Electronic Components

211 questions

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

In electronics design, technicians must evaluate semiconductor ratings to ensure circuit reliability and prevent component failure. Match each diode rating or parameter on the left with its correct operating definition on the right.

Click a left item, then click its matching right item

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Answer

Peak Inverse Voltage matches the maximum reverse-bias voltage before breakdown; Forward Voltage Drop matches the potential difference needed to overcome the internal barrier potential; Reverse Recovery Time matches the time required for stored charge to clear during switching; Maximum Forward Current matches the upper limit of continuous current before thermal damage occurs.

Each diode specification defines a critical operating limit or behavioral parameter: Peak Inverse Voltage (PIV) defines reverse breakdown tolerance, Forward Voltage Drop defines barrier potential overcoming voltage, Reverse Recovery Time defines switching speed between conduction and cutoff, and Maximum Forward Current defines the continuous forward conduction limit governed by thermal dissipation.

Step-by-Step Solution

Identify voltage ratings associated with diode polarity states.

Peak Inverse Voltage (PIV) pertains to reverse-bias breakdown limits, while Forward Voltage Drop (

V_F

) pertains to the forward-bias junction barrier potential.

Diodes behave as one-way valves with distinct electrical characteristics depending on whether bias polarity is positive (forward) or negative (reverse).

Distinguish between dynamic switching characteristics and steady-state current limits.

Reverse Recovery Time (

t_{rr}

) is a high-frequency switching parameter measuring carrier clearance time, whereas Maximum Forward Current (

I_{F(\max)}

) is a steady-state thermal safety boundary.

Proper component selection requires evaluating both high-speed switching performance and continuous power/current handling capability.

Pair each parameter with its corresponding functional definition.

All four parameters are mapped uniquely to their technical specifications.

Each rating addresses a specific operational constraint in semiconductor rectifiers and signal diodes.

Key Concept

Semiconductor Diode Ratings and Electrical Specifications

Estimated Time:1m 15s

Question 182Question →

In a basic electronic power supply circuit, an alternating current (AC) voltage source is connected in series with a standard semiconductor diode and a load resistor. Which of the following describes the electrical behavior of this circuit across a complete AC cycle?

It permits current to flow freely in both directions while stepping down the alternating voltage to a lower amplitude.

It stores electrical energy in an electrostatic field during the positive half-cycle and discharges it into the load during the negative half-cycle.

It conducts current only during the positive half-cycle when forward-biased, blocking current during the negative half-cycle to produce pulsating direct current.

It inverts the polarity of the negative half-cycle to convert the alternating waveform into continuous, perfectly smooth direct current.

Show answer & explanation

Answer: It conducts current only during the positive half-cycle when forward-biased, blocking current during the negative half-cycle to produce pulsating direct current.

Answer

The diode conducts current only during the positive half-cycle when forward-biased, blocking current during the negative half-cycle to produce pulsating direct current.

A semiconductor diode operates as a one-way electrical valve. During the positive half-cycle of an applied AC signal, the diode is forward-biased and conducts current through the load resistor. During the negative half-cycle, the diode is reverse-biased and blocks current flow. This action converts bidirectional AC into unidirectional pulsating DC through half-wave rectification.

Step-by-Step Solution

Analyze the diode biasing state during the positive half-cycle of the AC input voltage.

The anode becomes positive with respect to the cathode, placing the PN junction in forward bias.

Forward bias narrows the depletion region and allows current to pass through the diode and into the load resistor.

Analyze the diode biasing state during the negative half-cycle of the AC input voltage.

The anode becomes negative with respect to the cathode, placing the PN junction in reverse bias.

Reverse bias widens the depletion region, preventing current flow through the circuit during this half-cycle.

Determine the net output current and voltage waveform across the load resistor.

Current flows through the load only during the positive half-cycles in a single direction, resulting in pulsating direct current (DC).

This conversion of alternating current to pulsating direct current using a single diode is the foundation of half-wave rectification.

Key Concept

Half-wave rectification and unidirectional conduction of semiconductor diodes

Estimated Time:1m 0s

Question 183Question →

In a semiconductor PN junction diode, applying an external reverse-bias voltage causes the depletion region at the junction to widen, which prevents the conduction of majority charge carriers.

True

False

Show answer & explanation

Answer: True

Answer

True

Applying reverse bias attracts majority carriers away from the PN interface, widening the non-conductive depletion zone and preventing current conduction through the component.

Step-by-Step Solution

Identify the polarity of the applied external voltage across the diode terminals in reverse bias.

The positive terminal of the power source connects to the n-type cathode, and the negative terminal connects to the p-type anode.

Establishing the reverse-bias electrical configuration determines the direction of the electrostatic forces acting on internal charge carriers.

Determine how internal majority carriers respond to the applied electrostatic field.

Holes in the p-type region are pulled toward the negative terminal, and electrons in the n-type region are pulled toward the positive terminal.

Opposite electrical charges attract, pulling majority carriers away from the central PN boundary.

Evaluate the effect of charge carrier movement on the depletion layer and overall current flow.

The removal of mobile carriers from the junction widens the depletion layer of uncovered ionized atoms, preventing majority carrier flow and allowing only negligible minority leakage current.

The widened depletion layer acts as a high-resistance barrier to electric current, confirming the statement as true.

Key Concept

PN Junction Biasing and Depletion Layer Dynamics

Estimated Time:45s

Question 184Question →

In electronic circuit assembly and analysis, understanding diode terminology and physical structure is fundamental for proper circuit operation. Match each diode term or terminal to its correct operational description or physical characteristic.

Click a left item, then click its matching right item

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Answer

Anode pairs with the P-type semiconductor terminal requiring positive potential; Cathode pairs with the N-type terminal marked by a physical stripe; Depletion Layer pairs with the internal boundary zone depleted of mobile charge carriers; Forward Voltage Drop pairs with the fixed potential difference across the conducting PN junction.

Each diode term accurately matches its functional definition: the anode is the P-type terminal requiring higher positive voltage, the cathode is the N-type terminal identified by a package stripe, the depletion layer is the carrier-free junction zone, and the forward voltage drop is the barrier potential (approximately

0.7\text{ V}

for silicon).

Step-by-Step Solution

Identify the polarity and terminal designations of a semiconductor diode.

The anode corresponds to the P-type section (requiring positive bias), and the cathode corresponds to the N-type section (marked with a band on axial packages).

Correct terminal identification prevents reverse installation and ensures proper forward bias orientation in circuits.

Distinguish between junction phenomena and operating electrical characteristics.

The depletion layer is the carrier-free barrier region formed at the interface, while the forward voltage drop is the typical

0.7\text{ V}

threshold maintained during conduction.

Differentiating physical barrier structure from electrical conduction metrics clarifies semiconductor physics.

Key Concept

Semiconductor Diode Structure, Polarity Identification, and Electrical Characteristics

Estimated Time:1m 0s

Question 185Question →

When examining fixed through-hole resistors, what electrical property is directly governed by the component's physical dimensions, and what unit is used to express this capability? Complete the technical statement below.

Fill in the blanks below

Although the printed color bands on a standard carbon-composition resistor indicate its electrical resistance and tolerance, the physical size of the resistor body determines its rating, which is measured in .

Show answer & explanation

Answer

power (or wattage) and watts (W)

In electronic components, the physical package size of an axial-lead fixed resistor determines its ability to radiate heat into the surrounding air. Consequently, larger resistors have higher power (wattage) ratings, which are quantified in watts (W). The actual resistance value in ohms and the manufacturing tolerance percentage are represented by the color-coded bands printed on the component body.

Step-by-Step Solution

Identify the electrical property determined by a resistor's physical dimensions and surface area.

Power rating (or wattage / power dissipation capacity)

A larger physical package provides more surface area to safely dissipate thermal energy generated by current flow without overheating or damaging the component.

Identify the standard unit of measurement for electrical power dissipation.

Watts (W)

Resistor power ratings are specified in watts, with common standard through-hole ratings including

1/8\text{ W}

1/4\text{ W}

1/2\text{ W}

1\text{ W}

, and

2\text{ W}

Key Concept

Resistor physical package size correlates directly to power dissipation rating in watts rather than resistance value in ohms.

Estimated Time:45s

Question 186Question →

A direct-current (DC) solar charging circuit contains a standard silicon diode connected in series between the solar panel and a 12-volt battery. Which of the following best explains the primary function of this diode when the solar panel is not producing voltage at night?

It converts the direct current from the battery into alternating current to power the solar panel.

It prevents the battery from discharging backward through the dark solar panel.

It allows bidirectional current flow so the panel and battery maintain identical voltage levels.

It steps up the battery voltage to continuously supply power to the solar cells.

Show answer & explanation

Answer: It prevents the battery from discharging backward through the dark solar panel.

Answer

It prevents the battery from discharging backward through the dark solar panel.

A semiconductor diode permits electrical current to flow in only one direction (from anode to cathode under forward bias). When darkness falls, the solar panel's output drops to zero while the storage battery remains charged at 12 V. This higher potential at the cathode places the diode in reverse bias, creating an open circuit condition that prevents the battery from discharging through the solar panel.

Step-by-Step Solution

Analyze the circuit state under nighttime conditions.

The unilluminated solar panel produces 0 V, while the connected storage battery maintains approximately 12 V.

Determining the relative electric potentials at the diode terminals identifies its biasing condition.

Determine the bias state of the series diode.

The cathode connected to the battery positive terminal is at a higher potential than the anode connected to the panel, placing the diode in reverse bias.

A semiconductor diode conducts current only when its anode is more positive than its cathode by at least the forward barrier potential.

Identify the resulting circuit behavior.

The reverse-biased diode exhibits extremely high resistance, preventing current from leaving the battery.

This blocking action prevents the battery from draining energy into the internal resistance of the dark solar panel.

Key Concept

Unidirectional Current Flow and Reverse-Bias Blocking

Estimated Time:1m 0s

Question 187Question →

During a quality control inspection, a technician measures a resistor marked with the color bands Brown, Gray, Red, and Gold. For this resistor to meet its manufactured specification, what is the acceptable range of measured resistance?

171\ \Omega\text{ to } 189\ \Omega

1,710\ \Omega\text{ to } 1,890\ \Omega

1,620\ \Omega\text{ to } 1,980\ \Omega

172.9\ \Omega\text{ to } 191.1\ \Omega

Show answer & explanation

Answer: $1,710\ \Omega\text{ to } 1,890\ \Omega$

Answer

1,710\ \Omega\text{ to } 1,890\ \Omega

A four-band resistor with colors Brown (1), Gray (8), Red (

10^2 = 100

), and Gold (

\pm 5\%

) has a nominal value of

1,800\ \Omega

. With a

\pm 5\%

tolerance, the allowable variance is

1,800 \times 0.05 = 90\ \Omega

. Subtracting and adding

90\ \Omega

gives the acceptable operational range of

1,710\ \Omega

1,890\ \Omega

Step-by-Step Solution

Determine the nominal resistance from the first three color bands.

Brown = 1, Gray = 8, Red = multiplier of

10^2

(

100

). The nominal value is

18 \times 100 = 1,800\ \Omega

In a standard four-band resistor, the first two bands represent significant digits and the third band represents the power-of-ten multiplier.

Identify the tolerance percentage from the fourth band.

Gold band =

\pm 5\%

The fourth band specifies the manufacturing tolerance rating.

Calculate the acceptable tolerance deviation in ohms.

1,800\ \Omega \times 0.05 = 90\ \Omega

The allowable deviation is found by multiplying the nominal resistance by the tolerance percentage.

Calculate the minimum and maximum acceptable resistance boundaries.

Minimum:

1,800 - 90 = 1,710\ \Omega

; Maximum:

1,800 + 90 = 1,890\ \Omega

The acceptable range spans from nominal minus tolerance to nominal plus tolerance.

Key Concept

Decoding four-band resistor color codes and calculating nominal tolerance boundaries

Estimated Time:45s

Question 188Question →

The Peak Inverse Voltage (PIV) rating of a semiconductor diode specifies the maximum reverse-bias voltage that can be applied across its terminals before the PN junction breaks down and begins conducting in reverse.

True

False

Show answer & explanation

Answer: True

Answer

True: The Peak Inverse Voltage (PIV) rating defines the maximum reverse-bias voltage a diode is designed to withstand without entering junction breakdown.

The statement is true because the Peak Inverse Voltage (PIV) rating is the manufacturer's maximum allowable reverse-bias voltage. Operating beyond this limit causes avalanche breakdown, allowing destructive reverse current to surge through the component.

Step-by-Step Solution

Identify the standard definition of the Peak Inverse Voltage (PIV) specification for semiconductor diodes.

PIV represents the peak voltage experienced by the diode when it is reverse-biased (cathode positive relative to anode).

Understanding diode ratings is critical for selecting components capable of handling reverse voltage without failing.

Determine the physical consequence of exceeding the diode's PIV rating.

Exceeding the PIV rating forces the PN junction into avalanche or Zener breakdown, leading to sharp reverse conduction and potential permanent destruction.

Standard rectifier and switching diodes are not designed to sustain continuous operation in the reverse breakdown region.

Evaluate the accuracy of the stem statement against semiconductor theory.

The statement accurately reflects the definition and operational limit designated by the PIV parameter.

The PIV specification exists precisely to prevent unintentional reverse breakdown.

Key Concept

Diode Peak Inverse Voltage (PIV) Rating and Reverse Breakdown

Estimated Time:45s

Question 189Question →

Electronic circuits utilize various fixed and variable resistive components to control voltage, limit current, and calibrate operating points. Match each resistor type on the left with its primary functional description or application on the right. Which pairs correctly match each component to its function?

Click a left item, then click its matching right item

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Answer

Potentiometer matches the three-terminal voltage divider description; Rheostat matches the two-terminal series current regulator description; Trimmer Resistor matches the miniature calibration component description; Wirewound Resistor matches the high-power coiled-wire construction description.

Each component is mapped directly according to its terminal count, internal construction, and operational role: Potentiometers are three-terminal voltage dividers; Rheostats are two-terminal series current controllers; Trimmers (trimpots) are compact PCB calibration resistors; Wirewound resistors are heavy-duty components constructed with wire coils around a ceramic core for high-power dissipation.

Step-by-Step Solution

Differentiate between three-terminal and two-terminal variable resistor operations.

Identify the potentiometer as the three-terminal voltage divider and the rheostat as the two-terminal current limiter.

Potentiometers divide voltage across three terminals, whereas rheostats vary resistance in series using two active terminals.

Identify the distinct mechanical and physical packaging of precision alignment components versus power components.

Match the trimmer resistor (trimpot) to calibration/alignment and the wirewound resistor to high-power ceramic-core construction.

Trimpots are miniature preset controls adjusted during calibration, while wirewound resistors are constructed to handle significant thermal dissipation.

Key Concept

Classification, terminal configuration, and circuit applications of fixed and variable resistors.

Estimated Time:1m 0s

Question 190Question →

In a direct-current (DC) control circuit, a silicon rectifier diode is connected in parallel across the terminals of an electromechanical relay coil, with its cathode connected to the positive supply rail. What is the primary function of this diode in the circuit?

To convert the direct-current supply into alternating current needed to keep the relay energized

To conduct continuous forward-bias current around the coil while the circuit is actively powered

To safely dissipate high-voltage transient spikes generated by the collapsing magnetic field when the coil is de-energized

To step up the supply voltage across the coil terminals to increase contact closing speed

Show answer & explanation

Answer: To safely dissipate high-voltage transient spikes generated by the collapsing magnetic field when the coil is de-energized

Answer

To safely dissipate high-voltage transient spikes generated by the collapsing magnetic field when the coil is de-energized

An electromechanical relay coil is an inductor that stores energy in a magnetic field. When the control switch opens to turn off the relay, the sudden interruption of current causes the magnetic field to collapse rapidly. In accordance with Lenz's Law, this collapse generates a high-voltage back-EMF spike of reverse polarity. Because the diode is connected in anti-parallel (cathode to positive rail, anode to switched side), this reverse voltage places the diode into forward bias. The diode conducts, clamping the voltage spike to its forward drop (approximately

0.7\text{ V}

) and circulating the current through the coil until the energy safely dissipates as heat, protecting upstream transistors or switch contacts from electrical breakdown.

Step-by-Step Solution

Analyze the biasing condition of the diode during normal steady-state operation

The diode's cathode is connected to the positive supply rail and its anode to the switched ground side, placing it in reverse bias with negligible leakage current.

Understanding normal circuit operation confirms that the diode does not interfere with energizing the relay coil.

Analyze what happens when the circuit switch opens to de-energize the relay coil

The current suddenly drops to zero, causing the magnetic field around the inductive coil to collapse rapidly.

Inductors resist instantaneous changes in current by generating an induced electromotive force (back-EMF).

Determine the polarity of the induced voltage and the resulting state of the diode

The collapsing magnetic field reverses the polarity across the coil terminals, making the anode positive relative to the cathode and placing the diode into forward bias.

The forward-biased diode clamps the inductive spike to approximately

0.7\text{ V}

and circulates the decay current safely until the stored magnetic energy dissipates.

Key Concept

Flyback (freewheeling) diode protection across inductive loads

Estimated Time:1m 0s

Question 191Question →

In electronic systems, specialized diodes are engineered to leverage distinct semiconductor properties for control, sensing, and signaling. Match each specialized semiconductor diode with its primary operating characteristic or function.

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Answer

Zener diode matches with maintaining constant voltage in reverse breakdown; Light-emitting diode matches with emitting radiation under forward current; Photodiode matches with generating reverse current proportional to illumination; Varactor diode matches with functioning as a voltage-controlled variable capacitor.

Each diode type is engineered for a specific physical phenomenon: the Zener diode exploits controlled reverse breakdown for voltage stabilization; the LED converts electrical energy to light via forward-biased carrier recombination; the photodiode converts optical energy to current via reverse-biased photon absorption; and the varactor diode leverages the voltage-dependent width of the depletion layer to act as an adjustable capacitor in tuning circuits.

Step-by-Step Solution

Analyze the functional mechanism of a Zener diode

Identified as a device designed for stable reverse breakdown regulation

Zener diodes maintain a steady voltage rating when reverse biased beyond their breakdown threshold.

Analyze the functional mechanism of a Light-Emitting Diode (LED)

Identified as a forward-biased electroluminescent semiconductor

Forward current drives electron-hole recombination across the bandgap, releasing visible or infrared light.

Analyze the functional mechanism of a Photodiode

Identified as a reverse-biased light sensor

Incident light striking the depletion region creates free charge carriers, increasing reverse leakage current.

Analyze the functional mechanism of a Varactor diode

Identified as a voltage-tuned capacitive component

Varying the reverse-bias voltage changes the depletion layer thickness, which inversely varies junction capacitance.

Key Concept

Specialized semiconductor diode operational modes and functional characteristics

Estimated Time:1m 0s

Question 192Question →

A technician connects a digital multimeter across terminal 1 (the counter-clockwise outer terminal) and terminal 2 (the center wiper) of a $10\text{ k}\Omega$ rotary potentiometer. If the control shaft is rotated fully clockwise toward terminal 3, what resistance value should the multimeter display?

0\ \Omega

10\text{ k}\Omega

5\text{ k}\Omega

Infinite resistance

Show answer & explanation

Answer: $10\text{ k}\Omega$

Answer

The multimeter will display the full rated resistance of

10\text{ k}\Omega

A rotary potentiometer contains a continuous resistive track between two outer fixed terminals (terminals 1 and 3) and a movable wiper contact (terminal 2). When the shaft is rotated fully clockwise toward terminal 3, the wiper moves to the extreme end of the resistive track at terminal 3. Consequently, the multimeter connected across terminal 1 and terminal 2 spans the entire length of the resistive element, displaying the full nominal rating of

10\text{ k}\Omega

Step-by-Step Solution

Identify the internal construction and terminal functions of a standard three-terminal potentiometer.

Terminal 1 and terminal 3 connect to the opposite fixed ends of the internal

10\text{ k}\Omega

resistive strip, while terminal 2 connects to the movable wiper contact.

Understanding the internal layout establishes how rotation alters the length of resistive material between any two terminals.

Determine the physical position of the wiper relative to terminal 1 after rotating fully clockwise toward terminal 3.

Rotating clockwise shifts the wiper (terminal 2) all the way to terminal 3, placing the maximum possible distance of the resistive track between terminal 1 and terminal 2.

Electrical resistance is directly proportional to the length of resistive material between the two meter probes.

Calculate the measured resistance across terminal 1 and terminal 2.

The meter measures the complete length of the track, indicating

10\text{ k}\Omega

With the wiper at terminal 3, the resistance between terminal 1 and terminal 2 equals the total nominal resistance of the potentiometer (

10\text{ k}\Omega

), while the resistance between terminal 2 and terminal 3 drops to

0\ \Omega

Key Concept

Potentiometer terminal configuration and variable wiper resistance behavior

Estimated Time:45s

Question 193Question →

A dual-power electronic circuit uses two silicon diodes to connect a primary $12\text{ V}$ DC power supply and a secondary $9\text{ V}$ backup battery to a single shared load. The anode of each diode is connected to the positive terminal of its respective source, and both cathodes are connected together at the positive terminal of the load. When the $12\text{ V}$ primary supply is operating normally, which of the following best describes the condition of the diode connected to the $9\text{ V}$ battery?

Forward-biased, because current from the higher-voltage supply flows backward into the battery

Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltage

Forward-biased, because both voltage sources continuously force current simultaneously through the load

Reverse-biased, because the diode converts the direct-current input into alternating current to block it

Show answer & explanation

Answer: Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltage

Answer

Reverse-biased, because the cathode voltage established by the primary supply is higher than the battery anode voltage

A standard silicon diode becomes forward-biased only when its anode is roughly

0.7\text{ V}

more positive than its cathode. When the primary

12\text{ V}

DC supply conducts, the common cathode bus is held at approximately

12\text{ V} - 0.7\text{ V} = 11.3\text{ V}

. Because the backup battery diode has only

9\text{ V}

at its anode while its cathode is at

11.3\text{ V}

, the cathode is more positive than the anode, placing the backup diode into a reverse-biased (cutoff) state.

Step-by-Step Solution

Determine the voltage at the shared cathode connection using the active primary source.

The primary

12\text{ V}

supply forward-biases its diode, dropping approximately

0.7\text{ V}

and establishing a common cathode bus potential of approximately

11.3\text{ V}

Silicon diodes in the conducting state exhibit a standard forward voltage drop of roughly

0.7\text{ V}

Compare the voltage at the anode of the backup battery diode to the shared cathode voltage.

The backup diode anode is held at

+9\text{ V}

by the battery, while its cathode is held at

+11.3\text{ V}

A diode requires its anode to be more positive than its cathode by at least

0.7\text{ V}

to conduct.

Determine the biasing state of the backup diode.

Because the cathode potential (

11.3\text{ V}

) exceeds the anode potential (

9\text{ V}

), the backup diode is reverse-biased (in cutoff) and conducts negligible current.

When the cathode is at a higher electrical potential than the anode, the semiconductor PN junction depletion region widens, blocking current flow.

Key Concept

Diode Biasing and Power-OR Steering

Estimated Time:1m 0s

Question 194Question →

A rotary potentiometer has a total resistance of $50\text{ k}\Omega$ across its two outer fixed terminals (Terminals A and C). If the center wiper (Terminal B) is adjusted so that the resistance between Terminal A and Terminal B is $32\text{ k}\Omega$ , what is the resistance between Terminal B and Terminal C?

32\text{ k}\Omega

18\text{ k}\Omega

82\text{ k}\Omega

1.8\text{ k}\Omega

Show answer & explanation

Answer: $18\text{ k}\Omega$

Answer

The resistance between Terminal B and Terminal C is

18\text{ k}\Omega

A potentiometer contains a single continuous resistive track between its two outer terminals (

A

and

C

). The center wiper (

B

) contacts this track at an intermediate point, splitting the total resistance into two complementary series sections such that

R_{AC} = R_{AB} + R_{BC}

. Given a total resistance of

50\text{ k}\Omega

and a segment of

32\text{ k}\Omega

, the remaining segment between the wiper and the other outer terminal must be

50\text{ k}\Omega - 32\text{ k}\Omega = 18\text{ k}\Omega

Step-by-Step Solution

Identify the relationship between the three terminals of a potentiometer.

The total resistance between outer terminals A and C equals the sum of the two sub-segments created by wiper B:

R_{AC} = R_{AB} + R_{BC}

A potentiometer consists of a single continuous resistive element where the movable wiper divides the total track into two series-connected resistive portions.

Subtract the known segment resistance from the total end-to-end resistance to find

R_{BC}

R_{BC} = 50\text{ k}\Omega - 32\text{ k}\Omega = 18\text{ k}\Omega

Isolating

R_{BC}

algebraically provides the remaining resistance along the element to the opposite outer terminal.

Key Concept

Potentiometer resistive track division

Estimated Time:45s

Question 195Question →

A bench technician uses a $100\text{ k}\Omega$ linear potentiometer as an adjustable control element in a circuit. If the wiper is adjusted to $75\%$ of its full mechanical travel away from the lower reference terminal, what is the resulting resistance between the wiper and the upper terminal in kilo-ohms ( $\text{k}\Omega$ )?

Show answer & explanation

Answer: 25

Answer

25 kilo-ohms (

25\text{ k}\Omega

)

In a linear-taper potentiometer, the total end-to-end resistance (

100\text{ k}\Omega

) is distributed uniformly across the resistive track between the outer fixed terminals. When the wiper is moved

75\%

of the travel distance from the lower terminal, the remaining segment between the wiper and the upper terminal comprises

25\%

(

100\% - 75\%

) of the total resistance. Multiplying

100\text{ k}\Omega \times 0.25

gives

25\text{ k}\Omega

Step-by-Step Solution

Find the percentage of the resistive element situated between the wiper contact and the upper terminal.

100\% - 75\% = 25\%

The total resistive track spans

100\%

of the component's resistance between its two outer end terminals.

Calculate the resistance across this upper portion of the track.

100\text{ k}\Omega \times 0.25 = 25\text{ k}\Omega

Because the potentiometer is linear, resistance is directly proportional to the physical position of the wiper along the track.

Key Concept

Linear potentiometer terminal division and wiper position relationships

Alternative Method

Calculate the resistance from the lower terminal to the wiper first (

100\text{ k}\Omega \times 0.75 = 75\text{ k}\Omega

), and subtract that value from the total resistance:

100\text{ k}\Omega - 75\text{ k}\Omega = 25\text{ k}\Omega

Estimated Time:45s

Question 196Question →

In a direct-current circuit, applying a voltage of $+5.0\text{ V}$ to the anode and $+4.5\text{ V}$ to the cathode of a standard silicon PN junction diode is sufficient to overcome its barrier potential and place the diode into full forward conduction.

True

False

Show answer & explanation

Answer: False

Answer

False. The diode does not enter full forward conduction because the potential difference across its terminals (

0.5\text{ V}

) is below the

0.7\text{ V}

barrier potential threshold required for silicon.

For a standard silicon PN junction diode to conduct significant forward current, the anode potential must exceed the cathode potential by at least the internal barrier potential of approximately

0.7\text{ V}

. With

+5.0\text{ V}

on the anode and

+4.5\text{ V}

on the cathode, the net forward voltage is only

0.5\text{ V}

, leaving the diode in a non-conducting cutoff state.

Step-by-Step Solution

Calculate the forward voltage drop (

V_D

) across the diode from anode to cathode.

V_D = V_{\text{anode}} - V_{\text{cathode}} = 5.0\text{ V} - 4.5\text{ V} = +0.5\text{ V}

Diode biasing depends on the relative potential difference between the anode and cathode terminals.

Compare the applied forward voltage to the internal barrier potential of a silicon diode.

The applied voltage of

+0.5\text{ V}

is less than the required silicon barrier potential threshold of approximately

+0.7\text{ V}

A silicon PN junction requires approximately

0.7\text{ V}

of forward bias to overcome the depletion region barrier and permit significant forward current flow.

Determine the operating state of the diode.

The diode remains in the cutoff/non-conducting region with only negligible leakage current.

Since the threshold voltage is not reached, the statement claiming full forward conduction is false.

Key Concept

Silicon Diode Barrier Potential and Forward Conduction Threshold

Estimated Time:45s

Question 197Question →

A technician connects a standard silicon diode in series with a $220\,\Omega$ resistor and a $12\text{ V}$ DC power supply. The cathode terminal of the diode is connected to the positive supply terminal, and the anode terminal is connected through the resistor to the negative supply terminal. What is the expected voltage drop across the $220\,\Omega$ resistor?

11.3\text{ V}

0\text{ V}

0.7\text{ V}

12.0\text{ V}

Show answer & explanation

Answer: $0\text{ V}$

Answer

The voltage drop across the resistor is

0\text{ V}

because the diode is reverse-biased and prevents current from flowing through the circuit.

A diode is a polarized semiconductor component that permits current flow in only one direction. When the cathode terminal is connected to the positive supply side and the anode is connected toward the negative side, the diode is reverse-biased. In this state, the diode acts as an open circuit with extremely high resistance, preventing current from circulating (

I = 0\text{ A}

). By Ohm's Law (

V = I \times R

), with zero current flowing through the

220\,\Omega

resistor, the voltage drop across the resistor is

0\text{ V}

Step-by-Step Solution

Determine the biasing state of the diode based on the terminal connections.

The cathode is connected to the positive potential (

+12\text{ V}

) and the anode is connected toward the negative potential (

0\text{ V}

), meaning

V_{\text{anode}} < V_{\text{cathode}}

, which places the diode in reverse bias.

A PN junction diode only conducts when the anode is positive relative to the cathode by at least its forward threshold voltage.

Determine the circuit current under reverse-bias conditions.

The reverse-biased diode acts like an open switch, restricting current flow to negligible leakage levels (

I \approx 0\text{ A}

The reverse voltage widens the internal depletion region, blocking majority carrier flow.

Apply Ohm's Law (

V = I \times R

) to calculate the voltage drop across the

220\,\Omega

resistor.

V_R = 0\text{ A} \times 220\,\Omega = 0\text{ V}

Without current flowing through the resistor, no electromotive force is dropped across its resistance, and the full

12\text{ V}

source potential appears across the open diode.

Key Concept

Diode Biasing and One-Way Current Flow

Estimated Time:1m 0s

Question 198Question →

Complete the statement describing how variable resistive components are configured and classified in basic electronic circuits. Fill in the blanks with the correct component names.

Fill in the blanks below

A three-terminal variable resistor utilized to provide an adjustable voltage output is called a , whereas a two-terminal variable resistor used primarily to regulate current flow in a circuit is known as a .

Show answer & explanation

Answer

Blank 1: potentiometer; Blank 2: rheostat

A potentiometer uses three terminals to create an adjustable voltage divider, whereas a rheostat uses two terminals connected in series to vary resistance and regulate circuit current.

Step-by-Step Solution

Identify the component type and function for a three-terminal variable resistor used for voltage control

A potentiometer connects across all three terminals (two outer ends and a center wiper) to function as a variable voltage divider.

The wiper taps off an adjustable fraction of the total voltage applied across the outer resistive track.

Identify the component type and function for a two-terminal variable resistor used for current control

A rheostat uses only two connection points (one end and the wiper) wired in series with the load to vary total circuit resistance.

By changing series resistance directly, the rheostat regulates the magnitude of current flow according to Ohm's law (

I = V/R

Key Concept

Functional and structural distinction between three-terminal potentiometers (voltage division) and two-terminal rheostats (current limiting)

Estimated Time:45s

Question 199Question →

An electronics technician is replacing a scorched four-band resistor on a circuit board. The original component has the color sequence Blue, Gray, Brown, and Gold. What is the nominal resistance and tolerance of this resistor?

6.8\text{ k}\Omega \pm 5\%

68\ \Omega \pm 5\%

680\ \Omega \pm 5\%

570\ \Omega \pm 5\%

Show answer & explanation

Answer: $680\ \Omega \pm 5\%$

Answer

680\ \Omega \pm 5\%

According to standard electronic color code conventions, blue represents the digit 6, gray represents 8, brown denotes a multiplier of 10 (or

10^1

), and gold denotes a

\pm 5\%

tolerance. Combining these yields

68 \times 10 = 680\ \Omega \pm 5\%

Step-by-Step Solution

Determine the first two significant digits from the first and second color bands.

The first band (Blue) is

6

and the second band (Gray) is

8

, forming the number

68

In a standard four-band resistor code, the first two bands represent significant digits.

Apply the decimal multiplier from the third color band.

The third band (Brown) represents a multiplier of

10^1 = 10

, yielding

68 \times 10 = 680\ \Omega

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

Identify the tolerance percentage from the fourth color band.

The fourth band (Gold) designates a tolerance of

\pm 5\%

Gold in the fourth position standardly denotes a

\pm 5\%

manufacturer tolerance rating.

Key Concept

Four-band resistor color code decoding and tolerance interpretation

Estimated Time:35s

Question 200Question →

In a basic power supply circuit, an alternating current (AC) source is connected in series with a single PN-junction diode and a resistive load. Which of the following best describes the waveform and polarity of the voltage developed across the load resistor?

Continuous direct current (DC) that maintains a constant, ripple-free voltage level across both halves of the input cycle

Alternating current (AC) that flows in both directions with reduced amplitude across both halves of the cycle

Pulsating direct current (DC) that appears only during the half-cycles when the diode is forward-biased

Alternating voltage that is amplified into high-frequency oscillations by converting current into magnetic energy

Show answer & explanation

Answer: Pulsating direct current (DC) that appears only during the half-cycles when the diode is forward-biased

Answer

Pulsating direct current (DC) that appears only during the half-cycles when the diode is forward-biased

A standard PN-junction diode conducts current only when it is forward-biased (anode more positive than cathode) and prevents current flow when reverse-biased. In an AC circuit, this unidirectional conduction allows current to pass through the load during only one half of each alternating cycle, converting the bidirectional AC input into a pulsating direct current (DC) output.

Step-by-Step Solution

Analyze the one-way conduction property of the PN-junction diode.

The diode allows current to pass only when its anode is positive relative to its cathode (forward bias) and blocks current when polarity reverses (reverse bias).

Semiconductor PN junctions function as directional electrical check valves.

Evaluate the circuit response across both halves of the sinusoidal AC cycle.

During the positive half-cycle, the diode conducts and produces a voltage across the load resistor; during the negative half-cycle, the diode blocks conduction, dropping current to zero.

This process constitutes half-wave rectification.

Classify the resulting load voltage waveform.

The voltage across the load is unidirectional and periodic, known as pulsating DC.

Because current flows in only one direction in pulses without filter smoothing, the output is pulsating DC.

Key Concept

Diode rectification and unidirectional current conduction

Estimated Time:1m 0s

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