An electrical test setup uses a calibration wire of length L and cross-sectional area A wi… — Armed Services Vocational Aptitude Battery (ASVAB)

An electrical test setup uses a calibration wire of length $L$ and cross-sectional area $A$ with a measured resistance of $60\ \Omega$ . A technician needs to build a replacement wire of the same material that has a cross-sectional area of $2A$ and a target resistance of $15\ \Omega$ . What length must the replacement wire have?

A
$2L$
$\frac{1}{2}L$ Answer
C
$8L$
D
$\frac{1}{8}L$

Answer

The replacement wire must have a length of

\frac{1}{2}L

According to the physical resistance formula

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

, resistance is directly proportional to length and inversely proportional to cross-sectional area. The target resistance of

15\ \Omega

is one-fourth of the initial

60\ \Omega

. Since the cross-sectional area is doubled to

2A

, this area change alone halves the resistance to

30\ \Omega

. To achieve the final target of

15\ \Omega

, the length must be halved to

\frac{1}{2}L

, giving

R_2 = \rho \frac{0.5L}{2A} = 0.25 \left(\rho \frac{L}{A}\right) = 0.25 \times 60\ \Omega = 15\ \Omega

Step-by-Step Solution

Express the initial resistance in terms of conductor dimensions.

R_1 = \rho \frac{L}{A} = 60\ \Omega

The resistance of any uniform conductor is governed by

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

, where

\rho

is resistivity,

L

is length, and

A

is cross-sectional area.

Set up the equation for the replacement wire's target resistance.

R_2 = \rho \frac{L_{new}}{2A} = 15\ \Omega

The replacement wire uses the same material (constant

\rho

) and has a cross-sectional area of

2A

with a target resistance of

15\ \Omega

Solve for

L_{new}

relative to the original length

L

\frac{1}{2} \left(\rho \frac{L_{new}}{A}\right) = 15 \implies \rho \frac{L_{new}}{A} = 30\ \Omega

. Since

\rho \frac{L}{A} = 60\ \Omega

L_{new} = \frac{30}{60}L = \frac{1}{2}L

Dividing the required resistance relationship by the baseline formula yields the exact length ratio.

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

Question

Answer

Step-by-Step Solution

Key Concept

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