Electronic Components Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 4

Question 61Question →

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 62Question →

An automotive audio technician installs two filter capacitors connected in parallel across a power line to suppress voltage ripple. If the individual capacitor values are $47\text{ }\mu\text{F}$ and $33\text{ }\mu\text{F}$ , what is the total equivalent capacitance of this parallel combination in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 80

Answer

80 microfarads (

80\text{ }\mu\text{F}

)

When capacitors are wired in parallel, the total plate area increases, allowing more charge to be stored at a given voltage. Therefore, the equivalent capacitance is the simple arithmetic sum of the individual capacitances:

47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the network connection type and select the appropriate rule.

Capacitors are connected in parallel.

Connecting capacitors in parallel increases the total effective plate area, so their capacitances add directly.

2

Calculate the total capacitance using the sum formula

C_{\text{total}} = C_1 + C_2

.

C_{\text{total}} = 47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}

Direct addition of the individual capacitance ratings yields the total circuit capacitance.

Key Concept

Parallel Capacitance Calculation

Estimated Time:45s

Question 63Question →

A Single-Pole, Single-Throw (SPST) switch is capable of opening and closing two separate, electrically isolated circuits simultaneously.

True

False

Show answer & explanation

Answer: False

Answer

False. A Single-Pole, Single-Throw (SPST) switch controls only one individual circuit.

The statement is false because the pole count specifies how many distinct circuits a switch controls. A Single-Pole (SP) switch controls exactly one circuit path. Operating two separate, electrically isolated circuits simultaneously requires a Double-Pole (DP) switch, such as a DPST switch.

Step-by-Step Solution

1

Define the meaning of 'pole' in switch terminology.

The number of poles indicates how many separate, isolated electrical circuits the switch can control at one time.

Understanding switch nomenclature requires distinguishing between poles (circuits controlled) and throws (contact positions per pole).

2

Evaluate the capability of a Single-Pole (SP) switch against the statement.

A single-pole switch has only one set of contacts and can only control a single circuit.

To simultaneously switch two independent circuits, a double-pole (DP) configuration with a shared mechanical actuator is required.

Key Concept

Switch Poles and Throws Nomenclature

Estimated Time:30s

Question 64Question →

An automotive direct-current (DC) blower motor circuit with a continuous operating current of $5\text{ A}$ experiences a brief transient inrush current of $18\text{ A}$ for $150\text{ ms}$ each time the motor starts. When a technician installs a standard $6\text{ A}$ fast-acting fuse, the fuse opens immediately upon starting the motor despite no short circuit or wiring fault being present. To resolve this nuisance tripping while still providing reliable overcurrent and thermal protection for the continuous load and wiring, which protective device should be installed?

A

20\text{ A}

fast-acting fuse connected in series with the motor feed conductor

A

6\text{ A}

slow-blow (time-delay) fuse connected in series with the motor feed conductor

A

6\text{ A}

fast-acting fuse connected in parallel across the motor power terminals

A metal-oxide varistor (MOV) rated for

12\text{ V}

connected in series with the motor feed conductor

Show answer & explanation

Answer: A $6\text{ A}$ slow-blow (time-delay) fuse connected in series with the motor feed conductor

Answer

A

6\text{ A}

slow-blow (time-delay) fuse connected in series with the motor feed conductor

Inductive loads such as electric motors draw high transient inrush currents upon startup before counter-electromotive force (CEMF) develops. A slow-blow (time-delay) fuse utilizes a dual-element or thermal lag construction designed specifically to pass short-duration overcurrent surges without melting, while still opening quickly under prolonged overloads or dead short circuits. Sizing the slow-blow fuse at

6\text{ A}

and placing it in series provides full circuit protection matched to the

5\text{ A}

running current without nuisance tripping.

Step-by-Step Solution

1

Analyze the load characteristics and failure mode

The DC motor requires

5\text{ A}

continuous running current but produces a temporary inductive inrush surge of

18\text{ A}

(

3.6\times

rated current) lasting

150\text{ ms}

. Fast-acting fuses react instantaneously to current spikes, causing nuisance blowing during normal motor starts.

Inductive and capacitive loads naturally draw transient inrush currents that must be accommodated without blowing the protective element under non-fault conditions.

2

Determine the appropriate fuse operating characteristic and rating

A slow-blow (time-delay) fuse rated at

6\text{ A}

contains a dual-element or spring-loaded thermal design that absorbs brief transient current spikes without melting, yet reliably opens if current exceeds

6\text{ A}

continuously.

Maintaining the current rating near the continuous operating load (

6\text{ A}

for a

5\text{ A}

circuit) ensures proper thermal overload protection for the wiring and motor windings.

3

Verify proper circuit topology for overcurrent protection

The protective device must be placed in series with the ungrounded supply conductor so that opening the fuse disconnects the entire current path to the load.

Series placement ensures the total circuit current passes through the protective device.

Key Concept

Selection and application of time-delay (slow-blow) fuses for inductive inrush loads

Estimated Time:1m 15s

Question 65Question →

In an electronic filter circuit, two inductors with values $L_1 = 30\text{ mH}$ and $L_2 = 20\text{ mH}$ are connected in parallel. This parallel pair is connected in series with a third inductor $L_3 = 18\text{ mH}$ and a fourth inductor $L_4 = 10\text{ mH}$ . Assuming negligible mutual inductance between the coils, what is the total equivalent inductance of the circuit in millihenries ( $\text{mH}$ )?

Show answer & explanation

Answer: 40

Answer

40 mH

The total inductance is

40\text{ mH}

. Inductors follow the same combination rules as resistors: the parallel pair (

30\text{ mH}

and

20\text{ mH}

) yields an equivalent inductance of

12\text{ mH}

, which when added to the series inductors (

18\text{ mH}

and

10\text{ mH}

) gives

12 + 18 + 10 = 40\text{ mH}

.

Step-by-Step Solution

1

Calculate the equivalent inductance of the two parallel inductors

L_1

and

L_2

.

L_p = \frac{L_1 \times L_2}{L_1 + L_2} = \frac{30\text{ mH} \times 20\text{ mH}}{30\text{ mH} + 20\text{ mH}} = \frac{600}{50} = 12\text{ mH}

Inductors in parallel combine using reciprocal addition, identically to resistors in parallel.

2

Calculate the total circuit inductance by summing the parallel combination with the remaining series inductors

L_3

and

L_4

.

L_{total} = L_p + L_3 + L_4 = 12\text{ mH} + 18\text{ mH} + 10\text{ mH} = 40\text{ mH}

Inductors in series combine via direct addition because the magnetic fields and induced voltages add along the common current path.

Key Concept

Inductors combine like resistors: series inductances add directly, while parallel inductances combine using reciprocal addition.

Estimated Time:1m 30s

Question 66Question →

An audio technician connects a capacitor in series with a high-frequency speaker (tweeter) inside a multi-speaker enclosure. What is the primary electrical reason for using a capacitor in this configuration?

It stores energy in a magnetic field to boost low-frequency bass signals delivered to the speaker.

It easily passes high-frequency signals due to lower capacitive reactance while blocking low frequencies.

It blocks high-frequency signals and allows direct current and low audio frequencies to pass freely.

It acts as a permanent insulator that converts excess electrical voltage into heat at all audio frequencies.

Show answer & explanation

Answer: It easily passes high-frequency signals due to lower capacitive reactance while blocking low frequencies.

Answer

The capacitor is used because it easily passes high-frequency signals due to lower capacitive reactance while blocking low frequencies.

Capacitive reactance (

X_C = \frac{1}{2\pi f C}

) is inversely proportional to the frequency of the electrical signal. For high-frequency signals directed to a tweeter, the capacitor offers very low opposition (low reactance), allowing the high-frequency audio to pass smoothly. At low frequencies or DC, the reactance becomes extremely high, blocking low-frequency bass currents that could overheat or damage the delicate tweeter voice coil.

Step-by-Step Solution

1

Analyze the relationship between frequency and capacitive reactance (

X_C

).

Capacitive reactance is inversely proportional to frequency, governed by the formula

X_C = \frac{1}{2\pi f C}

.

Understanding how reactance changes with frequency is essential for determining how AC audio signals behave across the component.

2

Determine the effect on high versus low audio frequencies.

At high frequencies (treble),

X_C

is very low, presenting minimal opposition to current; at low frequencies (bass) and DC (

f = 0

),

X_C

becomes extremely large, effectively blocking the signal.

Tweeters are designed to handle only high-frequency signals, so placing a capacitor in series forms a high-pass filter that protects the tweeter from damaging low frequencies.

Key Concept

Capacitors exhibit capacitive reactance inversely proportional to frequency, acting as high-pass filters in series audio crossover circuits.

Estimated Time:1m 0s

Question 67Question →

In advanced circuit protection, different devices are engineered to respond to specific types of faults using unique physical properties. Based on their operating principles, match each specialized electronic protective device to its precise operational mechanism.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Magnetic Circuit Breaker matches the electromagnetic coil mechanism. Thermal Circuit Breaker matches the bimetallic strip mechanism. Slow-Blow Fuse matches the inrush current dual-element design. MOV matches the impedance-dropping clamping mechanism. Crowbar Circuit matches the deliberate shorting mechanism.

Each protective component relies on a distinct physical property: electromagnetism for magnetic breakers, thermodynamics for thermal breakers, thermal inertia for slow-blow fuses, non-linear resistance for MOVs, and intentional fault-creation for crowbar circuits.

Step-by-Step Solution

1

Analyze devices designed for overcurrent protection versus overvoltage protection.

Magnetic breakers, thermal breakers, and slow-blow fuses handle overcurrent. MOVs and crowbar circuits handle overvoltage.

Separating the components by their primary fault protection type narrows down the possible operating mechanisms.

2

Differentiate between the overcurrent devices based on their physical triggers.

Thermal breakers use heat (bimetallic strip). Magnetic breakers use electromagnetism (coils). Slow-blow fuses use dual-elements to delay blowing during temporary inrush currents.

Each overcurrent device is selected based on load characteristics (e.g., motors need slow-blow; precise instant tripping needs magnetic).

3

Differentiate between the overvoltage devices based on how they clear the fault.

MOVs passively clamp voltage by changing their internal resistance. Crowbar circuits actively short the circuit to force a fuse to blow.

Understanding the difference between passive clamping (varistors) and active destructive diversion (crowbar) confirms the final two matches.

Key Concept

Advanced circuit protection mechanisms (Overcurrent vs. Overvoltage strategies)

Question 68Question →

For a bipolar junction transistor (BJT) to operate in the active amplification mode, the base-emitter junction must be forward-biased while the base-collector junction is reverse-biased.

True

False

Show answer & explanation

Answer: True

Answer

True. Operating a bipolar junction transistor in the active amplification region requires a forward-biased base-emitter junction and a reverse-biased base-collector junction.

The statement is correct because linear amplification in a bipolar junction transistor strictly requires the input base-emitter diode junction to be forward-biased to initiate current flow, and the output collector-base junction to be reverse-biased to sweep and gather charge carriers into the collector circuit.

Step-by-Step Solution

1

Identify the specified operating state of the transistor

The transistor is designated to operate in active (linear) amplification mode.

A transistor exhibits distinct junction biasing configurations depending on whether it functions in cutoff, saturation, or active mode.

2

Determine the required bias across the base-emitter junction

The base-emitter junction must be forward-biased.

Forward-biasing this junction overcomes the internal barrier potential (approximately 0.7 V for silicon) and injects carriers into the base.

3

Determine the required bias across the base-collector junction

The base-collector junction must be reverse-biased.

Reverse-biasing this junction establishes a wide depletion region with an electric field that sweeps the injected carriers into the collector.

4

Evaluate the accuracy of the statement

The statement correctly describes both required junction conditions.

The combination of a forward-biased base-emitter junction and a reverse-biased base-collector junction precisely defines active region operation.

Key Concept

Bipolar Junction Transistor Biasing and Operating Modes

Estimated Time:45s

Question 69Question →

Which of the following statements correctly identifies the primary function and operating principle of a capacitor in an electronic circuit?

It stores energy in a magnetic field to oppose sudden changes in electrical current.

It converts alternating current into direct current by permitting electron flow in only one direction.

It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

It allows continuous direct current flow across its internal dielectric layer to dissipate energy as heat.

Show answer & explanation

Answer: It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

Answer

It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

A capacitor is designed with two conductive plates separated by an insulating dielectric material. When a voltage is applied, opposite charges accumulate on the plates, establishing an electrostatic field that stores electrical energy. Because charge accumulation requires finite time, a capacitor naturally opposes any sudden change in voltage across its terminals.

Step-by-Step Solution

1

Identify the fundamental physical construction and energy storage mechanism of a capacitor.

A capacitor consists of two conducting plates separated by an insulating dielectric, which accumulates opposite charges to establish an electrostatic field.

Capacitors store potential electrical energy in an electrostatic field rather than a magnetic field.

2

Determine how a capacitor interacts with circuit voltage and current over time.

Because electrical charge requires finite time to build up or dissipate across the plates (

q = C \cdot V

), the component naturally resists instantaneous changes in voltage.

This property makes capacitors essential for filtering voltage ripples in power supplies and setting time constants in electronic circuits.

Key Concept

Capacitor Energy Storage and Voltage Regulation

Estimated Time:45s

Question 70Question →

An automotive starter circuit uses an electromechanical relay between the ignition switch and the starter motor. What is the primary purpose of using a relay in this circuit?

To allow a low-current switch on the dashboard to safely control the high-current starter motor

To convert direct current from the vehicle battery into alternating current for the starter motor

To permanently break the circuit by melting an internal metal strip during an overcurrent event

To step up the battery voltage to a higher potential to overcome heavy wire resistance

Show answer & explanation

Answer: To allow a low-current switch on the dashboard to safely control the high-current starter motor

Answer

Allowing a low-current switch on the dashboard to safely control the high-current starter motor

An electromechanical relay uses a small control current flowing through an electromagnet coil to close heavy-duty contacts. In an automotive starting system, this enables a driver to use a small, safe ignition switch to engage the hundreds of amperes demanded by the starter motor without running heavy cables through the passenger cabin.

Step-by-Step Solution

1

Identify the basic operating principle of an electromechanical relay

A small control current through a wire coil generates a magnetic field that moves an armature to open or close heavy-duty load contacts.

Relays provide electrical isolation and current amplification between the control circuit and the load circuit.

2

Apply the relay function to the automotive starter scenario

The starter motor requires hundreds of amperes of current, which would overheat standard dashboard toggle switches and wiring if routed directly.

Placing the relay near the battery and starter allows short, heavy-gauge cables for the high-current path while using lightweight wiring for the ignition switch.

Key Concept

Electromechanical Relay Operation and High-Current Load Switching

Estimated Time:40s

Question 71Question →

Complete the statement describing the dual-action operating mechanisms inside a standard industrial circuit breaker.

Fill in the blanks below

In a thermal-magnetic circuit breaker, sustained moderate overloads cause a strip to heat and deflect over time, whereas severe short-circuit currents create an intense magnetic field in an coil to trip the mechanism instantaneously.

Show answer & explanation

Answer

Blank 1: bimetallic (or bimetal); Blank 2: electromagnetic (or magnetic)

Thermal-magnetic circuit breakers provide coordinated protection using two distinct physical principles: a bimetallic strip that flexes under prolonged thermal stress to prevent wire insulation degradation, and an electromagnetic solenoid that provides virtually instantaneous tripping when massive fault currents occur.

Step-by-Step Solution

1

Identify the thermal overload component

Bimetallic strip

Prolonged overcurrent conditions produce resistive heat proportional to

I^2R \cdot t

. Two bonded metals with different coefficients of thermal expansion expand at unequal rates, causing the strip to bend and unlatch the breaker contacts after a calibrated time delay.

2

Identify the short-circuit protection component

Electromagnetic coil (solenoid)

High-magnitude fault currents (such as direct phase-to-ground or phase-to-phase shorts) generate an immediate, powerful magnetic field inside a wire coil/solenoid, pulling an internal armature to trip the breaker instantly before conductors reach destructive temperatures.

Key Concept

Dual-trip circuit breakers utilize a slow-responding bimetallic element for thermal overcurrent protection and a rapid-responding electromagnetic solenoid for short-circuit protection.

Estimated Time:1m 15s

Question 72Question →

An industrial electronics technician is servicing a timing circuit on an automated assembly line. The circuit network contains two $12\text{ }\mu\text{F}$ capacitors connected in series, and this combination is connected in parallel with a third capacitor rated at $14\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this circuit network in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

20\text{ }\mu\text{F}

For the series branch, two identical

12\text{ }\mu\text{F}

capacitors yield an equivalent capacitance of

6\text{ }\mu\text{F}

. When placed in parallel with the

14\text{ }\mu\text{F}

capacitor, the capacitances add directly (

6\text{ }\mu\text{F} + 14\text{ }\mu\text{F}

), giving a total equivalent capacitance of

20\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = 6\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

, or

C_s = \frac{C}{n} = \frac{12}{2} = 6\text{ }\mu\text{F}

.

2

Add the capacitance of the parallel branch to the series equivalent.

C_{\text{total}} = 20\text{ }\mu\text{F}

Capacitors in parallel add directly together:

C_{\text{total}} = C_s + C_3 = 6\text{ }\mu\text{F} + 14\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

.

Key Concept

Series and parallel capacitor combinations and equivalent capacitance calculations

Estimated Time:1m 0s

Question 73Question →

A power adapter uses a step-down transformer with $400\text{ turns}$ on the primary coil and $20\text{ turns}$ on the secondary coil. If an input voltage of $120\text{ V AC}$ is applied across the primary coil, what is the secondary output voltage in volts?

Show answer & explanation

Answer: 6

Answer

The secondary output voltage is

6\text{ V}

.

In a transformer, the ratio of secondary voltage to primary voltage equals the ratio of secondary turns to primary turns (

V_s / V_p = N_s / N_p

). With

400

primary turns and

20

secondary turns, the turns ratio is

20 / 400 = 1/20

. Multiplying the primary voltage (

120\text{ V}

) by

1/20

yields a secondary output voltage of

6\text{ V}

.

Step-by-Step Solution

1

Identify the given transformer parameters from the problem statement.

V_p = 120\text{ V}

,

N_p = 400\text{ turns}

, and

N_s = 20\text{ turns}

.

These parameters are required to determine the relationship between primary and secondary voltages.

2

Set up the transformer turns ratio equation relating primary and secondary voltages to coil turns.

\frac{V_s}{V_p} = \frac{N_s}{N_p}

The voltage induced across a transformer winding is directly proportional to the number of turns in that winding.

3

Substitute the known values into the equation and compute the secondary voltage

V_s

.

V_s = 120\text{ V} \times \left(\frac{20}{400}\right) = 120\text{ V} \times 0.05 = 6\text{ V}

Direct algebraic calculation yields the step-down output voltage.

Key Concept

Transformer Turns Ratio and Voltage Transformation

Estimated Time:45s

Question 74Question →

A step-up transformer in a specialized power supply is connected to a $120\text{ V AC}$ source on its primary side. The secondary circuit delivers a current of $24\text{ mA}$ across a $25\text{ k}\Omega$ resistive load. If the primary winding has $150\text{ turns}$ , what is the total number of turns on the secondary winding?

30\text{ turns}

75\text{ turns}

750\text{ turns}

18\text{ turns}

Show answer & explanation

Answer: $750\text{ turns}$

Answer

750\text{ turns}

The correct answer is

750\text{ turns}

. First, determine the secondary voltage using Ohm's Law:

V_s = I_s \times R_L = (0.024\text{ A}) \times (25{}000\,\Omega) = 600\text{ V}

. Then, apply the transformer ratio formula

\frac{N_s}{N_p} = \frac{V_s}{V_p}

, yielding

N_s = 150 \times \left(\frac{600}{120}\right) = 750\text{ turns}

.

Step-by-Step Solution

1

Calculate the secondary voltage (

V_s

) across the load resistor using Ohm's Law (

V = I \times R

).

V_s = (24\text{ mA}) \times (25\text{ k}\Omega) = (0.024\text{ A}) \times (25{}000\,\Omega) = 600\text{ V}

Determining the voltage across the secondary winding is necessary to establish the transformer's voltage transformation ratio.

2

Apply the transformer turns ratio equation that relates winding turns directly to voltage.

\frac{N_s}{N_p} = \frac{V_s}{V_p}

In an ideal transformer, the ratio of secondary-to-primary turns is directly proportional to the ratio of secondary-to-primary voltage.

3

Substitute the known values (

V_s = 600\text{ V}

,

V_p = 120\text{ V}

, and

N_p = 150\text{ turns}

) to solve for the secondary turns (

N_s

).

N_s = 150 \times \left(\frac{600\text{ V}}{120\text{ V}}\right) = 150 \times 5 = 750\text{ turns}

Multiplying the primary turns by the voltage scaling factor gives the required secondary turns.

Key Concept

Transformer turns ratio and multi-step AC secondary load calculations

Estimated Time:2m 0s

Question 75Question →

In a bipolar junction transistor (BJT) amplifier circuit operating in its active region, a base current ( $I_B$ ) of $2\text{ mA}$ is applied. If the transistor has a common-emitter DC current gain ( $\beta$ or $h_{FE}$ ) of $50$ , what is the resulting collector current ( $I_C$ ) flowing through the transistor?

25\text{ mA}

52\text{ mA}

100\text{ mA}

0.04\text{ mA}

Show answer & explanation

Answer: $100\text{ mA}$

Answer

The collector current (

I_C

) is

100\text{ mA}

.

In the active region of a bipolar junction transistor, the collector current is directly proportional to the base current by the factor of DC current gain:

I_C = \beta \times I_B

. Multiplying the base current of

2\text{ mA}

by the gain of

50

yields

100\text{ mA}

.

Step-by-Step Solution

1

Identify the given values and formula for transistor DC current gain in the active region.

Given

I_B = 2\text{ mA}

and

\beta = 50

. The governing formula is

I_C = \beta \times I_B

.

Current gain (

\beta

) represents the amplification factor between the controlling base current and the controlled collector current.

2

Substitute the measured base current and gain into the current equation.

I_C = 50 \times 2\text{ mA} = 100\text{ mA}

.

Calculating the product gives the total current flowing through the collector terminal.

Key Concept

Bipolar Junction Transistor DC Current Gain (

\beta

or

h_{FE}

)

Estimated Time:1m 0s

Question 76Question →

Match each switch configuration and electromechanical relay component with its primary function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

SPST Switch matches with basic single-circuit on-off operation; SPDT Switch matches with directing one common terminal between two output paths; Relay Coil matches with generating a magnetic field to move the armature; Normally Open (NO) Contact matches with maintaining an open circuit until energized.

Each component is correctly matched to its defined operational role: an SPST switch opens or closes one path; an SPDT switch alternates between two paths; a relay coil uses current to create an electromagnetic field; and an NO contact stays open until coil actuation occurs.

Step-by-Step Solution

1

Analyze switch pole and throw designations.

SPST (Single Pole, Single Throw) has one input and one output for simple on/off control. SPDT (Single Pole, Double Throw) routes one input to either of two outputs.

Poles define the number of separate circuits controlled, while throws define the number of output positions available per pole.

2

Analyze relay component functions and contact resting states.

The coil provides electromagnetic force to actuate the switch mechanism, while Normally Open contacts only close and allow current flow when this magnetic force is active.

Relay contacts are designated based on their state when the control coil is unpowered (de-energized).

Key Concept

Switch Nomenclature and Electromechanical Relay Operation

Estimated Time:45s

Question 77Question →

An ideal transformer has a primary coil with $600$ turns and a secondary coil with $150$ turns. The primary coil is connected to a $120\text{ V}$ AC source, and the secondary coil is connected to a load consisting of two $15\ \Omega$ resistors wired in parallel. What is the total current flowing through the primary coil?

1\text{ A}

16\text{ A}

4\text{ A}

0.25\text{ A}

Show answer & explanation

Answer: $1\text{ A}$

Answer

The total current flowing through the primary coil is

1\text{ A}

.

The correct answer synthesizes multiple principles. First, the transformer step-down ratio is

600:150

, or

4:1

. This drops the primary

120\text{ V}

to a secondary

30\text{ V}

. Next, the two

15\ \Omega

resistors in parallel create an equivalent resistance of

7.5\ \Omega

. Using Ohm's Law (

I = V/R

), the secondary current is

30\text{ V} / 7.5\ \Omega = 4\text{ A}

. Finally, because a transformer steps current inversely to voltage, a

4:1

voltage step-down means a

1:4

current step-up from primary to secondary. Therefore, the primary current is

4\text{ A} / 4 = 1\text{ A}

.

Step-by-Step Solution

1

Calculate the turns ratio to determine the secondary voltage.

N_p/N_s = 600/150 = 4

. Therefore,

V_s = 120\text{ V} / 4 = 30\text{ V}

.

In a transformer, the voltage ratio is directly proportional to the turns ratio.

2

Determine the equivalent resistance of the secondary load.

For two equal resistors in parallel,

R_{eq} = 15\ \Omega / 2 = 7.5\ \Omega

.

Parallel branches provide multiple paths for current, reducing the total resistance of the circuit.

3

Apply Ohm's law to find the total secondary current.

I_s = V_s / R_{eq} = 30\text{ V} / 7.5\ \Omega = 4\text{ A}

.

The total current drawn by the load depends on the secondary voltage and the equivalent resistance of the load.

4

Calculate the primary current using the inverse turns ratio.

I_p = I_s / (N_p / N_s) = 4\text{ A} / 4 = 1\text{ A}

.

In an ideal transformer, power is conserved (

P_p = P_s

). If voltage steps down by a factor of 4, the current must step up by a factor of 4 from primary to secondary, making the primary current 1/4 of the secondary current.

Key Concept

Transformer turns ratio scaling for voltage and current, combined with parallel circuit analysis.

Question 78Question →

When inspecting a standard 14-pin Dual In-line Package (DIP) integrated circuit with the reference notch oriented at the top, which statement correctly describes the standard pinout numbering convention used to identify pin 1 and pin 14?

Pin 1 is at the top-left beside the notch, and numbering progresses sequentially down the left side from pin 1 to 7, then restarts at the top-right with pin 8 and counts down to pin 14 at the bottom-right.

Pin 1 is at the top-left beside the notch, and numbering progresses counter-clockwise down the left side from pin 1 to 7, across to the bottom-right for pin 8, and up the right side to pin 14 at the top-right.

Pin 1 is at the top-right beside the notch, and numbering progresses clockwise down the right side from pin 1 to 7, across to the bottom-left for pin 8, and up the left side to pin 14 at the top-left.

Pin 1 is at the top-left, and numbering alternates horizontally across opposing pairs (pin 1 top-left, pin 2 top-right, pin 3 second-row left) down the package to pin 14 at the bottom-right.

Show answer & explanation

Answer: Pin 1 is at the top-left beside the notch, and numbering progresses counter-clockwise down the left side from pin 1 to 7, across to the bottom-right for pin 8, and up the right side to pin 14 at the top-right.

Answer

Pin 1 is located at the top-left immediately adjacent to the notch, and numbering continues counter-clockwise down the left column (pins 1 through 7) and upward along the right column (pins 8 through 14), placing pin 14 at the top-right directly opposite pin 1.

For all standard Dual In-line Package (DIP) integrated circuits, placing the index notch at the top establishes pin 1 at the top-left. Numbering then advances counter-clockwise down the left row (pins 1 to 7 for a 14-pin IC), transitions across to the bottom-right pin (pin 8), and ascends the right row to pin 14 at the top-right.

Step-by-Step Solution

1

Orient the integrated circuit package correctly.

The reference notch or alignment dot is positioned facing upward (top).

Standardized electronic manufacturing standards (JEDEC) require referencing the orientation marker before indexing lead pins.

2

Identify pin 1 based on the reference notch.

Pin 1 is the lead located immediately to the left of the notch in the top row.

By industry convention, pin 1 is always directly counter-clockwise adjacent to the top reference notch.

3

Trace the indexing sequence around the Dual In-line Package perimeter.

Pins 1 through 7 run down the left side, pin 8 is at the bottom-right, and pins 9 through 14 advance upward on the right side to pin 14 at the top-right.

DIP integrated circuits use a continuous counter-clockwise loop rather than parallel or alternating row numbering.

Key Concept

Dual In-line Package (DIP) Integrated Circuit Pin Numbering Convention

Estimated Time:1m 0s

Question 79Question →

Electronic systems utilize various solid-state semiconductor transistors tailored for specific switching, amplification, and sensing applications. Match each transistor type to its defining operating principle and physical characteristic.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Bipolar Junction Transistor (BJT) pairs with the current-controlled device using both electron and hole carriers; Junction Field-Effect Transistor (JFET) pairs with the voltage-controlled device using a reverse-biased PN junction; Metal-Oxide-Semiconductor FET (MOSFET) pairs with the insulated-gate component featuring high input impedance and an oxide dielectric; Phototransistor pairs with the optoelectronic component converting incident light into amplified collector current.

Each semiconductor device corresponds uniquely to its operating physics: BJTs are bipolar current-controlled devices; JFETs utilize reverse-biased PN junctions for channel modulation; MOSFETs employ an insulated dielectric gate for field-effect control; and phototransistors utilize optical stimulation to generate base-region carriers.

Step-by-Step Solution

1

Identify the carrier conduction and control mechanism of the BJT

The BJT is a current-controlled bipolar device relying on both majority and minority carriers (electrons and holes).

The term bipolar explicitly indicates the participation of both carrier polarities, controlled by base current.

2

Differentiate between the field-effect transistor junction types (JFET vs. MOSFET)

The JFET relies on a direct reverse-biased PN gate junction, while the MOSFET incorporates an insulated oxide layer that eliminates DC gate current entirely.

The metal-oxide insulator in a MOSFET provides near-infinite input impedance compared to the reverse-biased PN junction of a JFET.

3

Distinguish the optoelectronic sensor category (Phototransistor)

The phototransistor operates by replacing or augmenting electrical base current with photon-generated charge carriers.

Light striking the transparent casing generates electron-hole pairs at the junction to initiate conduction.

Key Concept

Classification, control mechanisms, and structural properties of major transistor families

Estimated Time:1m 0s

Question 80Question →

In a multi-stage radio receiver, an electronics technician places a coupling capacitor in series along the signal line between two amplifier stages. What is the primary function of this capacitor regarding the combined Direct Current (DC) bias and Alternating Current (AC) signal?

Allows the continuous DC bias to pass freely while completely stopping all fluctuating AC signal frequencies.

Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

Transforms the AC signal into a higher DC voltage by generating a strong magnetic field across its plates.

Conducts the steady DC voltage directly across its dielectric material while dissipating the AC signal as heat.

Show answer & explanation

Answer: Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

Answer

Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

A capacitor acts as an open circuit to steady Direct Current (DC) because the insulating dielectric stops continuous charge conduction once the plates reach the bias voltage level. Simultaneously, the alternating fluctuations of an AC signal continuously charge and discharge the plates, allowing the alternating signal to transfer cleanly to the next amplifier stage without shifting its DC bias point.

Step-by-Step Solution

1

Analyze capacitor behavior in a steady Direct Current (DC) state.

Because the dielectric layer is an electrical insulator, steady DC charges the plates until the potential difference equals the supply voltage, reducing steady-state current to zero (

X_C = \infty

at

0\text{ Hz}

). The capacitor acts as an open circuit to DC.

Determines how the DC operating bias of the first amplifier stage is isolated from the second stage.

2

Analyze capacitor behavior when exposed to an Alternating Current (AC) signal.

The alternating polarity causes continuous charging and discharging cycles, allowing the AC signal to transfer across the component with low capacitive reactance (

X_C = \frac{1}{2\pi f C}

).

Determines how dynamic information and audio/RF waveforms propagate across the circuit.

3

Evaluate the combined circuit function of a series coupling capacitor.

The component blocks the steady DC bias voltage while coupling the alternating signal waveform directly to the next amplifier stage.

Confirms the core operational role of coupling capacitors in electronic signal chains.

Key Concept

Capacitor DC Blocking and AC Coupling

Estimated Time:1m 0s

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