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

Match each capacitor specification or parameter with its correct electrical description.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Dielectric matches the insulating material between conductive plates; Working Voltage (WVDC) matches the maximum continuous direct-current potential difference before breakdown; Capacitive Reactance matches the frequency-dependent opposition to alternating current; and Farad matches the base unit of electrostatic capacity storing one coulomb per volt.

Each capacitor term corresponds to its fundamental physical or operational definition: the dielectric is the insulating layer, working voltage is the maximum safe direct voltage before breakdown, capacitive reactance is the frequency-dependent opposition to alternating current, and the farad is the SI unit representing charge storage per volt.

Step-by-Step Solution

1

Identify the physical insulating component inside a capacitor.

The dielectric is the insulating material placed between conductive plates that establishes permittivity and prevents direct conduction.

Capacitors store energy in an electrostatic field across an insulator, which is termed the dielectric.

2

Identify the maximum safe operating voltage parameter.

Working Voltage (WVDC) indicates the maximum continuous direct voltage the component can withstand without dielectric puncture.

Exceeding this rating causes dielectric breakdown, arcing, and component destruction.

3

Identify the parameter defining AC opposition.

Capacitive Reactance (

X_C

) is the opposition offered to alternating current, calculated as

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

.

Capacitors impede alternating current depending inversely on the signal frequency and capacitance value.

4

Identify the primary electrical unit of capacitance.

The Farad (

F

) is the standard unit of capacitance, where

1\text{ F} = 1\text{ C/V}

.

Capacitance (

C

) equals stored electrostatic charge (

Q

) divided by the voltage difference (

V

).

Key Concept

Capacitor Construction, Ratings, Units, and AC Behavior

Estimated Time:1m 0s

Question 2Question →

An industrial machine uses a safety interlock limit switch on an access guard door. When the guard door is fully closed, it pushes down a spring-loaded plunger to complete the control circuit. When the door is opened, the plunger returns to its unactuated resting position, breaking the circuit and cutting power. Which switch configuration describes this device?

A normally open (NO) momentary switch that completes the circuit only when mechanically actuated

A normally closed (NC) momentary switch that maintains electrical continuity in its resting position

A double-pole, double-throw (DPDT) toggle switch configured for polarity reversal

A thermal bimetallic breaker switch that opens automatically when current exceeds a set limit

Show answer & explanation

Answer: A normally open (NO) momentary switch that completes the circuit only when mechanically actuated

Answer

A normally open (NO) momentary switch that completes the circuit only when mechanically actuated

The 'normal' state of any electrical switch or relay contact is defined as its resting condition when no external mechanical force or coil energization is applied. In this access guard configuration, opening the door releases the plunger to its resting position, opening the circuit (infinite resistance, stopping current flow). Closing the door applies the mechanical force necessary to hold the contacts closed. Therefore, the device is a normally open (NO) momentary switch.

Step-by-Step Solution

1

Identify the unactuated (resting) electrical state of the switch contacts.

When the guard door opens, the plunger is released back to its resting state, opening the electrical circuit.

By industry definition, the designation of a switch (Normally Open vs. Normally Closed) is determined by its state when no external force or actuation is present.

2

Determine the switch classification corresponding to this resting state and action.

Because the switch is open in its resting position and only conducts when held down by the door, it is a Normally Open (NO) momentary switch.

Momentary NO contacts require continuous physical force to maintain continuity and spring back to an open circuit when that force is removed.

Key Concept

Switch Contact Normal States (NO vs. NC) and Momentary Interlock Principles

Estimated Time:45s

Question 3Question →

A technician is assembling a low-pass filter circuit and connects two inductors in series with values of $45\text{ mH}$ and $80\text{ mH}$ . Assuming there is no mutual magnetic coupling between the coils, what is the total equivalent inductance of the circuit in millihenries ( $\text{mH}$ )?

Show answer & explanation

Answer: 125

Answer

125

When inductors are connected in series without mutual inductance, their magnetic opposition to current changes accumulates directly, meaning the total inductance is simply the sum of the individual inductances (

45\text{ mH} + 80\text{ mH} = 125\text{ mH}

).

Step-by-Step Solution

1

Recall the formula for combining series inductances without mutual magnetic coupling.

L_{\text{total}} = L_1 + L_2

Like series resistors, total series inductance is the direct sum of individual inductances when there is no magnetic interaction.

2

Add the two individual inductance values together.

45\text{ mH} + 80\text{ mH} = 125\text{ mH}

Direct addition provides the equivalent total inductance in millihenries.

Key Concept

Inductors in series combine by direct addition, identical to the rule for resistors in series.

Estimated Time:45s

Question 4Question →

A technician is troubleshooting a power supply circuit featuring a full-wave bridge rectifier composed of four standard silicon diodes. During each single half-cycle of the alternating current (AC) input signal, how many of the four diodes conduct current simultaneously through the load?

Only one diode conducts while the remaining three diodes stay idle.

Two diodes conduct while the remaining two diodes are reverse-biased.

All four diodes conduct simultaneously in parallel to share the load current.

Three diodes conduct in series while one diode blocks the return path.

Show answer & explanation

Answer: Two diodes conduct while the remaining two diodes are reverse-biased.

Answer

Two diodes conduct while the remaining two diodes are reverse-biased.

In a standard four-diode bridge rectifier, alternating current (AC) is converted to pulsating direct current (DC) by routing current through pairs of diodes. During any given half-cycle, two diagonally opposite diodes are forward-biased and conduct current through the load in a single direction, while the other two diodes are reverse-biased and block conduction.

Step-by-Step Solution

1

Analyze the bridge rectifier circuit topology under an AC input half-cycle

One AC input terminal is positive with respect to the other terminal.

AC voltage alternates polarity every half-cycle, determining the bias conditions of each diode in the bridge.

2

Trace the current flow through the forward-biased diodes and the load

Current enters the bridge, passes through one forward-biased diode, travels through the load resistor, and returns through a second forward-biased diode to the opposite AC line.

A complete circuit path requires a forward and return path through the bridge network.

3

Determine the state of the remaining two diodes

The opposite pair of diodes experiences reverse bias and blocks current flow.

Diodes only conduct when their anode is more positive than their cathode.

Key Concept

Full-Wave Bridge Rectification and Diode Conduction Pairs

Estimated Time:1m 0s

Question 5Question →

In electronic systems, standard PN-junction and specialized diodes exhibit distinct electrical behaviors depending on bias polarity and operating conditions. Match each diode operating condition with its characteristic electrical behavior.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Standard silicon diode in forward bias matches conducting current readily with an approximate

0.7\text{ V}

potential drop; Standard silicon diode in reverse bias matches blocking current flow with high resistance and negligible leakage; Zener diode in reverse breakdown matches maintaining a constant reference voltage across a range of currents; Light-emitting diode in forward bias matches releasing optical energy through electron-hole recombination.

Each diode state corresponds directly to its semiconductor junction physics: a forward-biased silicon diode exhibits a characteristic

0.7\text{ V}

barrier drop during conduction; a reverse-biased PN junction widens its depletion region to block current; a Zener diode exploits controlled reverse breakdown for voltage stabilization; and an LED emits light via radiative recombination under forward bias.

Step-by-Step Solution

1

Analyze standard forward bias characteristics of silicon diodes.

Identified that overcoming the silicon barrier potential (

0.7\text{ V}

) permits large forward current conduction with a steady

0.7\text{ V}

drop.

Forward bias narrows the depletion region once the threshold voltage is reached.

2

Analyze standard reverse bias behavior below breakdown limits.

Identified that reverse bias widens the depletion layer, preventing majority carrier flow and resulting in high resistance.

A reverse-biased PN junction acts essentially as an open circuit.

3

Analyze specialized Zener diode reverse breakdown operation.

Identified that reverse-biased Zener diodes hold a stable breakdown voltage across varying currents.

Controlled avalanche/Zener breakdown enables voltage regulation applications.

4

Analyze light-emitting diode (LED) forward conduction mechanics.

Identified that forward current injection causes radiative recombination emitting photons.

Electrons dropping across the direct bandgap release energy as light.

Key Concept

Diode Operating States and Biasing Characteristics

Estimated Time:1m 0s

Question 6Question →

A circuit requires a component that can store electrical energy in an electrostatic field and block direct current (DC) while permitting alternating current (AC) to pass. Which electronic component should be used?

Inductor

Capacitor

Resistor

Diode

Show answer & explanation

Answer: Capacitor

Answer

Capacitor

A capacitor is constructed of two conductive plates separated by an insulating dielectric material. When voltage is applied, electrical charge accumulates on the plates, creating an electrostatic field that stores energy. Because the dielectric prevents electrons from crossing directly between the plates, steady direct current is blocked. However, alternating current continuously charges and discharges the plates with alternating polarity, effectively allowing AC signals to pass.

Step-by-Step Solution

1

Identify the energy storage mechanism described in the circuit requirement.

Energy is stored in an electrostatic field created by separated opposite charges on conductive plates.

Capacitors store potential energy in an electric field across a dielectric, whereas inductors store energy in a magnetic field.

2

Evaluate the component behavior with DC and AC signals.

The insulating dielectric layer stops direct current once the plates are charged, while continuous alternating voltage polarity allows alternating current to pass through capacitive action.

This fundamental characteristic makes capacitors standard for DC blocking and AC coupling applications.

Key Concept

Capacitor electrostatic energy storage and DC blocking/AC passing characteristics

Estimated Time:45s

Question 7Question →

An electronics technician is analyzing an NPN bipolar junction transistor (BJT) configured as an amplifier in a common-emitter circuit. For the transistor to operate properly in its linear active region, which combination of internal junction biasing conditions must be established?

Forward-biasing both the base-emitter junction and the collector-base junction

Forward-biasing the base-emitter junction and reverse-biasing the collector-base junction

Reverse-biasing the base-emitter junction and forward-biasing the collector-base junction

Reverse-biasing both the base-emitter junction and the collector-base junction

Show answer & explanation

Answer: Forward-biasing the base-emitter junction and reverse-biasing the collector-base junction

Answer

Forward-biasing the base-emitter junction and reverse-biasing the collector-base junction

For any bipolar junction transistor (BJT) to function as a linear amplifier in the active region, the input base-emitter PN junction must be forward-biased to permit carrier injection, and the output collector-base PN junction must be reverse-biased to attract and collect these carriers across the junction.

Step-by-Step Solution

1

Identify the operational requirement for linear amplification in a BJT

Linear amplification requires the transistor to operate strictly in the active mode.

Operating outside the active region causes either signal clipping in saturation or complete signal cutoff.

2

Determine the required bias across the base-emitter (input) junction

The base-emitter junction must be forward-biased (positive base relative to emitter in an NPN transistor).

Forward bias reduces the potential barrier across the base-emitter junction, allowing majority carriers (electrons) to be injected from the emitter into the thin base layer.

3

Determine the required bias across the collector-base (output) junction

The collector-base junction must be reverse-biased (positive collector relative to base in an NPN transistor).

Reverse bias widens the collector-base depletion region and sets up a strong electric field that collects the injected electrons passing through the base into the collector.

Key Concept

BJT Active Region Biasing Conditions

Estimated Time:1m 15s

Question 8Question →

In an electronics schematic, a bipolar junction transistor (BJT) symbol features an arrow on the emitter lead pointing outward, away from the base. Which type of transistor is represented by this symbol?

PNP transistor

NPN transistor

Junction field-effect transistor (JFET)

Zener diode

Show answer & explanation

Answer: NPN transistor

Answer

An NPN transistor is represented by a BJT schematic symbol where the emitter arrow points outward, away from the base.

In electronics schematic diagrams, the arrow on the emitter terminal indicates the direction of conventional current flow. For an NPN transistor, this arrow points outward (away from the base). A common memory aid is 'NPN = Not Pointing iN'.

Step-by-Step Solution

1

Identify the component family and terminal structure.

The component is a Bipolar Junction Transistor (BJT) with three terminals: Collector, Base, and Emitter.

Recognizing the core three-terminal configuration establishes the category of semiconductor device.

2

Analyze the direction of the arrow located on the emitter terminal.

The arrow points outward (away from the base terminal).

In standard schematic conventions, the arrow on the emitter indicates the direction of conventional current flow during forward bias.

3

Match the arrow direction to the correct transistor polarity type.

An outward-pointing arrow corresponds to an NPN transistor ('Not Pointing iN'), whereas an inward-pointing arrow corresponds to a PNP transistor ('Pointing iN Proudly').

Differentiating between NPN and PNP configurations is essential for determining circuit polarity and proper biasing.

Key Concept

BJT schematic symbols use an arrow on the emitter terminal to distinguish between NPN (arrow pointing out) and PNP (arrow pointing in) configurations.

Estimated Time:45s

Question 9Question →

In a bipolar junction transistor (BJT) used in a common-emitter circuit, which terminal receives the small control current required to regulate the larger current flowing between the other two terminals?

Collector

Base

Emitter

Gate

Show answer & explanation

Answer: Base

Answer

The base terminal receives the small control current that regulates current flow between the collector and emitter.

In a bipolar junction transistor (BJT), the base is the central control terminal. A small current entering the base allows a significantly larger current to conduct between the collector and emitter, functioning as an amplifier or an electronic switch.

Step-by-Step Solution

1

Identify the terminal architecture of a bipolar junction transistor (BJT)

A standard BJT has three functional leads: the Base, the Collector, and the Emitter.

Establishes the available terminal functions specific to current-driven bipolar transistors.

2

Determine the control mechanism in a common-emitter configuration

Applying a small base current (

I_B

) modulates a larger collector current (

I_C

) according to the transistor gain (

\beta

), yielding

I_C = \beta \times I_B

.

Identifies the base as the designated input control lead in BJT switching and amplification circuits.

Key Concept

BJT Terminal Roles and Current Control

Estimated Time:45s

Question 10Question →

A circuit in a timing module contains two $30\text{ }\mu\text{F}$ capacitors connected in parallel. This parallel pair is then connected in series with a third $15\text{ }\mu\text{F}$ capacitor. What is the total equivalent capacitance of this network?

75\text{ }\mu\text{F}

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

30\text{ }\mu\text{F}

0.012\text{ }\mu\text{F}

Show answer & explanation

Answer: $12\text{ }\mu\text{F}$

Answer

The total equivalent capacitance of the network is

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

.

For capacitors in parallel, the total capacitance is the sum of the individual values:

30\text{ }\mu\text{F} + 30\text{ }\mu\text{F} = 60\text{ }\mu\text{F}

. When this

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

equivalent combination is connected in series with a

15\text{ }\mu\text{F}

capacitor, the product-over-sum rule is applied:

(60 \times 15) / (60 + 15) = 900 / 75 = 12\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two parallel capacitors (

C_1 = 30\text{ }\mu\text{F}

and

C_2 = 30\text{ }\mu\text{F}

).

C_p = C_1 + C_2 = 30\text{ }\mu\text{F} + 30\text{ }\mu\text{F} = 60\text{ }\mu\text{F}

Capacitors in parallel add their plate surface areas together, so their capacitances add directly.

2

Calculate the total capacitance when

C_p = 60\text{ }\mu\text{F}

is placed in series with

C_3 = 15\text{ }\mu\text{F}

.

C_{\text{total}} = \frac{C_p \times C_3}{C_p + C_3} = \frac{60 \times 15}{60 + 15} = \frac{900}{75} = 12\text{ }\mu\text{F}

Capacitors in series combine inversely using the reciprocal formula, identical to the formula used for parallel resistors.

Key Concept

Equivalent Capacitance in Series-Parallel Combinations

Estimated Time:1m 15s

Question 11Question →

An electronics technician is repairing a power supply filter network. The circuit requires replacing a section containing two $12\text{ }\mu\text{F}$ capacitors connected in series, which are arranged in parallel with a single $6\text{ }\mu\text{F}$ capacitor. What is the total equivalent capacitance of this entire combination?

4.8\text{ }\mu\text{F}

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

30\text{ }\mu\text{F}

12\text{ nF}

Show answer & explanation

Answer: $12\text{ }\mu\text{F}$

Answer

The total equivalent capacitance of the network is

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

.

To find the total equivalent capacitance, solve the series combination first: two

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

capacitors in series provide an equivalent capacitance of

\frac{12 \times 12}{12 + 12} = 6\text{ }\mu\text{F}

. Next, add the parallel branch value directly:

6\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 12\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two series-connected capacitors.

C_{\text{series}} = 6\text{ }\mu\text{F}

Capacitors in series combine reciprocally using the formula

C_{\text{series}} = \frac{C_1 \times C_2}{C_1 + C_2} = \frac{12 \times 12}{12 + 12} = \frac{144}{24} = 6\text{ }\mu\text{F}

.

2

Add the equivalent series capacitance to the parallel capacitor.

C_{\text{total}} = 12\text{ }\mu\text{F}

Capacitors in parallel add directly together:

C_{\text{total}} = C_{\text{series}} + C_3 = 6\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 12\text{ }\mu\text{F}

.

Key Concept

Capacitors in series combine reciprocally (reducing total capacitance), whereas capacitors in parallel add directly together (increasing total capacitance).

Alternative Method

Since the two series capacitors are identical, their equivalent capacitance is simply half of one capacitor's value:

\frac{12\text{ }\mu\text{F}}{2} = 6\text{ }\mu\text{F}

. Adding the identical

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

parallel branch doubles this value back to

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

.

Estimated Time:1m 30s

Question 12Question →

A technician connects a $10\text{ }\mu\text{F}$ capacitor in parallel with a $22\text{ }\mu\text{F}$ capacitor in a power supply filter circuit. What is the total equivalent capacitance of this combination in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 32

Answer

32 uF

When capacitors are connected in parallel, their total capacitance is the direct sum of the individual capacitances:

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

. Adding

10\text{ }\mu\text{F}

and

22\text{ }\mu\text{F}

results in a total equivalent capacitance of

32\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the circuit configuration and the relevant formula for equivalent capacitance.

The two capacitors are connected in parallel, meaning their total capacitance is the sum of their individual capacitances:

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

.

Connecting capacitors in parallel effectively increases the total plate surface area, allowing more electric charge to be stored at a given voltage.

2

Calculate the sum of the individual capacitances.

C_{\text{total}} = 10\text{ }\mu\text{F} + 22\text{ }\mu\text{F} = 32\text{ }\mu\text{F}

.

Direct addition yields the total equivalent capacitance.

Key Concept

Total capacitance in a parallel circuit is calculated by directly adding the individual capacitances (

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

).

Estimated Time:45s

Question 13Question →

A transistor can use a small electrical signal at one terminal to control a much larger current flowing through its other terminals.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because the defining characteristic of a transistor is its ability to govern a large current path using a relatively weak electrical control signal, making it effective for both amplification and switching.

Step-by-Step Solution

1

Identify the primary operational functions of a transistor in electronic circuits.

Transistors are solid-state components designed to act as electronic switches and signal amplifiers.

Recalling the basic definition and purpose of semiconductor devices establishes the core functional criteria.

2

Evaluate the relationship between the input control signal and the output circuit current.

Applying a relatively small signal to the control terminal modulates and directs a much larger current between the power terminals.

This control mechanism allows small electronic signals to switch heavy loads on or off and amplify weak audio or radio signals.

Key Concept

Transistor amplification and electronic switching principle

Estimated Time:30s

Question 14Question →

When a bipolar junction transistor operates in its cutoff region, the collector-to-emitter path behaves as an open switch, preventing significant current flow through the output circuit.

True

False

Show answer & explanation

Answer: True

Answer

True

In the cutoff region, the control junction receives negligible bias voltage, preventing base current from flowing. Consequently, the output conduction path between the collector and emitter remains non-conductive, functioning electrically like an open switch.

Step-by-Step Solution

1

Identify the specified operating state of the transistor

The statement specifies operation in the cutoff region.

A transistor's conductivity between its output terminals depends directly on its operating state: cutoff, active, or saturation.

2

Evaluate the current conduction characteristics in cutoff

Without base drive voltage, base current is zero, which halts collector current and creates an open-circuit condition.

In electronic switching applications, cutoff represents the 'off' or open-switch state where current cannot pass to the load.

Key Concept

Transistor Cutoff Region and Switching Operation

Estimated Time:45s

Question 15Question →

Unlike current-driven bipolar devices, an enhancement-mode metal-oxide-semiconductor field-effect transistor (MOSFET) conducts significant current between its drain and source terminals when its gate-to-source voltage is zero.

True

False

Show answer & explanation

Answer: False

Answer

False. Enhancement-mode MOSFETs are normally-off devices that require a gate-to-source voltage exceeding the threshold voltage to establish a channel and conduct current.

The evaluation 'False' is correct because enhancement-mode MOSFETs are normally-off solid-state devices. Without an applied gate-to-source voltage exceeding the device's threshold voltage (

V_{th}

), the channel between the drain and source remains non-conductive, preventing drain current from flowing.

Step-by-Step Solution

1

Identify the component classification and operating archetype specified in the statement.

The statement specifies an enhancement-mode metal-oxide-semiconductor field-effect transistor (MOSFET).

Different transistor families (bipolar vs. field-effect) and operational modes (enhancement vs. depletion) exhibit distinct default conduction states.

2

Analyze the channel status of an enhancement-mode device at zero gate-to-source voltage (

V_{GS} = 0\text{ V}

).

At

V_{GS} = 0\text{ V}

, no conductive channel connects the drain and source regions across the substrate, resulting in an open circuit where drain current (

I_D

) is approximately zero.

Enhancement-mode devices are manufactured without a pre-formed channel and operate as normally-off switches.

3

Determine what electrical condition is required to initiate conduction.

An electric field created by applying a gate-to-source voltage greater than the threshold voltage (

V_{GS} > V_{th}

) is mandatory to enhance/induce a channel.

Because current does not flow at zero bias, the assertion that it conducts significant current at zero gate voltage is incorrect.

Key Concept

Default conduction state and threshold voltage mechanics of enhancement-mode field-effect transistors

Alternative Method

Recall the root word 'enhance': the channel must be actively enhanced (created) by gate voltage, meaning at zero gate voltage, no channel exists and no current flows.

Estimated Time:45s

Question 16Question →

An audio receiver circuit contains three capacitors arranged in a sub-network. Two $8\text{ }\mu\text{F}$ capacitors are connected in series with each other, and this series combination is connected in parallel with a third capacitor rated at $16\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this sub-network in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

20 μF

The equivalent capacitance of two

8\text{ }\mu\text{F}

capacitors in series is

4\text{ }\mu\text{F}

because series capacitors follow the formula

\frac{1}{C_s} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}

, resulting in

C_s = 4\text{ }\mu\text{F}

. When this

4\text{ }\mu\text{F}

equivalent section is wired in parallel with the

16\text{ }\mu\text{F}

capacitor, their values add directly (

4\text{ }\mu\text{F} + 16\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

).

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two series-connected capacitors (

C_1 = 8\text{ }\mu\text{F}

and

C_2 = 8\text{ }\mu\text{F}

).

C_{\text{series}} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}\text{ }\mu\text{F}^{-1}

, which yields

C_{\text{series}} = 4\text{ }\mu\text{F}

.

2

Add the series branch equivalent capacitance (

4\text{ }\mu\text{F}

) to the parallel branch capacitor (

C_3 = 16\text{ }\mu\text{F}

).

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

Capacitors connected in parallel add directly:

C_{\text{total}} = C_{\text{series}} + C_3 = 4\text{ }\mu\text{F} + 16\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

.

Key Concept

Series and Parallel Capacitance Calculation

Estimated Time:1m 0s

Question 17Question →

A technician is troubleshooting a solid-state switching circuit that utilizes a silicon PNP bipolar junction transistor. The emitter terminal is connected directly to a +12V DC power source, and the collector is wired in series with a ground-referenced load. In order to drive this transistor into saturation and fully power the load, what specific voltage condition must exist at the base terminal, and what is the direction of conventional current flow through the device?

The base must be at a higher voltage potential than the emitter; conventional current flows from the collector to the emitter.

The base must be at a higher voltage potential than the emitter; conventional current flows from the emitter to the collector.

The base must be at a lower voltage potential than the emitter; conventional current flows from the emitter to the collector.

The base must be at a lower voltage potential than the emitter; conventional current flows from the collector to the emitter.

Show answer & explanation

Answer: The base must be at a lower voltage potential than the emitter; conventional current flows from the emitter to the collector.

Answer

The base must be at a lower voltage potential than the emitter; conventional current flows from the emitter to the collector.

The correct response accurately synthesizes both fundamental rules for a PNP transistor. First, to achieve saturation (full conduction), the base must be at a lower potential than the emitter to forward-bias the emitter-base junction. Second, unlike NPN devices, conventional current in a PNP transistor enters through the emitter and exits through the collector, following the direction of the emitter arrow on its schematic symbol.

Step-by-Step Solution

1

Identify the type of transistor and its fundamental biasing requirements.

The device is a PNP transistor. To turn it on (saturation), the base must be negative relative to the emitter.

PNP transistors operate with opposite polarities compared to NPN transistors. Forward-biasing the emitter-base junction requires the P-type emitter to be more positive than the N-type base.

2

Determine the direction of conventional current flow through the component.

Conventional current flows from the emitter to the collector.

The schematic symbol for a PNP transistor has an arrow on the emitter pointing inward (towards the base), which indicates the direction of conventional current flow from emitter to collector.

3

Synthesize the findings to select the correct operational state.

Base voltage must be lower than emitter voltage, with current flowing emitter-to-collector.

Combining the biasing rule (V_base < V_emitter) and the current path establishes the correct parameters for full conduction.

Key Concept

PNP Transistor Biasing and Current Flow

Question 18Question →

An avionics communication filter assembly contains a capacitor network designed for signal conditioning. The network consists of two $40\text{ }\mu\text{F}$ capacitors connected in series. This series combination is connected in parallel with an $80\text{ }\mu\text{F}$ capacitor. Finally, this entire parallel arrangement is connected in series with a single $25\text{ }\mu\text{F}$ capacitor. What is the total equivalent capacitance of the network, in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

The total equivalent capacitance of the network is

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

.

The correct equivalent capacitance is

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

. In capacitor networks, series components combine reciprocally (

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

) and parallel components add directly (

C_p = C_1 + C_2

). Resolving the initial series branch yields

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

, which in parallel with

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

produces

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

. Finally, combining

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

in series with

25\text{ }\mu\text{F}

results in

\frac{100 \times 25}{100 + 25} = 20\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the initial series branch consisting of two

40\text{ }\mu\text{F}

capacitors.

C_{s1} = 20\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{s1}} = \frac{1}{40\text{ }\mu\text{F}} + \frac{1}{40\text{ }\mu\text{F}} = \frac{2}{40\text{ }\mu\text{F}} = \frac{1}{20\text{ }\mu\text{F}}

.

2

Calculate the equivalent capacitance of the parallel combination formed by

C_{s1}

and the

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

capacitor.

C_p = 100\text{ }\mu\text{F}

Capacitors in parallel add directly:

C_p = C_{s1} + C_3 = 20\text{ }\mu\text{F} + 80\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

.

3

Calculate the total circuit capacitance by placing the parallel equivalent

C_p

in series with the

25\text{ }\mu\text{F}

capacitor.

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

Applying the series formula for the final combination:

\frac{1}{C_{\text{total}}} = \frac{1}{C_p} + \frac{1}{C_4} = \frac{1}{100\text{ }\mu\text{F}} + \frac{1}{25\text{ }\mu\text{F}} = \frac{1+4}{100\text{ }\mu\text{F}} = \frac{5}{100\text{ }\mu\text{F}} = \frac{1}{20\text{ }\mu\text{F}}

.

Key Concept

Calculation of total equivalent capacitance in compound series-parallel networks using reciprocal series addition and direct parallel addition rules.

Alternative Method

Use the product-over-sum shortcut for each two-component series reduction:

C_{s1} = \frac{40 \times 40}{40 + 40} = \frac{1600}{80} = 20\text{ }\mu\text{F}

. Next, add the parallel component directly:

C_p = 20 + 80 = 100\text{ }\mu\text{F}

. Finally, apply product-over-sum for the last series stage:

C_{\text{total}} = \frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}

.

Estimated Time:1m 30s

Question 19Question →

Which of the following correctly identifies the three terminal leads of a standard bipolar junction transistor (BJT)?

Source, gate, and drain

Anode, cathode, and gate

Emitter, base, and collector

Positive, negative, and ground

Show answer & explanation

Answer: Emitter, base, and collector

Answer

Emitter, base, and collector

A standard bipolar junction transistor (BJT) is a three-terminal semiconductor device consisting of the emitter (which emits charge carriers), the base (the thin middle region that controls current), and the collector (which gathers the charge carriers).

Step-by-Step Solution

1

Identify the component family and type specified in the question.

The component is a bipolar junction transistor (BJT).

Different transistor families utilize different lead naming conventions.

2

Recall the fundamental three-layer structure of a BJT.

A BJT is formed by two p-n junctions creating three regions: emitter, base, and collector.

The base controls current flow between the emitter and the collector.

Key Concept

Bipolar Junction Transistor (BJT) Terminals

Estimated Time:45s

Question 20Question →

In the physical construction of a standard fixed capacitor, what is the primary role of the dielectric material placed between the two conductive plates?

To create an electromagnetic field that opposes rapid changes in circuit current

To act as an insulating barrier that prevents direct electrical contact between the plates while supporting an electrostatic field

To provide a low-resistance path for direct current while completely blocking alternating current

To act as a conductive bridge that allows continuous electron flow across the gap between the plates

Show answer & explanation

Answer: To act as an insulating barrier that prevents direct electrical contact between the plates while supporting an electrostatic field

Answer

To act as an insulating barrier that prevents direct electrical contact between the plates while supporting an electrostatic field

The dielectric in a capacitor is an electrical insulator (such as ceramic, mica, paper, or air) placed between the conductive plates. Its primary purpose is to physically isolate the plates so direct current cannot cross the gap, thereby allowing equal and opposite charges to accumulate and establish an electrostatic field.

Step-by-Step Solution

1

Identify the basic construction of a capacitor

A capacitor consists of two conducting plates separated by an insulating layer known as the dielectric.

Understanding the physical structure establishes the role of each component.

2

Determine the electrical function of the dielectric material

The dielectric prevents charges on opposite plates from neutralizing each other through direct contact, allowing an electric field to be established and energy to be stored.

Insulators do not allow free electrons to flow through them, maintaining the potential difference between the plates.

Key Concept

Dielectric function and energy storage in capacitors

Estimated Time:45s

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