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

Explanation

Let `I_1,I_2,I_3,I_4` and `I_g` be the currents through the resistances, P,R,Q,S and G respectively as shown in figure.
Apply KVL to the mesh ABDA, we get
`I_1P+I_gG-I_2R=0` …(i)
Applying KVL to the mesh BCDB, we get
`I_3Q -I_4S-I_gG=0` ....(ii)
The network is balanced , if `I_g=0`, then
At B, `I_1=I_3`
At D, `I_2=I_4`
Equation (i) and (ii) becomes, ...(iii)
`I_3P=I_4R` ...(iv)
`I_3Q=I_4S` ...(v)
Equation (iv) and (v) , we get
`P/Q=R/S`