn Anderson's Bridge Balance equation.I dont know how this value is obtained
Why are they equating the voltage drops across branch one and the sum of the voltage drop across branch 2 and the one with the capacitor?
Please advice
Electronic – Obtaining Anderson’s Bridge Balance Equation
acbridgecircuit analysiskirchhoffs-laws
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Best Answer
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.