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

Explanation

Applying Kirchhoff.s current law to juction B.
` I _(1) - I _(g) - I _(3) =0" "…(i)`
Applying Kirchhoff.s current law to junction D
`I _(2) + I _(g) - I _(4) =0 " "…(ii)`
Applying Kirchhoff.s voltage law to closed path ABDA
`I _(1) P+ I _(3) G - I _(2) R =0 " "...(iii)`
Applying Kirchhoff.s voltage law to closed path ABCDA
`I _(1) P+ I _(3) Q - I _(4) S - I _(4) S - I _(2) R =0 " "...(iv)`
When the galvanometer shows zero deflection,k the points B and D are at same potential and I =0. Substituting `Ig =0` in equation. (i),(ii) and (iii)
`I_(1) = I _(3) " "...(v)`
`I _(2) = I _(4) " "...(vi)`
`I _(1) P = I _(2) R " "...(vii)`
Substituting the vlaues of (v) and (vi) in equation (iv)
`I _(1) P + I _(1) Q - I _(2) S- I _(2) R =0`
` I _(1) ( P+Q ) = I _(2) (R+S)" "...(viii)`
Dividing (viii) by (vii)
`(I _(1) (P+Q))/( I _(1) P ) = ( I _(2) ( R+S))/( I _(2) R )`
`therefore (P+Q)/( P ) = (R+S)/( R)`
`1 + Q/P =1 + S/R`
`therefore Q/P = S/R`
`or P/Q = S/R`