Obtaining condition for the balance Wheatstone bridgeApplying Kirchoff's loop rule to closed loop ADBA
$-{\mathit{I}}_1{\mathit{R}}_1+0+{\mathit{I}}_2{\mathit{R}}_2=0\left({\mathit{I}}_{\mathit{g}}=0\right)$ .......(i)
For loop CBDC
$-{\mathit{I}}_2{\mathit{R}}_4+0+{\mathit{I}}_1{\mathit{R}}_3=0$ .......(ii)
$\Rightarrow$ from equation (i)
$\frac{{\mathit{I}}_1}{{\mathit{I}}_2}=\frac{{\mathit{R}}_1}{{\mathit{R}}_2}$
From equation (ii)
$\frac{{\mathit{I}}_1}{{\mathit{I}}_2}=\frac{{\mathit{R}}_4}{{\mathit{R}}_3}$
$\therefore \frac{{\mathit{R}}_1}{{\mathit{R}}_2}=\frac{{\mathit{R}}_4}{{\mathit{R}}_3}$