Derive the balancing condition of a Wheatstone bridge

by

Wheatstone bridge (balanced condition) – Let i be the current flowing from battery E. At point A , current i1 flows through resistance P and current i−i1 flows through R. In balanced condition, no current flows through BD , hence point B and D are at same potential. Therefore, current i and i1 flows through resistance Q also and current i−i1 flows through S.

Applying Kirchoff’s current loop rule in closed circuit ABDA , i1P−(i−i1 )R=0 or      i1P=(i−i1)R  ………………..eq1

 In closed mesh BCDB ,

         i1−Q(i−i1)S=0

or      i1Q=(i−i 1)S  …………………eq2

Dividing eq1 from eq2 , we get

         P/Q=R/S    

This is the condition for balance in a Wheatstone’s bridge. This is the equation of whitsone bridge which is proved by kirchoffs rule. 

Leave a Comment