Question

A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resistance

Answer 1

Answer:

The resistance will be 2×R

Explanation:

We note that the resistivity of a cylindrical wire is given by the following relation;

$\rho = \frac{RA}{L}$

Where:

ρ = Resistivity of the wire

R = The wire resistance

A = Cross sectional area of the wire = π·D²/4

L = Length  of the wire

Rearranging, we have;

$R= \frac{\rho L}{A}$

If the length and the diameter are both cut in half, we have;

L₂ = L/2

A₂ =π·D₂²/4 = $\pi \cdot \left (\frac{D}{2} \right )^{2} \times \frac{1}{4} = \pi \cdot \frac{D^{2}}{16} = A/4$

Therefore, the new resistance, R₂ can be expressed as follows;

$R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times \frac{\rho L}{A}$

Hence, the new resistance R₂ =  2×R, that is the resistance will be doubled.