Electronic Components

211 questions

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

A field-effect transistor (FET) regulates current flow through a semiconductor channel using an electrostatic field. Which terminal receives the controlling input voltage to modulate this current?

Drain

Gate

Source

Cathode

Show answer & explanation

Answer: Gate

Answer

The gate terminal is the input control electrode of a field-effect transistor that receives voltage to regulate channel current.

In a field-effect transistor (FET), the gate terminal functions as the control element. An applied input voltage at the gate generates an electrostatic field that modulates the resistance and carrier flow through the channel connecting the source and drain.

Step-by-Step Solution

Identify the standard three terminals of a field-effect transistor (FET).

A field-effect transistor consists of the source, drain, and gate terminals.

Identifying the terminal designations is the first step in determining their functional roles.

Differentiate between the current-carrying terminals and the control terminal.

Current travels through the channel between the source and drain terminals, while voltage applied to the gate terminal establishes an electric field that widens or restricts the channel.

The gate provides voltage-based control over current conduction without requiring significant input current.

Key Concept

Field-Effect Transistor (FET) Terminal Roles and Electrostatic Control

Estimated Time:45s

Question 22Question →

An NPN bipolar junction transistor circuit is configured with voltage-divider biasing from a single $15\text{ V}$ DC power supply. The voltage divider consists of an upper resistor $R_1 = 20\text{ k}\Omega$ connected to the positive supply and a lower resistor $R_2 = 10\text{ k}\Omega$ connected to ground. The emitter terminal is connected to ground through an emitter resistor $R_E = 1\text{ k}\Omega$ . Assuming a standard silicon base-to-emitter forward voltage drop of $V_{BE} = 0.7\text{ V}$ and negligible base current, what is the approximate DC emitter current ( $I_E$ ) flowing through the transistor?

5.0\text{ mA}

4.3\text{ mA}

5.7\text{ mA}

10.0\text{ mA}

Show answer & explanation

Answer: $4.3\text{ mA}$

Answer

4.3\text{ mA}

In an NPN voltage-divider biased circuit, the base voltage relative to ground is determined by the divider ratio:

V_B = 15\text{ V} \times (10\text{ k}\Omega / 30\text{ k}\Omega) = 5.0\text{ V}

. Because the base-emitter junction is forward-biased, it drops

0.7\text{ V}

, establishing an emitter voltage of

V_E = 5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}

. Dividing this emitter voltage by the

1\text{ k}\Omega

emitter resistor yields an emitter current of

4.3\text{ mA}

Step-by-Step Solution

Calculate the DC voltage at the base terminal (

V_B

) using the voltage divider rule

V_B = 15\text{ V} \times \frac{10\text{ k}\Omega}{20\text{ k}\Omega + 10\text{ k}\Omega} = 15\text{ V} \times \frac{1}{3} = 5.0\text{ V}

The base voltage is established by the ratio of the lower resistor to the total resistance across the supply when base current loading is negligible.

Determine the DC voltage at the emitter terminal (

V_E

) by accounting for the base-emitter forward voltage drop

V_E = V_B - V_{BE} = 5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}

In an active NPN transistor, the base-emitter junction is forward-biased, dropping approximately 0.7 V from base to emitter.

Apply Ohm's law across the emitter resistor (

R_E

) to calculate the emitter current (

I_E

)

I_E = \frac{V_E}{R_E} = \frac{4.3\text{ V}}{1\text{ k}\Omega} = 4.3\text{ mA}

The emitter current equals the voltage developed across the emitter resistor divided by its resistance.

Key Concept

Transistor Voltage-Divider DC Biasing and Emitter Current Calculation

Estimated Time:1m 30s

Question 23Question →

A technician needs to modify a parallel-plate capacitor to increase its capacitance for an audio filtering application. Which design change will accomplish this?

Inserting a solid iron core between the plates to concentrate magnetic flux

Increasing the surface area of the conductive plates while decreasing the separation distance between them

Replacing the insulating dielectric spacer with a conductive aluminum strip

Widening the plate separation gap to permit steady direct current to flow through the component

Show answer & explanation

Answer: Increasing the surface area of the conductive plates while decreasing the separation distance between them

Answer

Increasing the surface area of the conductive plates while decreasing the separation distance between them

Capacitance (

C

) of a parallel-plate capacitor is given by the formula

C = \frac{\varepsilon A}{d}

, where

\varepsilon

represents the permittivity of the dielectric,

A

is the surface area of the plates, and

d

is the distance between them. Increasing the plate surface area and decreasing the separation distance between the plates directly increases the capacitance.

Step-by-Step Solution

Identify the formula governing the capacitance of a parallel-plate capacitor.

The capacitance formula is

C = \frac{\varepsilon A}{d}

, where

C

is capacitance,

\varepsilon

is the permittivity of the dielectric material,

A

is the plate surface area, and

d

is the plate separation distance.

Capacitance is determined by the physical geometry of the plates and the insulating properties of the dielectric.

Determine which physical modifications increase the value of

C

Increasing the plate area (

A

) and decreasing the separation distance (

d

) both increase the total capacitance.

A larger plate area accommodates more electric charge at a given voltage, while a smaller gap strengthens the electrostatic field between the plates.

Key Concept

Physical factors affecting parallel-plate capacitance

Estimated Time:1m 0s

Question 24Question →

Match each type of capacitor to its primary characteristic or typical circuit application.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrolytic capacitor matches polarized component offering high capacitance values for power supply filtering; Ceramic disc capacitor matches non-polarized component with low capacitance for high-frequency bypass circuits; Variable capacitor matches adjustable capacitance device for radio frequency tuning circuits.

Each capacitor type is tailored to specific circuit needs: electrolytic capacitors provide high capacitance values and require proper polarity orientation for power supply smoothing; ceramic disc capacitors are non-polarized and offer low capacitance values optimal for RF bypass; variable capacitors provide adjustable capacitance to tune receivers to target frequencies.

Step-by-Step Solution

Identify the construction and primary application of electrolytic capacitors.

Electrolytic capacitors use a thin chemical oxide layer as a dielectric, providing high capacitance in a compact size, but they must be connected with correct polarity (+/-) and are widely used to smooth ripple voltage in DC power supplies.

Polarity sensitivity and large capacitance are the defining operational traits of electrolytic capacitors.

Identify the construction and primary application of ceramic disc capacitors.

Ceramic disc capacitors have a solid ceramic dielectric, are non-polarized, and feature low capacitance ratings, making them stable and low-loss at high frequencies for bypass and filter tasks.

Non-polarity and low parasitic inductance suit ceramic capacitors for high-frequency AC bypassing.

Identify the construction and primary application of variable capacitors.

Variable capacitors allow the user to change plate overlap area or spacing, changing capacitance to tune resonant tank circuits to specific radio broadcast frequencies.

Adjustable capacitance directly shifts the resonant frequency in RF tuning circuits.

Key Concept

Capacitor construction types, polarity requirements, and their respective circuit applications.

Estimated Time:45s

Question 25Question →

An electronics technician is analyzing an AC circuit that consists of a capacitor connected directly across an alternating current signal generator. If the generator's operating frequency is doubled while the source voltage remains constant, which of the following statements correctly describes the effect on the capacitive reactance and the circuit current?

The capacitive reactance doubles, which causes the circuit current to decrease by half.

The capacitive reactance decreases by half, which causes the circuit current to double.

The capacitive reactance remains constant, which results in no change to the circuit current.

The capacitive reactance drops to zero, because the dielectric begins conducting continuous current at higher frequencies.

Show answer & explanation

Answer: The capacitive reactance decreases by half, which causes the circuit current to double.

Answer

The capacitive reactance decreases by half, which causes the circuit current to double.

Capacitive reactance is calculated using

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

, which shows that reactance is inversely proportional to the AC frequency. When the frequency is doubled, the opposition to AC current is halved. According to Ohm's law for AC circuits (

I = \frac{V}{X_C}

), reducing the reactance by half while holding the voltage constant doubles the current flowing through the circuit.

Step-by-Step Solution

Identify the formula for capacitive reactance

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

Capacitive reactance (

X_C

) measures the opposition a capacitor presents to alternating current and depends inversely on frequency (

f

) and capacitance (

C

Determine the change in capacitive reactance when frequency doubles

X_{C,\text{new}} = \frac{1}{2\pi (2f) C} = \frac{1}{2} X_C

Because frequency is in the denominator, doubling

f

halves the total capacitive reactance.

Apply Ohm's law for AC circuits to find the change in current

I_{\text{new}} = \frac{V}{X_{C,\text{new}}} = \frac{V}{\frac{1}{2}X_C} = 2 \times \left(\frac{V}{X_C}\right) = 2I

With constant source voltage

V

, cutting the opposition to current (

X_C

) by half causes the circuit current to double.

Key Concept

Capacitive Reactance and AC Frequency Relationship

Estimated Time:1m 0s

Question 26Question →

In electronics, bipolar junction transistors (BJTs) rely on three specialized terminals to control and route current. Match each BJT terminal to its primary functional role in a circuit.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Base matches with serving as the thin control layer that regulates carrier flow; Emitter matches with introducing and supplying majority charge carriers; Collector matches with gathering charge carriers to deliver output current.

Each terminal of a BJT has a designated task: the emitter supplies majority carriers, the base controls the rate of carrier flow via a small control signal, and the collector gathers these carriers to route current to the load circuit.

Step-by-Step Solution

Identify the primary role of the emitter terminal

The emitter is heavily doped to emit (inject) electrons or holes into the transistor body.

This establishes the supply of charge carriers needed for current flow.

Identify the primary role of the base terminal

The base controls the current flow through a small biasing voltage and current.

Because it is narrow and lightly doped, small variations at the base gate the bulk of current flowing across the device.

Identify the primary role of the collector terminal

The collector collects charge carriers swept across the base-collector junction.

It handles the bulk of power dissipation and passes the resulting amplified or switched current to the load.

Key Concept

Bipolar Junction Transistor (BJT) terminal roles and charge carrier dynamics

Estimated Time:45s

Question 27Question →

An electronics technician is analyzing solid-state switching and amplification circuits. Match each transistor operating region to its corresponding junction biasing condition and functional circuit behavior.

Click a left item, then click its matching right item

Show answer & explanation

Answer

BJT Cutoff matches both junctions reverse-biased (open switch); BJT Saturation matches both junctions forward-biased (closed switch); BJT Forward-Active matches forward-biased base-emitter and reverse-biased base-collector (linear amplification); JFET Ohmic Region matches unpinched channel at low drain voltage (voltage-controlled resistor).

Transistor states are governed by internal PN junction bias levels and gate field controls: Cutoff has both BJT junctions reverse-biased (acting as an open switch), Saturation has both BJT junctions forward-biased (acting as a closed switch), Forward-Active utilizes a forward-biased base-emitter and reverse-biased base-collector junction for linear amplification, and the JFET Ohmic region operates prior to pinch-off as a voltage-controlled variable resistor.

Step-by-Step Solution

Determine the junction bias conditions for digital BJT switching states.

Cutoff requires reverse-biasing both the base-emitter and base-collector junctions (open switch), while saturation requires forward-biasing both junctions (closed switch).

Biasing conditions control whether carrier flow is completely blocked or fully saturated across the semiconductor junctions.

Determine the junction bias conditions for BJT linear amplification.

Forward-active operation requires the base-emitter junction to be forward-biased (

V_{BE} \approx 0.7\text{ V}

) and the base-collector junction to be reverse-biased.

This asymmetric bias configuration allows a small base current to modulate a substantially larger collector current proportionally.

Analyze the operational characteristics of a field-effect transistor below pinch-off.

In the ohmic (triode) region, the drain voltage is lower than the pinch-off threshold, so the channel acts as an electric-field-controlled resistor.

The gate-to-source voltage alters depletion layer width, varying the cross-sectional conductive area and resistance of the channel.

Key Concept

Transistor Operating States, Biasing Rules, and Terminal Relationships

Estimated Time:1m 30s

Question 28Question →

Match each transistor terminal lead with its primary operational function in solid-state circuits.

Click a left item, then click its matching right item

Show answer & explanation

Answer

BJT Base pairs with 'Supplies the small input control current that regulates the main current flow'; BJT Emitter pairs with 'Injects the primary charge carriers into the middle semiconductor region'; BJT Collector pairs with 'Gathers majority charge carriers to route output current through the load'; FET Gate pairs with 'Applies an electrostatic field to modulate channel conductivity without drawing steady DC input current'.

In bipolar junction transistors, the base regulates the conduction pathway through a small input current, the emitter emits/injects majority charge carriers, and the collector gathers those carriers to deliver current to the load. In field-effect transistors, the gate acts as an electrostatic control terminal that modulates channel conduction via voltage without drawing significant steady-state current.

Step-by-Step Solution

Determine the control and injection roles of BJT leads

The base acts as the control terminal requiring current, and the emitter supplies the charge carriers.

Bipolar junction transistors are current-driven devices where a small base current forward-biases the base-emitter junction to release carriers from the emitter.

Determine the output gathering role of the BJT collector

The collector collects carriers that successfully pass through the thin base layer.

The collector is physically designed with a larger area and reverse bias to capture carriers and handle heat dissipation from load current.

Determine the voltage-control role of the FET gate lead

The gate regulates current through an electrostatic field without drawing steady current.

Field-effect transistors have extremely high input impedance at the gate, relying on voltage rather than current to modulate channel resistance.

Key Concept

Transistor Terminal Functions and Operating Mechanisms

Estimated Time:1m 0s

Question 29Question →

For a silicon PNP bipolar junction transistor to operate correctly in the active region as a linear amplifier, the DC biasing circuit must ensure that the collector terminal is maintained at a higher positive voltage relative to the emitter terminal.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because in a PNP transistor, the emitter must be at a higher positive potential than the collector to ensure current flows correctly (from emitter to collector) and the junctions are properly biased for the active region.

Step-by-Step Solution

Identify the specific transistor type and its intended operating state.

The device is a PNP bipolar junction transistor operating in the active region (as a linear amplifier).

Different transistor types (NPN vs. PNP) require completely opposite voltage polarities for the same operating state.

Recall the structural composition of a PNP transistor.

A PNP transistor consists of a P-type emitter, an N-type base, and a P-type collector.

Understanding the semiconductor materials (P or N) dictates how junctions must be biased.

Determine the required junction biasing for the active region.

For any BJT to act as a linear amplifier (active region), its emitter-base junction must be forward-biased, and its collector-base junction must be reverse-biased.

This specific biasing arrangement allows a small base current to control a larger collector current without entering saturation or cutoff.

Apply the biasing rules to the PNP terminal voltages.

To forward-bias the P-N emitter-base junction, the P-type emitter must be more positive than the N-type base. To reverse-bias the P-N collector-base junction, the P-type collector must be more negative than the N-type base.

Forward biasing requires connecting positive potential to P-type material; reverse biasing requires connecting negative potential to P-type material.

Evaluate the statement based on the deduced voltage potentials.

The emitter is the most positive terminal in a properly biased PNP amplifier circuit. Therefore, the collector is at a lower voltage relative to the emitter, making the original statement false.

The statement claims the collector is at a higher positive voltage, which incorrectly describes an NPN transistor's biasing.

Key Concept

Transistor Biasing and Polarity Configuration

Estimated Time:1m 30s

Question 30Question →

A technician is analyzing a power supply smoothing circuit. The circuit contains three capacitors arranged in a specific configuration: Capacitor $C_1$ ( $30 \mu\text{F}$ ) and Capacitor $C_2$ ( $15 \mu\text{F}$ ) are connected in series. This entire series branch is then connected in parallel with Capacitor $C_3$ ( $90 \mu\text{F}$ ). What is the total equivalent capacitance of this capacitor network?

100 \mu\text{F}

30 \mu\text{F}

135 \mu\text{F}

9 \mu\text{F}

Show answer & explanation

Answer: $100 \mu\text{F}$

Answer

The correct equivalent capacitance is

100 \mu\text{F}

To find the total capacitance, first evaluate the series branch. Unlike resistors, capacitors in series use the inverse formula:

C_{12} = (30 \times 15) / (30 + 15) = 10 \mu\text{F}

. This equivalent capacitor is in parallel with

C_3

. Parallel capacitors simply add their values together:

10 \mu\text{F} + 90 \mu\text{F} = 100 \mu\text{F}

Step-by-Step Solution

Calculate the equivalent capacitance of the series branch containing

C_1

and

C_2

C_{12} = \frac{30 \times 15}{30 + 15} = \frac{450}{45} = 10 \mu\text{F}

Capacitors in series add inversely, which is the exact opposite of resistors. The product-over-sum formula applies to two capacitors in series.

Calculate the total equivalent capacitance by combining the series branch (

C_{12}

) in parallel with

C_3

C_{\text{total}} = 10 + 90 = 100 \mu\text{F}

Capacitors in parallel add directly together, as the effective plate area is increased.

Key Concept

Calculating equivalent capacitance in mixed series and parallel circuits.

Estimated Time:1m 30s

Question 31Question →

Match each type of capacitor with its distinctive construction characteristic or typical circuit application.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrolytic Capacitor matches with polarized high-capacitance design; Ceramic Disc Capacitor matches with non-polarized high-frequency bypass; Variable (Trimmer) Capacitor matches with mechanically adjustable capacitance for RF tuning; Film (Polyester) Capacitor matches with high stability, low loss, and high insulation resistance.

Each capacitor type is tailored to specific circuit needs based on dielectric properties: electrolytic capacitors deliver high capacitance with strict polarity; ceramic discs provide non-polarized high-frequency bypassing; variable capacitors allow adjustable RF tuning; and film capacitors deliver low loss and high thermal stability.

Step-by-Step Solution

Identify the polarity and capacitance range requirements of electrolytic capacitors.

Electrolytic capacitors are uniquely polarized and offer high capacitance values suitable for power supply smoothing.

The thin chemical dielectric layer creates high capacitance per volume, but reverse biasing breaks down this layer destructively.

Evaluate the properties of ceramic disc capacitors.

Ceramic capacitors are non-polarized with small capacitance values, ideal for high-frequency decoupling.

Ceramic materials provide low parasitic inductance and withstand fast transient frequencies.

Identify the mechanism and function of variable capacitors.

Variable capacitors permit mechanical rotation of plates to vary capacitance for tuning circuits.

Varying the effective plate overlap area directly shifts the resonant frequency in radio frequency circuits.

Examine the dielectric properties of film capacitors.

Film capacitors provide exceptional stability, low dielectric absorption, and low loss.

Plastic film dielectrics maintain precise electrical characteristics across temperature variations without significant leakage.

Key Concept

Classification of capacitor types by dielectric construction, polarity constraints, and functional applications in electronic circuits.

Estimated Time:1m 0s

Question 32Question →

When a voltage source is connected across a capacitor, in what form does the component store its electrical energy?

In an electromagnetic field created around a conductive wire coil

In an electrostatic field established between two conductive plates

Through chemical oxidation-reduction reactions between electrolyte materials

As thermal energy dissipated across a resistive element

Show answer & explanation

Answer: In an electrostatic field established between two conductive plates

Answer

In an electrostatic field established between two conductive plates

A capacitor is designed to store electrical charge. When connected to a voltage source, opposite charges accumulate on its conductive plates, setting up an electrostatic field across the dielectric layer that stores potential electrical energy.

Step-by-Step Solution

Identify the basic structure and operating principle of a capacitor.

A capacitor consists of two conductive plates separated by an insulating dielectric material.

Understanding the physical construction determines how the component interacts with electrical charges.

Analyze how energy is stored when voltage is applied across the plates.

Positive charges accumulate on one plate while negative charges accumulate on the opposite plate, creating an electric (electrostatic) field across the dielectric.

Unlike inductors that store energy magnetically or batteries that store energy chemically, capacitors store potential energy directly in this electrostatic field.

Key Concept

Capacitors store electrical potential energy in an electrostatic field between oppositely charged conductive plates.

Estimated Time:45s

Question 33Question →

In electronic systems, capacitors perform distinct functional roles depending on their placement and frequency response. Match each specific capacitor circuit application to its operational mechanism and electrical effect.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Bypass/Decoupling Capacitor matches with shunting high-frequency transient switching noise to ground; Coupling/DC Blocking Capacitor matches with passing time-varying AC signals while blocking DC bias voltages; Power Supply Reservoir/Filter Capacitor matches with storing charge at peak voltage to reduce rectified ripple; and RC Differentiator Capacitor matches with converting rapid step transitions into sharp timing trigger pulses.

Each capacitor application leverages the relationship

i = C \frac{dv}{dt}

and

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

: decoupling capacitors provide low-impedance paths to ground for high-frequency transients; coupling capacitors isolate DC bias while passing dynamic AC waveforms; filter capacitors smooth pulsating rectified DC by storing and releasing electrostatic charge; and differentiator capacitors convert steep voltage edges into timing spikes.

Step-by-Step Solution

Analyze the frequency-dependent reactance formula

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

for each circuit topology.

At DC (

f = 0

X_C \rightarrow \infty

(open circuit / DC blocking); at high frequencies (

f \gg 0

X_C \rightarrow 0

(short circuit / AC conducting).

Capacitor impedance is inversely proportional to frequency, governing how capacitors separate or condition signals.

Identify parallel-to-ground noise attenuation vs. parallel-to-load energy storage.

Parallel IC bypass capacitors dump high-frequency noise spikes to ground, whereas large parallel rectifier capacitors supply energy during voltage valleys to smooth ripple.

Both configurations are parallel, but bypass handles high-frequency transient diversion while filter/reservoir handles low-frequency energy reservoir smoothing.

Differentiate series signal coupling from series differentiating networks.

Series coupling uses a large time constant (

RC \gg T

) to preserve signal shape while blocking DC bias; series differentiation uses a short time constant (

RC \ll T

) so the output is proportional to

\frac{dV}{dt}

, creating sharp spikes.

The ratio of circuit time constant

\tau = RC

relative to signal period determines whether a series capacitor acts as a pure coupler or a waveform differentiator.

Key Concept

Frequency-Dependent Impedance and Functional Topologies of Capacitors

Estimated Time:1m 30s

Question 34Question →

Solid-state electronic systems rely heavily on transistors for signal processing and power control. What are the two primary functions that a transistor performs in an electronic circuit?

Converting alternating current to direct current through rectification

Amplifying electrical signals and acting as an electronic switch

Storing electrical energy within an electrostatic field

Stepping up or down alternating voltages using electromagnetic induction

Show answer & explanation

Answer: Amplifying electrical signals and acting as an electronic switch

Answer

Amplifying electrical signals and acting as an electronic switch

Transistors are three-terminal active semiconductor devices designed to amplify weak electrical signals (such as audio or radio frequencies) and to serve as high-speed solid-state switches in digital logic circuits and power controllers.

Step-by-Step Solution

Identify the fundamental operational modes of a transistor in electronic circuits.

Transistors operate linearly in the active region to boost signal strength and non-linearly in cutoff/saturation regions to turn currents on and off.

Recognizing the functional scope of active semiconductor devices distinguishes them from passive components.

Match the identified operational modes with the standard electronic roles.

The two primary roles are amplification (signal gain) and switching (binary on/off control).

Alternative functions like rectification belong to diodes, charge storage belongs to capacitors, and inductive transformation belongs to transformers.

Key Concept

Primary functions of transistors (amplification and switching)

Estimated Time:45s

Question 35Question →

An inductor with an inductance of $2\text{ H}$ is connected across an alternating current ( $\text{AC}$ ) signal generator. If the generator frequency is doubled while maintaining a constant output voltage, how will the inductive reactance of the coil and the circuit current change?

The inductive reactance is halved, and the circuit current doubles.

The inductive reactance doubles, and the circuit current is halved.

The inductive reactance doubles, and the circuit current doubles.

The inductive reactance remains constant, and the circuit current drops to zero.

Show answer & explanation

Answer: The inductive reactance doubles, and the circuit current is halved.

Answer

The inductive reactance doubles, and the circuit current is halved.

Inductive reactance is defined by the relationship

X_L = 2\pi f L

. When the AC frequency (

f

) is doubled, the inductive reactance doubles because of their direct linear relationship. According to Ohm's law for AC circuits (

I = \frac{V}{X_L}

), increasing the total opposition by a factor of

2

at a constant voltage reduces the alternating current to half its original value.

Step-by-Step Solution

Determine the effect of frequency on inductive reactance

The inductive reactance (

X_L

) doubles.

Inductive reactance is calculated using the formula

X_L = 2\pi f L

. Since

X_L

is directly proportional to frequency (

f

), multiplying the frequency by

2

doubles the inductive reactance.

Determine the effect of changed reactance on circuit current using Ohm's law for AC circuits

The current (

I

) is reduced to one-half of its initial value.

Current in an inductive AC circuit is governed by

I = \frac{V}{X_L}

. With voltage (

V

) held constant, current is inversely proportional to inductive reactance (

X_L

). Doubling the reactance cuts the current in half.

Key Concept

Inductive Reactance and Frequency Relationship in AC Circuits

Estimated Time:1m 0s

Question 36Question →

An electronics technician is troubleshooting an NPN bipolar junction transistor (BJT) configured as a solid-state switch. To transition the transistor from the cutoff state into saturation to energize an attached load, which electrical condition must be established at the base-emitter junction?

Reverse-biasing the base-emitter junction by making the base negative with respect to the emitter

Forward-biasing the base-emitter junction by applying a positive voltage to the base relative to the emitter

Maintaining zero volts across the base-emitter junction while reverse-biasing the collector-base junction

Applying a negative voltage to the collector while drawing conventional current out of the base terminal

Show answer & explanation

Answer: Forward-biasing the base-emitter junction by applying a positive voltage to the base relative to the emitter

Answer

Forward-biasing the base-emitter junction by applying a positive voltage to the base relative to the emitter causes the NPN transistor to transition into saturation.

In an NPN bipolar junction transistor, the base is composed of P-type semiconductor material while the emitter is N-type. To turn the transistor ON and enter saturation, the base-emitter junction must be forward-biased with a positive base voltage that exceeds the silicon barrier potential (

\approx 0.7\text{ V}

). This introduces base current, which allows heavy electron flow from emitter to collector, effectively closing the circuit across the load.

Step-by-Step Solution

Identify the semiconductor structure and terminal polarities for an NPN transistor.

An NPN bipolar junction transistor consists of an N-type emitter, a thin P-type base, and an N-type collector.

Understanding component internal structure determines the required polarity for current control.

Determine the required junction biasing to turn the transistor ON.

The base-emitter (P-N) junction must be forward-biased by establishing a base potential approximately

0.7\text{ V}

higher than the emitter.

Forward bias reduces the depletion region barrier, allowing majority carriers to flow across the junction.

Analyze the conditions necessary to reach the saturation operating state.

Supplying sufficient forward base current drives the collector-emitter voltage (

V_{CE}

) to its minimum near

0.2\text{ V}

, acting as a closed switch.

Saturation occurs when maximum collector current flows, turning the solid-state switch fully ON to power the load.

Key Concept

Transistor Biasing and Solid-State Switching States

Estimated Time:1m 0s

Question 37Question →

In a multi-stage audio amplifier circuit, a capacitor is placed in series between the output of one transistor stage and the input of the next stage. What is the primary purpose of this coupling capacitor?

To store energy within a magnetic field to oppose sudden changes in circuit current

To convert the direct current power supply voltage into an alternating current audio signal

To pass the alternating current audio signal while blocking the direct current bias voltage

To conduct direct current continuously while insulating against alternating signal frequencies

Show answer & explanation

Answer: To pass the alternating current audio signal while blocking the direct current bias voltage

Answer

To pass the alternating current audio signal while blocking the direct current bias voltage

In multi-stage amplifiers, each transistor stage requires specific direct current (DC) biasing voltages to operate in its linear region. A coupling capacitor placed in series between stages acts as an open circuit to steady DC (

X_C = \infty

0\text{ Hz}

), preventing the DC bias of one stage from disturbing the next, while exhibiting low capacitive reactance to alternating current (AC) audio signals, allowing the desired signal to transfer unimpeded.

Step-by-Step Solution

Analyze the capacitive reactance formula in relation to frequency

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

, meaning capacitive reactance is inversely proportional to frequency.

Determining how the component reacts to different signal types establishes its filtering and coupling characteristics.

Evaluate the behavior of the capacitor at zero frequency (direct current)

For DC (

f = 0\text{ Hz}

), reactance

X_C \to \infty

, effectively acting as an open circuit that blocks DC voltage.

Amplifier stages require separate DC operating bias voltages that must not interfere with adjacent stages.

Evaluate the behavior of the capacitor for audio frequencies (alternating current)

For AC (

f > 0\text{ Hz}

X_C

drops to a low value, allowing the alternating signal voltage to pass freely to the next stage.

Coupling requires the time-varying signal information to transfer smoothly across the circuit without attenuation.

Key Concept

DC Blocking and AC Coupling Behavior of Capacitors

Estimated Time:1m 0s

Question 38Question →

An automotive auxiliary circuit contains three capacitors connected in parallel with capacitance values of $12\text{ }\mu\text{F}$ , $18\text{ }\mu\text{F}$ , and $30\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this parallel combination in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 60

Answer

The total equivalent capacitance of the parallel circuit is

60\text{ }\mu\text{F}

In parallel circuits, capacitor values add together directly (

C_{\text{total}} = C_1 + C_2 + C_3

). Adding

12\text{ }\mu\text{F} + 18\text{ }\mu\text{F} + 30\text{ }\mu\text{F}

yields an equivalent total capacitance of

60\text{ }\mu\text{F}

Step-by-Step Solution

Identify the circuit configuration and the appropriate formula for parallel capacitors.

C_{\text{total}} = C_1 + C_2 + C_3

Connecting capacitors in parallel effectively combines their plate surface areas, meaning the total capacitance is found by direct summation.

Substitute the individual capacitor values into the formula and solve.

C_{\text{total}} = 12\text{ }\mu\text{F} + 18\text{ }\mu\text{F} + 30\text{ }\mu\text{F} = 60\text{ }\mu\text{F}

Performing the addition gives the total equivalent capacitance in microfarads.

Key Concept

Total capacitance in a parallel circuit is the sum of the individual capacitances.

Question 39Question →

A technician is testing an unlabeled three-terminal semiconductor component using a digital multimeter in diode-check mode. The meter indicates a forward voltage drop of approximately $0.7\text{ V}$ when the positive (red) meter lead is placed on Terminal 1 and the negative (black) lead is placed on either Terminal 2 or Terminal 3. All reverse probe combinations, as well as direct measurements between Terminal 2 and Terminal 3, show an open-circuit (over-limit) reading. Which device type and lead configuration does this component represent?

PNP bipolar junction transistor with Terminal 1 as the base

NPN bipolar junction transistor with Terminal 1 as the base

NPN bipolar junction transistor with Terminal 1 as the collector

PNP bipolar junction transistor with Terminal 1 as the emitter

Show answer & explanation

Answer: NPN bipolar junction transistor with Terminal 1 as the base

Answer

An NPN bipolar junction transistor with Terminal 1 as the base.

In a bipolar junction transistor (BJT), the base-emitter and base-collector regions behave as two separate PN junction diodes sharing a common base material. For an NPN transistor, the base is composed of P-type semiconductor material, while the collector and emitter are N-type. Applying a positive test lead to the P-type base (Terminal 1) and a negative lead to either N-type terminal (Terminals 2 and 3) forward-biases each junction, yielding the standard silicon forward drop of approximately

0.7\text{ V}

Step-by-Step Solution

Analyze the multimeter diode test behavior

Conduction occurs only when Terminal 1 is connected to the positive meter lead and Terminals 2 or 3 are connected to the negative lead.

A digital multimeter diode test applies a small forward voltage from the red (+) probe to the black (-) probe to measure the junction threshold drop.

Identify the semiconductor layer structure

Terminal 1 acts as a common P-type anode, while Terminals 2 and 3 act as separate N-type cathodes.

Silicon PN junctions drop approximately

0.7\text{ V}

when forward-biased (anode positive relative to cathode).

Determine the transistor type and lead identity

The component is an NPN transistor where Terminal 1 is the Base (P-region), Terminal 2 is the Collector/Emitter (N-region), and Terminal 3 is the Emitter/Collector (N-region).

An NPN transistor electrically models as two back-to-back diodes with anodes joined at the base.

Key Concept

BJT internal PN junction polarity and diode-test verification

Estimated Time:1m 15s

Question 40Question →

On a standard electronic schematic diagram, the arrow on the emitter terminal of an NPN bipolar junction transistor points outward, away from the base.

True

False

Show answer & explanation

Answer: True

Answer

True: The emitter arrow on an NPN transistor schematic symbol points outward, away from the base.

In standard schematic diagrams, all bipolar junction transistors place a directional arrow on the emitter lead. For an NPN transistor, this arrow points outward, away from the base terminal, representing conventional current leaving the device. A widely used technician mnemonic is 'NPN = Not Pointing iN'.

Step-by-Step Solution

Identify the component type and schematic lead conventions.

The component is an NPN bipolar junction transistor (BJT) consisting of Base, Collector, and Emitter leads.

Determining the transistor family and terminal configuration is required to verify the symbol markings.

Determine the arrow placement and orientation for the NPN configuration.

The arrow is always positioned on the emitter lead and points outward (away from the base line).

Schematic symbols use the emitter arrow to indicate the direction of conventional current flow, which flows out of the emitter in an NPN transistor (commonly recalled with the mnemonic 'Not Pointing iN').

Evaluate the accuracy of the provided statement.

The statement matches standard schematic conventions exactly.

An outward-pointing emitter arrow uniquely defines the NPN transistor symbol.

Key Concept

BJT Schematic Symbols and Emitter Arrow Orientation

Estimated Time:40s

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