Find the new resistance if its length and D are doubled

[Fatima Hasan]

Homework Statement

Homework Equations

ρl = RA
A = π r²

The Attempt at a Solution

l = RA (ρ is constant)
A₁ = π (D/2)²
= (πD²)/4
A₂ =π (2D/2)²
= πD²
= A₁ /4
l₂ = 2*l₁
$R=\frac{l}{A}$
R₂ = (2l₁)/(A₁/4)
= 8 l /A
= 8R₁

Could someone tell me where is my mistake ?

 

[Vanadium 50]

Could someone tell me where is my mistake ?

Trying to do this in one bite.

What happens if only its diameter is doubled? (Hint: the area goes up by 4, so it's like having 4 original resistors in parallel)
What happens if only its length is doubled? (Hint: it's like having 2 original resistors in series)
What happens if you do both together?

 

[Fatima Hasan]

Trying to do this in one bite.

What happens if only its diameter is doubled? (Hint: the area goes up by 4, so it's like having 4 original resistors in parallel)
What happens if only its length is doubled? (Hint: it's like having 2 original resistors in series)
What happens if you do both together?

(D) . Right ?

 

[Vanadium 50]

Are you asking me or telling me?
(i.e. I'm not going to give you the answer.)

 

[mjc123]

Look at your calculations again. Is A₂ = A₁/4?

 

[Fatima Hasan]

Look at your calculations again. Is A₂ = A₁/4?

It should be A₂ = A₁*4
R = $\frac{ρ*2l}{4A}$
R = $\frac{ρl}{2A}$
R₂ = R₁ /2