Question

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 2.0 mm. Conductor B is a hollow tube of outside diameter 2.0 mm and inside diameter 1.0 mm. What is the resistance ratio RA/RB, measured between their ends

Answer 1

Answer:

Therefore,

The resistance ratio RA/RB, measured between their ends is

$\dfrac{R_{A}}{R_{B}}=\dfrac{1}{4}=0.25$

Explanation:

Given:

Consider Two conductors are made of the same material (A & B)and have the same length.,

$L_{A}=L_{B}$

Diameters,

$d_{A}=2\ mm$

$d_{B}=2\ mm....outer\\d_{B}=1\ mm.....inner$

Radius is half of Diameter,

Therefore  

$r_{A}= \dfrac{2}{2}=1\ mm\\\\r_{B}=\dfrac{2-1}{2}...hollow\ tube\\r_{B}=0.5\ mm$

To Find:

$\dfrac{R_{A}}{R_{B}}=?$   (resistance ratio)

Solution:

Resistance for long wire with a Area of cross section is given by

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

Where,

R = Resistance

l= length

A = Area of cross section = πr²

$\rho=Resistivity\ of\ material$

i.e

When

material and length is same then Resistance is inversely proportional to Area,

$R\alpha \dfrac{1}{A}$

Hence,

$R_{A}\alpha \dfrac{1}{A_{A}}$  

And

$R_{B}\alpha \dfrac{1}{A_{B}}$

Equating we get

$\dfrac{R_{A}}{R_{B}}=\dfrac{A_{B}}{A_{A}}=\dfrac{\pi r_{B}^{2}}{\pi r_{A}^{2}}$

Substituting the values we get

$\dfrac{R_{A}}{R_{B}}=\dfrac{0.5^{2}}{1^{2}}=0.25=\dfrac{1}{4}$

Therefore,

The resistance ratio RA/RB, measured between their ends is

$\dfrac{R_{A}}{R_{B}}=\dfrac{1}{4}=0.25$