Derive the equation of the balanced …

Based on the circuit given in the figure , we have  three known resistance's R1​,R2​,R3​ and an unknown variable resistor RX​, a source of voltage, and a sensitive ammeter.
 Kirchhoff's first rule is applied to find the currents in junctions B and D:
I₃​−I_x​+I_g​=0
I₁​−I₂​−I_g​=0
Now Kirchoff's second rule is used to find the voltage in the loops ABD and BCD:
I_3​.R₃​−I_g​.R_g​−I₁​.R₁​=0
I_x​.R_X​−I₂​.R_2​+I_g​.R_g​=0
The bridge is balanced and I_g​ = 0, so the second set of equations can be rewritten as:
I₃​.R₃​=I₁​.R₁​
I_x​.R_X​=I₂​.R_2​
From first rule of Kirchoff.
I₃​=I_x​
I1​=I2​

{tex}\begin{aligned} & \frac{\mathrm{I}_3 \cdot \mathrm{R}_3}{\mathrm{I}_{\mathrm{n}} \cdot \mathrm{R}_{\mathrm{X}}}=\frac{\mathrm{I}_1 \cdot \mathrm{R}_1}{\mathrm{I}_2 \cdot \mathrm{R}_2} \\ & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{R}_3}{\mathrm{R}_{\mathrm{x}}} \end{aligned}{/tex}