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

Explanation

Applying the KCL to
Junction B and D ,
`I_(1) = I_(3) +I_(g)`…. (1)
` I_(4) = I_(2) +I_(g) …(2)`

Applying KVL to the mesh ABDA ,
` I_(1) P+I_(g) G-I_(2) R=0`
`i.e., I_(1) P +I_(g) G=I_(2) R ` ....(3)
Applying KVL to the mesh BCDB ,
`I_(3) Q-I_(4) S-I_(g) G=0`
for a balanced condition of wheatstone.s network ,`I_(g) =0. `
` therefore (1) and (2)` may be written as `I_(1) =I_(3) and I_(2) =I_(4)`
(3) and (4) may be written as
` I_(1) P= I_(2) R ` ...(5)
`and I_(3) Q= I_(4) S ` ...(6)
`(5 ) div (6) ` gives `(p)/(Q)=(R)/(S)`