A continuous spool of aluminum hookup wire has a total length of 80\text{ meters} and a me… — Armed Services Vocational Aptitude Battery (ASVAB)

A continuous spool of aluminum hookup wire has a total length of $80\text{ meters}$ and a measured electrical resistance of $2.4\ \Omega$ . If an electronics technician cuts a $20\text{-meter}$ piece from this spool for a project, what is the electrical resistance of the cut piece in ohms ( $\Omega$ )?

Answer: 0.6 Ω

Answer

0.6 Ω

The electrical resistance of a uniform wire is directly proportional to its length (

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

). Since

20\text{ m}

is one-fourth (

\frac{1}{4}

0.25

) of the original

80\text{ m}

length, the resistance of the cut piece is one-fourth of the total resistance:

2.4\ \Omega \times 0.25 = 0.6\ \Omega

Step-by-Step Solution

Identify the physical relationship governing wire resistance and length.

Resistance is directly proportional to length:

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

, which means

R_2 = R_1 \times \frac{L_2}{L_1}

Because the material (resistivity

\rho

) and cross-sectional area (

A

) remain constant, resistance scales directly with length.

Find the proportion of the new length relative to the original length.

\frac{L_{\text{cut}}}{L_{\text{total}}} = \frac{20\text{ m}}{80\text{ m}} = 0.25

This establishes what fraction of the total opposition to electron flow the shorter segment represents.

Calculate the resistance of the 20-meter wire segment.

R_{\text{cut}} = 2.4\ \Omega \times 0.25 = 0.6\ \Omega

Multiplying the total resistance by the length ratio yields the exact resistance of the cut section.

Key Concept

The electrical resistance of a conductor is directly proportional to its length (

R \propto L

) when cross-sectional area and material remain constant.

Estimated Time:45s

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