All practice questions

Question 681Question →

An electronics technician is analyzing solid-state circuit topologies for an audio and radio frequency transceiver. Match each transistor circuit configuration or operational mode with its defining electrical characteristic.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Common Emitter Configuration matches with significant voltage/current gain and

180^\circ

phase inversion; Common Collector Configuration matches with high input impedance, low output impedance, and near-unity voltage gain; Common Base Configuration matches with low input impedance, high voltage gain, and no phase inversion for high-frequency RF; Transistor Switch Mode matches with alternating between cutoff and saturation to control load power.

Each transistor configuration serves a specific role determined by which terminal is common to both input and output: Common Emitter is the primary inverting voltage/current amplifier; Common Collector serves as a non-inverting impedance buffer with unity voltage gain; Common Base provides low input impedance and high-frequency stability without phase shift; and Switch Mode drives the transistor between saturation and cutoff for binary power control.

Step-by-Step Solution

1

Identify the phase relationship and gain characteristics of the Common Emitter amplifier.

The Common Emitter takes input at the base and extracts output at the collector, yielding voltage gain, current gain, and a

180^\circ

phase inversion.

This is the signature standard amplifier configuration in bipolar transistor circuitry.

2

Analyze the impedance transformation and voltage gain of the Common Collector configuration.

The Common Collector takes input at the base and extracts output at the emitter, yielding a voltage gain of slightly less than 1 with high input impedance and low output impedance.

This configuration acts as a buffer or impedance matching stage, commonly termed an emitter follower.

3

Identify the high-frequency and impedance properties of the Common Base configuration.

The Common Base takes input at the emitter and output at the collector with the base grounded for AC signals, offering low input impedance and no signal inversion.

The grounded base shields input from output, mitigating the Miller effect in RF circuits.

4

Distinguish linear amplification from digital switching operation.

Switching mode avoids the active linear region and toggles strictly between cutoff (

I_C = 0

) and saturation (

V_{CE} \approx 0.2\text{ V}

).

Digital control and relay driving rely on binary on/off states rather than linear proportional amplification.

Key Concept

Transistor Circuit Configurations and Operating Modes

Estimated Time:1m 0s

Question 682Question →

Match each electrical power source or practical application on the left with its corresponding current characteristic and operational principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair the high-voltage transmission grid with AC transformer voltage stepping, the automotive lead-acid battery with unidirectional DC from chemical reactions, the residential wall receptacle with AC reversing at

60\text{ Hz}

, and the electroplating tank with unidirectional DC for steady metal ion deposition.

Direct current (DC) flows in one unchanging direction with steady polarity, which is produced by chemical batteries and required for electroplating. Alternating current (AC) periodically alternates its direction of flow, enabling transformers to step voltage levels efficiently for transmission and supplying standard

60\text{ Hz}

household power.

Step-by-Step Solution

1

Identify AC-dependent applications and their physical mechanisms

Recognize that power distribution grids and residential outlets utilize alternating current (AC). Grids depend on AC for transformer stepping, while outlets deliver cyclical

60\text{ Hz}

polarity reversals.

Transformers require time-varying current to induce secondary voltage, and utility generators produce sinusoidal alternating current.

2

Identify DC-dependent applications and their physical mechanisms

Recognize that lead-acid batteries and electroplating tanks operate on direct current (DC). Batteries generate unidirectional current chemically, while electroplating requires steady polarity to maintain ion migration.

Chemical cells produce static polarities, and reversing current in electroplating would dissolve rather than plate metal.

3

Match each application to its unique operational principle

Pair grid transmission to transformer stepping, battery to chemical unidirectional DC, wall outlet to

60\text{ Hz}

AC reversal, and electroplating to continuous ionic migration.

Each assignment links the fundamental behavior of alternating or direct current to its specific operational requirement.

Key Concept

Direct current (DC) maintains constant polarity and unidirectional electron flow, making it ideal for electrochemical storage (batteries) and electrodeposition. Alternating current (AC) reverses polarity periodically, enabling electromagnetic induction in transformers for long-distance power distribution and standard domestic utility delivery.

Estimated Time:1m 0s

Question 683Question →

During the compression stroke of a standard four-stroke internal combustion engine, what are the positions of the intake and exhaust valves, and what happens to the trapped air-fuel charge as the piston moves upward?

The intake valve stays open to regulate cylinder pressure while the air-fuel mixture is partially vented.

Both valves remain tightly closed while the piston compresses the air-fuel mixture into the combustion chamber.

The exhaust valve opens briefly to release residual exhaust gases as the air-fuel mixture expands.

Both valves stay open simultaneously to allow a richer fuel mixture to enter before ignition occurs.

Show answer & explanation

Answer: Both valves remain tightly closed while the piston compresses the air-fuel mixture into the combustion chamber.

Answer

Both valves remain tightly closed while the piston compresses the air-fuel mixture into the combustion chamber.

During the compression stroke of a four-stroke internal combustion engine, the crankshaft rotates to push the piston upward from Bottom Dead Center (BDC) toward Top Dead Center (TDC). To compress the trapped air-fuel mixture into the combustion chamber at the top of the cylinder, both the intake valve and exhaust valve must remain fully closed to maintain an airtight seal.

Step-by-Step Solution

1

Identify the four-stroke cycle phase described in the question.

The phase is the compression stroke, which occurs immediately after the intake stroke and immediately before the power stroke.

Determining the specific stroke defines the mechanical motion of the piston and the synchronized state of the valvetrain.

2

Determine the required valve positions for compression.

Both the intake valve and the exhaust valve must be closed and firmly seated against the cylinder head.

A sealed cylinder chamber is necessary to prevent the escape of the air-fuel charge and allow cylinder pressure to build as the piston moves upward from bottom dead center (BDC) to top dead center (TDC).

Key Concept

Four-stroke internal combustion engine cycle events and cylinder sealing during compression.

Estimated Time:45s

Question 684Question →

Inside a standard capacitor, what is the primary role of the dielectric material placed between the two conductive plates?

To conduct direct current continuously between the plates to complete the circuit

To produce an electromagnetic field that resists sudden changes in current

To prevent physical contact between the plates while supporting an electrostatic field

To convert incoming direct current directly into alternating current

Show answer & explanation

Answer: To prevent physical contact between the plates while supporting an electrostatic field

Answer

To prevent physical contact between the plates while supporting an electrostatic field

The correct answer accurately states that the dielectric serves as an insulating barrier that prevents the conductive plates from touching (which would cause a short circuit) while enabling an electrostatic field to establish between the accumulated charges.

Step-by-Step Solution

1

Identify the basic physical structure of a capacitor

A capacitor consists of two conducting plates separated by an insulating layer called a dielectric.

Understanding the physical layout helps identify the specific function of each layer.

2

Determine the electrical properties and purpose of the dielectric

The dielectric material is a non-conductive insulator that blocks direct electron flow between the plates, allowing positive and negative charges to accumulate and establish an electrostatic field.

Without an insulating dielectric, the plates would touch or conduct, creating a short circuit instead of storing energy.

Key Concept

Function and Properties of Capacitor Dielectric Materials

Estimated Time:45s

Question 685Question →

An industrial control panel supplies a constant $24\text{ V}$ DC to a sensor circuit. During normal operation, the sensor draws a current of $16\text{ mA}$ . Following a component degradation, the measured current rises to $64\text{ mA}$ at the same operating voltage. By how many ohms ( $\Omega$ ) has the circuit's total resistance decreased?

Show answer & explanation

Answer: 1125

Answer

The circuit's total resistance decreased by

1125\text{ }\Omega

.

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

R = \frac{V}{I}

). Converting the currents to amperes gives

0.016\text{ A}

and

0.064\text{ A}

. The initial resistance is

R_1 = \frac{24\text{ V}}{0.016\text{ A}} = 1500\text{ }\Omega

, and the final resistance is

R_2 = \frac{24\text{ V}}{0.064\text{ A}} = 375\text{ }\Omega

. The decrease in resistance is

1500\text{ }\Omega - 375\text{ }\Omega = 1125\text{ }\Omega

.

Step-by-Step Solution

1

Convert current quantities from milliamperes (mA) to amperes (A)

I_1 = 0.016\text{ A}

and

I_2 = 0.064\text{ A}

Ohm's law requires current to be expressed in standard SI units (amperes) to yield resistance in ohms.

2

Calculate the initial resistance using Ohm's law (

R = \frac{V}{I}

)

R_1 = \frac{24\text{ V}}{0.016\text{ A}} = 1500\text{ }\Omega

Determines the baseline operating resistance of the sensor module.

3

Calculate the degraded state resistance using Ohm's law (

R = \frac{V}{I}

)

R_2 = \frac{24\text{ V}}{0.064\text{ A}} = 375\text{ }\Omega

Determines the lower operating resistance corresponding to the increased current draw.

4

Calculate the net difference between initial and degraded resistance

\Delta R = 1500\text{ }\Omega - 375\text{ }\Omega = 1125\text{ }\Omega

Finds the exact reduction in total circuit resistance.

Key Concept

Ohm's Law (

R = \frac{V}{I}

) applied to determine resistance changes in DC circuits with metric prefix conversion.

Estimated Time:1m 30s

Question 686Question →

An ammeter connected in series with a direct-current (DC) circuit registers a steady current of $250\text{ mA}$ . Which of the following statements correctly describes both the rate of electric charge transfer through the conductor and the physical direction of electron flow?

A charge of

0.25\text{ coulombs}

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

The electromotive force drives

4\text{ coulombs}

of energy each second, with electron flow maintaining constant potential throughout the conductor.

A charge of

0.25\text{ coulombs}

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

A charge of

250\text{ coulombs}

alternates direction each second, causing electrons to oscillate back and forth without net directional movement.

Show answer & explanation

Answer: A charge of $0.25\text{ coulombs}$ passes through the cross-section each second, with electrons drifting from the negative terminal toward the positive terminal.

Answer

A charge of

0.25\text{ coulombs}

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

One ampere equals one coulomb of electrical charge passing a point per second (

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

). Therefore, a direct current of

250\text{ mA}

(

0.25\text{ A}

) represents an electric charge transfer rate of

0.25\text{ coulombs per second}

. Furthermore, because electrons are negatively charged particles, electrostatic repulsion and attraction cause them to drift from the negative terminal of the DC source, through the external circuit, and into the positive terminal.

Step-by-Step Solution

1

Convert the current measurement from milliamperes (

\text{mA}

) to standard SI units of amperes (

\text{A}

).

250\text{ mA} = 250 \times 10^{-3}\text{ A} = 0.25\text{ A}

The ampere is the standard unit of current where

1\text{ A} = 1000\text{ mA}

.

2

Relate the current in amperes to the rate of electric charge flow using the fundamental formula

I = \frac{Q}{t}

.

0.25\text{ A} = 0.25\text{ coulombs per second}

One ampere is defined strictly as the flow of one coulomb of electrical charge per second.

3

Determine the physical direction of electron flow in the external circuit.

Electrons flow from the negative terminal to the positive terminal.

Electrons carry a negative charge and are repelled by the negative terminal and attracted toward the positive terminal of the DC power source.

Key Concept

Electric current is the time rate of charge flow (

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

), and physical electron flow travels from negative to positive potential in a DC circuit.

Alternative Method

Calculate the total time required for a standard unit charge (

1\text{ C}

) to pass:

t = \frac{Q}{I} = \frac{1\text{ C}}{0.25\text{ A}} = 4\text{ seconds}

. This verifies that each second carries

\frac{1}{4}\text{ C} = 0.25\text{ C}

.

Estimated Time:45s

Question 687Question →

In a direct current (DC) electronic device, two resistors with values of $30\ \Omega$ and $60\ \Omega$ are connected in series. What is the total equivalent resistance of this circuit?

90\ \Omega

20\ \Omega

30\ \Omega

1800\ \Omega

Show answer & explanation

Answer: $90\ \Omega$

Answer

90\ \Omega

The total resistance of a series circuit is the sum of the individual component resistances (

R_T = R_1 + R_2

). Adding

30\ \Omega

and

60\ \Omega

gives

90\ \Omega

.

Step-by-Step Solution

1

Identify the formula for total resistance in a series circuit.

R_T = R_1 + R_2

In a series circuit, current flows through each resistor sequentially, meaning total opposition to current is the sum of all individual resistances.

2

Substitute the given resistor values into the series formula.

R_T = 30\ \Omega + 60\ \Omega = 90\ \Omega

Adding the individual resistor values produces the total equivalent resistance.

Key Concept

Total resistance in a series circuit equals the sum of all individual resistances (

R_T = R_1 + R_2 + \dots + R_n

).

Estimated Time:30s

Question 688Question →

A handheld emergency flashlight is powered by four $1.5\text{ V}$ alkaline cells connected in series aiding. What is the total electromotive force (EMF) supplied to the flashlight bulb?

1.5\text{ V}

6.0\text{ V}

6.0\text{ A}

0.375\text{ V}

Show answer & explanation

Answer: $6.0\text{ V}$

Answer

6.0\text{ V}

When DC voltage sources are connected in series aiding (positive terminal to negative terminal), their individual voltages add together directly. Four

1.5\text{ V}

cells in series provide a total electromotive force of

4 \times 1.5\text{ V} = 6.0\text{ V}

.

Step-by-Step Solution

1

Identify the circuit configuration and the formula for total EMF in series.

E_{\text{total}} = E_1 + E_2 + E_3 + E_4

When electrical power sources (cells) are wired in series aiding, their individual voltages add directly together to produce the total potential difference.

2

Sum the voltages of the four identical

1.5\text{ V}

cells.

E_{\text{total}} = 1.5\text{ V} + 1.5\text{ V} + 1.5\text{ V} + 1.5\text{ V} = 6.0\text{ V}

Multiplying the cell voltage by the total number of cells (

4 \times 1.5\text{ V}

) yields the total electromotive force.

Key Concept

Series Cell Electromotive Force (EMF) Addition

Estimated Time:45s

Question 689Question →

A technician is preparing to service a removed MacPherson strut assembly to replace a worn hydraulic damper unit. In what sequential order should the technician perform the disassembly process to ensure mechanical integrity and workshop safety?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Mount the complete strut assembly securely in a fixture, compress the coil spring until tension is relieved from the upper seat, remove the center upper piston rod locknut, and lift off the upper mount and compressed spring components from the damper body.

The correct sequence begins with securing the strut assembly in a holding fixture to ensure stability. Next, the coil spring must be compressed using a specialized spring compressor to neutralize the substantial preload energy acting on the upper spring seat. Only after all spring force is relieved from the top plate can the center damper piston rod locknut be safely unthreaded. Finally, with the fastener removed, the upper mount, bearing assembly, and compressed spring can be lifted off the damper body.

Step-by-Step Solution

1

Secure the strut assembly in a dedicated holding fixture.

The strut is firmly anchored, preventing unexpected movement during heavy tool usage.

Safe disassembly requires stabilization to prevent slipping when torque and compression tools are applied.

2

Apply a coil spring compressor and tighten evenly until the spring is free from the upper mount.

Stored mechanical spring force is contained within the tool rather than acting against the upper mount.

MacPherson strut springs are under heavy preload; removing the fastener while under tension can cause severe injury.

3

Loosen and remove the top piston rod retaining nut while preventing rod rotation.

The mechanical attachment between the upper mounting plate and the damper shaft is disconnected.

The retaining nut holds the top seat to the damper shaft; loosening it without tension on the seat allows safe removal.

4

Disassemble the upper mount, bearing assembly, dust boot, and compressed spring from the strut body.

The damper unit is fully isolated and ready for replacement or inspection.

With the locknut removed, all upper hardware slides freely off the piston rod.

Key Concept

MacPherson Strut Safety and Disassembly Procedure

Estimated Time:1m 30s

Question 690Question →

A portable emergency radio beacon uses a power source consisting of three series-connected DC cells: a primary lithium cell with an electromotive force (EMF) of $3.6\text{ V}$ , a standard cell with an EMF of $1.5\text{ V}$ , and a photovoltaic booster cell with an EMF of $0.9\text{ V}$ . When all three cells are wired in a series-aiding configuration, what is the total electromotive force supplied to the beacon?

3.6\text{ V}

6.0\text{ V}

6.0\text{ A}

1.2\text{ V}

Show answer & explanation

Answer: $6.0\text{ V}$

Answer

The total electromotive force supplied to the beacon is

6.0\text{ V}

.

When multiple direct-current sources are connected in a series-aiding configuration, their individual electromotive forces (EMFs) add algebraically. Summing the three cell ratings gives

3.6\text{ V} + 1.5\text{ V} + 0.9\text{ V} = 6.0\text{ V}

.

Step-by-Step Solution

1

Identify the individual electromotive force (EMF) values of the series-connected sources.

E_1 = 3.6\text{ V}

,

E_2 = 1.5\text{ V}

, and

E_3 = 0.9\text{ V}

.

Extracting the component voltages is necessary to determine the overall potential difference across the power source.

2

Apply the rule for DC voltage sources connected in series-aiding polarity (

E_{\text{total}} = E_1 + E_2 + E_3

).

E_{\text{total}} = 3.6\text{ V} + 1.5\text{ V} + 0.9\text{ V} = 6.0\text{ V}

.

When cells are wired series-aiding (positive terminal of one cell connected to the negative terminal of the next), their individual electromotive forces combine additively.

Key Concept

Series-Aiding Electromotive Force (EMF) Summation

Estimated Time:1m 0s

Question 691Question →

A technician is evaluating a light truck's rear solid axle suspension system. With a static 1,000-pound payload placed directly over the rear axle, the vehicle's rear ride height sags severely, dropping three inches below the manufacturer's specified minimum loaded height. However, during a subsequent road test over a series of sharp pavement transitions, the chassis stabilizes immediately after each initial impact without exhibiting any continuous cyclical rebounding. Based on this combination of symptoms, which of the following is the most accurate diagnostic conclusion?

The rear leaf springs have fatigued and lost their spring rate, failing to support the vehicle's static load.

The internal piston seals in the rear shock absorbers have bypassed, eliminating their ability to hold up the payload.

The rear sway bar has lost its torsional rigidity, causing the static weight to collapse the suspension.

The jounce bumpers have deteriorated entirely, removing the primary load-bearing component of the rear axle.

Show answer & explanation

Answer: The rear leaf springs have fatigued and lost their spring rate, failing to support the vehicle's static load.

Answer

The diagnostic conclusion is that the leaf springs are fatigued and failing to support the static load, while the shock absorbers remain functional.

The correct answer accurately identifies that springs dictate static ride height and load capacity. The symptom of extreme sag points directly to fatigued springs. Furthermore, the absence of continuous bouncing after hitting bumps confirms that the shock absorbers are successfully performing their distinct job of damping oscillation, definitively ruling them out as the culprit.

Step-by-Step Solution

1

Analyze the first symptom: severe static sag under a 1,000-pound payload.

This indicates a failure in the primary load-bearing components of the suspension.

The springs (coil, leaf, or torsion bar) are the only components responsible for maintaining ride height and supporting the physical weight of the vehicle and payload.

2

Analyze the second symptom: the chassis stabilizes immediately after impacts without cyclical rebounding.

This indicates that the damping components of the suspension are functioning properly.

Shock absorbers (dampers) convert the kinetic energy of the moving springs into thermal energy. If they were faulty, the vehicle would bounce up and down multiple times after a bump.

3

Synthesize the findings to isolate the failed component.

The distinct separation of symptoms points exclusively to spring fatigue, ruling out shock absorber failure.

Isolating the function of energy storage/load capacity from energy dissipation allows for precise mechanical diagnosis without replacing unnecessary parts.

Key Concept

Differentiating the distinct mechanical roles of springs (load support and energy storage) versus shock absorbers (oscillation damping and energy dissipation).

Estimated Time:2m 0s

Question 692Question →

Match each electrical protective device to its primary operating characteristic and designated protection function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fast-acting fuse matches with rapid semiconductor protection; Slow-blow fuse matches with absorbing motor startup inrush currents; Metal-Oxide Varistor matches with clamping transient overvoltage surges; Thermal-magnetic circuit breaker matches with combining a bimetallic strip and electromagnetic coil.

Each protective component is matched to its specific physical mechanism and purpose: fast-acting fuses open immediately to protect solid-state devices; slow-blow fuses incorporate intentional delay for inductive motor inrush; MOVs clamp transient overvoltage spikes via non-linear resistance; and thermal-magnetic breakers utilize dual thermal and magnetic elements for coordinated overload and short-circuit clearance.

Step-by-Step Solution

1

Identify the protective role and operating speed of the fast-acting fuse.

The fast-acting fuse clears within milliseconds under overcurrent conditions.

Solid-state semiconductors lack the thermal mass to survive prolonged overcurrent, necessitating instantaneous circuit clearing.

2

Determine the operational necessity of the slow-blow (time-delay) fuse.

The slow-blow fuse tolerates transient current spikes without opening.

Inductive loads like motors pull substantial inrush current during initial startup, which would cause nuisance blowing in standard fuses.

3

Differentiate the voltage-clamping behavior of a Metal-Oxide Varistor (MOV) from overcurrent devices.

The MOV decreases its internal resistance when voltage exceeds a specific threshold.

Overvoltage protection requires shunting excess transient voltage away from downstream circuitry in parallel rather than breaking series current.

4

Analyze the dual-tripping mechanisms of a thermal-magnetic breaker.

It pairs a slow bimetal deflection mechanism with a fast electromagnetic solenoid.

This provides inverse-time response for minor thermal overloads while ensuring instantaneous clearing of dead short circuits.

Key Concept

Operating principles, response times, and application criteria of overcurrent and overvoltage protective devices.

Estimated Time:1m 30s

Question 693Question →

Electronics technicians must select the proper capacitor type for specific circuit requirements. Match each capacitor type on the left with its characteristic construction or typical circuit application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ceramic Disc Capacitor pairs with non-polarized low-capacitance RF bypassing; Aluminum Electrolytic Capacitor pairs with polarized high-capacitance power supply filtering; Tantalum Capacitor pairs with polarized low-leakage precision low-voltage applications; Trimmer (Variable) Capacitor pairs with adjustable capacitance for tuning resonant circuits.

Each capacitor type is tailored to specific circuit roles based on dielectric material, capacitance range, polarity, and frequency response: ceramic disc for high-frequency bypassing, aluminum electrolytic for bulk power smoothing, tantalum for stable low-leakage applications, and trimmer capacitors for variable circuit calibration.

Step-by-Step Solution

1

Identify the non-polarized capacitor designed for high frequencies.

Ceramic disc capacitors have low capacitance and excellent high-frequency response, matching the description for RF bypassing.

Ceramic dielectrics maintain stability at high frequencies without polarity constraints.

2

Identify the high-capacity polarized capacitor used in bulk power filtering.

Aluminum electrolytic capacitors provide the large microfarad values necessary to smooth DC ripple in rectifiers.

An electrochemical oxide layer allows very high capacitance per unit volume, but requires strict polarity observation.

3

Identify the polarized capacitor known for stability, low leakage, and high density.

Tantalum capacitors provide high capacitance in small packages with minimal leakage current for precision circuits.

Tantalum pentoxide dielectric layers are extremely thin and uniform compared to standard liquid electrolytes.

4

Identify the adjustable capacitor used for calibration.

Trimmer capacitors are variable capacitors used to tune resonant and RF circuits.

Movable plates or screw-adjusted dielectric spacing allow fine calibration of the total capacitance.

Key Concept

Classification, dielectric characteristics, polarity, and circuit roles of different capacitor types.

Estimated Time:1m 0s

Question 694Question →

In a field-effect transistor (FET), which terminal serves as the control element that regulates the current passing between the source and the drain?

Base

Gate

Collector

Cathode

Show answer & explanation

Answer: Gate

Answer

Gate

In a field-effect transistor (FET), the gate terminal applies an electric field that alters the conductivity of the semiconductor channel, thereby regulating the flow of current between the source and the drain.

Step-by-Step Solution

1

Identify the transistor family specified in the question.

The device is a field-effect transistor (FET), which utilizes Gate, Source, and Drain terminals.

Field-effect transistors use different terminal designations and operational principles compared to bipolar junction transistors.

2

Determine the functional role of each FET terminal.

Current enters at the source and leaves through the drain, while voltage applied to the gate controls channel conductivity.

The gate terminal acts as the primary control electrode that modulates current flow between the source and drain.

Key Concept

Field-Effect Transistor (FET) Terminal Designations and Roles

Estimated Time:45s

Question 695Question →

An iron-core step-down transformer connected to a $240\text{ V AC}$ power line has $1{}200\text{ turns}$ in its primary winding and $150\text{ turns}$ in its secondary winding. The secondary winding is connected across a $15\ \Omega$ resistive load. Assuming an ideal transformer with no power losses, what is the current drawn by the primary winding from the AC source?

16.0\text{ A}

0.25\text{ A}

2.00\text{ A}

0.025\text{ A}

Show answer & explanation

Answer: $0.25\text{ A}$

Answer

0.25\text{ A}

The transformer has an

8:1

step-down turns ratio (

1{}200 / 150 = 8

). The secondary voltage is

240\text{ V} / 8 = 30\text{ V}

. Connecting a

15\ \Omega

resistor across

30\text{ V}

produces a secondary current of

30\text{ V} / 15\ \Omega = 2.0\text{ A}

. Since power in an ideal transformer is conserved (

P_p = P_s

), the primary current is stepped down by the same ratio:

2.0\text{ A} / 8 = 0.25\text{ A}

.

Step-by-Step Solution

1

Calculate the turns ratio (

a

) of the transformer.

a = \frac{N_p}{N_s} = \frac{1{}200\text{ turns}}{150\text{ turns}} = 8

The ratio of primary to secondary turns governs the voltage step-down and current step-up proportions.

2

Calculate the secondary voltage (

V_s

) across the load.

V_s = \frac{V_p}{a} = \frac{240\text{ V}}{8} = 30\text{ V}

In a step-down transformer, secondary voltage equals primary voltage divided by the turns ratio.

3

Calculate the secondary current (

I_s

) through the load resistor using Ohm's Law.

I_s = \frac{V_s}{R_L} = \frac{30\text{ V}}{15\ \Omega} = 2.0\text{ A}

Current through a resistive load is directly proportional to voltage and inversely proportional to resistance.

4

Calculate the primary current (

I_p

) drawn from the AC source.

I_p = \frac{I_s}{a} = \frac{2.0\text{ A}}{8} = 0.25\text{ A}

Conservation of power in an ideal transformer requires that

V_p I_p = V_s I_s

, meaning primary current is the secondary current divided by the turns ratio.

Key Concept

Ideal transformer transformation ratios (

N_p / N_s = V_p / V_s = I_s / I_p

)

Alternative Method

Using reflected impedance: The load resistance reflected to the primary side is

R_p' = a^2 \times R_L = 8^2 \times 15\ \Omega = 64 \times 15\ \Omega = 960\ \Omega

. The primary current is then directly calculated using Ohm's Law:

I_p = \frac{V_p}{R_p'} = \frac{240\text{ V}}{960\ \Omega} = 0.25\text{ A}

.

Estimated Time:2m 0s

Question 696Question →

A technician connects a $24\text{ V}$ direct current (DC) power source across an auxiliary heating element with a measured resistance of $8\text{ }\Omega$ . What is the amount of current flowing through the heating element?

192\text{ A}

0.33\text{ A}

3\text{ A}

24\text{ A}

Show answer & explanation

Answer: $3\text{ A}$

Answer

3\text{ A}

According to Ohm's law, current (

I

) is calculated by dividing voltage (

V

) by resistance (

R

). For a

24\text{ V}

supply connected across an

8\text{ }\Omega

load, the calculation is

I = \frac{24\text{ V}}{8\text{ }\Omega} = 3\text{ A}

.

Step-by-Step Solution

1

Identify the given values from the problem statement.

Voltage (

V

) =

24\text{ V}

, Resistance (

R

) =

8\text{ }\Omega

.

Ohm's law relates voltage, current, and resistance in a DC circuit.

2

Select the appropriate form of Ohm's law to solve for current (

I

).

I = \frac{V}{R}

Current is directly proportional to voltage and inversely proportional to resistance.

3

Substitute the known values into the equation and compute the result.

I = \frac{24\text{ V}}{8\text{ }\Omega} = 3\text{ A}

Dividing

24

by

8

gives the circuit current in amperes.

Key Concept

Ohm's Law relationship between voltage, current, and resistance (

I = \frac{V}{R}

)

Estimated Time:45s

Question 697Question →

While standard bipolar junction transistors (BJTs) operate as current-controlled components, field-effect transistors (FETs) are commonly used in high-impedance input stages. What is the primary operating mechanism of a field-effect transistor?

Controlling collector current by drawing a continuous input current through a forward-biased base junction

Regulating current between the source and drain using an electric field established by a gate voltage

Directing electron flow by altering magnetic flux across a series of reverse-biased diode barriers

Modulating channel resistance by requiring a high input control current at the gate terminal

Show answer & explanation

Answer: Regulating current between the source and drain using an electric field established by a gate voltage

Answer

Regulating current between the source and drain using an electric field established by a gate voltage

A field-effect transistor (FET) controls the flow of electrical charge carriers between the source and drain terminals by varying the strength of an electrostatic field within the semiconductor channel. This field is established by applying a voltage to the gate terminal, allowing the device to modulate conductivity with almost zero input gate current.

Step-by-Step Solution

1

Identify the fundamental category and terminals of a Field-Effect Transistor (FET)

FETs have three primary terminals: Gate, Drain, and Source.

Recognizing terminal designations separates FET functionality from BJT terminals (Base, Collector, Emitter).

2

Determine the control mechanism of a FET

An electric field created by voltage applied to the gate controls the conductive channel between the source and drain.

FETs are voltage-controlled solid-state devices that draw negligible gate current due to extremely high input resistance.

Key Concept

Field-Effect Transistor (FET) Voltage-Controlled Operation

Estimated Time:1m 0s

Question 698Question →

A culinary instructor is reviewing the final exam scores of six students to determine the middle performance level of the class. The scores recorded are $84$ , $72$ , $90$ , $76$ , $88$ , and $64$ . What is the median score of this group of students?

79

80

83

118

Show answer & explanation

Answer: $80$

Answer

The median score is

80

.

To find the median, the values must first be arranged in numerical order:

64

,

72

,

76

,

84

,

88

,

90

. Because there are six numbers (an even amount), the median is the average of the two central numbers,

76

and

84

. Adding them together yields

160

, and dividing by

2

gives a median of

80

.

Step-by-Step Solution

1

Sort the given scores from least to greatest.

64

,

72

,

76

,

84

,

88

,

90

The median is the exact middle value of an ordered dataset, making sorting the mandatory first step.

2

Identify the middle value(s) of the sorted list.

The middle two values are

76

and

84

.

Since there is an even number of data points (

6

), there is no single middle value. The median position falls exactly between the 3rd and 4th values.

3

Calculate the average of the two middle values.

\frac{76 + 84}{2} = \frac{160}{2} = 80

The median of a dataset with an even number of values is the arithmetic mean of the two central values.

Key Concept

Median of a Dataset with an Even Number of Values

Estimated Time:1m 0s

Question 699Question →

Match each electrical current term with its correct definition or technical description.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electron Flow matches the movement of negative charge carriers from negative to positive; Conventional Current matches the flow modeled from positive to negative; Ampere matches the unit defined as one coulomb per second; Ammeter matches the instrument placed in series to measure charge flow rate.

Each electrical term is paired with its precise technical role: electron flow represents real electron drift from negative to positive; conventional current is the established positive-to-negative analysis standard; the ampere represents a charge flow rate of one coulomb per second; and the ammeter is the tool configured in series to monitor that current.

Step-by-Step Solution

1

Identify physical charge transport direction versus convention.

Electrons carry negative charge and move from negative to positive (Electron Flow), whereas engineering circuit analysis traditionally assumes positive-to-negative flow (Conventional Current).

Differentiating between physical particle drift and standardized schematic conventions is fundamental to electronics.

2

Identify the unit of rate of charge flow.

Current (

I

) is measured in amperes (

A

), where

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

.

Current is defined quantitatively as the rate of charge transfer over time (

I = Q/t

).

3

Identify the measurement device.

An ammeter measures current and is placed in series so the circuit current passes through it.

Measuring current requires intercepting the stream of charge carriers in a series configuration.

Key Concept

Fundamental definitions and flow conventions of electrical current

Estimated Time:45s

Question 700Question →

When an automobile travels over an uneven road surface, which suspension component prevents the chassis from continuously bouncing by dissipating the energy stored in the springs?

Coil spring

Control arm

Shock absorber

Stabilizer bar

Show answer & explanation

Answer: Shock absorber

Answer

Shock absorber

The correct answer is the shock absorber. While springs compress to absorb shocks from road irregularities, they naturally recoil and continue oscillating. The shock absorber dampens this movement by forcing hydraulic fluid through restricted passages, converting mechanical energy into heat and stabilizing the vehicle.

Step-by-Step Solution

1

Identify the primary suspension action required by the problem.

The target component must dampen oscillations and dissipate energy stored during spring compression.

Springs absorb impact through elastic deformation, but without damping, they will continue to oscillate.

2

Distinguish between energy storage and energy dissipation components.

The shock absorber (damper) specifically converts oscillation energy into heat via hydraulic fluid flow.

While springs carry the load, shock absorbers control the speed and rebound cycles of the spring movement.

Key Concept

Damping of spring oscillations by shock absorbers

Estimated Time:45s

Answer

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