Question

How does resistance change with thickness of the wire?

Answer 1

$\large{RESISTANCE}$

▶Resistance is the obstruction offered in the path of current.

▶Resistance of wire depends on the collision of the flowing electrons with protons and neutrons .When they collides heat is produced and the resistance is increases.

$\bold{DEPENDENCE \: OF \: RESISTANCE}$

◼AREA OF CROSS SECTION

When the area of cross section increases, a larger path will be made for the flow of free electrons. Therefore, it proved that more the area of cross section, less will be the resistance.

So,

$\boxed{Resistance \propto \dfrac{1}{Area \: of \: cross \: section}}$

So if area of cross section is increased to double then resistance will be one fourth.

$\large{Prove}$

Let the resistance be R.

Length be l.

Area of cross section be a.

$\boxed{Resistance = \rho \dfrac{length}{Area}}$

$\mathsf{R = \rho \dfrac{l}{a}}$ ----(1)

If area of cross section is doubled by doubling wire on itself then,

Let the resistance be $R_2$

Length be $\dfrac{l}{2}$

Area of cross section be 2a

$\boxed{Resistance = \rho \dfrac{length}{Area}}$

$\mathsf{R_2 = \rho \dfrac{\dfrac{l}{2}}{2a}}$

$\mathsf{R_2 = \rho \dfrac{l}{4a}}$ -------(2)

On dividing (1) by (2)

$\mathsf{\dfrac{R}{R_2} = \rho \dfrac{l}{a} \div \rho \dfrac{l}{4a}}$

$\mathsf{\dfrac{R}{R_2} = \rho \dfrac{l}{a} \times \rho \dfrac{4a}{l}}$

On solving

$\mathsf{\dfrac{R}{R_2} =4}$

$\mathsf{R=4R_2}$

$\mathsf{\dfrac{R}{4} =R_2}$

Hence proved that the new resistance will be one fourth of the original resistance.