Question

A copper wire has a diameter of 0.5mm and resistivity of 1.6×10^-18m.
1. what will be the length of the wire to make its resistance 10 ohm.
2. how much does the resistance change if the diameter is doubled.

Answer 1

Answer:

The length is the $1.23\times10^{12}\ m$ and the new resistance will be one fourth of the original resistance.

Explanation:

Given that,

Diameter d = 0.5 mm

Resistivity $\rho=1.6\times10^{-18}\ m$

Resistance R = 10 ohm

We know that,

The resistance of the wire is the defined as:

$R = \dfrac{\rho l}{A}$

(1). The length of the wire is

$l=\dfrac{RA}{\rho}$

$l = \dfrac{10\times3.14\times0.25\times0.25\times10^{-6}}{1.6\times10^{-18}}$

$l = 1.23\times10^{12}\ m$

(2). If the diameter is doubled then the resistance will be

We know that,

The resistance is inversely proportional to the area of cross section.

$R\propto\dfrac{1}{A}$....(I)

$R' \propto\dfrac{1}{4A}$....(II)

On dividing equation (I) by equation (II)\

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

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

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

If the diameter is double then the new resistance will be one fourth of the original resistance.

Hence, The length is the $1.23\times10^{12}\ m$ and the new resistance will be one fourth of the original resistance.