A solid copper conductor with a length of 150\text{ meters} and a uniform cross-sectional … — Armed Services Vocational Aptitude Battery (ASVAB)

A solid copper conductor with a length of $150\text{ meters}$ and a uniform cross-sectional area of $3.0\text{ mm}^2$ has a measured resistance of $0.85\ \Omega$ . An electrical technician replaces this conductor with another copper conductor of the same composition having a length of $90\text{ meters}$ and a cross-sectional area of $1.5\text{ mm}^2$ . What is the electrical resistance of the replacement conductor, in ohms ( $\Omega$ )?

Answer: 1.02 \Omega

Answer

1.02 ohms (

\Omega

)

Resistance depends directly on conductor length and inversely on cross-sectional area. The length decreases by a factor of

\frac{90}{150} = 0.60

, while the cross-sectional area decreases by a factor of

\frac{1.5}{3.0} = 0.50

, which doubles the resistance contribution (factor of

2.0

). Multiplying the original resistance by both factors gives

0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega

Step-by-Step Solution

State the governing formula for electrical resistance of a uniform conductor.

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

, where

\rho

is material resistivity,

L

is conductor length, and

A

is cross-sectional area.

Since both conductors are made of identical copper, resistivity

\rho

remains constant, making resistance directly proportional to length and inversely proportional to cross-sectional area.

Determine the scaling factor resulting from the change in length.

Length factor

= \frac{L_2}{L_1} = \frac{90\text{ m}}{150\text{ m}} = 0.60

Decreasing conductor length proportionally reduces resistance to 60% of its initial value.

Determine the scaling factor resulting from the change in cross-sectional area.

Area factor

= \frac{A_1}{A_2} = \frac{3.0\text{ mm}^2}{1.5\text{ mm}^2} = 2.0

Halving the cross-sectional area doubles the resistance due to inverse proportionality.

Calculate the final resistance of the replacement wire.

R_2 = 0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega

Combine the independent effects of length and cross-sectional area changes on the original resistance.

Key Concept

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

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

Estimated Time:1m 0s

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