A copper feed wire in a naval radar subsystem has a length of 50\text{ m}, a circular cros… — Armed Services Vocational Aptitude Battery (ASVAB)

A copper feed wire in a naval radar subsystem has a length of $50\text{ m}$ , a circular cross-sectional diameter of $2.0\text{ mm}$ , and a measured DC resistance of $0.27\ \Omega$ . During maintenance, this line is replaced with an alloy wire that has a resistivity $4.0$ times that of copper, a total length of $75\text{ m}$ , and a diameter of $3.0\text{ mm}$ . What is the electrical resistance of the replacement wire in ohms?

Answer: 0.72 ohms

Answer

The electrical resistance of the replacement wire is 0.72 ohms.

Electrical resistance is given by the formula

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

. Comparing the replacement wire to the original: resistivity increases by a factor of

4.0

, length increases by a factor of

\frac{75}{50} = 1.5

, and cross-sectional area increases by a factor of

\left(\frac{3.0}{2.0}\right)^2 = 2.25

. Combining these factors gives an overall resistance change factor of

\frac{4.0 \times 1.5}{2.25} = \frac{8}{3}

. Multiplying the baseline resistance of

0.27\ \Omega

\frac{8}{3}

gives exactly

0.72\ \Omega

Step-by-Step Solution

Relate electrical resistance to material resistivity and conductor dimensions.

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

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

Determine the relative scaling factors for resistivity, length, and diameter.

Resistivity factor

= 4.0

; Length factor

= \frac{75}{50} = 1.5

; Diameter factor

= \frac{3.0}{2.0} = 1.5

Proportional scaling allows direct calculation without calculating intermediate microscopic values.

Calculate the combined scaling factor for resistance.

\frac{R_2}{R_1} = \frac{\rho_2}{\rho_1} \times \frac{L_2}{L_1} \times \left(\frac{d_1}{d_2}\right)^2 = 4.0 \times 1.5 \times \left(\frac{1}{1.5}\right)^2 = \frac{4.0}{1.5} = \frac{8}{3}

Area scales quadratically with diameter, so diameter increase reduces resistance by a factor of

1.5^2 = 2.25

Calculate the final resistance of the replacement wire.

R_2 = 0.27\ \Omega \times \frac{8}{3} = 0.72\ \Omega

Multiplying the original resistance by the overall scaling factor yields the new total resistance.

Key Concept

Dependence of electrical resistance on resistivity, conductor length, and cross-sectional area

Alternative Method

Calculate the resistivity of copper first using

\rho_{\text{Cu}} = \frac{R_1 A_1}{L_1} = \frac{0.27 \times \pi (0.001)^2}{50} \approx 1.696 \times 10^{-8}\ \Omega\cdot\text{m}

. Then calculate

\rho_{\text{alloy}} = 4.0 \times \rho_{\text{Cu}} \approx 6.786 \times 10^{-8}\ \Omega\cdot\text{m}

, and compute

R_2 = \frac{\rho_{\text{alloy}} \times 75}{\pi (0.0015)^2} = 0.72\ \Omega

Estimated Time:1m 30s

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