Derive the necessity condition for wheatstone …

Balancing condition of Wheatstone bridge:

Consider four resistances P, Q, R and S connected to form a quadrilateral ABCD as shown in the figure below. A galvanometer G is connected between B and D and, BD is typically called the galvanometr arm. A battery is connected between A and C. The resistances are so adjusted that no current flows in the galvanometer G. The same current will flow in arms AB and BC. Similarly current I₂ flows in arms AD and DC.

Applying Kirchhoff’s second law for mesh ABCD,
I₁P – I₂R = 0

or, I₁P = I₂R           ...(i)

For mesh BCDB,
I₁Q – I₂S = 0

or, I₁Q = I₂S           ...(ii) 

Dividing (i) by (ii), we get



This is the balanced condition of the Wheatstone bridge.

Measurement of specific resistance: 

Slide wire or meter bridge is a practical form of Wheatstone bridge.


In the figure above, X is an unknown resistor and R.B is resistance box. After inserting the key k (circuit is closed), jockey is moved along the wire AC till galvanometer shows no deflection (point B). 
If k is the resistance per unit length of wire AC.
then,          P = resistance of AB = kl
                 Q = resistance of BC = k(100 - l) 

           

or,                
Hence, we can find the value of the unknown resistance.

Resistivity is derived as follows:

If 'r' is the radius of wire and l be its length, then its resistivity will be

             

Precautions:
(i) The null point should lie in the middle of the wire.
(ii) The current should not be allowed to flow in the wire for a long time or else the wire will get damaged.