Question

With the help of labelled diagram, show that the balancing condition of Wheatstone bridge is:

$\frac{R_1}{R_2}=\frac{R_3}{R_4}$ where the terms have their usual meaning.

Answer 1

In the arrangement, four resistances are so connected as to form a parallelogram. In one diagonal of this parallelogram is connected a galvanometer and in other diagonal a cell. The resistances $R_1,R_2,R_{3}and{R}_4$ are so adjusted that on pressing the key $K_2$ there is no deflection in the galvanometer G. That is, there is no current in the diagonal BD.
$V_B=V_D;i_g=0$
Applying Kirchhoff's seconds law for the closed loop ABDA, we have
$i_1{R}_1-i_2{R}_3=0$
$i_1{R}_1=i_2{R}_3$ ....(i)
Again on applying KVL in the closed loop BCDB
$i_1{R}_2-i_2{R}_4=0$
$i_1{R}_2=i_2{R}_4$ ....(ii)
Dividing eq. (i) by eq. (ii), we have

$\frac{1_1{R}_1}{i_1{R}_2}=\frac{i_2{R}_3}{i_2{R}_4}$

$\frac{R_1}{R_2}=\frac{R_3}{R_4}$