Using Kirchoff's law, derive the condition for the balance of a Wheatstone bridge circuit.

Explanation

Wheatstone bridge : Wheatstone bridge consists of four resistances P, Q, R and S arranged so as to form a quadrilateral ABCD as shown in fig. A cell E and a key K are connected between points A and C. A galvanometer G and a tapping key `K_(1)` are connected between points B and D. On closing the key K first and then key `K_(1)`, if galvanometer shows no deflection, then bridge is said to be balanced. In that case, `I_(g)=0` and

`(P)/(Q)= (R )/(S)`
Knowing P, Q and R, we can calculate unknown resistance S.
Proof Let E= emf of the cell
I = Total current given by the cell.
At junction A, current I is divided into two parts, `I_(1)` (flowing through arm (AD) and `I_(2)` (flowing through arm AB). At junction B, current `I_(2)` is again divided into two parts `I_(g)` (through BD) and `I_(2)- I_(g)` (through BC). At D, current `I_(1) and I_(g)` combine and hence current flowing through DC is `I_(1) + I_(g)`. At C, current `(I_(2)- I_(g)) and (I_(1) + I_(g))` combine and `I_(1) +I_(2)= I` flows out and completes the circuit. Using Kirchhoff.s Second law in closed mesh ABDA.
`I_(2)P + I_(g) G-I_(1) R= 0` ...(I)
where G= Resistance of galvanometer
Using Kirchhoff.s second law in closed circuit BDCB, `I_(g) G + (I_(1)+ I_(g)) S- (I_(2)-I_(g)) Q= 0` ...(II)
When bridge is balanced, i.e., when no current flows through the galvanometer (this can be done by adjusting the value of resistance R), then `I_(g)= 0`
Putting `I_(g)= 0`, in eqns (I) and (II) we get
`I_(2) P + 0 - I_(1)R= 0`
or `I_(2)P= I_(10R` ...(III)
and `0 + (I_(1) + 0) S- (I_(2)- 0) Q = 0`
`I_(2)S- I_(2) Q= 0`
`i_(2) Q= I_(1)S` ...(IV)
Dividing (III) by (IV),
`(I_(2)P)/(I_(2)Q)= (I_(1)R)/(I_(1)S)`
`rArr (P)/(Q)= (R )/(S)`
Hence the proof.