A cylindrical metal conductor with length L and diameter d has an electrical resistance of… — Armed Services Vocational Aptitude Battery (ASVAB)

A cylindrical metal conductor with length $L$ and diameter $d$ has an electrical resistance of $16\ \Omega$ . If a replacement conductor made of the identical material has its length doubled ( $2L$ ) and its diameter doubled ( $2d$ ), what is the electrical resistance of the replacement conductor?

A
$32\ \Omega$
$8\ \Omega$ Answer
C
$16\ \Omega$
D
$4\ \Omega$

Answer

The resistance of the replacement conductor is

8\ \Omega

The electrical resistance of a conductor is given by the formula

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

, where

A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}

. When the length is doubled (

2L

), the resistance tends to double. However, when the diameter is doubled (

2d

), the cross-sectional area quadruples (

4A

), which divides the resistance by

4

. Combining both changes yields a net factor of

\frac{2}{4} = \frac{1}{2}

. Thus, the new resistance is

\frac{1}{2} \times 16\ \Omega = 8\ \Omega

Step-by-Step Solution

Express the relationship between resistance, length, and diameter for a cylindrical conductor.

R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4\rho L}{\pi d^2} = 16\ \Omega

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

Determine the scaling factor for the new conductor with length

2L

and diameter

2d

R_{new} = \frac{4\rho (2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{1}{2} \left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R

Doubling the diameter increases the cross-sectional area by a factor of

2^2 = 4

, while doubling length increases resistance by a factor of

2

Calculate the numerical resistance value of the replacement conductor.

R_{new} = \frac{1}{2} \times 16\ \Omega = 8\ \Omega

Multiplying the original resistance value by the net proportionality factor of

0.5

yields the final resistance.

Key Concept

The resistance of a uniform conductor is directly proportional to its length and inversely proportional to the square of its diameter (cross-sectional area).

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