Resistance and Resistivity Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 2

Soru 21Soru →

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.

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

Cevap

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.

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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.

Anahtar Kavram

Distinction between extrinsic resistance (

R

) and intrinsic resistivity (

\rho

)

Tahmini Süre:45s

Soru 22Soru →

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.

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Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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).

Tahmini Süre:1m 15s

Soru 23Soru →

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

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

Cevap

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.

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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.

Anahtar Kavram

Conductor Resistance and Material Resistivity

Tahmini Süre:1m 0s

Soru 24Soru →

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 ( $\Omega$ )?

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Cevap: 12.5

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

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

R = \rho L / A

.

Tahmini Süre:1m 15s

Soru 25Soru →

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

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Cevap: $0.2\ \Omega$

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

R = \rho L / A

).

Tahmini Süre:1m 0s

Resistance and Resistivity

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