An electronics technician measures a baseline electrical resistance of 12\ \Omega for a so… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician measures a baseline electrical resistance of $12\ \Omega$ for a solid cylindrical wire lead of length $L$ , diameter $d$ , and resistivity $\rho$ . Match each physical modification of the wire lead to its resulting electrical resistance.

Tripling the wire length while keeping the cross-sectional area and material constant $36\ \Omega$
Doubling the wire diameter while keeping the length and material constant $3\ \Omega$
Halving the wire length while simultaneously tripling its cross-sectional area $2\ \Omega$
Using an alloy conductor with twice the resistivity and twice the length, with unchanged cross-sectional area $48\ \Omega$

Answer

Tripling length matches with

36\ \Omega

; doubling diameter matches with

3\ \Omega

; halving length while tripling area matches with

2\ \Omega

; doubling resistivity and doubling length matches with

48\ \Omega

Each pair correctly applies the mathematical relationship

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

and

A = \frac{\pi d^2}{4}

. Direct scaling by length yields

36\ \Omega

; inverse-square scaling by diameter yields

3\ \Omega

; combined reduction from shorter length and larger area yields

2\ \Omega

; and compound increases in resistivity and length yield

48\ \Omega

Step-by-Step Solution

Recall the governing physical equation for the resistance of a uniform cylindrical conductor.

R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4\rho L}{\pi d^2}

with initial baseline

R_0 = 12\ \Omega

This formula relates resistance to material resistivity (

\rho

), length (

L

), cross-sectional area (

A

), and diameter (

d

Evaluate the effect of tripling length with constant area and material.

R_1 = 3 R_0 = 3 \times 12\ \Omega = 36\ \Omega

Resistance varies directly with conductor length.

Evaluate the effect of doubling wire diameter with constant length and material.

R_2 = \frac{R_0}{2^2} = \frac{12\ \Omega}{4} = 3\ \Omega

Area increases with the square of diameter (

A' = 4A

), causing resistance to drop to one-fourth of its initial value.

Evaluate the simultaneous change of halving length (

0.5L

) and tripling cross-sectional area (

3A

R_3 = \frac{0.5}{3} R_0 = \frac{1}{6} \times 12\ \Omega = 2\ \Omega

Both modifications reduce resistance simultaneously.

Evaluate doubling both resistivity (

2\rho

) and length (

2L

) with constant area.

R_4 = (2 \times 2) R_0 = 4 \times 12\ \Omega = 48\ \Omega

Resistance scales proportionally with both resistivity and length.

Key Concept

The electrical resistance of a conductor is determined by

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

, where resistance is directly proportional to resistivity and length, and inversely proportional to the cross-sectional area (which is proportional to the square of diameter).

Estimated Time:1m 15s

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