Question

Two dimetrically opposite points of a metal ring are connected to two terminals of the left gap of metrebridge. The resistance of 11 $\Omega$ is connected in right gap. If null point is obtained at a distance of 45 cm from the left end, find the resistance of metal ring.

Answer 1

Let 2R be the resistance of the metal ring. As the diametrically opposite points are connected in the left gap.

The resistance of each half is R and these two halves are connected in parallel.

Thus resistance in left gap $=\frac{(R\times R)}{(R+R)}=\frac{R^2}{2R}=\frac{R}{2}\Omega$

Resistance in right gap $=11\Omega$

$l=45cm$ and $100-l=100-45=55cm$

For balanced meter bridge

$\frac{Resistanceintheleftgap}{Resistanceintherightgap}=\frac{l}{100-l}$

$\frac{R/2}{11}=\frac{45}{55}$

$\frac{R}{2}=\frac{45}{5}=9$

$\therefore R=18\Omega$

Now resistance of ring $=2R=2\times 18=36\Omega$