Electronic Components

211 questions

All types211 Multiple choice107 True / False24 Numeric27 Matching43 Fill in the blank10

An electronics technician is assembling an energy-storage circuit for an industrial sensor module. The circuit consists of two $100\text{ }\mu\text{F}$ capacitors connected in series, and this entire combination is wired in parallel with a single $150\text{ }\mu\text{F}$ capacitor. What is the total equivalent capacitance of this circuit in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 200

Answer

The total equivalent capacitance of the circuit is

200\text{ }\mu\text{F}

For capacitors in series, total capacitance decreases and is calculated as

\frac{1}{C_{\text{series}}} = \frac{1}{100\text{ }\mu\text{F}} + \frac{1}{100\text{ }\mu\text{F}} = \frac{2}{100\text{ }\mu\text{F}}

, giving

C_{\text{series}} = 50\text{ }\mu\text{F}

. When this combination is connected in parallel with the

150\text{ }\mu\text{F}

capacitor, parallel capacitances add directly:

C_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}

Step-by-Step Solution

Calculate the equivalent capacitance of the two

100\text{ }\mu\text{F}

capacitors connected in series using the reciprocal formula

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}

C_{\text{series}} = 50\text{ }\mu\text{F}

Capacitors in series combine inversely, meaning two identical capacitors in series have half the capacitance of one capacitor.

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

C_{\text{total}} = C_{\text{series}} + C_3

C_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}

Capacitors connected in parallel add directly because their effective plate areas combine, increasing overall charge storage capacity.

Key Concept

Series and parallel equivalent capacitance rules (series capacitors combine inversely; parallel capacitors add directly).

Estimated Time:1m 0s

Question 122Question →

A $10\text{ k}\Omega$ linear potentiometer is wired as a variable rheostat by tying the center wiper (terminal 2) directly to outside terminal 3. If the control shaft is set to $25\%$ of its total travel starting from terminal 1 toward terminal 3, what is the effective resistance measured between terminal 1 and the combined terminal 2/3 connection?

7.5\text{ k}\Omega

2.5\text{ k}\Omega

1.875\text{ k}\Omega

25\text{ k}\Omega

Show answer & explanation

Answer: $2.5\text{ k}\Omega$

Answer

2.5\text{ k}\Omega

A potentiometer wired with its wiper connected to one outer terminal functions as a two-terminal variable resistor (rheostat). Setting the shaft to

25\%

rotation from terminal 1 creates a

2.5\text{ k}\Omega

resistance path between terminal 1 and terminal 2. Because terminal 2 is strapped to terminal 3, the remaining

7.5\text{ k}\Omega

section of the resistive element is completely bypassed by the short, leaving an effective resistance of

2.5\text{ k}\Omega

Step-by-Step Solution

Determine the resistance of the active resistive track between terminal 1 and the wiper (terminal 2).

R_{1-2} = 0.25 \times 10\text{ k}\Omega = 2.5\text{ k}\Omega

Linear potentiometers divide resistance directly in proportion to shaft travel percentage from the reference end.

Analyze the state of the remaining track segment between the wiper (terminal 2) and terminal 3.

R_{2-3} = 7.5\text{ k}\Omega

is short-circuited.

Connecting terminal 2 directly to terminal 3 creates a zero-resistance path across this section, bypassing it entirely.

Calculate the total effective resistance between terminal 1 and the joined 2/3 terminal.

R_{\text{eff}} = 2.5\text{ k}\Omega

Current flows through the unshorted

R_{1-2}

segment and exits via the combined wiper/terminal 3 connection.

Key Concept

Rheostat Wiring and Potentiometer Track Division

Estimated Time:45s

Question 123Question →

A semiconductor diode is a two-terminal electronic component that acts as a one-way valve for electric current. For current to conduct freely through a standard diode in a direct current (DC) circuit, how must the diode be connected relative to the power source?

The cathode must be connected toward the positive voltage terminal and the anode toward the negative terminal.

The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

Both the anode and cathode must be connected to the positive terminal of the power source.

The diode must be connected to an alternating current source so voltage polarities constantly reverse.

Show answer & explanation

Answer: The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

Answer

The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

A diode conducts electric current when it is placed in forward bias. In a direct current circuit, forward bias is established by connecting the anode toward the positive terminal of the power supply and the cathode toward the negative terminal.

Step-by-Step Solution

Identify the two terminals of a standard diode and the condition needed for conduction.

A diode has an anode (p-type material) and a cathode (n-type material), and it conducts current only when forward-biased.

Understanding diode terminal polarity is necessary to determine proper circuit orientation.

Determine the voltage orientation required to forward-bias the diode.

The anode must be at a higher electrical potential than the cathode.

Applying a positive voltage to the anode and a negative voltage to the cathode overcomes the internal barrier potential, narrowing the depletion region and permitting current flow.

Key Concept

A semiconductor diode allows current to flow when it is forward-biased, which occurs when the anode is connected toward the positive supply voltage and the cathode toward the negative side.

Estimated Time:45s

Question 124Question →

An AC step-down transformer has $600\text{ turns}$ on its primary winding and $50\text{ turns}$ on its secondary winding. The primary winding is connected to a $120\text{ V}$ AC source, and the secondary circuit delivers current to a load drawing $2.4\text{ A}$ . What is the current flowing in the primary winding?

28.8\text{ A}

20\text{ mA}

0.2\text{ A}

0\text{ A}

Show answer & explanation

Answer: $0.2\text{ A}$

Answer

0.2\text{ A}

In an ideal transformer, the ratio of primary current to secondary current is inversely proportional to the ratio of primary turns to secondary turns (

I_p / I_s = N_s / N_p

). Since the transformer has a

12:1

step-down turns ratio (

600/50 = 12

), the secondary voltage decreases by a factor of 12 while the primary current must be 12 times smaller than the secondary current:

I_p = 2.4\text{ A} / 12 = 0.2\text{ A}

Step-by-Step Solution

Determine the turns ratio between the primary and secondary windings

\frac{N_p}{N_s} = \frac{600}{50} = 12

The turns ratio determines both the voltage transformation ratio and the inverse current transformation ratio.

Apply the transformer current relationship to solve for primary current (

I_p

)

\frac{I_p}{I_s} = \frac{N_s}{N_p} \implies I_p = 2.4\text{ A} \times \frac{50}{600} = \frac{2.4\text{ A}}{12} = 0.2\text{ A}

In an ideal transformer, power in equals power out (

V_p I_p = V_s I_s

), which means current is inversely proportional to the number of turns.

Key Concept

Transformer Current and Turns Ratio Relationship

Estimated Time:1m 0s

Question 125Question →

Match each specialized switching or relay device to its corresponding operational characteristic and actuation method.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Reed switch matches ferromagnetic flexible metal reeds sealed in glass; Rotary switch matches rotating shaft and wiper contact selecting between three or more positions; Contactor matches heavy-duty electromechanical device for high-current power loads; Thermal overload relay matches monitoring sustained overcurrent using a bimetallic strip.

Each switching component is matched according to its fundamental physics of operation: the reed switch relies on magnetic field induction on ferrous contacts; the rotary switch provides multi-position manual circuit routing; the contactor uses an electromagnetic solenoid for high-amperage switching; and the thermal overload relay uses thermal deflection of a bimetallic strip for motor overload protection.

Step-by-Step Solution

Identify the actuation mechanism of a reed switch.

Recognize that reed switches operate magnetically via enclosed flexible ferrous blades.

Reed switches rely on external magnetic flux rather than manual toggling or internal thermal deflection.

Analyze the mechanical multi-position selection function of a rotary switch.

Associate the rotary switch with angular shaft rotation routing a signal to multiple switch poles/throws.

Rotary switches provide multi-throw circuit selection through rotational movement.

Distinguish between standard low-power control relays and contactors.

Connect contactors to high-current power switching controlled via electromagnetic coils.

Contactors feature heavy-duty contacts built specifically to handle high current without contact welding.

Determine the protective operation of a thermal overload relay.

Pair the thermal overload relay with bimetallic thermal expansion sensing excessive motor current.

Thermal overload relays protect motor windings by converting persistent current into heat that mechanically triggers auxiliary contacts.

Key Concept

Specialized switch and relay actuation mechanisms and circuit roles

Estimated Time:1m 0s

Question 126Question →

In order for a standard PN junction diode to conduct conventional electric current, its cathode must be connected to a higher positive electrical potential than its anode.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because forward biasing a PN junction diode requires the anode (P-type material) to be at a higher positive potential than the cathode (N-type material). Applying a higher positive potential to the cathode places the diode in reverse bias, which widens the depletion layer and prevents current flow.

Step-by-Step Solution

Identify the relationship between diode terminals and semiconductor materials.

The anode is connected to the P-type semiconductor region, and the cathode is connected to the N-type semiconductor region.

Understanding terminal orientation is essential to determining how external voltage affects the internal PN junction barrier.

Evaluate the biasing condition described in the statement.

Connecting the cathode to a more positive potential than the anode creates a reverse-bias condition, widening the depletion region and preventing conventional current conduction.

For conduction to occur, the diode must be forward-biased by applying a more positive potential to the anode relative to the cathode.

Key Concept

Diode Biasing and Polarity

Estimated Time:45s

Question 127Question →

An electronics technician connects three inductors in parallel across an AC filter circuit. The inductors have inductance values of $60\text{ mH}$ , $30\text{ mH}$ , and $20\text{ mH}$ , with negligible mutual inductance between them. What is the total equivalent inductance of the circuit in millihenries (mH)?

Show answer & explanation

Answer: 10

Answer

The total equivalent inductance of the parallel circuit is

10\text{ mH}

Inductors in parallel without mutual coupling combine according to the reciprocal formula

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

. Computing

\frac{1}{60} + \frac{1}{30} + \frac{1}{20} = \frac{6}{60} = \frac{1}{10}

, and inverting yields

10\text{ mH}

Step-by-Step Solution

State the reciprocal formula for inductors connected in parallel

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

When inductors are in parallel and uncoupled, the total rate of current change is the sum of the individual branch rates, resulting in a reciprocal combination rule identical to parallel resistors.

Substitute the individual inductance values

\frac{1}{L_{\text{total}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}

Direct substitution of the known branch values:

L_1 = 60\text{ mH}

L_2 = 30\text{ mH}

, and

L_3 = 20\text{ mH}

Find the common denominator and sum the terms

\frac{1}{L_{\text{total}}} = \frac{1 + 2 + 3}{60} = \frac{6}{60} = \frac{1}{10}\text{ mH}^{-1}

Using the lowest common denominator (

60

) to perform fractional addition.

Invert the reciprocal to obtain equivalent inductance

L_{\text{total}} = 10\text{ mH}

Taking the reciprocal of

\frac{1}{10}\text{ mH}^{-1}

gives

10\text{ mH}

Key Concept

Parallel Inductance Calculation

Estimated Time:1m 0s

Question 128Question →

A quality assurance technician is inspecting a four-band carbon-composition resistor with the color bands in order: Yellow, Violet, Red, and Silver. To meet quality standards, the resistor must operate within its specified manufacturer tolerance. What is the minimum acceptable resistance value, in ohms ( $\Omega$ ), for this component?

Show answer & explanation

Answer: 4230

Answer

The minimum acceptable resistance value is

4230\text{ }\Omega

The first two bands (Yellow = 4, Violet = 7) establish the significant figures of 47. The third band (Red) indicates a multiplier of

10^2 = 100

, giving a nominal rating of

4700\text{ }\Omega

. The fourth band (Silver) specifies a

\pm 10\%

tolerance. Subtracting

10\%

4700\text{ }\Omega

(

470\text{ }\Omega

) from the nominal rating gives the minimum allowable threshold of

4230\text{ }\Omega

Step-by-Step Solution

Decode the first two color bands to obtain the significant numerical digits.

Yellow represents the digit

4

, and Violet represents the digit

7

, forming the base value of

47

Standard EIA four-band resistor color coding assigns the first two bands as significant figures of the resistance value.

Multiply the base significant figures by the power of ten indicated by the third band.

Red indicates a multiplier of

10^2

(

100

), so the nominal resistance is

47 \times 100 = 4700\text{ }\Omega

The third band functions as the decimal multiplier for four-band fixed resistors.

Identify the tolerance rating from the fourth band.

Silver represents a tolerance of

\pm 10\%

The fourth band specifies the allowable deviation range from the nominal resistance value.

Calculate the lower allowable resistance threshold.

Tolerance amount is

4700\text{ }\Omega \times 0.10 = 470\text{ }\Omega

. Minimum acceptable value is

4700\text{ }\Omega - 470\text{ }\Omega = 4230\text{ }\Omega

The minimum acceptable value represents the nominal value minus the total tolerance deviation.

Key Concept

Four-Band Resistor Color Code Decoding and Tolerance Range Calculation

Alternative Method

Multiply the nominal resistance directly by the minimum percentage factor:

4700\text{ }\Omega \times (1.00 - 0.10) = 4700\text{ }\Omega \times 0.90 = 4230\text{ }\Omega

Estimated Time:45s

Question 129Question →

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

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

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 130Question →

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

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.

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 131Question →

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

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.

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 132Question →

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

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.

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.

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 133Question →

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

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.

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

Question 134Question →

A step-down transformer has a $10:1$ turns ratio between its primary and secondary windings. If the primary winding is connected across a steady $12\text{ V}$ direct current (DC) power supply, what voltage is measured across the secondary winding under steady-state conditions?

1.2\text{ V}

12\text{ V}

0\text{ V}

120\text{ V}

Show answer & explanation

Answer: $0\text{ V}$

Answer

0\text{ V}

Transformers operate through mutual electromagnetic induction, requiring an alternating current (AC) to generate an expanding and collapsing magnetic field. A steady direct current (DC) produces a static, unvarying magnetic field once the circuit reaches steady state. Because Faraday's Law requires a changing magnetic field (

\frac{\Delta\Phi}{\Delta t} \neq 0

) to induce an electromotive force, the secondary winding produces exactly

0\text{ V}

Step-by-Step Solution

Identify the type of input voltage applied to the transformer primary winding.

The input source is a steady

12\text{ V}

direct current (DC) supply.

Transformers rely on mutual electromagnetic induction, which fundamentally requires a time-varying magnetic field.

Analyze the magnetic flux created in the core during steady-state DC operation.

A steady DC current produces a stationary, unchanging magnetic flux (

\frac{\Delta\Phi}{\Delta t} = 0

According to Faraday's Law of Induction, voltage is only induced in a conductor when magnetic flux lines expand or collapse across the winding turns.

Determine the resulting secondary output voltage.

The induced secondary voltage is

0\text{ V}

With zero rate of change of magnetic flux in steady state, no electromotive force (EMF) is induced in the secondary winding.

Key Concept

Transformers require alternating current (AC) to induce a secondary voltage via changing magnetic flux; steady direct current (DC) produces no secondary voltage.

Estimated Time:1m 0s

Question 135Question →

In an automotive electrical system, an electromechanical relay is installed between a dashboard switch and a high-current cooling fan. What is the primary purpose of using the relay in this circuit configuration?

It acts as a sacrificial protective device that permanently melts open to interrupt excessive circuit current.

It allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.

It steps up the source voltage to deliver a higher electromotive force across the fan motor terminals.

It converts direct current from the battery into alternating current before reaching the fan motor.

Show answer & explanation

Answer: It allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.

Answer

The correct answer states that it allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.

An electromechanical relay utilizes a low current passing through an internal coil to generate a magnetic field. This magnetic field pulls an armature to close heavy-duty contacts, safely powering high-current loads like cooling fans without subjecting delicate dashboard switches to high current or requiring thick, heavy wiring runs through the passenger compartment.

Step-by-Step Solution

Identify the distinct circuit sections established by the relay.

The dashboard switch operates within a low-current control circuit connected to the relay coil, while the cooling fan operates within a separate high-current load circuit connected through the relay contacts.

Relays provide electrical isolation and remote switching between control circuits and power circuits.

Evaluate the technical necessity of isolating high-current loads from cabin switches.

Routing substantial current directly through the dashboard switch would require thick gauge wiring into the vehicle interior, cause significant voltage drop over long distances, and degrade manual switch contacts due to arcing.

Electromechanical relays permit small control currents to safely actuate high-amperage contacts placed near the battery and load.

Match the identified operational principle to the corresponding functional description.

The relay serves as a remote electromechanical switch allowing low-current input to control high-current output.

Relays do not act as fuses, voltage transformers, or DC-to-AC inverters.

Key Concept

Electromechanical relay function in high-current DC switching

Estimated Time:1m 0s

Question 136Question →

On the cylindrical casing of a standard axial-lead semiconductor diode, what feature is identified by the painted stripe or band near one of the ends?

The cathode terminal

The anode terminal

The nominal resistance rating in ohms

The maximum operating alternating current (AC) frequency

Show answer & explanation

Answer: The cathode terminal

Answer

The cathode terminal

On standard through-hole axial diodes, the painted stripe marks the cathode terminal. In schematic diagrams, this corresponds to the vertical bar toward which the triangle points.

Step-by-Step Solution

Identify the standard packaging markings used on axial-lead semiconductor diodes.

Standard cylindrical diodes (such as the 1N4001 series) use a single painted band at one end to distinguish polarity.

Because diodes are polarized components that conduct current primarily in one direction, physical orientation must be identifiable.

Correlate the physical stripe with the corresponding electronic terminal.

The stripe aligns with the cathode (negative lead during forward conduction), matching the vertical bar in the standard schematic symbol.

Correct alignment ensures the diode is wired in forward bias rather than accidental reverse bias.

Key Concept

Physical diode lead identification and polarity marking

Estimated Time:45s

Question 137Question →

In electronic power and communication systems, various inductor and transformer configurations serve specialized functions. Match each inductive component on the left with its primary operational characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

RF Choke matches with presenting high opposition to high-frequency AC signals while allowing DC to pass; Isolation Transformer matches with providing a 1:1 turns ratio to electrically decouple circuits for safety; Air-Core Inductor matches with eliminating magnetic core losses at ultra-high frequencies; Step-Up Transformer matches with increasing alternating voltage while decreasing output current proportionally.

Each inductive device is matched according to its fundamental electromagnetic function: an RF choke blocks AC via inductive reactance while passing DC; an isolation transformer provides 1:1 galvanic isolation; an air-core inductor eliminates ferromagnetic core losses at high frequencies; and a step-up transformer increases voltage while proportionally reducing current.

Step-by-Step Solution

Identify the primary behavior of an RF choke

Recognized as a high-frequency AC blocking inductor that freely passes DC.

Inductive reactance (

X_L = 2\pi f L

) increases directly with frequency, creating significant impedance at radio frequencies.

Identify the function of an isolation transformer

Recognized as a 1:1 turns ratio transformer used for safety and ground-loop isolation.

Equal winding counts maintain input voltage on the secondary side while eliminating direct conductive connections.

Determine the main design advantage of an air-core inductor

Recognized as eliminating ferromagnetic core losses.

Without a solid iron or magnetic core, there are zero eddy currents or magnetic hysteresis losses at high frequencies.

Evaluate the voltage and current relationship of a step-up transformer

Recognized as increasing voltage while decreasing current proportionally.

Due to the principle of power conservation (

V_p I_p \approx V_s I_s

), increasing secondary voltage requires a corresponding drop in secondary current.

Key Concept

Operational characteristics and applications of inductors and transformers, including frequency-dependent choke behavior, core losses, galvanic isolation, and turns ratio power relationships.

Estimated Time:1m 0s

Question 138Question →

An electronics technician selects a four-band carbon-film resistor with the color sequence Brown, Green, Red, and Gold. What is the nominal resistance and tolerance of this resistor?

150\ \Omega\text{ with } \pm 5\%\text{ tolerance}

1.5\text{ k}\Omega\text{ with } \pm 5\%\text{ tolerance}

1.5\text{ k}\Omega\text{ with } \pm 10\%\text{ tolerance}

15\text{ k}\Omega\text{ with } \pm 5\%\text{ tolerance}

Show answer & explanation

Answer: $1.5\text{ k}\Omega\text{ with } \pm 5\%\text{ tolerance}$

Answer

1.5\text{ k}\Omega\text{ with } \pm 5\%\text{ tolerance}

(

1,500\ \Omega \pm 5\%

)

Decoding the four color bands in standard order gives: Brown = 1 (first digit), Green = 5 (second digit), Red =

10^2

\times 100

(multiplier), and Gold =

\pm 5\%

(tolerance). Calculating

15 \times 100\ \Omega

yields

1,500\ \Omega

, which converts to

1.5\text{ k}\Omega

with a

\pm 5\%

tolerance.

Step-by-Step Solution

Identify the first two significant digit color bands.

Brown represents the digit 1, and Green represents the digit 5, forming the number 15.

In a standard four-band resistor, the first two bands establish the base significant figures.

Determine the multiplier from the third band.

Red represents a multiplier of

10^2

(

100

The third band indicates the power of ten by which the significant digits are multiplied.

Calculate the nominal resistance value.

15 \times 100 = 1,500\ \Omega = 1.5\text{ k}\Omega

Multiplying the base digits by the multiplier yields the total nominal resistance.

Identify the tolerance from the fourth band.

Gold indicates a tolerance of

\pm 5\%

The fourth band specifies the acceptable manufacturing variation percentage.

Key Concept

Four-band resistor color code decoding and tolerance identification

Estimated Time:35s

Question 139Question →

In electrical and electronic control systems, electromechanical relays and multi-contact switches rely on specialized internal components and contact configurations to direct current and automate circuits. Match each switch or relay component to its primary mechanical or electrical operational function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Common (COM) Terminal matches the shared conductive connection point that routes current to selectable paths; Relay Armature matches the movable ferromagnetic component attracted by magnetic flux; Normally Closed (NC) Contacts match the contact pair that maintains a closed conductive path in the resting de-energized state and opens when actuated; Relay Solenoid Coil matches the insulated wire winding that produces an electromagnetic pulling force.

Each component corresponds directly to its mechanical or electrical role in relay operation: the Common terminal serves as the shared current pathway, the armature provides the physical motion via magnetic attraction, Normally Closed contacts provide resting continuity, and the solenoid coil generates the magnetic force from the control circuit.

Step-by-Step Solution

Identify the function of the Common (COM) Terminal

Matched with the shared conductive connection point that routes current to one of multiple selectable contact paths.

In single-pole double-throw (SPDT) configurations and standard relays, the COM terminal is the central node connected internally to either the NO or NC contact.

Identify the function of the Relay Armature

Matched with the movable ferromagnetic component that physically transfers motion to open or close contacts.

The armature is the mechanical piece of iron that pivots when drawn toward the core by electromagnetic attraction.

Identify the function of the Normally Closed (NC) Contacts

Matched with the contact pair that maintains continuity during the resting de-energized state and opens upon actuation.

'Normally' designates the shelf or unpowered state, meaning NC contacts conduct electricity until an active force separates them.

Identify the function of the Relay Solenoid Coil

Matched with the insulated wire winding that produces an electromagnetic pulling force when control current flows through it.

The coil converts electrical input energy into a magnetic field via electromagnetic induction to drive the switching mechanism.

Key Concept

Electromechanical Relay Structure and Contact Nomenclature

Alternative Method

Categorize each item first by its physical domain: identify the electrical input terminal (COM), the mechanical moving element (Armature), the resting electrical state (NC contacts), and the electromagnetic actuator (Coil).

Estimated Time:1m 0s

Question 140Question →

A light-emitting diode (LED) produces light when it is connected in a reverse-bias orientation within a direct current (DC) circuit.

True

False

Show answer & explanation

Answer: False

Answer

False. A light-emitting diode (LED) produces light only when forward-biased; in a reverse-bias configuration, it blocks current flow and stays dark.

The statement is false because a light-emitting diode (LED) must be forward-biased (anode connected to the positive potential and cathode connected to the negative potential) to conduct electrical current and release energy in the form of light. In a reverse-bias orientation, the diode blocks current and produces no light.

Step-by-Step Solution

Identify the operating requirement of a semiconductor diode and LED.

Diodes conduct electric current in only one direction (forward bias) and block current in the opposite direction (reverse bias).

An LED functions as a standard PN-junction diode with the specialized property of emitting light when electric current conducts through it.

Determine the state of the LED in a reverse-bias orientation.

Under reverse bias, the internal depletion region widens, preventing current flow across the junction.

Because no charge carriers cross the junction under reverse bias, no electron-hole recombination occurs, preventing any light emission.

Key Concept

Light-emitting diodes (LEDs) require forward biasing (anode connected to a higher positive potential than cathode) to conduct current and emit light.

Estimated Time:45s

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