Consider the diagram as shown below. The distribution of current in accordance with Kirchhoff’s rules is shown. Applying Kirchhoff’s loop rule to the closed loop ABDA, we have
-I1P - IgG+(I - I1)R = 0 ...(1)
Again, applying Kirchhoff's loop rule o closed loop BCDE, we have
-(I1 - Ig)Q + (I - I1 + Ig)X + IgG = 0 ...(2)
This value of P,O,R and X are so adjusted that the galvanometer gives zero deflection i.e., Ig = 0.
This situations is the balanced state of the Wheatstone bride. Putting Ig = 0 in equations (1) and (2), We have
-I1P + (I - I1)R = 0 .......(3)
and -I1Q+(I - I1)X = 0 ......(4)
Rewriting the above two equations, we have
I1P = (I - I1)R .....(5)
and I1Q = (I - I1)X ....(6)
Divinding the above equations, we have
\(\frac{P}{Q} = \frac{R}{X}\) ...(7)
The above expression gives the condition for the bance of a wheatstone bridge.