A technician tests a solid aluminum ground wire and measures its electrical resistance as … — Armed Services Vocational Aptitude Battery (ASVAB)

A technician tests a solid aluminum ground wire and measures its electrical resistance as $1.2\ \Omega$ . If the wire is replaced with an aluminum wire of the same alloy that is three times as long and has twice the cross-sectional area, what is the electrical resistance of the replacement wire?

A
$0.8\ \Omega$
$1.8\ \Omega$ Answer
C
$7.2\ \Omega$
D
$0.2\ \Omega$

Answer

The resistance of the replacement wire is

1.8\ \Omega

The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area (

R = \rho L / A

). Increasing the length by a factor of

3

increases resistance by

3\times

, while doubling the cross-sectional area reduces resistance by half (

/2

). Combining these effects yields a net multiplier of

3/2 = 1.5

. Multiplying the initial

1.2\ \Omega

1.5

results in

1.8\ \Omega

Step-by-Step Solution

Identify the relationship governing conductor resistance based on material and physical geometry.

The resistance formula is

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

, where

\rho

is resistivity,

L

is wire length, and

A

is cross-sectional area.

Electrical resistance is directly proportional to conductor length and inversely proportional to cross-sectional area.

Apply the scaling factors for the new conductor dimensions.

Length changes to

L' = 3L

and area changes to

A' = 2A

. The new resistance is

R' = \rho \frac{3L}{2A} = \frac{3}{2} \left(\rho \frac{L}{A}\right) = 1.5 R

Since the material remains identical, resistivity

\rho

is unchanged.

Calculate the numerical value of the new resistance.

R' = 1.5 \times 1.2\ \Omega = 1.8\ \Omega

Multiplying the baseline resistance by the scaling factor gives the final resistance.

Key Concept

Conductor resistance scales directly with length and inversely with cross-sectional area (

R = \rho L / A

Estimated Time:1m 0s

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