Question

To find the resistance of a gold bangle, two diametrically opposite points of the bangle are connected to the two terminals of the left gap of a meter bridge. A resistance of 4 Ω is introduced in the right gap. What is the resistance of the bangle if the null point is at 20 cm from the left end?

Options

• 2Ω

• 4Ω

• 8Ω

• 16Ω

Answer 1

4Ω

Explanation:

Given:

Resistance in the right gap of the meter bridge, R = 4Ω

The null point is at l = 20 cm from the left end.

The total length of the wire = 100 cm.

The gold bangle is connected across the left gap between two diametrically opposite points.

The equivalent resistance between the two opposite points is:

`R_(eq) = "R/2 × R/2"/"R/2 + R/2" = R/4`

Apply the Meter Bridge Principle

`R_"left"/R_"right" = l/(100-l)`

R_left ​= R_eq​ = `R/4`

R_right​ = 4Ω

l = 20 cm

`"R/4"/4 = 20/80`

`R/15 = 1/4`

Multiply both sides by 16:

R = 4Ω