Derive the condition of balance for Wheatstone bridge.

Explanation

Circuit arrangement for a Wheatstone bridge is being shown in the adjoining figure. In balance condition of bridge no current flows through galvanometer and it gives no deflection. Currents in various branches of network as per Kirchhoff.s first law are, thus, represented in figure.

Applying Kirchhoff.s second law to mesh ABDA, we have
`-P*I_(1)+R*(I-I_(1))=0`
or `PI_(1)=R(I-I_(1)) " " ...(i)`
Again for mesh BCDB, we have
`-Q*I_(1)+S*(I-I_(1))=0 or QI_(1)=S(I-I_(1)) " " ...(ii)`
Dividing (i) by (ii), we get
`(P)/(Q)=(R )/(S)`
which is the balance condition of Wheatstone.s bridge.