Deduce the condition for balance in a Wheatstone bridge. Using the principle of Wheatstone bridge, describe the method to determine the specific resistance of a wire in the laboratory. Draw the circuit diagram and write the formula used. Write any two important precautions you would observe while performing the experiment. 

 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

$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}$

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) 

$\therefore$           $\frac{\mathrm{R}}{\mathrm{X}} = \frac{\mathrm{P}}{\mathrm{Q}} = \frac{\mathrm{kl}}{\mathrm{k}(100-\mathrm{l})}$

or,               $\mathrm{X} = \frac{(100-\mathrm{l}) \mathrm{R}}{\mathrm{l}}$ 
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

             $\mathrm{\rho } = \frac{\mathrm{{\rm X}A}}{\mathrm{l}} = \frac{{\mathrm{\pi r}}^{2 }\mathrm{X}}{\mathrm{l}}$

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.