Resistance and Resistivity Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 2

Question 21Question →

A technician cuts a solid copper grounding rod exactly in half to create a shorter lead. How do the electrical resistance and the material resistivity of one of the shortened pieces compare to the original rod?

Both the electrical resistance and the resistivity are halved.

The electrical resistance is halved, while the resistivity remains unchanged.

The electrical resistance is doubled, while the resistivity is halved.

The electrical resistance remains unchanged, while the resistivity is halved.

Show answer & explanation

Answer: The electrical resistance is halved, while the resistivity remains unchanged.

Answer

The electrical resistance is halved, while the resistivity remains unchanged.

Electrical resistance (

R

) depends on geometry (

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

), so cutting the length in half directly halves the resistance. Resistivity (

\rho

) is an intrinsic property of the material (copper) at a given temperature, so altering the rod's length does not change its resistivity.

Step-by-Step Solution

1

Analyze how physical dimensions affect total electrical resistance (

R

).

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

, showing resistance is directly proportional to length (

L

).

Electrical resistance is an extrinsic property determined by the conductor's geometry and material.

2

Calculate the new resistance after halving the length.

R_{\text{new}} = \rho \frac{L/2}{A} = \frac{1}{2} R_{\text{original}}

.

Halving the conductor length reduces the opposition encountered by flowing electrons by half.

3

Determine the effect on resistivity (

\rho

).

\rho_{\text{new}} = \rho_{\text{original}}

.

Resistivity is an intrinsic physical property of copper determined by its atomic structure, independent of size or shape.

Key Concept

Distinction between extrinsic resistance (

R

) and intrinsic resistivity (

\rho

)

Estimated Time:45s

Question 22Question →

An electronics technician measures a baseline electrical resistance of $12\ \Omega$ for a solid cylindrical wire lead of length $L$ , diameter $d$ , and resistivity $\rho$ . Match each physical modification of the wire lead to its resulting electrical resistance.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Tripling length matches with

36\ \Omega

; doubling diameter matches with

3\ \Omega

; halving length while tripling area matches with

2\ \Omega

; doubling resistivity and doubling length matches with

48\ \Omega

.

Each pair correctly applies the mathematical relationship

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

and

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

. Direct scaling by length yields

36\ \Omega

; inverse-square scaling by diameter yields

3\ \Omega

; combined reduction from shorter length and larger area yields

2\ \Omega

; and compound increases in resistivity and length yield

48\ \Omega

.

Step-by-Step Solution

1

Recall the governing physical equation for the resistance of a uniform cylindrical conductor.

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

with initial baseline

R_0 = 12\ \Omega

.

This formula relates resistance to material resistivity (

\rho

), length (

L

), cross-sectional area (

A

), and diameter (

d

).

2

Evaluate the effect of tripling length with constant area and material.

R_1 = 3 R_0 = 3 \times 12\ \Omega = 36\ \Omega

.

Resistance varies directly with conductor length.

3

Evaluate the effect of doubling wire diameter with constant length and material.

R_2 = \frac{R_0}{2^2} = \frac{12\ \Omega}{4} = 3\ \Omega

.

Area increases with the square of diameter (

A' = 4A

), causing resistance to drop to one-fourth of its initial value.

4

Evaluate the simultaneous change of halving length (

0.5L

) and tripling cross-sectional area (

3A

).

R_3 = \frac{0.5}{3} R_0 = \frac{1}{6} \times 12\ \Omega = 2\ \Omega

.

Both modifications reduce resistance simultaneously.

5

Evaluate doubling both resistivity (

2\rho

) and length (

2L

) with constant area.

R_4 = (2 \times 2) R_0 = 4 \times 12\ \Omega = 48\ \Omega

.

Resistance scales proportionally with both resistivity and length.

Key Concept

The electrical resistance of a conductor is determined by

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

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

Estimated Time:1m 15s

Question 23Question →

A technician needs to replace a copper feeder cable with an aluminum cable of identical length while maintaining the exact same total electrical resistance. Since aluminum has a higher electrical resistivity than copper ( $\rho_{\text{Al}} > \rho_{\text{Cu}}$ ), which of the following physical characteristics must the replacement aluminum cable have?

A smaller cross-sectional area than the original copper cable

A larger cross-sectional area than the original copper cable

A higher American Wire Gauge (AWG) number than the original copper cable

An applied potential difference inversely proportional to the material resistivity

Show answer & explanation

Answer: A larger cross-sectional area than the original copper cable

Answer

A larger cross-sectional area than the original copper cable

The electrical resistance of a conductor is given by the formula

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

. When replacing a material with one having a higher resistivity (

\rho

), the cross-sectional area (

A

) must be increased proportionally to maintain the same total resistance (

R

) over an identical length (

L

). Therefore, a thicker conductor with a larger cross-sectional area is required.

Step-by-Step Solution

1

Identify the mathematical relationship governing conductor resistance

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

where

R

is resistance,

\rho

is resistivity,

L

is length, and

A

is cross-sectional area

Electrical resistance depends directly on material resistivity and length, and inversely on cross-sectional area.

2

Analyze the constraints given in the problem

R_{\text{Al}} = R_{\text{Cu}}

,

L_{\text{Al}} = L_{\text{Cu}}

, and

\rho_{\text{Al}} > \rho_{\text{Cu}}

The replacement wire must preserve equal total resistance and length despite the change in material.

3

Determine the required adjustment to cross-sectional area

A_{\text{Al}} = \frac{\rho_{\text{Al}}}{\rho_{\text{Cu}}} \cdot A_{\text{Cu}} > A_{\text{Cu}}

To counterbalance the higher resistivity

\rho_{\text{Al}}

, the denominator

A

must increase proportionally.

Key Concept

Conductor Resistance and Material Resistivity

Estimated Time:1m 0s

Question 24Question →

A specialized alloy heating wire has a length of $40\text{ meters}$ , a uniform cross-sectional area of $0.8\text{ mm}^2$ , and a material resistivity of $2.5 \times 10^{-7}\ \Omega\cdot\text{m}$ . What is the total electrical resistance of this wire in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 12.5

Answer

12.5 Ω

The electrical resistance of a uniform conductor is determined by

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

. Converting

0.8\text{ mm}^2

to

8 \times 10^{-7}\text{ m}^2

and substituting

L = 40\text{ m}

along with

\rho = 2.5 \times 10^{-7}\ \Omega\cdot\text{m}

yields

\frac{(2.5 \times 10^{-7})(40)}{8 \times 10^{-7}} = 12.5\ \Omega

.

Step-by-Step Solution

1

Convert the cross-sectional area from square millimeters to square meters.

A = 0.8\text{ mm}^2 = 0.8 \times 10^{-6}\text{ m}^2 = 8 \times 10^{-7}\text{ m}^2

Resistivity is given in ohm-meters (

\Omega\cdot\text{m}

), requiring all geometric dimensions to be in standard SI units (meters and square meters).

2

Apply the fundamental resistance formula relating resistivity, length, and cross-sectional area.

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

The total resistance of a uniform conductor is directly proportional to its resistivity and length, and inversely proportional to its cross-sectional area.

3

Substitute the known values into the equation and solve for resistance.

R = \frac{(2.5 \times 10^{-7}\ \Omega\cdot\text{m})(40\text{ m})}{8 \times 10^{-7}\text{ m}^2} = \frac{100 \times 10^{-7}}{8 \times 10^{-7}} = 12.5\ \Omega

Canceling the common power factor

10^{-7}

simplifies the expression to

100 / 8 = 12.5\ \Omega

.

Key Concept

Calculating conductor resistance from material resistivity, length, and cross-sectional area using

R = \rho L / A

.

Estimated Time:1m 15s

Question 25Question →

An avionics technician replaces a copper ground wire in an aircraft instrument panel. The replacement copper wire is twice as long and has twice the diameter of the original wire. If the original wire had a measured resistance of $0.4\ \Omega$ , what is the electrical resistance of the replacement wire?

0.8\ \Omega

0.2\ \Omega

0.4\ \Omega

1.6\ \Omega

Show answer & explanation

Answer: $0.2\ \Omega$

Answer

The electrical resistance of the replacement wire is

0.2\ \Omega

.

The electrical resistance of a conductor is determined by the formula

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

, where

\rho

is resistivity,

L

is length, and

A

is cross-sectional area. Because the wire cross-section is circular,

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

, meaning area scales with the square of diameter (

d^2

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

2^2 = 4

. Doubling the length increases resistance by a factor of

2

. Combining these gives a new resistance of

\frac{2}{4} \times 0.4\ \Omega = 0.5 \times 0.4\ \Omega = 0.2\ \Omega

.

Step-by-Step Solution

1

Identify the relationship between conductor dimensions and resistance

The resistance formula is

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

, where cross-sectional area is

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

.

Resistance varies directly with length

L

and inversely with cross-sectional area

A

.

2

Determine the change in cross-sectional area when diameter doubles

Since

A \propto d^2

, doubling the diameter (

2d

) increases cross-sectional area by a factor of

2^2 = 4

.

Area is proportional to the square of the diameter for a circular cross-section.

3

Calculate the scaling factor for total resistance

The new resistance scales by

\frac{\text{length factor}}{\text{area factor}} = \frac{2}{4} = 0.5

.

Doubling length multiplies resistance by

2

, and quadrupling area divides resistance by

4

.

4

Calculate the final resistance of the replacement wire

R_{\text{new}} = 0.5 \times 0.4\ \Omega = 0.2\ \Omega

.

Multiplying the baseline resistance by the net geometric scaling factor yields the new resistance value.

Key Concept

Conductor resistance is directly proportional to length and inversely proportional to the square of its diameter (

R = \rho L / A

).

Estimated Time:1m 0s

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