Resistance and Resistivity Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

An automotive technician is replacing a damaged section of wiring in a 12-volt vehicle circuit. If the replacement copper wire has the same length as the original wire but a smaller American Wire Gauge (AWG) number, what effect will this have on the wire's electrical resistance?

The electrical resistance will increase because a lower gauge number corresponds to a narrower wire.

The electrical resistance will decrease because a lower gauge number corresponds to a larger cross-sectional area.

The electrical resistance will remain constant because resistance depends solely on conductor length.

The electrical resistance will drop to zero because changing gauge eliminates all opposition to current.

Show answer & explanation

Answer: The electrical resistance will decrease because a lower gauge number corresponds to a larger cross-sectional area.

Answer

The electrical resistance will decrease because a lower gauge number corresponds to a larger cross-sectional area.

In the American Wire Gauge (AWG) standard, gauge numbers are inversely related to physical wire diameter. A conductor with a smaller AWG number has a larger diameter and therefore a greater cross-sectional area. According to the formula

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

, resistance is inversely proportional to cross-sectional area when material resistivity and length are constant. Thus, replacing a wire with one of a smaller AWG number results in lower electrical resistance.

Step-by-Step Solution

1

Identify the relationship between the American Wire Gauge (AWG) scale and conductor diameter.

In the AWG standard, smaller numerical gauge numbers designate thicker conductors with larger cross-sectional areas (

A

).

The AWG numbering system is inverse; as the gauge number decreases, the physical diameter of the wire increases.

2

Apply the physical resistance formula:

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

.

Because conductor length (

L

) and material resistivity (

\rho

) remain unchanged, increasing cross-sectional area (

A

) reduces total electrical resistance (

R

).

Electrical resistance is inversely proportional to the cross-sectional area of the conductor.

Key Concept

Relationship between American Wire Gauge (AWG), cross-sectional area, and electrical resistance (

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

)

Question 2Question →

An avionics technician is evaluating how specific physical modifications and environmental conditions affect the electrical resistance of solid cylindrical conductors. Match each physical change on the left with its corresponding effect on electrical resistance on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Uniformly stretching a wire to double length at constant volume pairs with resistance quadrupling (

4R

); doubling wire diameter pairs with resistance decreasing to one-fourth (

\frac{1}{4}R

); heating a metallic conductor pairs with resistance increasing due to thermal lattice vibrations; and substituting a lower-resistivity material pairs with resistance decreasing proportionally.

Each physical modification maps directly to the governing formula

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

and thermal properties of conductors: stretching at constant volume doubles length while halving area (

2 / 0.5 = 4

); doubling diameter quadruples area reducing resistance to one-fourth; increasing temperature increases electron scattering in metals; and lower intrinsic material resistivity directly lowers resistance for identical dimensions.

Step-by-Step Solution

1

Analyze the volume-conserving elongation of a wire

Stretching to length

2L

at constant volume

V = A \cdot L

forces the cross-sectional area to become

A/2

, producing resistance

R' = \rho \frac{2L}{A/2} = 4R

.

Volume conservation requires that axial stretching produces a proportional lateral contraction.

2

Determine the effect of doubling wire diameter at fixed length

Cross-sectional area scales with diameter squared (

A \propto d^2

), so doubling diameter increases area by a factor of 4, reducing resistance to

R' = \frac{1}{4}R

.

Resistance is inversely proportional to cross-sectional area.

3

Evaluate the temperature dependence of resistance in metals

Elevated temperature increases lattice ion vibrations, which impedes electron drift and increases resistance.

Pure metallic conductors have a positive temperature coefficient of resistance.

4

Examine the direct role of material resistivity for identical geometry

From

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

, decreasing material resistivity

\rho

directly decreases resistance

R

.

Resistivity is the intrinsic property defining resistance per unit length and cross-sectional area.

Key Concept

Physical and material determinants of electrical resistance governed by

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

and temperature effects.

Estimated Time:1m 15s

Question 3Question →

Consider the physical principles governing electrical resistance ( $R = \rho \frac{L}{A}$ ) and material resistivity. Match each specific physical modification of a cylindrical wire conductor with its resulting net effect on the wire's total electrical resistance. Which pairings accurately reflect these proportional and thermal relationships?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Stretching a wire to double length at constant volume pairs with increasing resistance by a factor of 4; doubling wire diameter pairs with decreasing resistance to one-fourth; doubling both length and radius pairs with decreasing resistance to one-half; and heating a copper conductor pairs with increasing resistance due to higher resistivity.

Each match strictly obeys the physical governing equation for resistance

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

. For cylindrical geometry,

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

. Uniform stretching alters length and cross-sectional area inversely (

2L \times 0.5A \to 4R

). Diameter changes scale quadratically (

2d \to 4A \to 0.25R

). Simultaneous doubling of length and radius yields a ratio of

2/4 = 0.5R

. Thermal elevation in conductors increases resistivity (

\rho

) due to enhanced electron-phonon collisions.

Step-by-Step Solution

1

Analyze volume conservation under mechanical stretching

Stretching to double length (

L' = 2L

) causes cross-sectional area to halve (

A' = A/2

), giving

R' = \rho \frac{2L}{A/2} = 4R

.

Since volume

V = A \cdot L

remains constant, stretching alters both length and cross-sectional area simultaneously.

2

Calculate the geometric effect of doubling diameter

Doubling diameter (

d' = 2d

) quadruples area (

A' = 4A

), resulting in

R' = \rho \frac{L}{4A} = 0.25R

.

Cross-sectional area varies quadratically with diameter (

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

).

3

Calculate simultaneous scaling of length and radius

Doubling length (

2\times

) and radius (

4\times

area) produces

R' = \frac{2}{4}R = 0.5R

.

The length factor of 2 in the numerator is divided by the area factor of 4 in the denominator.

4

Evaluate the thermal effect on metallic conductors

Increasing temperature increases copper's resistivity

\rho

, raising resistance without altering geometry.

Metals possess a positive temperature coefficient of resistance where lattice vibrations hinder electron drift.

Key Concept

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

R = \rho \frac{L}{\pi r^2}

), with metallic resistivity increasing as temperature rises.

Estimated Time:1m 30s

Question 4Question →

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 $d$ ?

2\,d

\sqrt{2}\,d

\frac{1}{2}\,d

4\,d

Show answer & explanation

Answer: $\sqrt{2}\,d$

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

1

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}

.

2

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.

3

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 5Question →

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?

Show answer & explanation

Answer: 0.72

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

by

\frac{8}{3}

gives exactly

0.72\ \Omega

.

Step-by-Step Solution

1

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).

2

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.

3

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

.

4

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

Question 6Question →

A solid copper wire used in an industrial application has a length of $200\text{ meters}$ and a cross-sectional area of $0.85 \, mm^2$ . The resistivity of copper at $20^\circ C$ is $1.7 \times 10^{-8} \, \Omega\cdot m$ , and its temperature coefficient of resistance is $0.004 \text{ per } ^\circ C$ . What is the resistance of this wire, in ohms, when its operating temperature reaches $70^\circ C$ ?

Show answer & explanation

Answer: 4.8

Answer

4.8

The solution requires a synthesis of two core resistance principles. First, the baseline resistance is found by dividing the length by the area (converted to square meters) and multiplying by the resistivity, yielding 4.0 ohms. Second, the 50°C temperature increase is multiplied by the coefficient (0.004) to find a 20% increase in total resistance, raising the final value to 4.8 ohms.

Step-by-Step Solution

1

Convert the cross-sectional area from

mm^2

to

m^2

.

0.85 \, mm^2 = 0.85 \times 10^{-6} \, m^2

Standard SI units must be used to cancel with the meters in the resistivity value.

2

Calculate the resistance at

20^\circ C

(

R_{20}

) using Pouillet's law (

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

).

R_{20} = (1.7 \times 10^{-8}) \times \frac{200}{0.85 \times 10^{-6}} = 4.0 \, \Omega

This establishes the baseline resistance of the wire before temperature effects are considered.

3

Calculate the change in temperature (

\Delta T

).

\Delta T = 70^\circ C - 20^\circ C = 50^\circ C

The temperature coefficient is applied to the difference between the operating temperature and the reference temperature.

4

Calculate the new resistance at

70^\circ C

using the temperature coefficient formula:

R_{new} = R_{ref}[1 + \alpha(\Delta T)]

.

R_{70} = 4.0 \times [1 + 0.004 \times 50] = 4.0 \times [1 + 0.2] = 4.0 \times 1.2 = 4.8 \, \Omega

As temperature increases, the resistance of a standard conductor like copper increases proportionally.

Key Concept

Calculating absolute resistance using dimensions and resistivity, followed by applying temperature coefficients.

Estimated Time:3m 0s

Question 7Question →

In electrical circuits, the physical characteristics of a conductor govern its opposition to current flow. Match each conductor modification on the left with its resulting effect on electrical resistance on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Extending length matches with increasing resistance due to a longer path; increasing diameter matches with decreasing resistance due to a wider cross-section; elevating temperature matches with increasing resistance due to increased atomic vibrations; replacing copper with nichrome matches with increasing resistance due to higher intrinsic resistivity.

Each modification is correctly paired according to the fundamental resistance formula

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

and the positive temperature coefficient of metallic conductors: increasing length increases

R

, increasing cross-sectional area decreases

R

, heating the metal increases

R

, and selecting a higher resistivity material increases

R

.

Step-by-Step Solution

1

Analyze the relationship between wire length and resistance using

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

Resistance increases linearly as length increases.

Electrons encounter more resistive material over a longer distance.

2

Analyze the relationship between wire cross-sectional area (thickness) and resistance

Resistance decreases as area increases.

A greater cross-sectional area allows more electrons to pass simultaneously with fewer collisions.

3

Determine the effect of thermal energy on metallic conductors

Resistance increases as temperature rises.

Thermal energy causes metal lattice ions to vibrate more vigorously, impeding the drift of free electrons.

4

Compare material resistivity (

\rho

) between copper and nichrome

Nichrome creates greater resistance than copper of identical dimensions.

Nichrome is a high-resistivity alloy specifically designed to impede current flow and generate heat.

Key Concept

The resistance of a conductor is determined by its physical dimensions (length and cross-sectional area), material resistivity, and operating temperature.

Estimated Time:45s

Question 8Question →

An electrical technician wants to reduce the electrical resistance of a standard copper wire run without changing the metal type or operating temperature. Which of the following modifications will decrease the resistance of the wire?

Increasing the total length of the wire

Increasing the cross-sectional area of the wire

Operating the wire at a significantly elevated temperature

Selecting a wire with a smaller circular diameter

Show answer & explanation

Answer: Increasing the cross-sectional area of the wire

Answer

Increasing the cross-sectional area of the wire decreases its electrical resistance.

Electrical resistance in a conductor is inversely proportional to its cross-sectional area (

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

). Increasing the thickness or cross-sectional area provides more conductive pathways for free electrons, thereby lowering the total resistance.

Step-by-Step Solution

1

Recall the fundamental physical relationship between conductor dimensions and electrical resistance

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

, where

R

is resistance,

\rho

is material resistivity,

L

is conductor length, and

A

is cross-sectional area

This formula defines how physical geometry directly influences opposition to electron flow.

2

Analyze how changing the cross-sectional area (

A

) impacts resistance (

R

)

Because cross-sectional area (

A

) is in the denominator, resistance is inversely proportional to area (

R \propto \frac{1}{A}

)

Increasing the area (

A

) reduces the calculated resistance (

R

), allowing electrons to flow more easily.

Key Concept

Physical factors governing conductor resistance

Estimated Time:45s

Question 9Question →

When comparing two electrical wires of identical length, diameter, and temperature, which material property determines which wire has lower electrical resistance?

Applied voltage

Resistivity

Dielectric strength

Inductance

Show answer & explanation

Answer: Resistivity

Answer

Resistivity is the material property that determines the electrical resistance of equal-dimensioned conductors.

Resistivity (represented by

\rho

) is an intrinsic physical property of a material that measures its natural opposition to electric current flow. According to the formula

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

, when two conductors share identical length (

L

) and cross-sectional area (

A

) at constant temperature, the wire made of a material with lower resistivity will have lower total electrical resistance.

Step-by-Step Solution

1

Apply the fundamental formula for conductor resistance:

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

.

Total resistance (

R

) is directly proportional to conductor length (

L

), inversely proportional to cross-sectional area (

A

), and directly proportional to resistivity (

\rho

).

Because length, diameter (area), and temperature are held constant between the two wires, any difference in total resistance is strictly governed by the material's resistivity.

2

Identify the property representing the material's specific opposition to current.

Resistivity (

\rho

) is the intrinsic characteristic of the substance.

A material with lower resistivity (such as copper compared to iron) will always yield a lower resistance when shaped into identical physical dimensions.

Key Concept

Resistance and Resistivity

Estimated Time:45s

Question 10Question →

A piece of copper hookup wire with a length of $15\text{ m}$ has an electrical resistance of $0.6\ \Omega$ . If the wire is replaced with a $45\text{ m}$ piece of the exact same wire type and gauge at the same temperature, what is the resistance of the new wire in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 1.8

Answer

1.8 \Omega

According to the fundamental resistance relationship

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

, resistance is directly proportional to the length

L

of the conductor when resistivity

\rho

and cross-sectional area

A

remain unchanged. Tripling the wire length from

15\text{ m}

to

45\text{ m}

multiplies the resistance by a factor of 3, resulting in

0.6\ \Omega \times 3 = 1.8\ \Omega

.

Step-by-Step Solution

1

Calculate the proportional change in conductor length.

The new length is 3 times the original length (

45\text{ m} \div 15\text{ m} = 3

).

Identifying the scaling factor allows direct application of the resistance formula.

2

Apply the direct proportionality between conductor length and electrical resistance (

R \propto L

).

R_{\text{new}} = R_{\text{old}} \times 3 = 0.6\ \Omega \times 3 = 1.8\ \Omega

.

Since material resistivity (

\rho

) and cross-sectional area (

A

) are held constant, tripling the length triples the total resistance.

Key Concept

The electrical resistance of a uniform conductor is directly proportional to its length (

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

).

Estimated Time:45s

Question 11Question →

In electrical and electronic systems, understanding the fundamental physical properties governing current flow is essential. Match each resistance-related concept on the left with its correct physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrical Resistance matches the total opposition of a specific component in ohms; Material Resistivity matches the intrinsic geometry-independent material property; American Wire Gauge matches the inverse sizing scale where smaller numbers mean thicker wire; and Positive Temperature Coefficient matches the increase in resistance as temperature rises.

Each electrical concept matches its core definition: Electrical Resistance (

R

) is the total opposition to current flow in ohms; Material Resistivity (

\rho

) is the intrinsic material property; American Wire Gauge (AWG) is the inverse sizing standard where smaller numbers indicate thicker conductors; and a Positive Temperature Coefficient defines the increase of resistance with rising temperature.

Step-by-Step Solution

1

Distinguish between total electrical resistance and intrinsic material resistivity.

Electrical resistance (

R

) is the overall opposition in a specific conductor measured in ohms (

\Omega

), while resistivity (

\rho

) is the inherent property of the material itself.

Total resistance depends on both the material's resistivity and its physical dimensions according to

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

.

2

Analyze standard conductor sizing and thermal behavior principles.

AWG numbers inversely scale with conductor diameter (smaller number = thicker wire), and a positive temperature coefficient means resistance rises with heat.

Wire gauge dictates cross-sectional area, and thermal excitation in metals increases electron collisions, raising resistance.

Key Concept

Core physical factors and definitions governing electrical resistance and resistivity in conductors

Estimated Time:45s

Question 12Question →

Which of the following factors causes an increase in the electrical resistance of a solid metallic wire?

Increasing the cross-sectional area of the wire

Decreasing the overall length of the wire

Increasing the operating temperature of the wire

Using a wire material with higher electrical conductivity

Show answer & explanation

Answer: Increasing the operating temperature of the wire

Answer

Increasing the operating temperature of the wire causes an increase in its electrical resistance.

In metallic conductors, electrical resistance increases as temperature rises. The added thermal energy causes the atoms in the metal lattice to vibrate more vigorously, resulting in more frequent collisions with moving free electrons and hindering the flow of electric current.

Step-by-Step Solution

1

Identify the relationship between temperature and electrical resistance in metallic conductors.

Standard metals have a positive temperature coefficient of resistance, meaning resistance rises as temperature increases.

Thermal energy causes atoms in the metal lattice to vibrate more intensely, increasing collisions with flowing electrons.

2

Evaluate the effects of wire geometry and material properties using the resistance formula.

Electrical resistance is given by

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

. Resistance increases with greater length (

L

) and higher resistivity (

\rho

), but decreases with larger cross-sectional area (

A

) or higher conductivity.

Analyzing all choices confirms that among the options provided, only increasing the temperature results in higher resistance.

Key Concept

Physical factors governing conductor resistance (temperature coefficient, geometry, and material resistivity)

Estimated Time:45s

Question 13Question →

A continuous spool of aluminum hookup wire has a total length of $80\text{ meters}$ and a measured electrical resistance of $2.4\ \Omega$ . If an electronics technician cuts a $20\text{-meter}$ piece from this spool for a project, what is the electrical resistance of the cut piece in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 0.6

Answer

0.6 Ω

The electrical resistance of a uniform wire is directly proportional to its length (

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

). Since

20\text{ m}

is one-fourth (

\frac{1}{4}

or

0.25

) of the original

80\text{ m}

length, the resistance of the cut piece is one-fourth of the total resistance:

2.4\ \Omega \times 0.25 = 0.6\ \Omega

.

Step-by-Step Solution

1

Identify the physical relationship governing wire resistance and length.

Resistance is directly proportional to length:

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

, which means

R_2 = R_1 \times \frac{L_2}{L_1}

.

Because the material (resistivity

\rho

) and cross-sectional area (

A

) remain constant, resistance scales directly with length.

2

Find the proportion of the new length relative to the original length.

\frac{L_{\text{cut}}}{L_{\text{total}}} = \frac{20\text{ m}}{80\text{ m}} = 0.25

This establishes what fraction of the total opposition to electron flow the shorter segment represents.

3

Calculate the resistance of the 20-meter wire segment.

R_{\text{cut}} = 2.4\ \Omega \times 0.25 = 0.6\ \Omega

Multiplying the total resistance by the length ratio yields the exact resistance of the cut section.

Key Concept

The electrical resistance of a conductor is directly proportional to its length (

R \propto L

) when cross-sectional area and material remain constant.

Estimated Time:45s

Question 14Question →

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?

32\ \Omega

8\ \Omega

16\ \Omega

4\ \Omega

Show answer & explanation

Answer: $8\ \Omega$

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

1

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.

2

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

.

3

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).

Estimated Time:1m 0s

Question 15Question →

An electronics technician evaluates how different physical modifications alter the electrical resistance of solid wire conductors based on the fundamental relationship $R = \rho \frac{L}{A}$ . Match each conductor modification on the left with its exact resulting effect on electrical resistance on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Doubling conductor length doubles resistance; doubling diameter reduces resistance to one-fourth; replacing copper with nichrome increases resistance due to higher resistivity; increasing metal temperature increases resistance due to thermal lattice vibrations.

Each physical conductor property maps directly to the governing formula

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

and temperature coefficient behavior. Length is directly proportional (

2L \rightarrow 2R

), diameter has an inverse-square relationship (

2d \rightarrow 4A \rightarrow \frac{1}{4}R

), material change alters

\rho

, and thermal excitation increases atomic collisions in metals.

Step-by-Step Solution

1

Analyze the effect of length change using

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

Since

R \propto L

, multiplying length by

2

multiplies resistance by

2

.

Electrons must traverse twice as much resistive medium, doubling opposition to flow.

2

Analyze the effect of diameter change using cross-sectional area

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

Doubling

d

yields

(2d)^2 = 4A

. Since

R \propto \frac{1}{A}

, resistance becomes

\frac{R}{4}

.

Resistance varies inversely with the square of the conductor diameter.

3

Analyze material substitution from copper to nichrome

Nichrome has a higher resistivity (

\rho

) than copper.

Resistivity is an intrinsic material property that directly scales total resistance.

4

Analyze thermal effects on metallic conductors

Resistance increases with temperature in standard metals.

Thermal energy increases lattice atom vibrations, creating more frequent collisions with moving valence electrons.

Key Concept

Physical factors determining conductor resistance: length, cross-sectional area, resistivity, and temperature

Estimated Time:1m 15s

Question 16Question →

A 10-meter copper conductor has a measured resistance of $0.05\ \Omega$ . A technician needs to fabricate a replacement wire of the exact same length and resistance using an alloy with 4 times the resistivity of copper. How must the cross-sectional area of the replacement wire compare to the original copper wire?

The cross-sectional area must be reduced to 1/4 of the original wire.

The cross-sectional area must remain exactly the same as the original wire.

The cross-sectional area must be 4 times larger than the original wire.

The cross-sectional area must be 16 times larger than the original wire.

Show answer & explanation

Answer: The cross-sectional area must be 4 times larger than the original wire.

Answer

The cross-sectional area must be 4 times larger than the original wire.

Based on the resistance formula

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

, resistance is directly proportional to resistivity

\rho

and inversely proportional to cross-sectional area

A

. When the replacement wire material has 4 times the resistivity (

\rho' = 4\rho

) but must maintain identical length

L

and total resistance

R

, the cross-sectional area must increase by a factor of 4 (

A' = 4A

) to counterbalance the higher material resistivity.

Step-by-Step Solution

1

Identify the relationship between electrical resistance, material resistivity, length, and cross-sectional area.

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

, where

R

is resistance,

\rho

is resistivity,

L

is wire length, and

A

is cross-sectional area.

This formula defines how physical geometry and material properties determine electrical resistance.

2

Rearrange the equation to express cross-sectional area as a function of resistivity, length, and resistance.

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

Isolating cross-sectional area enables direct evaluation of how area must adjust when resistivity changes while length and resistance remain constant.

3

Substitute the scaled resistivity

\rho' = 4\rho

into the rearranged formula with constant length

L

and constant resistance

R

.

A' = \frac{(4\rho)L}{R} = 4\left(\frac{\rho L}{R}\right) = 4A

Because cross-sectional area is directly proportional to resistivity under constant length and resistance, quadrupling resistivity requires quadrupling the area.

Key Concept

Resistance and Resistivity Proportionality

Estimated Time:1m 0s

Question 17Question →

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?

Show answer & explanation

Answer: 33

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

1

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.

2

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.

3

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

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

Dividing

6.60

by

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

Question 18Question →

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?

2L

\frac{1}{2}L

8L

\frac{1}{8}L

Show answer & explanation

Answer: $\frac{1}{2}L$

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

1

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.

2

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

.

3

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 19Question →

A technician tests a solid aluminum ground wire and measures its electrical resistance as $1.2\ \Omega$ . If the wire is replaced with an aluminum wire of the same alloy that is three times as long and has twice the cross-sectional area, what is the electrical resistance of the replacement wire?

0.8\ \Omega

1.8\ \Omega

7.2\ \Omega

0.2\ \Omega

Show answer & explanation

Answer: $1.8\ \Omega$

Answer

The resistance of the replacement wire is

1.8\ \Omega

.

The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area (

R = \rho L / A

). Increasing the length by a factor of

3

increases resistance by

3\times

, while doubling the cross-sectional area reduces resistance by half (

/2

). Combining these effects yields a net multiplier of

3/2 = 1.5

. Multiplying the initial

1.2\ \Omega

by

1.5

results in

1.8\ \Omega

.

Step-by-Step Solution

1

Identify the relationship governing conductor resistance based on material and physical geometry.

The resistance formula is

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

, where

\rho

is resistivity,

L

is wire length, and

A

is cross-sectional area.

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

2

Apply the scaling factors for the new conductor dimensions.

Length changes to

L' = 3L

and area changes to

A' = 2A

. The new resistance is

R' = \rho \frac{3L}{2A} = \frac{3}{2} \left(\rho \frac{L}{A}\right) = 1.5 R

.

Since the material remains identical, resistivity

\rho

is unchanged.

3

Calculate the numerical value of the new resistance.

R' = 1.5 \times 1.2\ \Omega = 1.8\ \Omega

.

Multiplying the baseline resistance by the scaling factor gives the final resistance.

Key Concept

Conductor resistance scales directly with length and inversely with cross-sectional area (

R = \rho L / A

).

Estimated Time:1m 0s

Question 20Question →

A solid copper conductor with a length of $150\text{ meters}$ and a uniform cross-sectional area of $3.0\text{ mm}^2$ has a measured resistance of $0.85\ \Omega$ . An electrical technician replaces this conductor with another copper conductor of the same composition having a length of $90\text{ meters}$ and a cross-sectional area of $1.5\text{ mm}^2$ . What is the electrical resistance of the replacement conductor, in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 1.02

Answer

1.02 ohms (

\Omega

)

Resistance depends directly on conductor length and inversely on cross-sectional area. The length decreases by a factor of

\frac{90}{150} = 0.60

, while the cross-sectional area decreases by a factor of

\frac{1.5}{3.0} = 0.50

, which doubles the resistance contribution (factor of

2.0

). Multiplying the original resistance by both factors gives

0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega

.

Step-by-Step Solution

1

State the governing formula for electrical resistance of a uniform conductor.

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

, where

\rho

is material resistivity,

L

is conductor length, and

A

is cross-sectional area.

Since both conductors are made of identical copper, resistivity

\rho

remains constant, making resistance directly proportional to length and inversely proportional to cross-sectional area.

2

Determine the scaling factor resulting from the change in length.

Length factor

= \frac{L_2}{L_1} = \frac{90\text{ m}}{150\text{ m}} = 0.60

.

Decreasing conductor length proportionally reduces resistance to 60% of its initial value.

3

Determine the scaling factor resulting from the change in cross-sectional area.

Area factor

= \frac{A_1}{A_2} = \frac{3.0\text{ mm}^2}{1.5\text{ mm}^2} = 2.0

.

Halving the cross-sectional area doubles the resistance due to inverse proportionality.

4

Calculate the final resistance of the replacement wire.

R_2 = 0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega

.

Combine the independent effects of length and cross-sectional area changes on the original resistance.

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

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