A potential difference of 10 V is placed across a certain so

A potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If everything stays the same but the diameter of this resistor is now tripled, what is the new value of the current?

Solution

V =10 v , I₁ = 2 A , and we know R = (resistivity x length) / area . here I denote resistivity by \'p\' and length by L

here let us take initial radius as r₁ = d /2 and final radius r₂ = 3d /2

area = ( pi x r² ) where pi = 3.14

so initial area A₁ = pi x (d /2)²   and final area A₂ = pi x (3d / 2)²

now as we know V = I R

here V = 10 v so

I₁ R₁ = I₂ R₂   ------> I₂ = ( I₁ R₁ ) / R₂

R₁ = (4p L ) / (pi x d² ) and R₂   = (4pL) / (9 pi x d²)

so I₁ x ( 4pL / pi d² ) = I₂ x ( 4pL / 9 pi d² )

solving we get I₂ = 9 I₁  = 9 x 2 = 18 A

I₂ = 18 A