Question

A wire of resistance 4R is bent in the form of a circle. what is the effective resistance between the ends of the diameter.

Answer 1

if we bent sire in a circular form ,
then
length of wire = perimeter of circle
let length of wire is L
then ,

L = π× diameter of circle

diameter = L/π

now , semiperimer of circle =π/2× diameter = L/2

we know,
resitance depends upon length and cross sectional area of wire .
e.g R = pl/A
where p is resistivity .
but here only length change e.g half so,
resistance exist in two semiperimeter is half e.g 2R

now ,

let A and B is the end of diameter then combination of both above and below , wire is In parallel combination .

so, effective resistance = Req

1/Req = 1/R1 + 1/R2
= 1/2R + 1/2R
= 1/R

Req = R

hence effective resistance = R

Answer 2

Let 'l' be the length of wire and 'd' be the diameter.

Length of the wire = Circumference of circle
l = 2 x d/2 x π
l = dπ
d = l/π -----(i)

Now ,
Half the circumference = π/2 x d = l/2

Since resistance depends on both length and the area of cross section so the resistance will be 2R ( length is halved)

Now the 2 resistances are in parallel combination.

1/Rp= 1/R1 + 1/R2
1/Rp = 1/2R + 1/2R
1/Rp = 1/R = R

So the required effective resistance is R.

Hope This Helps You!