Deduce the condition for balance of a wheatstone's bridge using Kirchoffs rules .

Explanation

It is the combination of four resistances P,Q,R and S in the form of quadrilateral ABCD. A galvanometer is connected between junctions B and D. The line BD is called galvanometer arm. A battery is connected between junctions A and C. This AC is called the battery arm.
On applying Kirchhoff.s loop rule to the loop ABDA, we get
`I_(1)P+I_(g)G-I_(2)R=0`
But for balance of network `I_(g)=0`
`I_(1)P-I_(2)R=0`
`I_(1)P=I_(2)R" "_________(1)`
On applying Kirchhoff.s loop rule to the loop BCDB, we get
`(I_(1)-I_(g))-(I_(2)+I_(g))S-I_(g)G=0`
`I_(1)Q-I_(g)Q-I_(2)S-I_(g)S-I_(g)G=0`
For balance of network `I_(g)=0`
`I_(1)Q-I_(2)S=0`
`I_(1)Q=I_(2)S" "_________(2)`
`((1))/((2))implies(I_(1)P)/(I_(1)Q)=(I_(2)R)/(I_(2)S)`
`(P)/(Q)=( R )/(S)`
This is the condition for balance wheatstone bridge.