Respuesta :
Answer:
The resistance will be 2×R
Explanation:
We note that the resistivity of a cylindrical wire is given by the following relation;
[tex]\rho = \frac{RA}{L}[/tex]
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;
[tex]R= \frac{\rho L}{A}[/tex]
If the length and the diameter are both cut in half, we have;
L₂ = L/2
A₂ =π·D₂²/4 = [tex]\pi \cdot \left (\frac{D}{2} \right )^{2} \times \frac{1}{4} = \pi \cdot \frac{D^{2}}{16} = A/4[/tex]
Therefore, the new resistance, R₂ can be expressed as follows;
[tex]R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times \frac{\rho L}{A}[/tex]
Hence, the new resistance R₂ = 2×R, that is the resistance will be doubled.
If the length and diameter are BOTH cut in half, then the resistance of the wire will be double.
What is resistance?
The resistance is the opposition that is been offered to electrons when flowing through the wire.
We know that the resistivity of a cylindrical wire is given by the formula,
[tex]\rho = \dfrac{RA}{L}[/tex]
where R is the resistance, A is the cross-sectional area of the wire, and L is the length of the wire.
As given to us that the length and the diameter, both are cut in half, therefore,
[tex]\rho = \dfrac{RA}{L}\\\\\dfrac{rho \times L}{A}=R_1[/tex]
Substitute the value, L = L/2, and d=d/2,
[tex]\rho = \dfrac{R_2\times \dfrac{\pi}{4}(\dfrac{d}{2})^2}{\dfrac{L}{2}}\\\\\\\rho = {R_2\times \dfrac{\pi}{4}\times \dfrac {d^2}{4}}\times {\dfrac{2}{L}}\\\\\\\rho = \dfrac{R_2\times A}{L}\times {\dfrac{2}{4}}\\\\\\\rho = \dfrac{R_2\times A}{L}\times\dfrac{1}{2}}\\\\\dfrac{2 \times \rho L}{A} = R_2\\\\R_2 = 2 \times R_1[/tex]
Hence, if the length and diameter are BOTH cut in half, then the resistance of the wire will be double.
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