An avionics technician replaces a solid copper grounding wire of length L and circular dia… — Armed Services Vocational Aptitude Battery (ASVAB)

An avionics technician replaces a solid copper grounding wire of length $L$ and circular diameter $d$ with a replacement wire made of a resistive alloy. The alloy has $4$ times the resistivity of copper ( $\rho_{\text{alloy}} = 4\rho_{\text{copper}}$ ), and the replacement wire is cut to half the original length ( $\frac{1}{2}L$ ). If the replacement wire must have the exact same electrical resistance as the original copper wire, what must be the diameter of the replacement wire in terms of $d$ ?

A
$2\,d$
$\sqrt{2}\,d$ Answer
C
$\frac{1}{2}\,d$
D
$4\,d$

Answer

\sqrt{2}\,d

Resistance depends on material resistivity and physical dimensions according to

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

. Multiplying resistivity by

4

and multiplying length by

\frac{1}{2}

produces a numerator factor of

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

. To keep the overall resistance constant, the denominator area must also increase by a factor of

2

. Because cross-sectional area is proportional to the square of diameter (

A \propto d^2

), the diameter must increase by a factor of

\sqrt{2}

, giving

\sqrt{2}\,d

Step-by-Step Solution

Express the electrical resistance of the original copper wire using its physical dimensions and resistivity.

R_{\text{orig}} = \rho_{\text{copper}} \frac{L}{A_1} = \rho_{\text{copper}} \frac{L}{\frac{\pi d^2}{4}} = \frac{4\rho_{\text{copper}}L}{\pi d^2}

The resistance of a cylindrical conductor is governed by

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

, where circular cross-sectional area is

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

Express the electrical resistance of the replacement alloy wire in terms of its new parameters

\rho_{\text{alloy}} = 4\rho_{\text{copper}}

L_{\text{new}} = 0.5L

, and unknown diameter

d_{\text{new}}

R_{\text{new}} = (4\rho_{\text{copper}}) \frac{0.5L}{\frac{\pi d_{\text{new}}^2}{4}} = \frac{4(2\rho_{\text{copper}}L)}{\pi d_{\text{new}}^2} = \frac{8\rho_{\text{copper}}L}{\pi d_{\text{new}}^2}

Substitute the scaled values of resistivity and length into the conductor resistance equation.

Equate the original resistance to the replacement resistance and solve for

d_{\text{new}}

\frac{4\rho_{\text{copper}}L}{\pi d^2} = \frac{8\rho_{\text{copper}}L}{\pi d_{\text{new}}^2} \implies \frac{1}{d^2} = \frac{2}{d_{\text{new}}^2} \implies d_{\text{new}}^2 = 2d^2 \implies d_{\text{new}} = \sqrt{2}\,d

The replacement specification mandates identical resistance (

R_{\text{orig}} = R_{\text{new}}

Key Concept

Conductor resistance is directly proportional to resistivity and length, and inversely proportional to the square of conductor diameter (

R \propto \frac{\rho L}{d^2}

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic