An electronics technician is fabricating a heating coil using nichrome resistance wire. Th… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician is fabricating a heating coil using nichrome resistance wire. The wire has a cross-sectional area of $0.2\text{ mm}^2$ , a total length of $6\text{ meters}$ , and a material resistivity of $1.10\ \Omega\cdot\text{mm}^2/\text{m}$ . What is the total electrical resistance of the heating coil in ohms ( $\Omega$ ) across its entire length?

Answer: 33 ohms

Answer

The total resistance of the nichrome wire heating coil is

33\ \Omega

The total resistance of any conductor is governed by

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

. Substituting the length of

6\text{ m}

, cross-sectional area of

0.2\text{ mm}^2

, and resistivity of

1.10\ \Omega\cdot\text{mm}^2/\text{m}

yields

R = \frac{1.10 \times 6}{0.2} = 33\ \Omega

Step-by-Step Solution

State the governing physical formula relating electrical resistance to conductor dimensions and material resistivity.

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

, where

R

is resistance in ohms,

\rho

is resistivity,

L

is length, and

A

is cross-sectional area.

Electrical resistance is directly proportional to wire length and resistivity, and inversely proportional to cross-sectional area.

Substitute the given numerical values into the formula with compatible units.

R = \frac{1.10\ \Omega\cdot\text{mm}^2/\text{m} \times 6\text{ m}}{0.2\text{ mm}^2}

The unit of resistivity (

\Omega\cdot\text{mm}^2/\text{m}

) aligns directly with meters for length and square millimeters for area, canceling out spatial units.

Calculate the numerator product and divide by the cross-sectional area.

R = \frac{6.60}{0.2} = 33\ \Omega

Dividing

6.60

0.2

gives the final total resistance in ohms.

Key Concept

Calculation of electrical resistance from material resistivity, conductor length, and cross-sectional area (

R = \rho L / A

)

Estimated Time:1m 0s

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