Basic Electrical Concepts Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 8

Question 141Question →

In electronics manufacturing, materials are categorized by how easily they conduct electricity. Which of the following elements has four valence electrons in its outer shell and is widely used as a semiconductor?

Copper

Silicon

Rubber

Silver

Show answer & explanation

Answer: Silicon

Answer

Silicon is the element with four valence electrons that functions as a semiconductor.

Silicon contains four valence electrons in its outer shell and exhibits electrical conductivity between that of a conductor and an insulator, making it the primary material for solid-state electronics like diodes and transistors.

Step-by-Step Solution

1

Identify the defining atomic property of semiconductor materials.

Semiconductors possess four valence electrons in their outermost atomic shell (Group IV elements).

This atomic structure gives them an intermediate level of conductivity between conductors and insulators.

2

Evaluate the listed materials by their valence electron structure and electrical classification.

Silicon has four valence electrons (semiconductor), copper and silver have one valence electron (conductors), and rubber is a compound insulator.

Matching the criteria of four valence electrons identifies silicon as the correct semiconductor material.

Key Concept

Classification of materials by atomic valence structure into conductors (1-3 valence electrons), semiconductors (4 valence electrons), and insulators (5-8 valence electrons).

Estimated Time:45s

Question 142Question →

A technician is inspecting a steady direct-current (DC) auxiliary circuit in a vehicle. Which of the following statements correctly describes the physical movement of electric charge carriers through the copper conductor?

Free electrons drift through the conductor away from the negative terminal and toward the positive terminal.

Positively charged atomic nuclei drift through the crystal lattice from the positive terminal toward the negative terminal.

Electromotive potential builds up statically across the wire without any physical displacement of particles.

Electrons rapidly alternate their direction of travel back and forth with no net forward displacement through the wire.

Show answer & explanation

Answer: Free electrons drift through the conductor away from the negative terminal and toward the positive terminal.

Answer

Free electrons drift through the conductor away from the negative terminal and toward the positive terminal.

In solid metal conductors such as copper wiring, electric current is the physical flow of negatively charged free electrons. According to electrostatic principles, negative electrons are repelled by the negative terminal of the DC power source and attracted toward the positive terminal, establishing a continuous unidirectional drift.

Step-by-Step Solution

1

Identify the mobile charge carriers in a solid metallic conductor.

In metals such as copper, loosely bound valence electrons are free to move, while positive atomic nuclei remain locked in the crystal lattice.

Determining which particles are mobile is the first step in analyzing physical current flow.

2

Determine the direction of electron drift under direct-current (DC) conditions.

Negatively charged electrons are repelled by the negative terminal and attracted toward the positive terminal, creating a continuous net directional drift.

Fundamental electrostatic laws dictate that opposite charges attract and like charges repel.

Key Concept

In metallic conductors carrying direct current, electric current is physically constituted by the net drift of negatively charged free electrons from the negative potential toward the positive potential.

Estimated Time:45s

Question 143Question →

In a solid copper conductor connected to a direct-current (DC) power source, the physical drift velocity of individual conduction electrons through the wire occurs at nearly the speed of light.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because free electrons in a metallic conductor undergo frequent collisions with copper atoms, resulting in a net drift velocity of only fractions of a millimeter per second, even though the electrical impulse travels across the circuit near the speed of light.

Step-by-Step Solution

1

Differentiate between the speed of electric field propagation and the drift velocity of charge carriers.

The electric field sets up almost instantaneously across the conductor at near light speed, while charge carriers collide repeatedly with lattice atoms.

Understanding the physical mechanism of electrical current requires separating the transmission of electromagnetic energy from the physical motion of electrons.

2

Evaluate the actual physical movement rate of electrons in typical DC conductor currents.

Free electrons undergo constant random thermal collisions, resulting in a net directional drift velocity of only fractions of a millimeter to a few millimeters per second.

Frequent scattering against stationary copper ions severely impedes direct linear acceleration of individual electrons.

3

Determine the validity of the statement based on the evaluated speeds.

The statement claiming that electron drift velocity occurs at nearly the speed of light is false.

The statement falsely equates signal propagation speed with individual particle drift speed.

Key Concept

Distinction between electron drift velocity and electric field propagation speed in electrical current

Estimated Time:45s

Question 144Question →

Match each electrical material to its correct classification and physical property.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Silver matches with the conductor having high free electron concentration; Silicon matches with the semiconductor having four valence electrons; Rubber matches with the flexible insulator; Ceramic matches with the rigid, heat-resistant insulator.

Materials are classified by their atomic structure and ability to carry electrical current. Silver contains loosely bound electrons making it an excellent conductor; Silicon contains four valence electrons making it a semiconductor; Rubber has tightly bound electrons making it a flexible insulator; Ceramic has tightly bound ionic and covalent bonds making it a rigid, heat-resistant insulator.

Step-by-Step Solution

1

Identify the conductor among the listed materials.

Silver is a metal with a high density of free valence electrons, allowing electrical current to pass with minimal resistance.

Conductors readily allow electron flow due to loosely bound valence electrons.

2

Identify the semiconductor among the listed materials.

Silicon contains four valence electrons, giving it an intermediate conductivity level between conductors and insulators.

Group IV elements like silicon and germanium serve as the foundational materials for solid-state electronics.

3

Distinguish between the two insulating materials by their mechanical and thermal properties.

Rubber is a flexible insulator commonly used on cables, while ceramic is a rigid, heat-resistant insulator used in high-voltage applications.

Both materials possess tightly bound electrons that resist current flow, but serve distinct structural applications.

Key Concept

Classification of materials into conductors, semiconductors, and insulators based on electron mobility and atomic structure

Estimated Time:45s

Question 145Question →

A technician is troubleshooting an industrial robotics controller that delivers direct current (DC) to a solid-state indicator module. The module has an internal resistance of $1.5\text{ k}\Omega$ . When the circuit is energized, a voltmeter measures a potential difference of $24\text{ V}$ across the module. What is the current flowing through the indicator module?

16\text{ A}

36\text{ mA}

16\text{ mA}

62.5\text{ mA}

Show answer & explanation

Answer: $16\text{ mA}$

Answer

The current flowing through the indicator module is

16\text{ mA}

.

According to Ohm's Law, current equals voltage divided by resistance (

I = \frac{V}{R}

). Converting the resistance of

1.5\text{ k}\Omega

to base ohms gives

1{}500\text{ }\Omega

. Dividing the voltage of

24\text{ V}

by

1{}500\text{ }\Omega

yields

0.016\text{ A}

, which converts to

16\text{ mA}

.

Step-by-Step Solution

1

Convert the resistance from kilo-ohms to base ohms

R = 1.5\text{ k}\Omega = 1.5 \times 10^3\text{ }\Omega = 1{}500\text{ }\Omega

Standard electrical calculations using Ohm's Law require values in fundamental SI units (volts, amperes, ohms).

2

Apply Ohm's Law to calculate the current in amperes

I = \frac{V}{R} = \frac{24\text{ V}}{1{}500\text{ }\Omega} = 0.016\text{ A}

Ohm's Law states that current equals electromotive force (voltage) divided by resistance.

3

Convert the resulting current from amperes to milliamperes

I = 0.016\text{ A} \times 1{}000\text{ mA/A} = 16\text{ mA}

Expressing fractional amperes in milliamperes aligns with standard instrumentation and option specifications.

Key Concept

Ohm's Law (

I = \frac{V}{R}

) and metric prefix conversion (kilo- and milli-).

Estimated Time:1m 0s

Question 146Question →

Standard electrical cables typically consist of an inner copper wire surrounded by a flexible rubber or plastic jacket. What is the primary electrical purpose of this outer jacket?

To act as a secondary conductor that increases total current capacity

To function as a semiconductor that regulates voltage fluctuations

To serve as an insulator that prevents electric current from escaping the wire

To reverse the direction of electron flow along the surface of the copper core

Show answer & explanation

Answer: To serve as an insulator that prevents electric current from escaping the wire

Answer

To serve as an insulator that prevents electric current from escaping the wire

The rubber or plastic jacket is an electrical insulator. Because insulators have extremely high resistance and tightly bound valence electrons, they block the flow of electric current. This ensures that electrical current remains confined within the copper conductor, preventing accidental contact, short circuits, and electrical shock.

Step-by-Step Solution

1

Identify the electrical classification of the outer jacket material (rubber or plastic).

Rubber and plastic are non-metallic materials classified as electrical insulators.

Insulators have tightly bound valence electrons and extremely high resistance to electrical current.

2

Determine the functional role of an insulator surrounding a metallic conductor.

The insulator confines the flow of electric charges inside the copper conductor.

Preventing current leakage protects people from electric shock and prevents dangerous short circuits.

Key Concept

Insulators have high electrical resistance and tightly bound valence electrons, preventing current from leaking out of conductors.

Estimated Time:45s

Question 147Question →

An avionics technician tests a direct-current (DC) instrument subsystem and measures a steady current of $2.5\text{ amperes}$ through a supply conductor. Which of the following statements correctly describes the movement of electrical charge through this conductor?

The direction of electron drift continuously alternates back and forth across the conductor at a rate of

2.5\text{ cycles}

per second.

A total charge of

2.5\text{ coulombs}

passes a cross-section of the conductor each second, with electrons drifting from the negative terminal toward the positive terminal.

A total charge of

2.5\text{ coulombs}

passes a cross-section of the conductor each second, with electrons physically moving from the positive terminal toward the negative terminal.

An electromotive force of

2.5\text{ volts}

is consumed every second to push static charges through the conductor without net transfer of charge.

Show answer & explanation

Answer: A total charge of $2.5\text{ coulombs}$ passes a cross-section of the conductor each second, with electrons drifting from the negative terminal toward the positive terminal.

Answer

A total charge of

2.5\text{ coulombs}

passes a cross-section of the conductor each second, with electrons drifting from the negative terminal toward the positive terminal.

One ampere is formally defined as one coulomb of electric charge passing a given point in a circuit per second (

I = Q/t

). For a steady current of

2.5\text{ amperes}

,

2.5\text{ coulombs}

of charge pass through the conductor each second. Furthermore, in solid metallic wires, current is carried by negatively charged electrons that physically drift away from the negative polarity terminal toward the positive polarity terminal.

Step-by-Step Solution

1

Define electric current in terms of fundamental electrical units

Current (

I

) is defined as the rate of charge flow:

I = \frac{Q}{t}

, where

1\text{ ampere} = 1\text{ coulomb per second}

. Therefore,

2.5\text{ A} = 2.5\text{ C/s}

.

Establishing the relationship between amperes and coulombs per second confirms the rate of charge transfer.

2

Determine the physical direction of charge carriers in a DC conductor

In solid metal conductors, charge carriers are negatively charged free electrons that are repelled by the negative power supply terminal and attracted to the positive terminal.

Identifying the electron flow model distinguishes physical charge motion from conventional current flow.

Key Concept

Definition of the ampere as coulombs per second and the direction of physical electron drift in DC circuits

Estimated Time:40s

Question 148Question →

Match each fundamental electrical current concept on the left with its correct technical definition or physical characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electron Flow Theory matches the physical movement of negative charges from negative to positive; Conventional Current Flow matches the circuit analysis model from positive to negative; One Ampere matches one coulomb per second; Free Conduction Electrons match the mobile valence particles drifting through the metal lattice.

Each concept accurately pairs with its physical definition: Electron flow physically travels from negative to positive; conventional current travels from positive to negative; the ampere measures the rate of charge flow in coulombs per second; and free conduction electrons serve as the mobile charge carriers in metallic conductors.

Step-by-Step Solution

1

Identify the physical mechanism of charge carrier drift in solid conductors.

Free conduction electrons are the charge carriers in metals, and their physical drift toward the positive potential defines electron flow theory.

Electrons carry negative charge and are repelled by negative potential and attracted to positive potential.

2

Distinguish between conventional current and electron flow directional models.

Conventional current assumes positive-to-negative flow, while electron flow describes physical negative-to-positive movement.

Both conventions are standard in electronics and electrical engineering documentation.

3

Apply the fundamental SI definition of current magnitude.

Current (

I

) is the rate of charge (

Q

) flow over time (

t

), where

1\text{ A} = 1\text{ C/s}

.

The unit ampere measures rate of charge transfer, not static charge quantity or voltage.

Key Concept

Distinction between directional current models, charge carriers, and the SI rate definition of current.

Estimated Time:45s

Question 149Question →

Materials are categorized by their ability to facilitate or oppose the movement of electrical charge. Arrange the following substances in order from greatest electrical conductivity (easiest electron flow) to least electrical conductivity (highest electrical resistance).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Aluminum wire, Silicon wafer, Glass sheet

The correct sequence begins with Aluminum wire because metals are good conductors with low resistivity due to their sea of free electrons. Silicon wafer comes next because it is a semiconductor with four valence electrons, exhibiting moderate electrical conductivity. Glass sheet comes last because it is an insulator with a wide bandgap and tightly bound electrons, preventing current flow under normal operating conditions.

Step-by-Step Solution

1

Identify the electrical classification of each listed material.

Aluminum wire is a conductor, Silicon wafer is a semiconductor, and Glass sheet is an insulator.

Electrical classification defines how easily electrons move through a material's atomic structure.

2

Rank the materials according to their relative ability to conduct electric current from highest to lowest.

Conductor (Aluminum wire) > Semiconductor (Silicon wafer) > Insulator (Glass sheet).

Conductors have freely moving valence electrons, semiconductors have moderate carrier concentrations, and insulators hold their valence electrons tightly to impede electron flow.

Key Concept

Relative electrical conductivity of conductors, semiconductors, and insulators based on atomic structure.

Estimated Time:45s

Question 150Question →

A direct-current (DC) power supply maintains a constant potential difference of $15\text{ V}$ across a circuit resistor that dissipates $4.5\text{ W}$ of electrical power. What is the electrical resistance of the resistor in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 50

Answer

The resistance of the resistor is

50\ \Omega

.

According to Watt's Law, power dissipated by a resistor with a known voltage across it is given by

P = \frac{V^2}{R}

. Solving for resistance gives

R = \frac{V^2}{P} = \frac{15^2}{4.5} = \frac{225}{4.5} = 50\ \Omega

.

Step-by-Step Solution

1

State the known values and identify the target variable.

Voltage

V = 15\text{ V}

, Power

P = 4.5\text{ W}

, Resistance

R = ?

Establishing the known electrical parameters determines the appropriate form of Watt's Law to apply.

2

Select and rearrange the Watt's Law power equation that relates power, voltage, and resistance.

R = \frac{V^2}{P}

Since voltage and power are known, using

P = \frac{V^2}{R}

allows direct solving for resistance.

3

Calculate the squared voltage.

V^2 = 15^2 = 225\text{ V}^2

Evaluating the numerator is the first calculation step in solving for

R

.

4

Divide the squared voltage by the power dissipation.

R = \frac{225}{4.5} = 50\ \Omega

Carrying out the division yields the final resistance value in ohms.

Key Concept

Watt's Law relates electrical power, voltage, and resistance through the formula

P = \frac{V^2}{R}

, which rearranges algebraically to

R = \frac{V^2}{P}

.

Alternative Method

First find the circuit current using

I = \frac{P}{V} = \frac{4.5\text{ W}}{15\text{ V}} = 0.3\text{ A}

. Then apply Ohm's Law to determine the resistance:

R = \frac{V}{I} = \frac{15\text{ V}}{0.3\text{ A}} = 50\ \Omega

.

Estimated Time:1m 0s

Question 151Question →

Which of the following atomic characteristics explains why metals such as copper and silver are classified as good electrical conductors?

They have tightly bound outer electrons that cannot migrate between atoms.

They possess an outer shell with exactly four valence electrons in a crystal grid.

They have loosely held valence electrons that move freely from atom to atom.

They exhibit high electrical resistivity that opposes the flow of electric charge.

Show answer & explanation

Answer: They have loosely held valence electrons that move freely from atom to atom.

Answer

They have loosely held valence electrons that move freely from atom to atom.

Electrical conductors such as copper, silver, and gold have few electrons in their outermost shell that are held loosely by the nucleus. These free electrons move easily from one atom to another when a voltage is applied, permitting electric current to flow with minimal resistance.

Step-by-Step Solution

1

Analyze the atomic structure of electrical conductors.

Conductors like copper, silver, and aluminum typically have only one or two valence electrons in their outermost shell.

Understanding atomic structure determines how easily charges can move through a material.

2

Determine how these valence electrons behave under an applied electrical potential.

Because these outer electrons are loosely bound to the nucleus, small electromotive forces cause them to drift freely across the lattice as electric current.

This free electron mobility is what defines high electrical conductivity.

Key Concept

Conductors, Insulators, and Semiconductors

Estimated Time:45s

Question 152Question →

During a maintenance check on an emergency backup power system, an ammeter registers a continuous reading of $4\text{ amperes}$ through a primary feeder wire. How many coulombs of electrical charge pass through any given cross-section of this feeder wire in $15\text{ seconds}$ ?

3.75\text{ coulombs}

60\text{ coulombs}

0.27\text{ coulombs}

60\text{ volts}

Show answer & explanation

Answer: $60\text{ coulombs}$

Answer

60\text{ coulombs}

of electrical charge pass through the cross-section.

Electric current in amperes measures the rate of charge movement where

1\text{ ampere} = 1\text{ coulomb per second}

. Multiplying the steady current (

4\text{ A}

) by the elapsed time (

15\text{ s}

) gives the total charge transfer:

4\text{ C/s} \times 15\text{ s} = 60\text{ coulombs}

.

Step-by-Step Solution

1

Identify the relationship between electric current, charge, and time

I = \frac{Q}{t}

, where

I

is current in amperes,

Q

is charge in coulombs, and

t

is time in seconds

One ampere is defined as the flow of one coulomb of charge per second (

1\text{ A} = 1\text{ C/s}

).

2

Rearrange the formula to solve for total charge

Q

Q = I \times t

Multiplying both sides of the rate equation by time isolates the total charge quantity.

3

Substitute the given values into the equation and compute the result

Q = 4\text{ A} \times 15\text{ s} = 60\text{ C}

Evaluating the product yields the total accumulated charge that traversed the wire.

Key Concept

Electric current is the time rate of electric charge transfer through a conductor (

1\text{ A} = 1\text{ C/s}

).

Estimated Time:45s

Question 153Question →

An electronics technician measures a power dissipation of $20\text{ W}$ across a $500\ \Omega$ test resistor in a direct-current circuit. What is the electrical current flowing through this resistor in milliamperes ( $\text{mA}$ )?

40\text{ mA}

200\text{ mA}

0.2\text{ mA}

100\text{ mA}

Show answer & explanation

Answer: $200\text{ mA}$

Answer

200\text{ mA}

According to Watt's Law, power dissipated in a resistive load is

P = I^2 R

. Rearranging to solve for current gives

I = \sqrt{\frac{P}{R}} = \sqrt{\frac{20\text{ W}}{500\ \Omega}} = \sqrt{0.04} = 0.2\text{ A}

. Converting amperes to milliamperes gives

0.2\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}

.

Step-by-Step Solution

1

Identify known parameters and select the Watt's Law formula relating power, resistance, and current.

P = 20\text{ W}

,

R = 500\ \Omega

, and

P = I^2 R

The circuit power and resistance are given, and current is the unknown quantity.

2

Rearrange the formula to solve for current (

I

) and substitute the given values.

I = \sqrt{\frac{P}{R}} = \sqrt{\frac{20\text{ W}}{500\ \Omega}} = \sqrt{0.04} = 0.2\text{ A}

Isolating

I

requires dividing power by resistance and taking the square root.

3

Convert the calculated current from amperes to milliamperes.

0.2\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}

The question asks for the final answer in milliamperes.

Key Concept

Watt's Law relates power, current, and resistance via

P = I^2 R

, which rearranges to

I = \sqrt{\frac{P}{R}}

.

Alternative Method

First compute the voltage drop across the resistor using

V = \sqrt{P \times R} = \sqrt{20\text{ W} \times 500\ \Omega} = \sqrt{10{}000} = 100\text{ V}

. Then apply Ohm's Law to find current:

I = \frac{V}{R} = \frac{100\text{ V}}{500\ \Omega} = 0.2\text{ A} = 200\text{ mA}

.

Estimated Time:1m 0s

Question 154Question →

In an operating direct-current (DC) circuit, conventional current is defined as moving from the positive terminal to the negative terminal through the external circuit, which is opposite to the physical drift direction of electrons.

True

False

Show answer & explanation

Answer: True

Answer

True: Conventional current flows from positive to negative in the external circuit, while electron flow moves from negative to positive.

The statement is true because standard electrical convention defines current as positive charge moving from positive to negative, whereas atomic theory proves that negative electrons physically flow from negative to positive in the external circuit.

Step-by-Step Solution

1

Identify the historical definition of conventional current flow.

Conventional current assumes positive charge movement from the positive terminal (+) through the external circuit to the negative terminal (-).

Benjamin Franklin and early physicists established current flow under the convention that positive charges were the moving carriers.

2

Identify the physical direction of electron flow in solid conductors.

Electrons carry a negative charge and travel from the negative terminal (-) through the external circuit toward the positive terminal (+).

Electrostatic laws state that like charges repel and opposite charges attract; hence, negative electrons are repelled by the negative battery terminal.

3

Compare the two flow directions to evaluate the statement.

Conventional current and electron flow describe opposite directions through the external circuit.

Since the statement accurately defines both directions as being opposite to one another, the statement is true.

Key Concept

Directional distinction between conventional current and electron flow in DC circuits

Estimated Time:40s

Question 155Question →

A technician evaluates four direct-current (DC) electrical components operating under steady-state conditions. Match each component and operating condition on the left with its corresponding dissipated electrical power on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The automotive lamp (

12\text{ V}, 750\text{ mA}

) matches

9\text{ W}

; the heating element (

50\ \Omega, 0.6\text{ A}

) matches

18\text{ W}

; the instrumentation sensor (

600\ \Omega, 200\text{ mA}

) matches

24\text{ W}

; and the relay coil (

100\ \Omega, 60\text{ V}

) matches

36\text{ W}

.

Each electrical component's power dissipation is determined by selecting the appropriate algebraic form of Watt's Law based on the known parameters:

P = V \times I

for the indicator lamp (

12\text{ V} \times 0.75\text{ A} = 9\text{ W}

),

P = I^2 \times R

for the heating element (

(0.6\text{ A})^2 \times 50\ \Omega = 18\text{ W}

) and sensor (

(0.2\text{ A})^2 \times 600\ \Omega = 24\text{ W}

), and

P = \frac{V^2}{R}

for the relay coil (

(60\text{ V})^2 / 100\ \Omega = 36\text{ W}

).

Step-by-Step Solution

1

Calculate power for the automotive indicator lamp using voltage and current

P = 12\text{ V} \times 0.75\text{ A} = 9\text{ W}

Convert milliamperes to amperes (

750\text{ mA} = 0.75\text{ A}

) and apply the fundamental Watt's Law equation

P = V \times I

.

2

Calculate power for the heating element using current and resistance

P = (0.6\text{ A})^2 \times 50\ \Omega = 0.36 \times 50 = 18\text{ W}

Apply Watt's Law in the form

P = I^2 \times R

when current and resistance are known.

3

Calculate power for the instrumentation sensor using current and resistance

P = (0.2\text{ A})^2 \times 600\ \Omega = 0.04 \times 600 = 24\text{ W}

Convert

200\text{ mA}

to

0.2\text{ A}

and calculate power using

P = I^2 \times R

.

4

Calculate power for the communications relay coil using voltage and resistance

P = \frac{(60\text{ V})^2}{100\ \Omega} = \frac{3600}{100} = 36\text{ W}

Apply Watt's Law in the form

P = \frac{V^2}{R}

when potential difference and resistance are known.

Key Concept

Watt's Law and Electrical Power Dissipation

Estimated Time:1m 15s

Question 156Question →

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 157Question →

A direct-current (DC) relay coil has an internal resistance of $80\ \Omega$ and draws a steady current of $250\text{ mA}$ . How much electrical power is dissipated by the coil?

20\text{ W}

5{}000\text{ W}

5\text{ W}

320\text{ W}

Show answer & explanation

Answer: $5\text{ W}$

Answer

The electrical power dissipated by the coil is

5\text{ W}

.

To find the power dissipated when current and resistance are known, apply Watt's Law in the form

P = I^2 R

. First, convert

250\text{ mA}

to

0.25\text{ A}

. Squaring the current gives

(0.25\text{ A})^2 = 0.0625\text{ A}^2

. Multiplying by the resistance yields

0.0625\text{ A}^2 \times 80\ \Omega = 5\text{ W}

.

Step-by-Step Solution

1

Convert the current from milliamperes (mA) to amperes (A).

I = 250\text{ mA} = 0.25\text{ A}

Standard electrical formulas require current in base units of amperes.

2

Calculate the power dissipation using Watt's Law (

P = I^2 R

).

P = (0.25\text{ A})^2 \times 80\ \Omega = 0.0625\text{ A}^2 \times 80\ \Omega = 5\text{ W}

When current and resistance are known, power equals the square of current multiplied by resistance.

Key Concept

Calculating electrical power using current and resistance (

P = I^2 R

) with metric prefix conversion.

Estimated Time:1m 0s

Question 158Question →

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 159Question →

An industrial direct-current (DC) indicator lamp dissipates $1.8\text{ W}$ of power when a current of $150\text{ mA}$ flows through it. What is the voltage drop across the lamp, in volts?

Show answer & explanation

Answer: 12

Answer

The voltage drop across the indicator lamp is

12\text{ V}

.

According to Watt's Law, electrical power is the product of voltage and current (

P = V \times I

). Rearranging the formula to solve for voltage gives

V = \frac{P}{I}

. Converting

150\text{ mA}

to

0.15\text{ A}

and evaluating

\frac{1.8\text{ W}}{0.15\text{ A}}

yields

12\text{ V}

.

Step-by-Step Solution

1

Convert the current from milliamperes to amperes.

I = 150\text{ mA} = 0.15\text{ A}

Standard electrical formulas require current in base units of amperes (A).

2

Rearrange Watt's Law to solve for voltage.

V = \frac{P}{I}

Watt's Law defines power as

P = V \times I

, so isolating voltage yields

V = \frac{P}{I}

.

3

Substitute the known values into the equation to find the voltage.

V = \frac{1.8\text{ W}}{0.15\text{ A}} = 12\text{ V}

Dividing the power dissipation (

1.8\text{ W}

) by the current (

0.15\text{ A}

) gives the operating voltage.

Key Concept

Watt's Law describes the mathematical relationship between electrical power, voltage, and current:

P = V \times I

. When power and current are known, voltage is calculated using

V = \frac{P}{I}

, provided current is expressed in amperes.

Estimated Time:1m 0s

Question 160Question →

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

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