Question

Two diametrically opposite points of a metal ring are connected to terminals of the left gap of meter bridge. In the right gap, resistance of $15\Omega$ is introduced. If the null point is obtained at a distance of $40cm$ from the left end, find resistance of the wire that is bent in the shape of the ring.

Answer 1

Here we will use the principle of meter bridge

So, $\frac{P}{Q}=\frac{R}{S}$

where,

$P=40\rho$ [$\rho$ is resistance/unit length]

$Q=60\rho$

$S=15\Omega$

$R=10\Omega$

Now, two points are connected to the diameter of a ring.Therefore we have two parallel path.

If the resistance of both the path is $r$

Then equivalent resistance $=\frac{r}{2}$

$\frac{r}{2}=10$

$r=20\Omega$