Electronic Components

211 questions

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

An uncharged capacitor is connected in series with a resistor and a DC voltage source through an open switch. When the switch is suddenly closed, how do the voltage drop across the capacitor and the circuit current behave from the initial instant ( $t = 0$ ) to steady-state condition ( $t \ge 5\tau$ )?

Initial capacitor voltage equals the source voltage with zero circuit current; steady-state capacitor voltage is zero with maximum circuit current.

Initial and steady-state capacitor voltage both remain at zero while continuous DC current flows indefinitely through the circuit.

Initial capacitor voltage is zero with maximum circuit current; steady-state capacitor voltage equals the source voltage with zero circuit current.

Initial capacitor voltage equals the source voltage with maximum current; steady-state voltage and current both drop to zero.

Show answer & explanation

Answer: Initial capacitor voltage is zero with maximum circuit current; steady-state capacitor voltage equals the source voltage with zero circuit current.

Answer

At the initial instant (

t = 0

), the uncharged capacitor has zero voltage drop and allows maximum circuit current, while at steady state (

t \ge 5\tau

), the capacitor voltage reaches the full source voltage and circuit current drops to zero.

Because a capacitor stores energy in an electrostatic field between its plates, its voltage cannot change instantaneously. When connected to a DC source at

t = 0

, an uncharged capacitor exhibits zero potential difference (

0\text{ V}

) across its plates and behaves momentarily like a short circuit, allowing peak current (

I = V_{\text{source}} / R

) to flow through the series circuit. As charge accumulates over five time constants (

5\tau

), the potential difference across the capacitor rises to match the source voltage (

V_C = V_{\text{source}}

), completely balancing the applied EMF. At this steady state, the capacitor acts as an open circuit, reducing direct current flow to zero.

Step-by-Step Solution

Analyze the electrical state of an uncharged capacitor at the exact instant the DC switch is closed (

t = 0

Because charge cannot accumulate instantaneously (

Q = C \times V

), the initial capacitor voltage is

V_C(0) = 0\text{ V}

. The capacitor acts momentarily as a short circuit, allowing maximum current determined entirely by the series resistor:

I_{\text{max}} = \frac{V_{\text{source}}}{R}

Capacitors oppose instantaneous changes in voltage.

Analyze the transient charging phase over time (

0 < t < 5\tau

As electrons accumulate on the negative plate and depart the positive plate, an opposing electric field develops across the dielectric. The voltage

V_C(t)

increases exponentially according to

V_C(t) = V_{\text{source}}(1 - e^{-t/RC})

, while current decreases exponentially:

I(t) = I_{\text{max}} e^{-t/RC}

The accumulating charge creates a counter-voltage that reduces the net potential driving current through the series resistor.

Determine the steady-state circuit values after five time constants (

t \ge 5\tau

The capacitor reaches approximately

99.3\%

of full charge. The voltage across the capacitor equals the source voltage (

V_C = V_{\text{source}}

), and the net voltage across the resistor becomes

0\text{ V}

, causing circuit current to cease entirely (

I = 0\text{ A}

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

Once the opposing electrostatic potential equals the applied DC source potential, no further charge transfer can occur.

Key Concept

DC Transient and Steady-State Behavior of Capacitors in RC Circuits

Estimated Time:1m 30s

Question 42Question →

In standard electronic schematic symbols for bipolar junction transistors (BJTs), the directional arrow indicating current flow is positioned on the emitter terminal.

True

False

Show answer & explanation

Answer: True

Answer

True. The directional arrow in a BJT schematic symbol is always located on the emitter terminal.

The statement is true because electronic drafting standards universally position the current direction arrow on the emitter terminal of a BJT. This arrow serves two primary purposes: it distinguishes the emitter lead from the collector lead, and its direction (pointing out for NPN, pointing in for PNP) identifies the polarity type of the transistor.

Step-by-Step Solution

Identify the standard terminals of a bipolar junction transistor (BJT).

A BJT has three terminals: base, collector, and emitter.

Recognizing the physical and schematic lead names is necessary to evaluate the statement.

Recall the schematic drawing conventions used to represent BJTs.

The base is drawn as a vertical bar, the collector is an un-arrowed diagonal branch, and the emitter is a diagonal branch with an arrow.

Standard electronics symbology uses the arrow to indicate the direction of conventional current flow through the device.

Verify the terminal location of the identifying arrow.

The arrow is placed exclusively on the emitter terminal for both NPN and PNP configurations.

The arrow differentiates the emitter from the collector and establishes the transistor type (pointing outward for NPN, inward for PNP).

Key Concept

BJT Schematic Symbol Conventions

Estimated Time:45s

Question 43Question →

A pulse-generator circuit contains two $50\text{ }\mu\text{F}$ capacitors connected in parallel. This parallel pair is then connected in series with a third capacitor rated at $25\text{ }\mu\text{F}$ . What is the total equivalent capacitance of the circuit in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

The total equivalent capacitance of the circuit is

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

The correct answer is

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

. When two

50\text{ }\mu\text{F}

capacitors are wired in parallel, their total capacitance is

50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

. When this

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

equivalent capacitance is connected in series with the

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

capacitor, the product-over-sum formula yields

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

Step-by-Step Solution

Calculate the equivalent capacitance of the two parallel capacitors (

C_1

and

C_2

C_p = 50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

Capacitors in parallel add their plate surface areas directly, so their individual capacitance values are summed (

C_p = C_1 + C_2

Calculate the total series capacitance combining the

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

parallel equivalent with the

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

third capacitor (

C_3

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

Capacitors in series combine reciprocally according to the product-over-sum rule for two components:

C_{\text{total}} = \frac{C_p \cdot C_3}{C_p + C_3}

Key Concept

Capacitors combine opposite to resistors: parallel capacitors add directly (

C_p = C_1 + C_2

), whereas series capacitors combine using reciprocal formulas (

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

Alternative Method

Use the standard reciprocal formula for the series stage:

\frac{1}{C_{\text{total}}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{100} + \frac{1}{25} = \frac{1}{100} + \frac{4}{100} = \frac{5}{100} = \frac{1}{20}

, hence

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

Estimated Time:1m 0s

Question 44Question →

In basic electronics, the capacitance of a parallel-plate capacitor is determined by its physical construction and the dielectric material between its plates. Which of the following modifications will increase the capacitance of a parallel-plate capacitor?

Increasing the distance separating the two conductive plates

Increasing the surface area of the conductive plates

Winding the conductive material into tight coils to store energy in a magnetic field

Operating the component at higher AC frequencies to increase its Farad rating

Show answer & explanation

Answer: Increasing the surface area of the conductive plates

Answer

Increasing the surface area of the conductive plates increases the capacitance of a parallel-plate capacitor.

Capacitance (

C

) of a parallel-plate capacitor is given by

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

, where

\varepsilon

represents the permittivity of the dielectric material,

A

is the surface area of the plates, and

d

is the distance between the plates. Because capacitance is directly proportional to plate area, increasing the plate surface area allows more charge to be stored at any given voltage, directly increasing total capacitance.

Step-by-Step Solution

Recall the fundamental formula defining the capacitance of a parallel-plate capacitor.

The formula is

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

, where

C

is capacitance,

\varepsilon

is dielectric permittivity,

A

is the plate surface area, and

d

is the distance between plates.

Identifying the relationship between physical parameters and capacitance establishes which changes increase or decrease the value.

Analyze the proportional relationship between plate area and capacitance.

Capacitance is directly proportional to plate surface area (

C \propto A

) and inversely proportional to separation distance (

C \propto \frac{1}{d}

Because plate area

A

appears in the numerator of the formula, expanding

A

causes capacitance

C

to increase.

Identify the correct modification that increases capacitance.

Increasing the surface area of the conductive plates increases the total capacitance.

A larger plate area accommodates a greater quantity of charge for a given electric potential.

Key Concept

Physical factors affecting parallel-plate capacitance (

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

)

Estimated Time:1m 0s

Question 45Question →

Match each capacitor term or specification on the left with its correct electrical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Farad (F) matches with the standard unit used to quantify electrical capacitance; Dielectric matches with the insulating material separating the conductive plates; Electrolytic capacitor matches with a polarized capacitor type sensitive to connection polarity; Working voltage rating matches with the maximum continuous voltage that can safely be applied.

Each term accurately pairs with its foundational physical definition: the farad is the SI unit of capacitance, the dielectric is the insulating medium between plates, electrolytic capacitors are polarized devices requiring correct terminal orientation, and the working voltage rating defines the maximum safe operating voltage before dielectric puncture.

Step-by-Step Solution

Identify the standard unit of capacitance

Farad (F) is the metric unit measuring charge storage ability.

Capacitance (

C = Q/V

) is quantified in units of farads.

Identify the insulating component inside a capacitor

The dielectric is the insulating layer.

Dielectrics block steady direct current conduction across the plates while concentrating the electric field.

Identify the polarized capacitor category

Electrolytic capacitors have designated positive and negative terminals.

Reverse polarization disrupts the chemical oxide layer, potentially destroying the component.

Identify the safe maximum voltage parameter

Working voltage rating represents the dielectric breakdown threshold.

Operating above this threshold risks arcing and dielectric breakdown.

Key Concept

Fundamental capacitor terminology, units, internal construction, and operational ratings

Estimated Time:45s

Question 46Question →

Once a capacitor placed in a direct current (DC) circuit reaches a fully charged state, how does it affect the continuous flow of current in that branch?

It blocks direct current completely because the accumulated charge creates an opposing voltage equal to the supply voltage.

It allows direct current to conduct freely because the dielectric spacer functions as an electrical conductor.

It transforms the incoming direct current into an oscillating alternating current output.

It sustains ongoing current flow by continuously storing energy within a magnetic field.

Show answer & explanation

Answer: It blocks direct current completely because the accumulated charge creates an opposing voltage equal to the supply voltage.

Answer

A fully charged capacitor completely blocks the continuous flow of direct current by acting as an open circuit.

The correct choice explains that as a capacitor charges, its plate voltage rises until it matches the applied direct current source voltage. At this equilibrium point, no net voltage difference exists to move additional electrons through the circuit branch, completely stopping current flow and causing the capacitor to act as an open circuit to steady DC.

Step-by-Step Solution

Identify the charging behavior of a capacitor connected to a DC source.

Current flows initially as electrons accumulate on one plate and leave the other, creating a potential difference across the dielectric.

Capacitors respond to changes in voltage by charging until their terminal voltage matches the applied source voltage.

Determine the state of the circuit when the capacitor is fully charged.

The voltage across the capacitor equals the supply voltage (

V_C = V_S

), so the net voltage driving current becomes zero (

I = 0

With no potential difference remaining across the circuit resistance, current ceases entirely, meaning the fully charged capacitor acts as an open circuit to DC.

Key Concept

DC Blocking Behavior of Capacitors

Estimated Time:45s

Question 47Question →

In an electromechanical relay, energizing the coil produces a magnetic field that pulls the armature, which causes the normally closed (NC) contacts to establish circuit continuity and the normally open (NO) contacts to break circuit continuity.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because the descriptor 'normally' refers to the unenergized resting state of the relay. When current energizes the electromagnetic coil, the armature moves to close normally open (NO) contacts—completing that circuit—and open normally closed (NC) contacts—interrupting that circuit.

Step-by-Step Solution

Identify the baseline resting (de-energized) state of relay contacts

Normally open (NO) contacts are open (no continuity) and normally closed (NC) contacts are closed (continuity present) when the coil carries no current.

The term 'normally' specifies the default mechanical condition of the contacts prior to coil energization.

Analyze the mechanical effect of coil energization

Current flowing through the coil generates an electromagnetic field that pulls the movable armature toward the core.

Electromagnetic attraction provides the physical force required to switch contact positions.

Determine the resulting contact states when energized

Armature displacement forces NO contacts to close (establishing continuity) and NC contacts to open (breaking continuity).

Energization transitions all contact pairs to the state opposite their resting configuration.

Compare the derived operation with the given statement

The statement asserts that energization causes NC contacts to establish continuity and NO contacts to break continuity, which is the exact inverse of actual relay operation.

Because the statement inverts the functional behavior of NO and NC contacts during coil activation, it is false.

Key Concept

Electromechanical Relay Contact States (NO vs. NC) During Coil Energization

Estimated Time:1m 15s

Question 48Question →

In a standard bipolar junction transistor (BJT), which terminal receives the small input control current that regulates the larger current flow between the remaining two terminals?

Collector

Base

Emitter

Gate

Show answer & explanation

Answer: Base

Answer

The base terminal is the control electrode that receives the small input current in a bipolar junction transistor.

In a bipolar junction transistor (BJT), the base terminal is the thin middle semiconductor layer that controls the flow of charge carriers. Injecting a small base current allows a much larger current to flow from the collector to the emitter (or vice versa in PNP transistors), making the base the primary control terminal.

Step-by-Step Solution

Identify the three terminals of a bipolar junction transistor (BJT) and their basic functions.

A BJT consists of an emitter, a base, and a collector.

Recognizing the standard terminal configuration is essential for analyzing transistor operation.

Determine which terminal acts as the input control mechanism for current regulation.

The base is the thin, lightly doped central region where a small base current (

I_B

) directly controls a substantially larger collector current (

I_C

In current-controlled bipolar devices, variation in base current governs overall device conduction.

Key Concept

BJT Terminal Roles and Current Control Mechanism

Estimated Time:45s

Question 49Question →

An avionics technician constructs a filtering network using three capacitors. Two capacitors, rated at $0.05\text{ }\mu\text{F}$ and $50\text{ nF}$ , are connected in series with each other. This series combination is then connected in parallel with a third capacitor rated at $15\text{ nF}$ . What is the total equivalent capacitance of the network in nanofarads ( $\text{nF}$ )?

Show answer & explanation

Answer: 40

Answer

The total equivalent capacitance of the network is

40\text{ nF}

First, convert

0.05\text{ }\mu\text{F}

to nanofarads:

0.05\text{ }\mu\text{F} \times 1000\text{ nF}/\mu\text{F} = 50\text{ nF}

. Next, the two series capacitors (

50\text{ nF}

and

50\text{ nF}

) combine via the reciprocal formula to give

25\text{ nF}

. Finally, adding the parallel capacitor (

15\text{ nF}

) yields

25\text{ nF} + 15\text{ nF} = 40\text{ nF}

Step-by-Step Solution

Convert the microfarad rating to nanofarads to ensure uniform units across all calculations.

0.05\text{ }\mu\text{F} = 0.05 \times 10^3\text{ nF} = 50\text{ nF}

Capacitor combination formulas require all terms to share the same metric prefix.

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

C_1 = 50\text{ nF}

and

C_2 = 50\text{ nF}

C_{\text{series}} = \frac{1}{\frac{1}{50\text{ nF}} + \frac{1}{50\text{ nF}}} = 25\text{ nF}

Capacitors in series combine reciprocally, meaning two identical series capacitors yield half of an individual capacitor's value.

Add the parallel branch capacitance (

C_3 = 15\text{ nF}

) to the series combination equivalent.

C_{\text{total}} = C_{\text{series}} + C_3 = 25\text{ nF} + 15\text{ nF} = 40\text{ nF}

Capacitors in parallel add directly because their plate areas effectively combine to store more charge at a given voltage.

Key Concept

Series and parallel equivalent capacitance rules with metric prefix conversion

Alternative Method

Convert all components into microfarads first:

C_1 = 0.05\text{ }\mu\text{F}

C_2 = 0.05\text{ }\mu\text{F}

, and

C_3 = 0.015\text{ }\mu\text{F}

. The series pair gives

\frac{0.05}{2} = 0.025\text{ }\mu\text{F}

. Adding the parallel capacitor gives

0.025\text{ }\mu\text{F} + 0.015\text{ }\mu\text{F} = 0.040\text{ }\mu\text{F}

. Finally, converting

0.040\text{ }\mu\text{F}

back to nanofarads gives

0.040 \times 1000 = 40\text{ nF}

Estimated Time:1m 30s

Question 50Question →

In a bipolar junction transistor circuit, a small input current supplied to the base terminal is used to control a significantly larger current flowing between which two terminals?

The gate and the source

The collector and the emitter

The anode and the cathode

The base and the power supply rail

Show answer & explanation

Answer: The collector and the emitter

Answer

The collector and the emitter

A bipolar junction transistor functions as a current-controlled amplifier or switch. A relatively small base current (

I_B

) modulates a significantly larger current (

I_C

) flowing across the main conduction channel between the collector and the emitter.

Step-by-Step Solution

Identify the component type and its standard terminal leads.

A bipolar junction transistor (BJT) consists of three active terminals: the base (B), collector (C), and emitter (E).

Terminal designations distinguish BJTs from other semiconductor devices such as FETs or diodes.

Determine the functional relationship between the input control lead and the main output conduction channel.

A small current entering the base terminal forward-biases the base-emitter junction, permitting a much larger current (proportional by the gain factor beta) to flow between the collector and the emitter.

BJTs operate as current-controlled devices where the base-emitter circuit governs the primary collector-to-emitter current channel.

Key Concept

Bipolar Junction Transistor (BJT) terminal roles and current amplification principle

Estimated Time:45s

Question 51Question →

An industrial control circuit uses a relay (K1) to control a DC motor. The circuit is wired as follows:

- A normally closed (NC) 'Stop' pushbutton is wired in series with a momentary normally open (NO) 'Start' pushbutton and the K1 relay coil.
- A normally open (NO) contact from the K1 relay is wired in parallel across the 'Start' pushbutton.
- A second normally open (NO) contact from the K1 relay connects the motor to the power supply.

If a short circuit develops directly across the two terminals of the 'Stop' pushbutton, how will the circuit respond when an operator presses and releases the 'Start' button, and then later presses the 'Stop' button?

The motor will start and continue running, but pressing the Stop button will fail to stop it.

The circuit's protective device will trip immediately when power is applied because of the short circuit.

The motor will start when the Start button is pressed, but will immediately shut off when the Start button is released.

The motor will fail to start entirely because the shorted Stop button creates an open circuit condition.

Show answer & explanation

Answer: The motor will start and continue running, but pressing the Stop button will fail to stop it.

Answer

The motor will start and continue running, but pressing the Stop button will fail to stop it.

The circuit described is a classic motor starter latching circuit. The parallel NO contact acts as a 'seal-in' contact, keeping the relay coil energized after the momentary Start button is released. The NC Stop button is designed to break this latched path. However, a short circuit directly across the Stop button's terminals provides a permanent alternate path for the current to flow, entirely bypassing the button's internal mechanical contacts. When the operator presses Stop, the internal contacts open, but current continues to flow freely through the short circuit. As a result, the relay coil remains energized and the motor cannot be turned off using the intended control.

Step-by-Step Solution

Analyze the normal operation of the latching relay circuit.

Pressing 'Start' energizes K1, closing the parallel NO contact (latching the circuit) and the motor NO contact. The circuit stays energized through the NC 'Stop' button when 'Start' is released.

Understanding the baseline functionality is required before analyzing fault conditions.

Determine the electrical effect of a short circuit across the 'Stop' button terminals.

A short circuit creates a permanent zero-ohm bypass around the button's internal mechanical contacts.

To understand how the fault alters the available current paths.

Trace the circuit logic when 'Start' is pressed and released with the fault present.

Current flows through the bypassed 'Stop' button and the pressed 'Start' button, energizing the coil. When 'Start' is released, current continues flowing through the bypassed 'Stop' button and the latched K1 NO contact. The motor runs normally.

To determine the initial state of the motor after the start sequence.

Trace the circuit logic when 'Stop' is subsequently pressed.

The mechanical contacts of the 'Stop' button open, but the short circuit wire continues to provide a zero-ohm path around them. The relay coil remains energized, and the motor continues to run.

To evaluate if the intended stopping mechanism can overcome the fault.

Key Concept

Relay Latching Logic and Short Circuit Fault Analysis

Question 52Question →

A technician is designing a control system operated by a single manual toggle switch. When the switch is actuated, it must simultaneously disconnect an existing 12 V DC cooling fan and connect a separate, electrically isolated 120 V AC warning beacon to power. Which switch configuration is the minimum requirement to control both isolated circuits in opposite states with a single mechanical actuator?

Double-Pole, Single-Throw (DPST) switch

Single-Pole, Double-Throw (SPDT) switch

Double-Pole, Double-Throw (DPDT) switch

Single-Pole, Single-Throw (SPST) switch

Show answer & explanation

Answer: Double-Pole, Double-Throw (DPDT) switch

Answer

Double-Pole, Double-Throw (DPDT) switch

The Double-Pole, Double-Throw (DPDT) configuration provides two electrically separate poles (required for isolating the 12 V DC and 120 V AC systems) and two throws per pole. This allows Pole 1 to open the 12 V DC fan circuit while Pole 2 simultaneously closes the 120 V AC beacon circuit under a single mechanical toggle.

Step-by-Step Solution

Determine the required number of poles based on circuit isolation.

Two poles (Double-Pole) are required.

The system involves two distinct voltage sources (12 V DC and 120 V AC) that must remain electrically isolated to prevent dangerous short circuits or cross-voltage damage.

Determine the required number of throws based on contact operation.

Two throws (Double-Throw) are required.

When the switch is actuated, one load must disconnect (normally closed contact opening) while the other load connects (normally open contact closing), requiring two contact positions per pole.

Combine the pole and throw requirements.

DPDT configuration.

Two independent poles combined with two selectable throws per pole define a Double-Pole, Double-Throw (DPDT) switch.

Key Concept

Switch Pole and Throw Nomenclature and Electrical Isolation

Estimated Time:1m 15s

Question 53Question →

Match each fundamental transistor operating concept on the left with its correct electrical definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Amplification matches 'Using a small varying input signal to control a larger output signal'; Switching matches 'Alternating between full conduction and complete cutoff to control power to a load'; Biasing matches 'Applying preset DC voltages to establish a stable quiescent operating point'.

Each concept pairs directly with its fundamental role in electronics: Amplification scales a small input variation into a larger output signal; Switching toggles between full conduction and total cutoff to activate or deactivate loads; Biasing establishes the necessary static direct-current voltages across terminals for stable circuit operation.

Step-by-Step Solution

Identify the definition of signal amplification in solid-state electronics.

Amplification corresponds to increasing signal strength by using small input fluctuations to modulate larger power supplies.

In linear active mode, transistors amplify weak audio, radio, or sensor signals.

Identify the functional definition of electronic switching.

Switching corresponds to toggling between saturation (closed/on) and cutoff (open/off).

Digital and power control circuits use transistors as high-speed electronic switches without moving parts.

Identify the purpose and definition of transistor biasing.

Biasing corresponds to setting baseline DC voltage levels at the terminals.

Proper DC bias ensures the semiconductor junctions maintain the required operational state (forward-biased base-emitter, reverse-biased base-collector).

Key Concept

Core operational principles and functional modes of transistors

Estimated Time:45s

Question 54Question →

During the maintenance of an aerospace navigation system, an avionics technician isolates a filtering network to verify its total capacitance. The network is configured as follows: a $20 \mu\text{F}$ capacitor and a $5 \mu\text{F}$ capacitor are wired in series. This series combination is placed in parallel across a $36 \mu\text{F}$ capacitor. Finally, this entire grouped section is wired in series with a $120 \mu\text{F}$ capacitor. What is the total equivalent capacitance of this filtering network? Enter your answer as a number.

Show answer & explanation

Answer: 30

Answer

To find the total equivalent capacitance, you must simplify the circuit step-by-step applying the specific rules for capacitors, which are the exact reverse of resistors. First, the

20 \mu\text{F}

and

5 \mu\text{F}

series capacitors are combined using the product-over-sum method to yield

4 \mu\text{F}

. Next, this

4 \mu\text{F}

equivalent is added directly to the

36 \mu\text{F}

parallel capacitor, resulting in

40 \mu\text{F}

for the entire parallel section. Finally, this

40 \mu\text{F}

equivalent block is in series with the final

120 \mu\text{F}

capacitor. Using the product-over-sum method once more (

\frac{40 \times 120}{160}

), the final total equivalent capacitance equals

30 \mu\text{F}

Step-by-Step Solution

Calculate the equivalent capacitance of the

20 \mu\text{F}

and

5 \mu\text{F}

series pair.

4 \mu\text{F}

Capacitors in series are calculated using the reciprocal formula or product-over-sum rule:

\frac{20 \times 5}{20 + 5} = 4

Calculate the equivalent capacitance of the parallel section.

40 \mu\text{F}

The

4 \mu\text{F}

series equivalent is in parallel with the

36 \mu\text{F}

capacitor. Capacitors in parallel simply add together:

4 + 36 = 40

Calculate the final total capacitance of the entire network.

30 \mu\text{F}

The

40 \mu\text{F}

equivalent section is in series with the

120 \mu\text{F}

capacitor. Using the product-over-sum rule again:

\frac{40 \times 120}{40 + 120} = 30

Key Concept

Calculating equivalent capacitance in compound series-parallel networks

Estimated Time:3m 0s

Question 55Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

Common Emitter Configuration matches with significant voltage/current gain and

180^\circ

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

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

Step-by-Step Solution

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

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

180^\circ

phase inversion.

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

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

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

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

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

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

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

Distinguish linear amplification from digital switching operation.

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

I_C = 0

) and saturation (

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

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

Key Concept

Transistor Circuit Configurations and Operating Modes

Estimated Time:1m 0s

Question 56Question →

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

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

To produce an electromagnetic field that resists sudden changes in current

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

To convert incoming direct current directly into alternating current

Show answer & explanation

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

Answer

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

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

Step-by-Step Solution

Identify the basic physical structure of a capacitor

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

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

Determine the electrical properties and purpose of the dielectric

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

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

Key Concept

Function and Properties of Capacitor Dielectric Materials

Estimated Time:45s

Question 57Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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

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

Step-by-Step Solution

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

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

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

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

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

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

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

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

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

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

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

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

Key Concept

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

Estimated Time:1m 30s

Question 58Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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

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

Step-by-Step Solution

Identify the non-polarized capacitor designed for high frequencies.

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

Ceramic dielectrics maintain stability at high frequencies without polarity constraints.

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

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

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

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

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

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

Identify the adjustable capacitor used for calibration.

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

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

Key Concept

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

Estimated Time:1m 0s

Question 59Question →

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

Base

Gate

Collector

Cathode

Show answer & explanation

Answer: Gate

Answer

Gate

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

Step-by-Step Solution

Identify the transistor family specified in the question.

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

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

Determine the functional role of each FET terminal.

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

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

Key Concept

Field-Effect Transistor (FET) Terminal Designations and Roles

Estimated Time:45s

Question 60Question →

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

16.0\text{ A}

0.25\text{ A}

2.00\text{ A}

0.025\text{ A}

Show answer & explanation

Answer: $0.25\text{ A}$

Answer

0.25\text{ A}

The transformer has an

8:1

step-down turns ratio (

1{}200 / 150 = 8

). The secondary voltage is

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

. Connecting a

15\ \Omega

resistor across

30\text{ V}

produces a secondary current of

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

. Since power in an ideal transformer is conserved (

P_p = P_s

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

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

Step-by-Step Solution

Calculate the turns ratio (

a

) of the transformer.

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

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

Calculate the secondary voltage (

V_s

) across the load.

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

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

Calculate the secondary current (

I_s

) through the load resistor using Ohm's Law.

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

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

Calculate the primary current (

I_p

) drawn from the AC source.

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

Conservation of power in an ideal transformer requires that

V_p I_p = V_s I_s

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

Key Concept

Ideal transformer transformation ratios (

N_p / N_s = V_p / V_s = I_s / I_p

)

Alternative Method

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

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

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

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

Estimated Time:2m 0s

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