Basic Electrical Concepts Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 9

Question 161Question →

A direct-current (DC) soldering iron operates on a $24\text{ V}$ power supply and dissipates $72\text{ W}$ of power. What is the electrical resistance of the heating element?

0.33\ \Omega

8\ \Omega

3\ \Omega

24\ \Omega

Show answer & explanation

Answer: $8\ \Omega$

Answer

8\ \Omega

According to Watt's Law, power dissipated by a resistance is

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

. Solving for resistance gives

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

. Substituting

24\text{ V}

and

72\text{ W}

yields

R = \frac{24^2}{72} = \frac{576}{72} = 8\ \Omega

.

Step-by-Step Solution

1

Identify the given values and select the Watt's Law formula that relates voltage, power, and resistance.

V = 24\text{ V}

,

P = 72\text{ W}

, and

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

The circuit voltage and power dissipation are known, and the resistance is the target variable.

2

Rearrange the formula to isolate resistance (

R

).

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

Multiplying both sides by

R

and dividing by

P

provides the direct expression for resistance.

3

Square the voltage and divide by the power to find the resistance in ohms.

R = \frac{24^2}{72} = \frac{576}{72} = 8\ \Omega

Evaluating the numerical expression yields the final resistance value.

Key Concept

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

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

, which can be rearranged as

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

.

Estimated Time:1m 0s

Question 162Question →

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 163Question →

An electronics technician evaluates four direct-current (DC) circuit devices under steady-state conditions. Match each electrical device and operating parameter on the left with its correct calculated electrical power dissipation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The backlight (

12\text{ V}, 0.5\text{ A}

) matches

6\text{ W}

; the dummy load (

50\ \Omega, 0.4\text{ A}

) matches

8\text{ W}

; the defroster element (

60\text{ V}, 300\ \Omega

) matches

12\text{ W}

; and the bleeder resistor (

4\text{ k}\Omega, 100\text{ V}

) matches

2.5\text{ W}

.

Each load correctly maps to its power calculation derived from Watt's Law formulas (

P = V \times I

,

P = I^2 \times R

, and

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

) with proper unit conversions.

Step-by-Step Solution

1

Calculate the power of the instrument panel backlight using known voltage and current.

P = V \times I = 12\text{ V} \times 0.5\text{ A} = 6\text{ W}

When voltage (

V

) and current (

I

) are provided, the direct formula for electrical power is

P = V \times I

.

2

Calculate the power of the dummy load resistor using known current and resistance.

P = I^2 \times R = (0.4\text{ A})^2 \times 50\ \Omega = 0.16 \times 50 = 8\text{ W}

When current (

I

) and resistance (

R

) are known, substitute Ohm's Law (

V = I \times R

) into Watt's Law to get

P = I^2 \times R

.

3

Calculate the power of the defroster element using known voltage and resistance.

P = \frac{V^2}{R} = \frac{(60\text{ V})^2}{300\ \Omega} = \frac{3600}{300} = 12\text{ W}

When voltage (

V

) and resistance (

R

) are known, substitute

I = \frac{V}{R}

into Watt's Law to get

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

.

4

Convert the bleeder resistance from kilo-ohms to ohms and calculate its power dissipation.

R = 4\text{ k}\Omega = 4000\ \Omega

, then

P = \frac{V^2}{R} = \frac{(100\text{ V})^2}{4000\ \Omega} = \frac{10000}{4000} = 2.5\text{ W}

Metric prefixes must be converted to base units before applying Watt's Law formulas.

Key Concept

Electrical power in direct-current circuits is calculated using Watt's Law variations:

P = V \times I

,

P = I^2 \times R

, and

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

, depending on which two electrical parameters are known.

Estimated Time:1m 15s

Question 164Question →

A direct-current (DC) avionics heating resistor with a resistance of $200\ \Omega$ dissipates $18\text{ W}$ of electrical power under steady-state operating conditions. What is the magnitude of the current flowing through the resistor?

90\text{ mA}

300\text{ mA}

0.30\text{ mA}

60\text{ mA}

Show answer & explanation

Answer: $300\text{ mA}$

Answer

300\text{ mA}

According to Watt's Law, power dissipation in a resistive load is given by

P = I^2 R

. Rearranging to solve for current gives

I = \sqrt{\frac{P}{R}} = \sqrt{\frac{18\text{ W}}{200\ \Omega}} = \sqrt{0.09\text{ A}^2} = 0.3\text{ A}

. Multiplying by

1{}000

converts the current to

300\text{ mA}

.

Step-by-Step Solution

1

Select the appropriate Watt's Law formula that relates power (

P

), current (

I

), and resistance (

R

).

P = I^2 R

The known quantities are electrical power (

18\text{ W}

) and resistance (

200\ \Omega

), and the target quantity is current (

I

).

2

Rearrange the formula to solve for the squared current (

I^2

).

I^2 = \frac{P}{R} = \frac{18\text{ W}}{200\ \Omega} = 0.09\text{ A}^2

Isolating the variable term allows direct computation of

I^2

.

3

Take the square root of both sides to calculate the current in amperes (

A

).

I = \sqrt{0.09\text{ A}^2} = 0.3\text{ A}

Current is the square root of

I^2

.

4

Convert the current from amperes (

A

) to milliamperes (

\text{mA}

).

0.3\text{ A} \times 1{}000\text{ mA/A} = 300\text{ mA}

Expressing the value in standard sub-unit milliamperes.

Key Concept

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

P = I^2 R

, which can be rearranged as

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

.

Estimated Time:1m 0s

Question 165Question →

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 166Question →

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 167Question →

In an electromagnetic generator, a coil of wire rotates through a uniform magnetic field. If the rotational speed of the coil is doubled while all other physical parameters remain constant, what happens to the peak induced electromotive force (voltage)?

It remains unchanged, because peak voltage is determined solely by the electrical resistance of the wire.

It doubles, because the rate of cutting magnetic lines of force is doubled.

It is halved, because higher rotational speeds reduce the amount of direct current flowing into the coil.

It drops to zero, because alternating voltage can only be induced at low, steady frequencies.

Show answer & explanation

Answer: It doubles, because the rate of cutting magnetic lines of force is doubled.

Answer

The peak induced electromotive force doubles because the rate of cutting magnetic lines of force is doubled.

The peak electromotive force (EMF) induced in a rotating coil is directly proportional to the strength of the magnetic field, the number of turns in the coil, and the velocity at which the coil cuts through the magnetic flux lines. Doubling the rotational speed doubles the rate of flux cutting per unit time, which doubles the peak induced voltage.

Step-by-Step Solution

1

Identify the fundamental physical relationship governing induced voltage.

According to Faraday's law of electromagnetic induction, induced voltage is directly proportional to the rate of change of magnetic flux:

\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}

.

Establishes the governing formula relating coil movement to generated electromotive force.

2

Determine the effect of doubling the coil's rotational speed on the rate of flux change.

Doubling the rotational speed doubles the velocity at which the conductors pass through the magnetic field lines, doubling

\frac{\Delta \Phi}{\Delta t}

.

Connects rotational mechanical speed directly to the rate of cutting magnetic flux lines.

3

Calculate the resulting change in peak induced voltage.

Since the magnetic field strength and the number of turns (

N

) remain unchanged, doubling the rate of flux cutting doubles the peak induced voltage (

\ ext{EMF}_{\text{new}} = 2 \times \text{EMF}_{\text{original}}

).

Applies the proportional factor to determine the final output voltage.

Key Concept

Faraday's Law of Induction and rate of flux cutting

Estimated Time:1m 0s

Question 168Question →

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

Question 169Question →

In an electromagnetic coil, increasing the rate at which electrical current changes produces a greater magnitude of induced counter-electromotive force (back-EMF) opposing the change.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because the self-induced voltage (back-EMF) across an electromagnetic coil is directly proportional to the time rate of change of current (

V_L = L \frac{\Delta I}{\Delta t}

). As the current changes more rapidly, the expanding or collapsing magnetic flux induces a correspondingly larger opposing electromotive force.

Step-by-Step Solution

1

Identify the relationship between current flow through a coil and the resulting magnetic flux.

Current flowing through a coil generates a magnetic field proportional to the current magnitude; any change in current directly changes the magnetic flux linked with the coil.

Electromagnetic induction in an inductor depends directly on time-varying magnetic flux.

2

Apply Faraday's law of induction to determine the magnitude of the induced electromotive force.

The magnitude of the induced electromotive force is directly proportional to the time rate of change of magnetic flux (

V_L = L \frac{\Delta I}{\Delta t}

).

A higher rate of current change causes a faster rate of flux change, generating a greater induced voltage.

3

Apply Lenz's law to determine the direction and nature of the induced voltage.

The polarity of the induced voltage always opposes the change in current that produced it, acting as a counter-electromotive force (back-EMF).

Lenz's law establishes that the induced magnetic field always acts to oppose the change in magnetic flux.

Key Concept

Faraday's and Lenz's laws governing self-induction and counter-electromotive force (back-EMF) in electromagnetic coils.

Estimated Time:45s

Question 170Question →

An automotive technician modifies an electromagnetic solenoid by inserting a soft iron core into its center while keeping the applied direct current (DC) and the number of wire turns constant. What effect does inserting the iron core have on the strength of the magnetic field produced by the solenoid?

The magnetic field strength decreases because the iron core absorbs voltage from the coil windings.

The magnetic field strength increases because iron has high permeability and concentrates magnetic flux lines.

The magnetic field collapses because iron blocks magnetic field generation under direct current conditions.

The magnetic field polarity reverses because the core forces electron flow in the opposite direction through the wire.

Show answer & explanation

Answer: The magnetic field strength increases because iron has high permeability and concentrates magnetic flux lines.

Answer

The magnetic field strength increases because iron has high permeability and concentrates magnetic flux lines.

Inserting a ferromagnetic core such as soft iron into a coil increases the magnetic field strength because iron has a much higher magnetic permeability than air. This allows the magnetic lines of force (flux) to flow with far less reluctance, greatly concentrating the magnetic field.

Step-by-Step Solution

1

Identify the factors determining magnetic flux density inside a solenoid

The magnetic field strength is directly proportional to the core material's magnetic permeability (

\mu

), the number of turns (

N

), and the current (

I

).

The formula

B = \mu \frac{N I}{L}

defines the magnetic field strength inside a solenoid of length

L

.

2

Compare the magnetic properties of soft iron versus air

Soft iron is a ferromagnetic substance with a magnetic permeability hundreds to thousands of times higher than air.

Permeability measures a material's ability to support the formation of a magnetic field within itself.

3

Determine the resulting change in the solenoid's magnetic field

The soft iron core provides a path of low reluctance, channeling and concentrating the magnetic lines of force (flux).

Since the applied current and number of turns remain constant, replacing air with high-permeability iron significantly increases overall magnetic field intensity.

Key Concept

Magnetic permeability and core material effects on solenoid magnetic field strength

Estimated Time:45s

Question 171Question →

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 172Question →

Match each fundamental principle of electromagnetism and induction with its correct operational description.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Faraday's Law of Induction matches generating a voltage from changing magnetic flux; Lenz's Law matches the principle where induced polarity opposes the flux change; Conductor Magnetic Field Rule matches circular flux lines around a straight wire; Mutual Induction matches inducing a voltage in a secondary coil from primary current changes.

Each principle describes a specific cornerstone of electromagnetics: Faraday's Law links changing flux to induced electromotive force; Lenz's Law establishes that induced effects oppose their cause; the conductor field rule defines the circular magnetic field around current-carrying wires; and mutual induction explains voltage induction between coupled adjacent coils.

Step-by-Step Solution

1

Identify the basic electromagnetic field structure around a straight wire conductor.

Current flowing through a single conductor produces concentric circular magnetic lines of force perpendicular to the wire axis.

This isolates the Conductor Magnetic Field Rule from dynamic induction phenomena.

2

Distinguish between the primary generator of induced voltage and the directional opposition rule.

Faraday's Law governs the generation of electromotive force from varying flux, while Lenz's Law dictates that this induced voltage acts in opposition to the flux variation.

Faraday determines magnitude and existence of induction, whereas Lenz determines polarity and counter-action.

3

Identify multi-component coupling across separate windings.

Mutual induction links changing current in a primary winding to an induced voltage in an adjacent secondary winding.

This specifically characterizes the core operating mechanism of transformers and coupled inductors.

Key Concept

Fundamental Laws of Electromagnetism and Electromagnetic Induction

Estimated Time:1m 0s

Question 173Question →

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 174Question →

A straight electrical conductor carries a steady direct current vertically through a flat horizontal surface. Which pattern best describes the magnetic field lines formed around the conductor?

Straight magnetic lines that travel parallel alongside the wire in the direction of current flow

Concentric closed circles centered on the wire in planes perpendicular to the conductor

Radial field lines pointing straight outward from the wire due to expanding voltage pressure

Pulsating magnetic loops that appear only when alternating current flows through the wire

Show answer & explanation

Answer: Concentric closed circles centered on the wire in planes perpendicular to the conductor

Answer

Concentric closed circles centered on the wire in planes perpendicular to the conductor

When an electric current passes through a straight conductor, it generates a magnetic field consisting of concentric circular lines of magnetic flux that wrap around the wire in planes perpendicular to the conductor's axis.

Step-by-Step Solution

1

Identify the relationship between electric current and magnetic fields.

Every electrical current produces a magnetic field surrounding the conductor.

According to the fundamental principles of electromagnetism (Ampère's Law), moving electrical charges inherently generate a magnetic field.

2

Determine the geometry of the magnetic field around a straight wire.

The magnetic flux forms continuous concentric circular loops centered on the wire.

The magnetic field lines lie in planes perpendicular to the direction of current flow, following the right-hand rule.

Key Concept

A straight current-carrying conductor generates a circular magnetic field in planes perpendicular to the direction of the current.

Estimated Time:45s

Question 175Question →

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 176Question →

A straight copper conductor moved parallel to the lines of a uniform magnetic field develops an induced electromotive force across its terminals.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because electromagnetic induction requires a conductor to cut across magnetic lines of force. When the direction of motion is parallel to the magnetic flux lines, zero flux is cut, resulting in zero induced electromotive force.

Step-by-Step Solution

1

Identify the physical requirement for motional electromagnetic induction in a conductor.

Electromagnetic induction occurs when a conductor moves across magnetic lines of force, represented by the relationship

E = B \cdot l \cdot v \cdot \sin\theta

.

A magnetic force acts on charge carriers only when there is a component of velocity perpendicular to the magnetic field vector.

2

Determine the angle of motion relative to the magnetic field lines.

Since the conductor moves parallel to the magnetic field lines, the angle

\theta = 0^\circ

, and

\sin(0^\circ) = 0

.

Parallel motion means the conductor cuts zero magnetic flux lines per unit of time.

3

Calculate the resulting induced electromotive force (EMF).

The induced voltage is zero, confirming the statement is false.

Without cutting magnetic flux lines, no potential difference can be established between the conductor ends.

Key Concept

Motional electromagnetic induction requires cutting across magnetic flux lines; movement parallel to magnetic field lines produces zero induced voltage.

Estimated Time:45s

Question 177Question →

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 178Question →

A transformer is constructed with 500 turns on its primary coil and 100 turns on its secondary coil. If a technician connects a steady 12 V direct current (DC) source across the primary coil, what continuous voltage is measured across the open secondary terminals?

2.4 V

0 V

12 V

60 V

Show answer & explanation

Answer: 0 V

Answer

0 V

A continuous voltage of 0 V is measured because electromagnetic induction requires a changing magnetic flux (

\frac{d\Phi}{dt}

) to induce an electromotive force in a secondary conductor. Because a steady direct current creates a stationary, unchanging magnetic field, mutual induction cannot occur continuously.

Step-by-Step Solution

1

Analyze the type of electrical source applied to the primary coil

The input source provides a steady direct current (DC) of 12 V.

Electromagnetic induction depends fundamentally on whether the source produces a changing or constant current.

2

Evaluate the magnetic field produced by the steady DC input

The steady current creates a static, unchanging magnetic field within the transformer core.

According to Ampere's law, a constant current generates a constant magnetic flux (

\frac{d\Phi}{dt} = 0

).

3

Apply Faraday's Law of Electromagnetic Induction to determine the secondary voltage

The induced voltage across the secondary coil is

V_s = -N_s \frac{d\Phi}{dt} = 0\text{ V}

.

Faraday's law states that an induced electromotive force requires a changing magnetic flux through the coil.

Key Concept

Faraday's Law of Induction and Transformer Operation on DC vs. AC

Estimated Time:45s

Question 179Question →

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

Question 180Question →

During the assembly of an avionics chassis, a technician handles components made of various substances selected for their specific electrical transport properties. Arrange the following materials in order of their electrical resistivity at room temperature, from lowest resistivity (best conductor) to highest resistivity (best insulator).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Gold contact plating, followed by Intrinsic germanium crystal, followed by Porcelain standoff bushing

Gold contact plating is a metal with high electron mobility and the lowest resistivity. Intrinsic germanium is a semiconductor containing four valence electrons, providing an intermediate resistivity between conductors and non-conductors. Porcelain is a ceramic insulator whose valence electrons are tightly locked in ionic/covalent bonds, giving it the highest resistivity.

Step-by-Step Solution

1

Classify each material according to its electrical conduction group (conductor, semiconductor, or insulator).

Gold is a conductor, germanium is a semiconductor, and porcelain is an insulator.

Electrical resistivity directly corresponds to how tightly a material's atomic structure holds its valence electrons.

2

Compare the relative electrical resistivities of the three material classifications.

Conductors have the lowest resistivity, semiconductors have moderate/intermediate resistivity, and insulators have the highest resistivity.

Resistivity is the inverse of conductivity; materials with abundant free electrons oppose current flow the least.

3

Sequence the items from lowest resistivity to highest resistivity.

Gold contact plating

\rightarrow

Intrinsic germanium crystal

\rightarrow

Porcelain standoff bushing.

Matching the classified items to the required ordering sequence satisfies the question criteria.

Key Concept

Classification of materials by electrical resistivity and atomic valence properties (conductors vs. semiconductors vs. insulators)

Estimated Time:1m 0s

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