Question

A wire of length L and area of cross section is doubled on itself. find the ratio of the initial resistance to the final resistance

Answer 1

R1:R2 = 1:1
it is a correct answer.

Answer 2

Answer:

The ratio of the initial resistance to the final resistance is 4:1.

Explanation:

We know that,

The resistance of the wire is defied as:

$R = \dfrac{\rho L}{A}$  .....(I)

here, R = resistance of the wire

L = length of the wire

A = area of cross section

$\rho$ = resistivity of the wire

We know that,

Initial volume = final volume

$LA=L'A'$

If the length of the wire is L and area of cross section is doubled on itself.

$L=2L'$

$A'=2A$

The final resistance is

$R'=\dfrac{\rho L'}{A'}$  ....(II)

The ratio of the initial resistance to the final resistance is

$\dfrac{R}{R'}=\dfrac{\dfrac{\rho L}{A}}{\dfrac{\rho L}{2A}}$

$\dfrac{R}{R'}=\dfrac{\dfrac{\rho 2L'}{A}}{\dfrac{\rho L'}{2A}}$

$\dfrac{R}{R'}=\dfrac{4}{1}$

Hence, The ratio of the initial resistance to the final resistance is 4:1.