Question

when a diameter of a wire is doubled then its resistance become

Answer 1

Hey.

For a wire of length L , resistivity p and Area A = πd^2/4

Resistance , R = p L/A = p L/ (πd^2/4)

as the diameter d is doubled

the new resistance will be

R' = p L/ (π4d^2/4) = p L / 4 (πd^2/4)

or, R' = p L/ 4A = R/4

So, the new resistance will get reduced by 4 times .

Thanks.

Answer 2

Diameter of previous wire=d

Then, radius= d/2 =2r/2=r

Diameter of new wire=2d

Then, radius=2d/2=d=2r

Therefore, R'/R = rho(l)/pie (2r²)*pie r²/rho l

R'/R=r²/4r²

R'/R=1/4

R'=1/4R

Therefore , resistance becomes 1/4 of the actual resistance

Actually, doubling the resistance is same as doubling the radius