Answer: Kelvin’s bridge is a modification of Wheatstone’s bridge and always used in measurement of four-terminal low resistance. Is is also called as double-bridge as it consists of two set of ratio arms.The circuit diagram is shown below: 

One set of arms consists of the resistor $R_1 , \ R_2$  and other set of arm consists of the resistors $R_a , \ R_b$ . The unknown resistor $R_x$ is to be measured and $R_3$ is the adjustable resistor.All the resistors are non - inductive resistors . The galvanometer is connected between the points $k \ and \ p$.  The galvanomtere reading is zero when potential at points k and p are equal. $R_y$ is the resistor of connecting lead from $R_3 \ to \ R_x$ .

For zero deflection ,$E_{lk}=E_{lmp}$......................(1)

But $E_{lk}=\dfrac{E \times R_2}{R_1+R_2}$

Now , $E=I[R_x+(R_y \mid\mid(R_a+R_b))+R_3]$

$\therefore E_{lk}=I[R_x+\mid\mid(R_a+R_b)+R_3]\times \dfrac{R_2}{R_1+R_2}$

Similarly , $\therefore E_{lmp}=I[R_3+\dfrac{R_b}{R_a+R_b}(R_y\mid\mid (R_a+R_b))]$

Now by using 1 , we get

$I[R_x+\mid\mid(R_a+R_b)+R_3]\times \dfrac{R_2}{R_1+R_2}=I[R_3+\dfrac{R_b}{R_a+R_b}(R_y\mid\mid (R_a+R_b))]$

$R_x+\dfrac{(R_a+R_b)R_y}{R_a+R_b+R_y}+R_3=\dfrac{R_1+R_2}{R_2}[R_3+\dfrac{R_bR_y}{R_a+R_b+R_y}]$

$\therefore R_x=\dfrac{R_1R_3}{R_2}+\dfrac{R_1R_bR_y}{R_2(R_a+R_b+R_y)}+\dfrac{R_bR_y-R_aR_y-R_bR_y}{R_a+R_b+R_y}$

$\therefore R_x=\dfrac{R_1R_3}{R_2}+\dfrac{R_1R_bR_y}{R_2(R_a+R_b+R_y)}-\dfrac{R_aR_y}{R_a+R_b+R_y}$

$\therefore R_x=\dfrac{R_1R_3}{R_2}+\dfrac{R_bR_y}{R_a+R_b+R_y}(\dfrac{R_1}{R_2}-\dfrac{R_a}{R_b})$

But $\dfrac{R_1}{R_2}-\dfrac{R_a}{R_b}$

$\therefore R_x=\dfrac{R_1R_3}{R_2}$

The known resistance $R_3$ is adjusted such that the galvanometer indicates zero and then by using the above equation , we can calculate the unknown resistance .