All practice questions

Question 901Question →

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

Increasing the cross-sectional area of the wire

Decreasing the overall length of the wire

Increasing the operating temperature of the wire

Using a wire material with higher electrical conductivity

Show answer & explanation

Answer: Increasing the operating temperature of the wire

Answer

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

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

Step-by-Step Solution

1

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

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

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

2

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

Electrical resistance is given by

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

. Resistance increases with greater length (

L

) and higher resistivity (

\rho

), but decreases with larger cross-sectional area (

A

) or higher conductivity.

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

Key Concept

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

Estimated Time:45s

Question 902Question →

Monolithic integrated circuits contain microscopic electronic components such as transistors, diodes, and resistors fabricated directly onto a single thin piece of semiconductor material. What is this semiconductor base commonly called?

Insulating ceramic substrate

Silicon wafer

Magnetic ferrite core

Bipolar junction terminal

Show answer & explanation

Answer: Silicon wafer

Answer

Silicon wafer

A monolithic integrated circuit is produced by creating all necessary circuit elements—including transistors, diodes, and resistors—on a single thin slice of crystalline semiconductor material (most commonly silicon). In microelectronics manufacturing, this foundation slice is called a silicon wafer.

Step-by-Step Solution

1

Identify the structural foundation of a monolithic integrated circuit

Monolithic ICs are manufactured as a single, solid physical unit on a continuous semiconductor slice

The word 'monolithic' indicates that all circuit components share a single, unified semiconductor base rather than separate discrete parts

2

Apply standard electronics manufacturing terminology to the semiconductor base

A thin, polished slice of crystalline silicon used as the substrate is termed a wafer

Wafers provide the crystal foundation into which microscopic active and passive components are diffused, etched, and interconnected

Key Concept

Monolithic integrated circuit construction and semiconductor wafer substrate

Question 903Question →

In an electronic power supply circuit, what is the primary function of a rectifier diode?

To maintain a constant opposition to electric current regardless of circuit voltage polarity

To convert alternating current (AC) into direct current (DC) by allowing current to flow in only one direction

To store electrical charge in an electrostatic field to filter out voltage variations

To step up or step down alternating voltage levels through electromagnetic induction

Show answer & explanation

Answer: To convert alternating current (AC) into direct current (DC) by allowing current to flow in only one direction

Answer

To convert alternating current (AC) into direct current (DC) by allowing current to flow in only one direction

A semiconductor diode permits electrical current to flow easily in only one direction (when forward-biased) and blocks current flow in the reverse direction (when reverse-biased). In power supply circuits, this unidirectional property is utilized in rectification to convert alternating current (AC) into direct current (DC).

Step-by-Step Solution

1

Identify the fundamental operational characteristic of a semiconductor diode.

A diode acts as a one-way electrical valve, exhibiting very low resistance in the forward direction and extremely high resistance in the reverse direction.

Understanding the unidirectional conduction property of the PN junction is essential for determining its circuit application.

2

Connect this one-way conduction property to its purpose in a power supply circuit.

When alternating current (AC) is applied, the diode conducts only during the positive half-cycles, producing a pulsating direct current (DC) output.

Rectification is the process of converting bidirectional AC into unidirectional DC.

Key Concept

Unidirectional current conduction and AC-to-DC rectification by semiconductor diodes

Estimated Time:45s

Question 904Question →

A technician is troubleshooting a direct-current parallel lighting system. Match each circuit modification on the left with its direct electrical effect on the system on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The circuit modifications match their electrical effects as follows: adding an additional branch decreases total resistance and increases total supply current; disconnecting one branch increases total resistance while allowing remaining branches to operate normally; increasing resistance of a single branch reduces current only through that branch; creating a short circuit across one branch bypasses all branches and causes a dangerous surge in total current.

Each pair reflects core parallel circuit behavior: adding branches lowers equivalent resistance and raises total current; disconnecting a branch raises equivalent resistance without interrupting other branches; changing one branch resistance alters only that branch's current; and shorting across one branch shorts the entire common bus.

Step-by-Step Solution

1

Determine the impact of adding a parallel branch

Total equivalent resistance drops and total supply current rises (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

).

Adding a parallel branch creates an extra pathway for current flow.

2

Determine the impact of opening or disconnecting an active branch

Total equivalent resistance increases, but unaffected branches stay powered.

Removing a branch eliminates a current path while full source voltage remains across remaining branches.

3

Determine the impact of increasing a single branch's resistance

Current through that specific branch decreases (

I = V/R

) without changing the other branches.

Branch voltages in a parallel network are independent and equal to source voltage.

4

Determine the impact of short-circuiting one branch

The entire power source is shorted, bypassing all parallel loads.

All parallel branches connect directly to the same common positive and negative supply nodes.

Key Concept

Parallel Circuit Branch Behavior and Independence

Estimated Time:45s

Question 905Question →

A technician is servicing an off-grid solar telemetry station and measures the operating parameters of four distinct direct current (DC) load circuits. Match each circuit description on the left with its corresponding calculated electrical quantity on the right using Ohm's law.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The

24\text{ V}

fan at

120\text{ }\Omega

matches

0.20\text{ A}

; the

48\text{ V}

beacon at

16\text{ mA}

matches

3.0\text{ k}\Omega

; the

15\text{ V}

heater drawing

2.5\text{ A}

matches

6.0\text{ }\Omega

; and the

50\text{ mA}

indicator across

180\text{ }\Omega

matches

9.0\text{ V}

.

Each match correctly applies the appropriate algebraic variation of Ohm's law (

V = I \times R

,

I = \frac{V}{R}

, or

R = \frac{V}{I}

) while properly converting metric units (

1\text{ mA} = 0.001\text{ A}

and

1\text{ k}\Omega = 1000\text{ }\Omega

).

Step-by-Step Solution

1

Calculate the current for the cooling fan circuit

I = \frac{V}{R} = \frac{24\text{ V}}{120\text{ }\Omega} = 0.20\text{ A}

Ohm's law defines current as voltage divided by resistance.

2

Calculate the resistance for the remote sensor beacon

R = \frac{V}{I} = \frac{48\text{ V}}{0.016\text{ A}} = 3000\text{ }\Omega = 3.0\text{ k}\Omega

Convert milliamperes to amperes (

16\text{ mA} = 0.016\text{ A}

) and solve for resistance by dividing voltage by current.

3

Calculate the resistance for the environmental heating strip

R = \frac{V}{I} = \frac{15\text{ V}}{2.5\text{ A}} = 6.0\text{ }\Omega

Ohm's law defines resistance as voltage divided by current.

4

Calculate the voltage for the backup LED indicator branch

V = I \times R = 0.050\text{ A} \times 180\text{ }\Omega = 9.0\text{ V}

Convert milliamperes to amperes (

50\text{ mA} = 0.050\text{ A}

) and calculate voltage by multiplying current by resistance.

Key Concept

Ohm's Law states the fundamental relationship between voltage, current, and resistance in a direct current circuit:

V = I \times R

,

I = \frac{V}{R}

, and

R = \frac{V}{I}

.

Estimated Time:1m 15s

Question 906Question →

A technician is installing a master disconnect switch for a 240 V AC equipment circuit supplied by two active 120 V lines ( $L_1$ and $L_2$ ). To completely isolate the load by opening both hot conductors simultaneously using a single mechanical toggle, which type of switch is required?

Single-Pole, Single-Throw (SPST) switch

Double-Pole, Single-Throw (DPST) switch

Single-Pole, Double-Throw (SPDT) switch

Thermal overcurrent fuse

Show answer & explanation

Answer: Double-Pole, Single-Throw (DPST) switch

Answer

Double-Pole, Single-Throw (DPST) switch

A Double-Pole, Single-Throw (DPST) switch features two independent internal contact sets (poles) mechanically linked to move together under a single toggle. This enables simultaneous opening and closing of both

L_1

and

L_2

hot conductors to achieve complete electrical isolation of the equipment.

Step-by-Step Solution

1

Identify the number of isolated conductive paths (poles) required by the circuit.

Two distinct live conductors (

L_1

and

L_2

) must be controlled at the same time, which requires a Double-Pole (DP) configuration.

The term 'pole' defines the number of separate, electrically isolated circuits or lines the switch can open or close concurrently.

2

Determine the number of active output positions (throws) required for each pole.

The application requires simple ON/OFF isolation, which corresponds to a Single-Throw (ST) contact arrangement.

The term 'throw' defines the number of closed contact positions available per pole; a single-throw switch has one closed state (ON) and one open state (OFF).

Key Concept

Switch Pole and Throw Configurations (DPST vs. SPST/SPDT)

Estimated Time:1m 0s

Question 907Question →

A technician is troubleshooting a series circuit powered by a $12\text{ V}$ DC source that normally operates with three identical $10\,\Omega$ resistive loads ( $R_1$ , $R_2$ , and $R_3$ ). Match each specific multimeter reading or diagnostic symptom on the left to its corresponding circuit fault condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The multimeter readings correctly pair with the specific circuit faults as follows: measuring full supply voltage across a single component with zero current corresponds to an open circuit in that component; measuring zero volts across a component while current and remaining voltage drops increase corresponds to a shorted component; reduced voltage across all loads with an unexpected voltage drop at a junction corresponds to a high-resistance connection; and measuring zero ohms to chassis before loads corresponds to a direct short-to-ground fault.

Each fault produces a distinct electrical signature in a series circuit: an open component exhibits infinite resistance and drops the full source voltage with zero current; a shorted component exhibits zero resistance and zero voltage drop while increasing circuit current; a high-resistance connection adds unwanted series resistance that drops voltage and reduces current to all loads; and a short-to-ground creates a direct zero-ohm path to ground before the load network.

Step-by-Step Solution

1

Analyze the voltage and current behavior for the open circuit symptom

Full source voltage (

12\text{ V}

) across

R_2

with

0\text{ A}

current indicates an open circuit at

R_2

.

An open circuit creates infinite resistance, stopping all current (

I = 0

) so no voltage drops occur across intact series components, placing total source EMF across the open terminals.

2

Analyze the voltage and current behavior for the shorted component symptom

Zero volts across

R_2

with increased current from

0.4\text{ A}

to

0.6\text{ A}

indicates a short circuit across

R_2

.

A shorted component has

0\,\Omega

, dropping

0\text{ V}

while decreasing total circuit resistance (

R_{total} = 20\,\Omega

) and increasing current (

I = 12\text{ V} / 20\,\Omega = 0.6\text{ A}

).

3

Analyze the effect of an unwanted series resistance

Lower voltage drops across all loads combined with an unexpected drop across a terminal indicates a high-resistance connection.

Corrosion or a loose terminal adds resistance to the series string, lowering total current and taking a portion of the source voltage away from the loads.

4

Evaluate the de-energized ohmmeter reading to chassis

A reading of

0\,\Omega

from the power lead to chassis ground prior to the loads confirms a short-to-ground fault.

An ohmmeter reading continuity (

0\,\Omega

) to chassis ground before the loads shows an unintended low-resistance return path that bypasses the load network entirely.

Key Concept

Circuit Fault Diagnosis and Multimeter Reading Interpretation in DC Series Circuits

Estimated Time:1m 30s

Question 908Question →

A military supply depot receives a shipment of $400$ uniform jackets. Of these jackets, $25\%$ are size Small, $\frac{3}{8}$ are size Medium, and the remaining jackets are divided equally between size Large and size Extra-Large. If a quartermaster randomly selects one jacket from the entire shipment, what is the probability that it is a size Extra-Large jacket?

\frac{3}{16}

\frac{3}{13}

\frac{1}{3}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{3}{16}$

Answer

\frac{3}{16}

The probability of a single event is the number of favorable outcomes divided by the total number of possible outcomes. First, determine the number of Extra-Large jackets. There are

100

Small jackets (

25\%

of

400

) and

150

Medium jackets (

\frac{3}{8}

of

400

). This leaves

150

jackets (

400 - 100 - 150 = 150

). Since these remaining jackets are divided equally between Large and Extra-Large, there are exactly

75

Extra-Large jackets. The probability of randomly selecting an Extra-Large jacket is the number of Extra-Large jackets (

75

) over the total number of jackets (

400

). Simplifying the fraction

\frac{75}{400}

by dividing the numerator and denominator by

25

yields

\frac{3}{16}

.

Step-by-Step Solution

1

Calculate the number of Small jackets in the shipment.

25\%

of

400

is

100

jackets.

To find the remaining jackets, you first need to determine the exact counts of the known sizes.

2

Calculate the number of Medium jackets in the shipment.

\frac{3}{8}

of

400 = 3 \times 50 = 150

jackets.

Converting the fraction to an exact count is required to subtract from the total.

3

Determine the number of jackets remaining for the Large and Extra-Large sizes.

400 - (100 + 150) = 400 - 250 = 150

remaining jackets.

Subtracting the known sizes from the total gives the combined count of the remaining two sizes.

4

Calculate the number of Extra-Large jackets.

150 \div 2 = 75

Extra-Large jackets.

The problem states the remaining jackets are divided equally between the final two sizes.

5

Calculate the probability of selecting an Extra-Large jacket.

Probability

= \frac{75}{400}

.

Single event probability is the number of favorable outcomes divided by the total number of possible outcomes.

6

Simplify the probability fraction.

\frac{75}{400} = \frac{3}{16}

.

Standardized tests require probability fractions to be in simplest form.

Key Concept

Single Event Probability

Question 909Question →

Electromagnetism and induction are core principles governing how electric circuits interact with magnetic fields. Match each electromagnetic concept with its correct definition or operational description.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electromagnet matches the temporary magnet created by electric current through a coil; Electromagnetic induction matches the generation of voltage from a changing magnetic field; Mutual induction matches voltage induced in a nearby coil by varying current; Magnetic field lines match imaginary loops representing flux direction and strength.

Each electromagnetic term is matched to its precise definition: an electromagnet creates a magnetic field from electric current; electromagnetic induction creates voltage from a changing magnetic field; mutual induction describes two inductively coupled coils; and magnetic field lines visually represent flux patterns.

Step-by-Step Solution

1

Identify the basic construction and function of an electromagnet.

An electromagnet requires electric current flowing through a coil wound on a ferromagnetic core to produce magnetism.

This pairs 'Electromagnet' directly with the definition describing a temporary magnet formed by current through a wire coil around a core.

2

Define the fundamental principle of electromagnetic induction.

Discovered by Faraday, moving a conductor through a magnetic field or exposing it to changing flux creates an induced voltage.

This matches 'Electromagnetic induction' with the generation of voltage across a conductor experiencing changing magnetic flux.

3

Distinguish mutual induction from general induction.

Mutual induction specifically requires interaction between two distinct coils (primary and secondary).

This links 'Mutual induction' to the definition involving a varying current in one coil inducing voltage across a separate nearby coil.

4

Identify the visual concept used for magnetic flux.

Field lines illustrate lines of magnetic force and field intensity.

This pairs 'Magnetic field lines' with the representation of flux strength and direction.

Key Concept

Core terminology and physical principles of electromagnetism and induction in basic electrical systems.

Estimated Time:45s

Question 910Question →

An electrical circuit that transfers $180\text{ coulombs}$ of charge through a conductor over a span of $3\text{ minutes}$ maintains an electric current of $60\text{ amperes}$ .

True

False

Show answer & explanation

Answer: False

Answer

False. The actual current is

1\text{ A}

because time must be expressed in seconds (

3\text{ minutes} = 180\text{ seconds}

) when calculating current in amperes (

I = \frac{Q}{t}

).

The statement is false because the ampere represents charge transfer per second (

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

). A charge of

180\text{ coulombs}

flowing over

3\text{ minutes}

(

180\text{ seconds}

) yields an electric current of exactly

1\text{ ampere}

, not

60\text{ amperes}

.

Step-by-Step Solution

1

Convert the elapsed time from minutes to seconds

3\text{ minutes} \times 60\text{ seconds/minute} = 180\text{ seconds}

The base unit of electric current (ampere) is defined as coulombs per second (

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

).

2

Calculate the electric current using the charge rate formula

I = \frac{Q}{t}

I = \frac{180\text{ C}}{180\text{ s}} = 1\text{ A}

Electric current is the rate of charge flow through a cross-section of a conductor.

3

Compare the calculated value with the statement's value

The calculated current is

1\text{ A}

, whereas the statement asserts

60\text{ A}

.

The statement erroneously divides charge by time in minutes rather than seconds.

Key Concept

Electric current is defined as the rate of charge flow per unit time, where one ampere equals one coulomb per second (

I = \frac{Q}{t}

).

Estimated Time:45s

Question 911Question →

An electronic technician needs to measure the resistance of a resistor installed on a circuit board to verify its value. What is the correct sequence of steps to safely and accurately perform this measurement?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Disconnect power to the circuit

\rightarrow

Isolate at least one lead of the resistor

\rightarrow

Set the multimeter dial to resistance (

\Omega

)

\rightarrow

Place the test probes across the component leads and observe the display.

The correct sequence begins with de-energizing the circuit for safety and instrument protection, isolating at least one lead to eliminate parallel circuit paths, configuring the multimeter to the Ohms function, and finally contacting the leads across the component to observe the reading.

Step-by-Step Solution

1

Ensure all power to the circuit is disconnected.

The circuit is completely de-energized and safe to work on.

Multimeters measure resistance by supplying their own small internal voltage; external power will damage the meter and produce invalid readings.

2

Lift or disconnect at least one lead of the resistor from the circuit board.

The resistor is isolated from other components.

Parallel paths in the circuit network would provide alternative current loops, causing the meter to read lower than the true component value.

3

Turn the rotary switch on the multimeter to the resistance (

\Omega

) range.

The multimeter is configured in ohmmeter mode.

The meter must be in the correct mode before probes are applied to the device under test.

4

Connect the black and red test leads across the resistor terminals.

The resistance value appears on the digital readout.

The meter sends a small internal current through the component, measures the resulting voltage drop, and displays the calculated resistance.

Key Concept

Safely and accurately measuring component resistance requires de-energizing the circuit, isolating the component from parallel paths, selecting the Ohms mode, and applying the test probes.

Estimated Time:45s

Question 912Question →

A technician is troubleshooting a direct-current circuit that suddenly stopped operating. In what order should the technician perform the standard diagnostic steps to locate and resolve the fault safely?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct troubleshooting sequence begins with a visual inspection, followed by verifying supply voltage, then de-energizing the circuit, and finally measuring isolated component resistance.

Systematic troubleshooting follows a logical hierarchy: first perform a visual check for obvious physical damage; second, verify that input voltage is present; third, de-energize the circuit for safety; and fourth, perform resistance checks on isolated components to identify the specific faulty device.

Step-by-Step Solution

1

Inspect the physical condition of the circuit.

Identifies obvious signs of overheating, loose terminals, or mechanical damage without disturbing connections.

Visual checks are fast, safe, and can resolve obvious problems immediately.

2

Measure input voltage while the circuit is connected to power.

Confirms whether proper operational voltage reaches the circuit input terminals.

If supply voltage is missing, troubleshooting must shift to the upstream power source rather than the circuit itself.

3

Shut off and isolate the power supply.

Removes all live electrical energy from the circuit.

Safety protocol requires de-energizing the circuit before testing resistance or removing parts.

4

Use an ohmmeter to measure continuity and resistance across individual components.

Pinpoints open coils, shorted semiconductor junctions, or out-of-tolerance resistors.

Ohmmeters supply their own internal battery voltage and will be damaged if used on an active, energized circuit.

Key Concept

Standard systematic troubleshooting progresses from non-invasive visual inspection to live voltage verification, followed by complete power isolation before conducting ohmmeter resistance measurements.

Estimated Time:45s

Question 913Question →

A semiconductor diode is a fundamental component that controls current flow in electronic circuits. Match each diode term on the left with its corresponding characteristic or operational definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The Anode pairs with the P-type terminal represented by the triangle base, the Cathode pairs with the N-type terminal identified by the painted stripe, Forward bias pairs with the state allowing conduction when the P-type region is positive, and Reverse bias pairs with the state blocking conduction when the N-type region is positive.

A semiconductor diode consists of an anode (P-type side, represented by the base of the triangle symbol) and a cathode (N-type side, indicated by a band on the physical package and a vertical bar on the symbol). Connecting the positive supply to the anode creates forward bias, collapsing the depletion layer and allowing current conduction. Connecting the positive supply to the cathode creates reverse bias, expanding the depletion layer and insulating the circuit.

Step-by-Step Solution

1

Identify the two terminals of a PN junction diode

The anode represents the P-type material (triangle base on schematics), while the cathode represents the N-type material (vertical line on schematics and painted ring on physical components).

Correctly identifying component leads is essential for analyzing circuit orientation and polarity.

2

Determine the operating conditions for current conduction and cutoff

Applying positive voltage to the anode creates forward bias and allows conduction; applying positive voltage to the cathode creates reverse bias and halts conduction.

Forward bias shrinks the internal depletion layer, whereas reverse bias widens it to prevent electron movement.

3

Match each term to its exact operational and physical descriptor

All four terms are unambiguously paired with their respective terminal characteristics and biasing behaviors.

Verifies complete conceptual understanding of diode fundamentals.

Key Concept

Diode terminal identification (anode vs. cathode) and junction biasing states (forward vs. reverse bias)

Estimated Time:45s

Question 914Question →

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

Show answer & explanation

Answer: 0.6

Answer

0.6 Ω

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

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

). Since

20\text{ m}

is one-fourth (

\frac{1}{4}

or

0.25

) of the original

80\text{ m}

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

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

.

Step-by-Step Solution

1

Identify the physical relationship governing wire resistance and length.

Resistance is directly proportional to length:

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

, which means

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

.

Because the material (resistivity

\rho

) and cross-sectional area (

A

) remain constant, resistance scales directly with length.

2

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

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

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

3

Calculate the resistance of the 20-meter wire segment.

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

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

Key Concept

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

R \propto L

) when cross-sectional area and material remain constant.

Estimated Time:45s

Question 915Question →

In an alternating current (AC) generator, increasing the rotational speed of the armature coil within a constant magnetic field increases the output frequency but has no effect on the magnitude of the peak induced electromotive force (EMF).

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because induced electromotive force depends directly on how fast the magnetic field lines are cut. Spinning the armature coil faster increases both the frequency of the alternating current and the peak induced voltage.

Step-by-Step Solution

1

Identify the relationship between rotational speed and AC frequency in a generator.

Higher rotational speed increases the number of electrical cycles completed per second, directly increasing frequency (

f

).

Each mechanical revolution corresponds to a fixed number of electrical cycles based on the generator's pole pairs.

2

Apply Faraday's Law of Electromagnetic Induction to determine how rotational speed affects induced voltage amplitude.

Faster rotation increases the angular velocity (

\omega

), causing magnetic flux lines to be cut more rapidly over time (

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

).

Faraday's Law states that induced electromotive force is directly proportional to the rate of change of magnetic flux (

E_{\text{peak}} = N B A \omega

).

3

Evaluate the statement based on the combined physical effects.

Since rotational speed increases both output frequency and peak induced EMF, the claim that peak EMF remains unaffected is incorrect.

A generator produces higher peak voltage when spun faster because conductor velocity through the magnetic field increases.

Key Concept

Faraday's Law of Electromagnetic Induction in AC Generation

Estimated Time:1m 0s

Question 916Question →

A $60\text{ V}$ direct-current power supply feeds a circuit containing three parallel branches. The total current drawn from the supply is $10\text{ A}$ . If the first two branches have resistances of $20\ \Omega$ and $30\ \Omega$ , what is the resistance of the third branch?

6\ \Omega

12\ \Omega

18\ \Omega

50\ \Omega

Show answer & explanation

Answer: $12\ \Omega$

Answer

The resistance of the third branch is

12\ \Omega

.

Because the circuit is wired in parallel, the full supply voltage of

60\text{ V}

is applied across each branch. Using Ohm's Law (

I = V / R

), the first branch draws

60\text{ V} / 20\ \Omega = 3\text{ A}

and the second branch draws

60\text{ V} / 30\ \Omega = 2\text{ A}

. The sum of these two branch currents is

5\text{ A}

. Since the total source current is

10\text{ A}

, the third branch must carry the remaining

10\text{ A} - 5\text{ A} = 5\text{ A}

. Applying Ohm's Law to the third branch yields

R_3 = 60\text{ V} / 5\text{ A} = 12\ \Omega

.

Step-by-Step Solution

1

Determine the voltage across each parallel branch.

Each branch receives the full source voltage of

60\text{ V}

.

In a parallel circuit, the voltage across every parallel branch is identical to the source voltage.

2

Calculate the individual currents flowing through the first two known branches using Ohm's Law (

I = V / R

).

I_1 = 60\text{ V} / 20\ \Omega = 3\text{ A}

and

I_2 = 60\text{ V} / 30\ \Omega = 2\text{ A}

.

Branch currents must be determined to find how much of the total current is accounted for by the known branches.

3

Find the current flowing through the third branch by subtracting the known branch currents from the total supply current (

I_3 = I_{\text{total}} - I_1 - I_2

).

I_3 = 10\text{ A} - (3\text{ A} + 2\text{ A}) = 5\text{ A}

.

Total current in a parallel circuit equals the sum of the individual branch currents.

4

Calculate the resistance of the third branch using Ohm's Law (

R_3 = V / I_3

).

R_3 = 60\text{ V} / 5\text{ A} = 12\ \Omega

.

Applying Ohm's Law to the third branch gives its required resistance value.

Key Concept

In a parallel circuit, voltage remains constant across all branches, the total current equals the sum of all individual branch currents (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

), and each branch resistance satisfies Ohm's Law (

R_n = V / I_n

).

Estimated Time:1m 30s

Question 917Question →

Analyze the series circuit parameters under normal operation and after a short-circuit fault by filling in the correct numerical values (use decimal format without units, e.g., '0.5' or '10').

Fill in the blanks below

An aviation warning beacon is powered by a

120\text{ V}

DC supply connected in series with three resistive components:

R_1 = 100\ \Omega

,

R_2 = 150\ \Omega

, and

R_3 = 50\ \Omega

. Under normal operation, the total current flowing through the circuit is A and the voltage drop across

R_1

is V. If component

R_2

experiences a catastrophic failure and completely shorts out (its internal resistance drops to

0\ \Omega

), the equivalent resistance of the entire circuit drops to

\Omega

, causing the voltage drop across

R_3

to become V.

Show answer & explanation

Answer

Under normal operation, the current is 0.4 A, and the voltage drop across R1 is 40 V. After R2 shorts out, the total resistance drops to 150 ohms, and the increased current causes the voltage drop across R3 to become 40 V.

In a series circuit, total resistance is the linear sum of all resistive components. Current remains uniform throughout the entire loop and is determined by the total voltage divided by the total resistance. When a component fails via a short circuit, it effectively acts as a direct wire (zero ohms). This absence of resistance lowers the total circuit resistance, which consequently forces a higher current to flow from the constant voltage source. The increased current then results in proportionately higher voltage drops across the surviving resistors.

Step-by-Step Solution

1

Calculate the total equivalent resistance under normal operating conditions.

R_{total} = R_1 + R_2 + R_3 = 100 + 150 + 50 = 300\ \Omega

In a series circuit, the total resistance is the sum of all individual resistances.

2

Use Ohm's Law to find the total current flowing through the normal circuit.

I_{total} = \frac{V_{total}}{R_{total}} = \frac{120\text{ V}}{300\ \Omega} = 0.4\text{ A}

(This is blank 1).

Total current is equal to the source voltage divided by the total equivalent resistance.

3

Calculate the voltage drop across

R_1

under normal conditions.

V_{R1} = I_{total} \times R_1 = 0.4\text{ A} \times 100\ \Omega = 40\text{ V}

(This is blank 2).

The voltage drop across any single component in a series circuit is the total current multiplied by that component's resistance.

4

Determine the new total resistance after

R_2

shorts out.

R_{new\_total} = 100 + 0 + 50 = 150\ \Omega

(This is blank 3).

A shorted component has

0\ \Omega

resistance. The circuit remains closed, but the total resistance decreases.

5

Calculate the new total current and the resulting new voltage drop across

R_3

.

I_{new} = \frac{120\text{ V}}{150\ \Omega} = 0.8\text{ A}

. Then,

V_{R3} = 0.8\text{ A} \times 50\ \Omega = 40\text{ V}

(This is blank 4).

The lowered total resistance causes a higher total current to flow from the same voltage source, which proportionally increases the voltage drop across the remaining functional resistors.

Key Concept

Series Circuit Analysis and Component Fault (Short Circuit) Behavior

Question 918Question →

A structural mechanic is testing the tensile strength of a batch of aluminum rivets. The mean tensile strength of the first $4$ rivets tested is $410$ MPa. If the quality standard requires the mean strength of $5$ rivets to be exactly $415$ MPa, what tensile strength must the $5$ th rivet have?

Show answer & explanation

Answer: 435

Answer

The 5th rivet must have a tensile strength of 435 MPa.

The correct answer is derived by finding the total required sum for 5 rivets (2075) and subtracting the existing sum of the first 4 rivets (1640).

Step-by-Step Solution

1

Calculate the total tensile strength of the first 4 rivets.

1640 MPa

The sum of a data set is equal to its mean multiplied by the number of values.

2

Calculate the required total tensile strength for all 5 rivets.

2075 MPa

To achieve a target mean of 415 MPa with 5 items, the overall sum must equal the new mean multiplied by the new count.

3

Find the difference between the target total and the current total.

435 MPa

Subtracting the sum of the first 4 items from the required sum of all 5 items isolates the value needed for the 5th item.

Key Concept

Calculating a missing data value given an initial mean and a target mean.

Alternative Method

Use the deviation method: The first 4 rivets have a mean of 410 MPa, which is 5 MPa below the target mean of 415 MPa. This creates a total deficit of 4 × 5 = 20 MPa. The 5th rivet must provide enough strength to cover its own requirement (415 MPa) plus the 20 MPa deficit from the others. Therefore, 415 + 20 = 435 MPa.

Estimated Time:1m 0s

Question 919Question →

A DC cooling fan installed in an equipment enclosure has an internal resistance of $16\ \Omega$ and draws a steady current of $1.5\text{ A}$ during normal operation. According to Ohm's law, what is the voltage drop across the fan in volts ( $\text{V}$ )?

Show answer & explanation

Answer: 24

Answer

24 V

According to Ohm's law, voltage (

V

) is found by multiplying current (

I

) by resistance (

R

). With a current of

1.5\text{ A}

and a resistance of

16\ \Omega

, the voltage is

1.5 \times 16 = 24\text{ V}

.

Step-by-Step Solution

1

Identify the known electrical quantities from the problem statement.

I = 1.5\text{ A}

and

R = 16\ \Omega

Ohm's law requires current and resistance values to calculate voltage.

2

State Ohm's law formula arranged to solve for voltage (

V

).

V = I \times R

The potential difference across a resistive element is directly proportional to the product of current and resistance.

3

Multiply the current by the resistance to find the voltage.

V = 1.5\text{ A} \times 16\ \Omega = 24\text{ V}

Performing the multiplication yields the potential difference in volts.

Key Concept

Ohm's Law (

V = I \times R

)

Estimated Time:30s

Question 920Question →

A double-pole, single-throw (DPST) switch is designed to simultaneously open or close two electrically independent circuits using a single mechanical actuator.

True

False

Show answer & explanation

Answer: True

Answer

True. A DPST switch controls two isolated circuit paths simultaneously with a single actuator mechanism.

The statement accurately defines a DPST switch. In electrical engineering and electronics, 'pole' refers to the number of separate circuits controlled by the switch, and 'throw' refers to the number of closed contact positions available per pole. A DPST switch therefore controls two separate circuits with a single ON/OFF throw action.

Step-by-Step Solution

1

Analyze the switch terminology for poles and throws.

'Double-pole' (DP) signifies two separate, electrically isolated circuits or lines being controlled, while 'single-throw' (ST) indicates that each pole has only one active closed position (ON/OFF).

Understanding standard electrical switch nomenclature is necessary to identify contact arrangements.

2

Evaluate the mechanical and electrical operation of the switch.

The two internal sets of contacts move in unison via one toggle or lever, switching both independent circuits at the same instant.

A single mechanical linkage operates both poles together to ensure simultaneous opening or closing.

Key Concept

Switch Pole and Throw Nomenclature (DPST Operation)

Estimated Time:45s

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