Question

answer this this is 5 marks question ​

Answer 1

Answer:

Given:

3 conductors of same material have been supplied.

Their specifications are as follows :

• L and A
• 2L and A/2
• L/2 and 2A

To find:

Which wire has a greater resistance.

Concept:

First let's define Ohm's law :

Ohm's Law states that at constant temperature, the resistance of a conductor is directly proportional to the length and inversely proportional to the Area of cross-section of the conductor.

R = ρ(l/a)

where R => resistance , ρ => resistivity ,

l => length and a => area of cross section.

Calculation:

For 1st wire :

R1 = ρ(L/A).........(1)

For 2nd wire:

R2 = ρ{2L/(A/2)}

=> R2 = 4 ρ(L/A)........(2)

For 3rd wire:

R3 = ρ{(L/2)/2A}

=> R3 = ¼ ρ(L/A).........(3)

So final answer is max resistance is that of 2nd wire (2).

Answer 2

$<font color ="red">$GiVEN:-

$<font color ="orange">$SPECIFICATIONS ARE:-

$<font color ="blue">$1)L AND A

2)2L AND A/2.

3)L/2 AND 2 A.

$<font color ="red">$FIND:-

GREATER RESISTANCE.

$<font color ="red">$ANSWER:-

$<font color ="purple">$ USING THIS FORMULA,

$r =\rho ( \frac{l}{a} )$

HERE,

R IS RESISTANCE.

$\rho$ is resistivity .

L is a length.

a=area of section.

$<font color ="red">$ IN FIRST WIRE :-

1)R1=$\rho$(L/A)

HERE,

R1 IS FIRST RESISTANCE.

$\rho$ is resistivity .

L is a length.

a=area of section.

$<font color ="green">$IN SECOND WIRE:-

$2)r2 = \rho( \frac{2l}{ \frac{a}{2} } ) \\ r2 = 2 \times 2 \rho( \frac{l}{a}) \\.°. r2 = 4 \rho( \frac{l}{a} )$

HERE,

R2 IS SECOND RESISTANCE.

$\rho$ is resistivity .

L is a length.

a=area of section.

$<font color ="red">$IN THIRD WIRE:-

$3)r3 = \rho( \frac{2l}{ 2a} ) \\ r3 = \frac{1}{2 \times 2} \rho( \frac{l}{a} ) \\ .°. r3 = \frac{1}{4} \rho( \frac{l}{a} ) .$

$<font color ="green">$ FROM, EQUATION 1 & 2 OR 3.

HENCE,

$<font color ="orange ">$ THE MAX RESISTANCE IS IN 2nd. wire.

$<font color ="maroon">$THANKS.