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

Explanation

Circuit diagram with current direction
Applying KCL and KVL
Arriving final expression `(R_(1))/(R_(2))=`__
Detailed Answer:

Applying Kirchhoff.s loop law to the closed loop ABCD, we get
Applying Kirchhoff.s loop law to the closed loop BCDB, we get
`-(I_(2)-I_(g))Q+(I_(1)+I_(g))s+I_(g)G=0`
In case of balanced Wheatstone bridge, no current flows through the galvanometer i.e.,
Hence equation (1) becomes
`-I_(2)P+I_(1)R=0`
`I_(1)R=I_(2)P`
`(I_(1))/(I_(2))=(P)/(Q)`
Similarly equation (2) becomes `-I_(2)Q+I_(1)S=0`
`I_(1)S=I_(2)Q`
`(I_(1))/(I_(2))=(Q)/(S)`
For equation (3) and (4) we get
`(P)/(Q)=(Q)/(S)`
Or `(P)/(Q)=(R)/(S)`.