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Verdich [7]
2 years ago
8

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

tance
Physics
1 answer:
snow_lady [41]2 years ago
8 0

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.

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