Obtain the condition for bridge balance in Wheatstone's bridge.

Explanation

An important application of Kirchhoff's rule is Wheatstone's bridge.
(i) The bridge consist of four resistance P,Q R and S connected as shown in Figure. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is `I_(G)` and its resistance is G.
(ii) Applying Kirchhoff's current rule to junction B
`I_(1)-I_(G)-I_(3)=0 " "...(1)`
Applying Kirchhoff's current rule to junction D,
`I_(2)+I_(G)-I_(4)=0 " "....(2)`

(iii) Applying Kirchhoff's voltage rule to loop ABDA,
`I_(1)P+I_(G)-I_(2)R=0`
Applying Kirchhoff's voltage rule to loop ABCDA.
`I_(1)P+I_(3)Q-1_(4)S-I_(2)R=0`
(iv) When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through gavanometer (IG=0). Substituting `I_(G)=0` in equation (1),(2) and (2), we get
`I_(1)=I_(3) " "...(4)`
`I_(2)=I_(4)" "...(5)`
`I_(1)P=I_(2)R" "...(6)`
Substituting the equation (4) and (5) in eqution (6)
`I_(1)P+I_(1)Q-I_(2)S-I_(2)r=0 `
`I_(1)(P+Q)=I_(2)(R+S)" "...(7)`
Dividing equation (7) equation (6), we get
`(P+Q)/(P)=(R+S)/(R)`
`1+(Q)/(P)=1+(S)/(R)`
`(Q)/(P)=(S)/(R)`
`(P)/(Q)=(R)/(S)`
this is the bridge balance condition. Under this condition galvanometer shows null deflection.