Electronic Components Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 6

Question 101Question →

Electromechanical relays and manual switches use specific contact arrangements to govern how electrical circuits open, close, and divert current under various operating states. Which of the following correctly pairs each switch configuration or relay contact state on the left with its functional definition on the right?

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Normally Open (NO) Relay Contact matches with 'Remains open to block current when de-energized, closing to complete the circuit only when the coil is energized'; Normally Closed (NC) Relay Contact matches with 'Maintains continuity in the unpowered resting state and opens to interrupt current when the relay coil is energized'; Single-Pole, Double-Throw (SPDT) Switch matches with 'Routes current from a single common input terminal to either of two separate output paths depending on the actuator position'; and Double-Pole, Single-Throw (DPST) Switch matches with 'Simultaneously opens or closes two separate, electrically isolated circuits using a single mechanical actuator'.

Each contact type is paired based on standard electrical conventions: Normally Open contacts require coil excitation to establish continuity; Normally Closed contacts conduct by default until coil energization breaks continuity; SPDT switches route one source line between two alternate destinations; and DPST switches control two isolated lines in tandem.

Step-by-Step Solution

1

Analyze the Normally Open (NO) relay contact behavior.

Identified as an open circuit in the resting state that closes upon coil energization.

'Normal' in relay terminology denotes the de-energized or unpowered coil state; thus, an NO contact remains disconnected until magnetic attraction pulls the contacts shut.

2

Analyze the Normally Closed (NC) relay contact behavior.

Identified as a closed circuit in the resting state that opens when the coil is energized.

An NC contact provides continuous circuit conduction until an energized coil generates a magnetic field that pulls the armature away to break continuity.

3

Evaluate the Single-Pole, Double-Throw (SPDT) switch configuration.

Identified as one input pole with two selectable throw positions.

'Single-Pole' means one circuit path is controlled; 'Double-Throw' means the common terminal can connect to either of two output load contacts.

4

Evaluate the Double-Pole, Single-Throw (DPST) switch configuration.

Identified as two electrically isolated circuit poles that simultaneously turn on or off.

'Double-Pole' designates two separate conductive lines controlled by a unified toggle/actuator; 'Single-Throw' indicates simple open/closed switching for both.

Key Concept

Switch pole/throw nomenclature and electromechanical relay contact states

Estimated Time:1m 0s

Question 102Question →

A technician servicing a radio receiver needs to verify a fixed carbon-film resistor with four color bands: Brown, Black, Yellow, and Silver. Calculate the nominal resistance and the minimum allowable resistance under its tolerance rating, and complete the statement below.

Fill in the blanks below

The nominal resistance of the resistor is

\text{k}\Omega

, and its minimum allowable resistance within tolerance is

\text{k}\Omega

.

Show answer & explanation

Answer

The nominal resistance is

100\text{ k}\Omega

and the minimum allowable resistance is

90\text{ k}\Omega

.

According to the electronic color code system, Brown represents 1 and Black represents 0, forming the base number 10. The third band, Yellow, represents a multiplier of

10^4

(

10{}000

), giving a nominal resistance of

10 \times 10{}000 = 100{}000\ \Omega = 100\text{ k}\Omega

. The fourth band, Silver, designates a

\pm 10\%

tolerance. Ten percent of

100\text{ k}\Omega

is

10\text{ k}\Omega

, meaning the minimum allowable resistance under tolerance is

100\text{ k}\Omega - 10\text{ k}\Omega = 90\text{ k}\Omega

(and the maximum is

110\text{ k}\Omega

).

Step-by-Step Solution

1

Decode the significant digits from the first two color bands.

Band 1 (Brown) =

1

, Band 2 (Black) =

0

, giving significant digits

10

.

In standard 4-band EIA color coding, the first two bands represent the first two significant figures of the resistance value.

2

Apply the multiplier from the third color band and convert to kilohms.

Band 3 (Yellow) = multiplier

10^4

(

10{}000

). Nominal resistance

= 10 \times 10{}000 = 100{}000\ \Omega = 100\text{ k}\Omega

.

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

3

Determine the tolerance percentage from the fourth color band.

Band 4 (Silver) =

\pm 10\%

tolerance.

Silver indicates a manufacturing variation of plus or minus ten percent from nominal value.

4

Calculate the minimum allowable resistance limit.

Tolerance amount

= 100\text{ k}\Omega \times 0.10 = 10\text{ k}\Omega

. Minimum resistance

= 100\text{ k}\Omega - 10\text{ k}\Omega = 90\text{ k}\Omega

.

Subtracting the tolerance deviation from the nominal value establishes the lower acceptable boundary.

Key Concept

Standard 4-band resistor color decoding and tolerance range calculation

Estimated Time:45s

Question 103Question →

A technician connects a step-up transformer to a $30\text{ V}$ AC power supply. The transformer has $200\text{ turns}$ on its primary winding and $800\text{ turns}$ on its secondary winding. Assuming ideal operating conditions with no energy losses, what is the output voltage at the secondary winding?

7.5\text{ V}

12\text{ V}

120\text{ V}

0\text{ V}

Show answer & explanation

Answer: $120\text{ V}$

Answer

The secondary output voltage is

120\text{ V}

.

For an ideal transformer, secondary voltage is calculated using the formula

V_s = V_p \times \left(\frac{N_s}{N_p}\right)

. Given a primary voltage of

30\text{ V}

and a turns ratio of

\frac{800}{200} = 4

, the secondary voltage is

30\text{ V} \times 4 = 120\text{ V}

.

Step-by-Step Solution

1

Identify the known values and the transformer voltage-turns relationship formula.

V_p = 30\text{ V}

,

N_p = 200\text{ turns}

,

N_s = 800\text{ turns}

, using

\frac{V_s}{V_p} = \frac{N_s}{N_p}

.

In an ideal transformer, the ratio of the secondary voltage to the primary voltage equals the ratio of secondary turns to primary turns.

2

Calculate the turns ratio (

N_s / N_p

).

\frac{800}{200} = 4

.

Finding the ratio simplifies multiplying the primary voltage to determine secondary voltage.

3

Multiply the primary input voltage by the turns ratio to find the secondary voltage (

V_s

).

V_s = 30\text{ V} \times 4 = 120\text{ V}

.

Since the secondary winding has four times as many turns as the primary, the secondary voltage must be four times greater than the primary voltage.

Key Concept

Transformer Turns and Voltage Ratio

Estimated Time:45s

Question 104Question →

A standard glass cartridge fuse is rated for a continuous operating current of $3\text{ A}$ . What occurs within this protective device if an electrical short circuit causes a sustained current of $10\text{ A}$ to flow through it?

The internal element temporarily flexes open and automatically snaps closed once the overload clears.

The internal metal element overheats and melts, creating an open circuit that permanently stops current flow.

The internal element increases its electrical resistance to convert the extra current into additional supply voltage.

The device alters its polarity to force excess electron flow back toward the power source.

Show answer & explanation

Answer: The internal metal element overheats and melts, creating an open circuit that permanently stops current flow.

Answer

The internal metal element overheats and melts, creating an open circuit that permanently stops current flow.

A fuse is a one-time sacrificial protective device wired in series with a load. When the circuit current (

10\text{ A}

) exceeds the rated threshold (

3\text{ A}

), resistive heating (

I^2R

) rapidly raises the temperature of the internal metal link past its melting point. The link separates, creating an open circuit (infinite resistance) that permanently halts current flow to prevent wiring damage or fire.

Step-by-Step Solution

1

Identify the type and rating of the protective device.

The device is a standard cartridge fuse rated for

3\text{ A}

.

Determining the device type establishes whether it is a one-time sacrificial device or a resettable mechanism.

2

Compare the actual fault current to the device's rated current.

The fault current of

10\text{ A}

significantly exceeds the

3\text{ A}

rating.

Current passing through the fuse element generates heat proportional to

I^2R

(

P = I^2R

).

3

Determine the physical response of the fuse element under this overcurrent condition.

The excessive current heats the thin metal wire or strip beyond its melting point, burning it open.

Melting breaks electrical continuity, producing an open circuit with infinite resistance that stops all downstream current.

Key Concept

Operating principle of a standard overcurrent fuse

Estimated Time:45s

Question 105Question →

An electronics technician is testing an optical pulse circuit and measures that a capacitor stores an electrostatic charge of $450\text{ }\mu\text{C}$ when connected across a potential difference of $15\text{ V}$ . What is the capacitance of this component?

6{}750\text{ }\mu\text{F}

30\text{ }\mu\text{H}

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

0.03\text{ }\mu\text{F}

Show answer & explanation

Answer: $30\text{ }\mu\text{F}$

Answer

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

Capacitance is defined by the fundamental relationship

C = \frac{Q}{V}

. Dividing the stored charge of

450\text{ }\mu\text{C}

by the applied potential difference of

15\text{ V}

yields

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

.

Step-by-Step Solution

1

Identify the relationship between charge, capacitance, and voltage

The governing equation is

Q = C \times V

, which rearranges to

C = \frac{Q}{V}

.

Capacitance is defined as the amount of electric charge stored per unit of electrical potential difference.

2

Substitute the given values into the formula and solve for capacitance

C = \frac{450\text{ }\mu\text{C}}{15\text{ V}} = 30\text{ }\mu\text{F}

Dividing the charge in microcoulombs by the potential difference in volts directly yields the capacitance in microfarads.

Key Concept

The capacitance (

C

) of a component is the ratio of electric charge (

Q

) stored on its plates to the applied voltage (

V

), measured in Farads (F).

Estimated Time:1m 0s

Question 106Question →

Electromagnetic components are essential for controlling current and voltage levels in alternating current circuits. Which characteristic or operational principle correctly corresponds to each listed component or term?

Click a left item, then click its matching right item

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Answer

Step-up transformer matches a device with more secondary turns than primary turns that increases alternating voltage; Inductor matches a device that opposes rapid changes in electric current by storing energy in a magnetic field; Henry matches the standard unit of measurement for electrical inductance; Mutual induction matches the process by which a changing current in one coil induces an electromotive force in an adjacent coil.

Each item is matched to its exact operational role and definition: the step-up transformer increases AC voltage via a greater secondary turn count; the inductor resists current changes through magnetic storage; the henry serves as the unit of inductance; and mutual induction represents the principle of transferring energy across coupled coils.

Step-by-Step Solution

1

Identify the function of a step-up transformer.

A step-up transformer increases alternating voltage because its secondary winding has more turns than its primary winding (

N_s > N_p

).

According to the transformer turns ratio formula,

\frac{V_s}{V_p} = \frac{N_s}{N_p}

, higher secondary turns produce higher secondary voltage.

2

Identify the operating behavior of an inductor.

An inductor resists sudden current fluctuations by temporarily storing electrical energy in a concentrated magnetic field.

Lenz's law states that the induced back-EMF always acts to oppose the original change in current.

3

Identify the standard unit for inductance.

The henry (

\text{H}

) is the recognized unit of electrical inductance.

One henry of inductance induces an electromotive force of one volt when the current changes at a rate of one ampere per second.

4

Identify the definition of mutual induction.

Mutual induction is the magnetic coupling effect where current variations in a primary winding induce a voltage across a separate secondary winding.

This electromagnetic linkage allows electrical energy to transfer between isolated circuits without direct conductive contact.

Key Concept

Core definitions, units, and operational characteristics of inductors and transformers

Estimated Time:45s

Question 107Question →

Fill in the blanks to complete the statement describing how overcurrent protective devices are connected and restored in an electrical circuit.

Fill in the blanks below

To protect a circuit from excessive current, a protective device such as a fuse or circuit breaker must always be connected in with the load. Unlike a one-time fuse that melts its element and must be replaced, a standard circuit breaker can simply be after the fault is cleared.

Show answer & explanation

Answer

blank_1: series; blank_2: reset

Overcurrent protective devices like fuses and circuit breakers must always be installed in series with the load so that the full circuit current passes through them. If an overcurrent or short circuit occurs, opening the device breaks the single path for current flow, completely de-energizing the circuit. Additionally, while fuses contain a sacrificial metal strip that melts and requires complete replacement after opening, circuit breakers utilize reusable mechanical contacts that can be manually reset to restore normal operation.

Step-by-Step Solution

1

Determine the required wiring configuration for an overcurrent protection device.

Protective devices must be placed in series with the load.

Connecting the protective device in series ensures that all circuit current flows through it, allowing it to interrupt the entire current path when an overload occurs.

2

Identify the restoration method for a tripped circuit breaker compared to a blown fuse.

A circuit breaker can be mechanically reset.

Fuses rely on a sacrificial metal link that melts permanently when rated current is exceeded, whereas circuit breakers use reusable mechanical contacts that open and can be manually reset.

Key Concept

Overcurrent protective device circuit placement and operational restoration

Estimated Time:45s

Question 108Question →

A $100\text{ k}\Omega$ linear potentiometer is connected across a $12\text{ V}$ DC power supply, with terminal 1 connected to $+12\text{ V}$ , terminal 3 connected to ground ( $0\text{ V}$ ), and terminal 2 serving as the adjustable wiper. If the wiper is set to $40\%$ of its total travel distance away from terminal 3 toward terminal 1, what is the output voltage measured at terminal 2 relative to ground, and what is the resistance between terminal 1 and terminal 2?

7.2\text{ V}

and

40\text{ k}\Omega

4.8\text{ V}

and

60\text{ k}\Omega

4.8\text{ V}

and

40\text{ k}\Omega

4.8\text{ V}

and

600\ \Omega

Show answer & explanation

Answer: $4.8\text{ V}$ and $60\text{ k}\Omega$

Answer

4.8\text{ V}

output voltage and

60\text{ k}\Omega

resistance between terminal 1 and terminal 2

A linear potentiometer configured across a DC voltage source acts as a continuous voltage divider. With terminal 3 connected to ground (

0\text{ V}

) and the wiper at

40\%

travel from ground, the lower resistance segment

R_{2-3}

is

40\text{ k}\Omega

, yielding an output voltage of

V_{\text{out}} = 12\text{ V} \times (40\text{ k}\Omega / 100\text{ k}\Omega) = 4.8\text{ V}

. The upper resistance segment between terminal 1 (

+12\text{ V}

) and terminal 2 (wiper) is the remaining portion:

100\text{ k}\Omega - 40\text{ k}\Omega = 60\text{ k}\Omega

.

Step-by-Step Solution

1

Calculate the resistance of the lower track section between terminal 2 (wiper) and terminal 3 (ground).

R_{2-3} = 0.40 \times 100\text{ k}\Omega = 40\text{ k}\Omega

The wiper is positioned at

40\%

of total travel measured from the ground terminal.

2

Calculate the output voltage

V_{\text{out}}

at terminal 2 relative to ground using the voltage divider relationship.

V_{\text{out}} = V_{\text{in}} \times \left(\frac{R_{2-3}}{R_{\text{total}}}\right) = 12\text{ V} \times \left(\frac{40\text{ k}\Omega}{100\text{ k}\Omega}\right) = 4.8\text{ V}

A potentiometer connected across a voltage source functions as an adjustable voltage divider.

3

Determine the remaining resistance between terminal 1 (

+12\text{ V}

) and terminal 2 (wiper).

R_{1-2} = R_{\text{total}} - R_{2-3} = 100\text{ k}\Omega - 40\text{ k}\Omega = 60\text{ k}\Omega

The total resistance is divided into two series segments by the wiper contact.

Key Concept

Potentiometer voltage division and wiper segment resistance distribution

Estimated Time:45s

Question 109Question →

Two standard silicon rectifier diodes are arranged in a dual-input diode OR logic configuration. The anode of Diode 1 is connected to a $+5.0\text{ V}$ DC supply, and the anode of Diode 2 is connected to a $+2.0\text{ V}$ DC supply. Both cathodes connect to a common junction that ties to ground through a $2.2\text{ k}\Omega$ load resistor. Assuming each forward-biased silicon diode exhibits a nominal barrier potential drop of $0.7\text{ V}$ , what is the resulting output voltage measured at the common cathode junction?

1.3\text{ V}

4.3\text{ V}

5.0\text{ V}

6.3\text{ V}

Show answer & explanation

Answer: $4.3\text{ V}$

Answer

The voltage measured at the common output cathode junction is

4.3\text{ V}

.

The output potential is determined by the diode with the highest anode voltage. Diode 1 has

+5.0\text{ V}

applied to its anode, which forward-biases its PN junction and establishes a common cathode potential of

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

. Because Diode 2 has only

+2.0\text{ V}

at its anode while its cathode is held at

+4.3\text{ V}

, it experiences a net reverse bias of

-2.3\text{ V}

and remains completely non-conductive (OFF).

Step-by-Step Solution

1

Determine which diode enters forward conduction first by comparing anode potentials.

Diode 1 has an anode voltage of

+5.0\text{ V}

, which is higher than Diode 2's anode voltage of

+2.0\text{ V}

.

In a common-cathode configuration, the diode with the most positive anode potential forward-biases first and sets the output voltage.

2

Calculate the output node voltage established by the conducting diode.

V_{\text{out}} = 5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}

.

A conducting silicon diode drops approximately

0.7\text{ V}

across its PN junction.

3

Evaluate the biasing state of Diode 2 with the cathode node established at

4.3\text{ V}

.

V_{\text{anode2}} - V_{\text{cathode}} = 2.0\text{ V} - 4.3\text{ V} = -2.3\text{ V}

(Reverse Biased / OFF).

Because the cathode is more positive than the anode, Diode 2 is reverse-biased, acts as an open circuit, and contributes no current to the load.

Key Concept

Diode conduction states, forward barrier potential drop, and reverse-bias isolation in multi-diode circuits

Estimated Time:1m 30s

Question 110Question →

In an electromechanical relay, the normally closed (NC) contacts maintain electrical continuity and allow current to flow through the load circuit when the control coil is de-energized.

True

False

Show answer & explanation

Answer: True

Answer

True

Relay contact states are defined relative to the de-energized (unpowered) coil condition. Normally Closed (NC) contacts are held shut by a mechanical return spring when the coil has no current, thereby maintaining a complete circuit and allowing current to pass.

Step-by-Step Solution

1

Identify the operating definition of 'normal' in relay terminology.

The normal state refers to the resting condition of the relay when no voltage or current is applied to the electromagnetic coil.

All contact designations (Normally Open and Normally Closed) are standardized based on the unpowered coil state.

2

Determine the physical contact position of Normally Closed (NC) contacts in the unpowered state.

The internal return spring holds the movable armature firmly against the NC contact terminal, forming a continuous electrical path.

Mechanical spring tension keeps the circuit closed until magnetic force from an energized coil pulls the armature away.

3

Evaluate circuit continuity and current flow capability.

Because the physical contacts touch, electrical continuity exists and current can flow through the load circuit.

Closed contacts provide a low-resistance path allowing electrical current to circulate.

Key Concept

Normally Closed (NC) vs. Normally Open (NO) Relay Contact States

Estimated Time:45s

Question 111Question →

A standard circuit breaker trips and interrupts power to an appliance branch circuit due to an electrical overload. After the fault condition has been resolved, what action is required to restore power to the circuit?

Replace the unit with a new protective device because the internal conductor has permanently melted.

Increase the source voltage to force the internal sensing mechanism to disengage.

Manually reset the breaker operating handle back to the closed (ON) position.

Rewire the branch circuit so the protective device is connected in parallel with the load.

Show answer & explanation

Answer: Manually reset the breaker operating handle back to the closed (ON) position.

Answer

Manually reset the breaker operating handle back to the closed (ON) position.

Circuit breakers are reusable protective devices engineered with bimetallic strips or electromagnetic coils that open electrical contacts during overcurrent events. Unlike fuses that melt and must be discarded, a circuit breaker simply trips its mechanical latch and can be reset manually to the ON position once the electrical fault is cleared.

Step-by-Step Solution

1

Identify the type of protective device specified in the scenario.

The device is a circuit breaker, which is a reusable electromechanical switch.

Differentiating between sacrificial devices (fuses) and resettable devices (circuit breakers) determines the correct maintenance action.

2

Determine the procedure for restoring an open circuit breaker after a trip.

The internal trip mechanism mechanically separates the contacts, which can be re-engaged by toggling the external handle.

Circuit breakers do not destroy their internal elements during normal tripping and do not need to be replaced.

Key Concept

Circuit Breaker Operation and Resetting

Estimated Time:45s

Question 112Question →

Match each circuit protective device on the left with its primary operating characteristic or application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fast-acting fuse pairs with opening rapidly to protect sensitive semiconductor electronics; Slow-blow fuse pairs with tolerating brief inrush currents from inductive motors; Magnetic circuit breaker pairs with using an electromagnetic coil to trip during heavy short circuits; Metal-oxide varistor pairs with clamping transient high-voltage surges.

Each protective device is engineered for a specific fault type or application: fast-acting fuses react immediately to protect delicate electronics; slow-blow fuses accommodate brief inductive surges; magnetic breakers use electromagnetism for instant short-circuit tripping; and metal-oxide varistors suppress high-voltage spikes.

Step-by-Step Solution

1

Identify the characteristic of a fast-acting fuse

Matches with immediate element melting to protect delicate solid-state electronics from sudden overcurrent.

Semiconductors cannot withstand overcurrent for even brief durations.

2

Identify the characteristic of a slow-blow fuse

Matches with tolerating brief inrush currents during electric motor startup.

Inductive loads draw high starting current that would blow standard fuses prematurely.

3

Identify the characteristic of a magnetic circuit breaker

Matches with utilizing an electromagnetic coil to trip instantaneously during high-current short circuits.

Electromagnetic pull is proportional to current and trips the contact latch mechanism without relying on heat.

4

Identify the characteristic of a metal-oxide varistor (MOV)

Matches with clamping high-voltage transient spikes.

MOVs are voltage-dependent resistors designed specifically for surge suppression rather than direct overcurrent interruption.

Key Concept

Operating principles and applications of circuit protection devices including thermal, magnetic, time-delay, and voltage suppression elements

Estimated Time:45s

Question 113Question →

A circuit is constructed with a $10\text{ V}$ DC power supply connected in series with a $220\ \Omega$ current-limiting resistor. After the resistor, the circuit splits into two parallel branches that eventually reconnect to the negative terminal of the power supply. The first parallel branch contains a Zener diode (rated at $V_Z = 6.2\text{ V}$ ) installed in a reverse-biased orientation. The second parallel branch contains a standard silicon diode installed in a forward-biased orientation. Based on this configuration, what is the approximate voltage measured across the parallel diode combination, and which component dictates this voltage?

0.7\text{ V}

, dictated by the forward-biased standard silicon diode clamping the voltage.

6.2\text{ V}

, dictated by the reverse-biased Zener diode regulating the branch voltage.

6.9\text{ V}

, dictated by the combined additive voltage drops of both diodes in the circuit.

10\text{ V}

, dictated by the power supply because the reverse-biased Zener diode acts as an open circuit.

Show answer & explanation

Answer: $0.7\text{ V}$ , dictated by the forward-biased standard silicon diode clamping the voltage.

Answer

The correct answer is

0.7\text{ V}

, dictated by the forward-biased standard silicon diode clamping the voltage.

The correct answer accurately identifies that in a parallel circuit, the component with the lowest forward voltage drop or breakdown voltage dictates the voltage across the entire parallel block. Since the standard silicon diode is forward-biased, it drops approximately 0.7 V. This prevents the voltage across the parallel branches from ever reaching the 6.2 V required to cause the Zener diode to break down and conduct. The Zener diode remains effectively an open circuit, and the standard diode clamps the branch voltage at 0.7 V.

Step-by-Step Solution

1

Determine the biasing state of each diode in the parallel branches based on the power supply orientation.

The Zener diode is reverse-biased (requires 6.2 V to reach breakdown), and the standard silicon diode is forward-biased (requires ~0.7 V to conduct).

A diode's behavior depends fundamentally on whether it is forward or reverse-biased.

2

Analyze the voltage rules for parallel circuits.

The voltage across all components connected in parallel must be identical.

This fundamental circuit law means both the Zener diode branch and the standard diode branch must share the exact same voltage drop.

3

Determine the clamping voltage of the parallel combination.

The forward-biased silicon diode begins conducting heavily at 0.7 V, clamping the voltage across both branches at this level.

Because the silicon diode provides a low-resistance path at 0.7 V, the voltage cannot rise high enough to reach the Zener diode's 6.2 V breakdown threshold. The Zener diode remains effectively off (open circuit).

Key Concept

Diode Biasing and Parallel Circuit Voltage Clamping

Estimated Time:1m 30s

Question 114Question →

A power supply circuit utilizes a step-down transformer to convert an alternating current (AC) input of $100\text{ V}$ to an output of $25\text{ V}$ . If the primary winding contains $400\text{ turns}$ , how many turns are on the secondary winding?

1{}600\text{ turns}

100\text{ turns}

25\text{ turns}

10\text{ turns}

Show answer & explanation

Answer: $100\text{ turns}$

Answer

100\text{ turns}

The correct answer is

100\text{ turns}

. According to the transformer turns ratio law, the ratio of secondary voltage to primary voltage equals the ratio of secondary turns to primary turns:

\frac{V_s}{V_p} = \frac{N_s}{N_p}

. Rearranging to solve for the secondary turns yields

N_s = N_p \times \left(\frac{V_s}{V_p}\right) = 400 \times \left(\frac{25}{100}\right) = 100\text{ turns}

. Since this is a step-down transformer (

100\text{ V}

down to

25\text{ V}

, a

4:1

step-down ratio), the secondary coil must have one-fourth the number of turns of the primary coil.

Step-by-Step Solution

1

Identify the given transformer parameters

Primary voltage

V_p = 100\text{ V}

, secondary voltage

V_s = 25\text{ V}

, primary turns

N_p = 400\text{ turns}

Extracting the known values is necessary to set up the turns ratio equation.

2

State the transformer voltage and turns ratio formula

\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies N_s = N_p \times \left(\frac{V_s}{V_p}\right)

In an ideal transformer, the voltage across each winding is directly proportional to the number of turns.

3

Substitute the values and calculate the secondary winding turns

N_s = 400 \times \left(\frac{25\text{ V}}{100\text{ V}}\right) = 400 \times 0.25 = 100\text{ turns}

Performing the proportional calculation gives the exact number of secondary turns required.

Key Concept

Transformer Turns Ratio and Voltage Transformation

Estimated Time:45s

Question 115Question →

An electronics technician is replacing components on a circuit board and selects an electrolytic capacitor for the DC power supply filtering stage. Why must the technician observe strict polarity when installing an electrolytic capacitor into a DC circuit?

They store electrical energy in a magnetic field that collapses and damages surrounding components if polarity is reversed.

They possess an electrochemical insulating oxide layer that breaks down and can cause the component to overheat or rupture under reverse voltage.

They are constructed to allow direct current to flow continuously while blocking all alternating current signals.

They operate through direct electrical conduction between the two internal plates without an insulating dielectric barrier.

Show answer & explanation

Answer: They possess an electrochemical insulating oxide layer that breaks down and can cause the component to overheat or rupture under reverse voltage.

Answer

Electrolytic capacitors have an electrochemical oxide dielectric layer that breaks down and causes the component to overheat or rupture if connected with reversed polarity.

Electrolytic capacitors are polarized components because their extremely thin dielectric layer (aluminum or tantalum oxide) is formed and maintained electrochemically in one specific voltage direction. If connected in reverse polarity, this insulating layer breaks down rapidly, allowing heavy leakage current to generate boiling electrolyte gases, resulting in device failure, venting, or explosion.

Step-by-Step Solution

1

Identify the internal construction of electrolytic capacitors.

Electrolytic capacitors utilize an extremely thin dielectric film of aluminum oxide or tantalum oxide maintained by a chemical electrolyte.

Understanding the unique dielectric mechanism explains why electrolytic capacitors provide high capacitance in a compact size but possess polarity restrictions.

2

Analyze the electrical behavior of the dielectric layer under reverse polarity.

When reversed voltage is applied, the electrochemical reaction breaks down and de-forms the insulating oxide layer.

The formation and insulation of the oxide layer are unidirectional.

3

Evaluate the practical consequence of reverse installation in a circuit.

With the dielectric compromised, large leakage currents flow through the electrolyte, producing excessive heat, gas buildup, and catastrophic rupture or short circuit.

This necessitates strict observance of the positive and negative terminal markings during assembly and maintenance.

Key Concept

Electrolytic Capacitor Polarity and Dielectric Integrity

Estimated Time:1m 0s

Question 116Question →

A technician is troubleshooting a test circuit connected to a $32\text{V}$ DC power supply. The circuit consists of a fixed resistor $R_1$ in series with a parallel block. The parallel block contains a $12\text{ k}\Omega$ fixed resistor and a variable resistor (rheostat) $R_v$ . The resistor $R_1$ has four color bands: Green, Blue, Red, and Gold. The technician adjusts the rheostat $R_v$ until the total current drawn from the power supply is exactly $4\text{ mA}$ .

What is the resistance setting of the rheostat $R_v$ in kilohms (k $\Omega$ )?

Show answer & explanation

Answer: 3

Answer

3

The total resistance required by the circuit to draw

4\text{ mA}

from a

32\text{V}

source is

8\text{ k}\Omega

. The series resistor

R_1

(Green, Blue, Red) contributes

5.6\text{ k}\Omega

. This leaves exactly

2.4\text{ k}\Omega

required for the parallel section. For a

12\text{ k}\Omega

resistor in parallel with an unknown

R_v

to yield an equivalent resistance of

2.4\text{ k}\Omega

,

R_v

must be set to

3\text{ k}\Omega

.

Step-by-Step Solution

1

Decode the color code of resistor

R_1

.

R_1 = 5.6\text{ k}\Omega

The color bands are Green (5), Blue (6), and Red (multiplier of

10^2

or 100). Thus,

56 \times 100 = 5,600\ \Omega

, which is

5.6\text{ k}\Omega

. The Gold band indicates a 5% tolerance, which applies to real-world variance but does not alter our nominal calculation.

2

Calculate the total equivalent resistance (

R_{eq}

) of the circuit.

R_{eq} = 8\text{ k}\Omega

Using Ohm's Law (

R = \frac{V}{I}

), divide the source voltage (

32\text{V}

) by the total current (

4\text{ mA}

or

0.004\text{ A}

) to find the total resistance required to draw that current.

3

Determine the required equivalent resistance of the parallel block.

R_{parallel} = 2.4\text{ k}\Omega

Since

R_1

is in series with the parallel block, subtract

R_1

from the total circuit resistance to find the resistance of the parallel segment:

8\text{ k}\Omega - 5.6\text{ k}\Omega = 2.4\text{ k}\Omega

.

4

Solve for the rheostat's resistance (

R_v

).

R_v = 3\text{ k}\Omega

The parallel block contains a

12\text{ k}\Omega

resistor and

R_v

. Using the parallel resistance formula

\frac{1}{R_p} = \frac{1}{R_a} + \frac{1}{R_b}

, we get

\frac{1}{2.4} = \frac{1}{12} + \frac{1}{R_v}

. Rearranging gives

\frac{1}{R_v} = \frac{1}{2.4} - \frac{1}{12} = \frac{10}{24} - \frac{2}{24} = \frac{8}{24} = \frac{1}{3}

, meaning

R_v = 3\text{ k}\Omega

.

Key Concept

Integrating Ohm's law with series-parallel equivalent resistance and resistor color code decoding.

Estimated Time:3m 0s

Question 117Question →

Match each resistor configuration or color-coded component on the left with its corresponding electrical characteristic and operating principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The three-terminal potentiometer matches the voltage divider function; the rheostat matches the series current limiter with wiper-to-end protection; the Yellow-Violet-Orange-Gold resistor matches

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

; and the Brown-Gray-Green-Silver resistor matches

1.8\text{ M}\Omega \pm 10\%

.

Each item correctly links the component physical description or color code to its precise electrical definition: potentiometers provide variable potential division, rheostats adjust circuit current, Yellow-Violet-Orange-Gold evaluates to

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

, and Brown-Gray-Green-Silver evaluates to

1.8\text{ M}\Omega \pm 10\%

.

Step-by-Step Solution

1

Analyze the three-terminal potentiometer and rheostat wiring configurations

A potentiometer connected across a power source acts as a continuous voltage divider (

0\text{ V}

to

V_{\text{in}}

) at the wiper terminal. A rheostat uses a variable two-terminal configuration in series with the load to regulate current.

Variable resistors differ in circuit function based on whether they are wired as three-terminal potential dividers or two-terminal current adjusters.

2

Decode the four-band color code for Yellow-Violet-Orange-Gold

Yellow = 4, Violet = 7, Orange = multiplier

10^3 = 1,000

, Gold =

\pm 5\%

. Value

= 47 \times 1,000\ \Omega = 47\text{ k}\Omega \pm 5\%

.

Standard EIA 4-band color coding assigns significant digits to the first two bands, a decimal multiplier to the third band, and tolerance percentage to the fourth band.

3

Decode the four-band color code for Brown-Gray-Green-Silver

Brown = 1, Gray = 8, Green = multiplier

10^5 = 100,000

, Silver =

\pm 10\%

. Value

= 18 \times 100,000\ \Omega = 1.8\text{ M}\Omega \pm 10\%

.

Standard EIA 4-band color coding converts

1,800,000\ \Omega

using standard metric prefixes to

1.8\text{ M}\Omega

with a silver tolerance of

\pm 10\%

.

Key Concept

Operating principles and terminal configurations of fixed and variable resistors (potentiometers, rheostats, and standard color codes)

Alternative Method

To quickly decode 4-band resistors, write down the two digits (e.g., 47 for Yellow-Violet or 18 for Brown-Gray), append the number of zeros indicated by the multiplier color (3 zeros for Orange = 47,000, 5 zeros for Green = 1,800,000), and simplify to metric units (kilo or mega).

Estimated Time:1m 0s

Question 118Question →

When an individual internal transistor fails inside an integrated circuit, that specific component can be desoldered and replaced separately without replacing the whole package.

True

False

Show answer & explanation

Answer: False

Answer

False

Integrated circuits are fabricated on a single microscopic semiconductor die sealed inside plastic or ceramic, meaning internal components cannot be individually accessed or replaced.

Step-by-Step Solution

1

Identify the physical structure and manufacturing method of an integrated circuit.

An integrated circuit combines thousands or millions of microscopic components fabricated directly onto a single silicon chip sealed inside a protective housing.

Understanding IC construction determines whether internal elements can be accessed or individually serviced.

2

Evaluate the feasibility of repairing individual components within an integrated circuit.

Because the components are microscopic and permanently sealed on a monolithic die, individual parts cannot be desoldered or replaced.

Directly confirms that the entire IC package must be replaced when any internal component fails.

Key Concept

Monolithic construction and non-serviceable nature of integrated circuits

Estimated Time:45s

Question 119Question →

A technician is designing a control circuit to reverse the operating direction of a direct current (DC) motor by reversing the polarity of both supply leads simultaneously using a single manual switch. Which switch configuration is required to achieve this operation?

Single-Pole, Single-Throw (SPST)

Single-Pole, Double-Throw (SPDT)

Double-Pole, Double-Throw (DPDT)

Double-Pole, Single-Throw (DPST)

Show answer & explanation

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

Answer

Double-Pole, Double-Throw (DPDT)

A Double-Pole, Double-Throw (DPDT) switch contains two electrically isolated common terminals (poles) that simultaneously connect to one of two sets of output contacts (throws). By cross-wiring the two throw sets, toggling the switch swaps the positive and negative supply leads connected to the motor, reversing its rotational direction.

Step-by-Step Solution

1

Determine the number of isolated poles required

Two poles (Double-Pole) are needed

Reversing DC motor polarity requires controlling both the positive and negative supply lines simultaneously within isolated circuit paths.

2

Determine the number of positions/throws required per pole

Two throws (Double-Throw) are needed

Each pole must alternate between two different electrical connections (forward polarity and reversed polarity) rather than simply opening or closing.

3

Combine pole and throw specifications

A Double-Pole, Double-Throw (DPDT) switch is identified

When the outer terminals of a DPDT switch are cross-wired, flipping the switch reverses the voltage polarity applied across the motor terminals.

Key Concept

Switch nomenclature regarding poles (number of separate circuits controlled) and throws (number of output paths per pole) for polarity-reversal applications.

Estimated Time:1m 0s

Question 120Question →

In electronics, integrated circuits (ICs) are widely used to replace complex discrete component assemblies. Which of the following best describes the construction of a monolithic integrated circuit?

A protective device containing a metal strip that melts to interrupt excessive current flow

A complete electronic circuit composed of multiple interconnected components fabricated on a single semiconductor substrate

A non-conductive ceramic block designed strictly to isolate and prevent current from flowing between connections

A two-terminal semiconductor device that functions exclusively to permit current flow in only one direction

Show answer & explanation

Answer: A complete electronic circuit composed of multiple interconnected components fabricated on a single semiconductor substrate

Answer

A complete electronic circuit composed of multiple interconnected components fabricated on a single semiconductor substrate

A monolithic integrated circuit (IC), commonly called a microchip, consists of many microscopic active components (like transistors and diodes) and passive elements (like resistors and capacitors) manufactured together on a single semiconductor crystal substrate (silicon wafer).

Step-by-Step Solution

1

Identify the fundamental definition of an integrated circuit (IC)

An IC integrates multiple microscopic components (transistors, diodes, resistors, capacitors) into a functional system.

The prefix 'integrated' refers to combining individual circuit elements into a single package.

2

Determine the physical construction of a monolithic IC

Monolithic ICs are manufactured directly on a single piece of semiconductor material (substrate), typically silicon.

This contrasts with discrete component circuits where separate parts are wired together on a circuit board.

Key Concept

Monolithic integrated circuit construction and characteristics

Estimated Time:45s

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