All practice questions

Question 1041Question →

A technician is using a digital multimeter set to the resistance ( $Ω$ ) function to test a de-energized DC solenoid valve coil with a nominal rating of $24\ \Omega$ . Which multimeter display reading indicates that the solenoid coil has an internal open circuit?

0.0\ \Omega

24\ \Omega

\text{OL}

(over-limit) or infinite resistance

24\ \text{k}\Omega

Show answer & explanation

Answer: $\text{OL}$ (over-limit) or infinite resistance

Answer

An over-limit (OL) or infinite resistance reading on the digital multimeter display indicates an open circuit.

An open circuit occurs when there is a physical break in the conductive path of the coil winding. Because electric current cannot bridge the gap, the electrical resistance across the terminals becomes infinite (

\infty

). On a digital multimeter, infinite resistance that exceeds the instrument's measurement capability is displayed as 'OL' (over-limit).

Step-by-Step Solution

1

Identify the electrical state being tested on the de-energized coil.

An internal open circuit means the continuous wire winding has broken, creating an incomplete path.

When a conductor or winding breaks, current can no longer pass through the component.

2

Determine the theoretical resistance value of an open circuit.

An open circuit provides infinite resistance (

\infty\ \Omega

) to the multimeter's internal test current.

Air or physical gaps between broken wire ends have extremely high dielectric resistance.

3

Correlate infinite resistance with standard digital multimeter display conventions.

Digital multimeters display infinite resistance exceeding the measurement range as 'OL' (over-limit) or '

\infty

'.

Multimeters use 'OL' to indicate that the measured resistance exceeds the measurable range of the selected scale.

Key Concept

Interpreting digital multimeter resistance displays for open circuits versus normal continuity

Estimated Time:1m 0s

Question 1042Question →

In an operating direct-current (DC) circuit, conventional current is defined as moving from the positive terminal to the negative terminal through the external circuit, which is opposite to the physical drift direction of electrons.

True

False

Show answer & explanation

Answer: True

Answer

True: Conventional current flows from positive to negative in the external circuit, while electron flow moves from negative to positive.

The statement is true because standard electrical convention defines current as positive charge moving from positive to negative, whereas atomic theory proves that negative electrons physically flow from negative to positive in the external circuit.

Step-by-Step Solution

1

Identify the historical definition of conventional current flow.

Conventional current assumes positive charge movement from the positive terminal (+) through the external circuit to the negative terminal (-).

Benjamin Franklin and early physicists established current flow under the convention that positive charges were the moving carriers.

2

Identify the physical direction of electron flow in solid conductors.

Electrons carry a negative charge and travel from the negative terminal (-) through the external circuit toward the positive terminal (+).

Electrostatic laws state that like charges repel and opposite charges attract; hence, negative electrons are repelled by the negative battery terminal.

3

Compare the two flow directions to evaluate the statement.

Conventional current and electron flow describe opposite directions through the external circuit.

Since the statement accurately defines both directions as being opposite to one another, the statement is true.

Key Concept

Directional distinction between conventional current and electron flow in DC circuits

Estimated Time:40s

Question 1043Question →

A mobile radar subsystem is powered by a $24\text{ V}$ direct-current source. Its internal circuitry contains a parallel configuration with three distinct branches.

The total equivalent resistance of this parallel section is exactly $3\ \Omega$ .
A technician measures the current through the first branch to be $3\text{ A}$ .
The second branch has a known resistance of $24\ \Omega$ .

What is the resistance of the third branch in ohms?

Show answer & explanation

Answer: 6

Answer

The resistance of the third branch is

6\ \Omega

.

The correct value is derived by finding the total circuit current (

8\text{ A}

) and subtracting the known branch currents (

3\text{ A}

and

1\text{ A}

) to find the current in the third branch (

4\text{ A}

). Dividing the common

24\text{ V}

source by this

4\text{ A}

current yields

6\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit current using the total voltage and equivalent resistance.

I_{total} = 24\text{ V} / 3\ \Omega = 8\text{ A}

Ohm's law applies to the entire circuit, linking total voltage, total current, and equivalent resistance.

2

Determine the current flowing through the second branch.

I_2 = 24\text{ V} / 24\ \Omega = 1\text{ A}

In a parallel circuit, the full source voltage drops across every branch. Ohm's law (

I = V/R

) gives the branch current.

3

Calculate the current flowing through the third branch.

I_3 = 8\text{ A} - (3\text{ A} + 1\text{ A}) = 4\text{ A}

Kirchhoff's Current Law states that the total current is the sum of the individual branch currents (

I_{total} = I_1 + I_2 + I_3

).

4

Calculate the resistance of the third branch.

R_3 = 24\text{ V} / 4\text{ A} = 6\ \Omega

Apply Ohm's law (

R = V/I

) specifically to the third branch using the source voltage and the calculated branch current.

Key Concept

Applying Ohm's law to both individual branches and the entire circuit, combined with parallel circuit rules for voltage and current.

Alternative Method

Use the parallel resistance reciprocal formula:

1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3

. First, find

R_1 = V/I_1 = 24\text{ V} / 3\text{ A} = 8\ \Omega

. Then, substitute the known values:

1/3 = 1/8 + 1/24 + 1/R_3

. Finding a common denominator of 24 gives

8/24 = 3/24 + 1/24 + 1/R_3

, which simplifies to

8/24 = 4/24 + 1/R_3

. Therefore,

1/R_3 = 4/24 = 1/6

, meaning

R_3 = 6\ \Omega

.

Estimated Time:1m 30s

Question 1044Question →

An electric motor circuit experiences a momentary inrush current of 25 A when starting up, but drops to a steady operating current of 6 A during normal operation. Which protective device is specifically designed to tolerate this brief startup surge without opening the circuit while still protecting against prolonged overcurrent faults?

A fast-acting fuse

A slow-blow (time-delay) fuse

A parallel shunt resistor

A step-down transformer

Show answer & explanation

Answer: A slow-blow (time-delay) fuse

Answer

A slow-blow (time-delay) fuse

Slow-blow (time-delay) fuses are specifically engineered to accommodate temporary, harmless current surges—such as inductive motor starting currents—by incorporating a thermal delay element. If the overcurrent condition persists beyond the brief startup interval or reaches dangerous short-circuit levels, the fuse element melts to isolate and protect the circuit.

Step-by-Step Solution

1

Analyze the electrical load characteristics of the motor.

The motor draws a high transient inrush current (25 A) that significantly exceeds its continuous operating current (6 A).

Inductive loads like electric motors require substantial starting torque and draw temporary surge currents before establishing normal operating counter-electromotive force.

2

Determine the required response time characteristics for the protective device.

The device must absorb brief current spikes without nuisance tripping, yet still interrupt sustained overloads.

Protective devices must match circuit operational profiles to avoid unnecessary service interruptions while maintaining safety.

3

Identify the component designed with an intentional time lag mechanism.

A slow-blow (time-delay) fuse provides the necessary delay for inductive surges while maintaining full overload protection.

Time-delay fuses utilize dual-element or thermal delay construction that allows momentary overloads to pass harmlessly but melts if current remains high.

Key Concept

Time-delay protective devices and motor inrush current handling

Estimated Time:1m 0s

Question 1045Question →

An automobile's four-stroke gasoline engine operates at a steady speed of 2,400 RPM. For any single cylinder in this engine, how many power strokes occur each minute, and at what rotational speed does the camshaft turn?

2,400 power strokes per minute, with the camshaft turning at 2,400 RPM

1,200 power strokes per minute, with the camshaft turning at 1,200 RPM

1,200 power strokes per minute, with the camshaft turning at 4,800 RPM

2,400 power strokes per minute, with the camshaft turning at 1,200 RPM

Show answer & explanation

Answer: 1,200 power strokes per minute, with the camshaft turning at 1,200 RPM

Answer

1,200 power strokes per minute, with the camshaft turning at 1,200 RPM

In a four-stroke cycle, completing all four strokes (intake, compression, power, exhaust) requires two complete rotations (

720^\circ

) of the crankshaft. Consequently, each cylinder produces one power stroke every two revolutions, yielding

2,400 / 2 = 1,200

power strokes per minute. Because the intake and exhaust valves must each actuate only once per complete four-stroke cycle, the camshaft is geared to rotate at half crankshaft speed (

2,400 / 2 = 1,200\text{ RPM}

).

Step-by-Step Solution

1

Determine the relationship between crankshaft revolutions and power strokes in a four-stroke cycle.

A single cylinder completes one full four-stroke cycle (intake, compression, power, exhaust) every 2 crankshaft revolutions (

720^\circ

), producing exactly 1 power stroke per cycle.

Four distinct piston strokes require two complete rotations of the crankshaft.

2

Calculate the number of power strokes per minute for a single cylinder at 2,400 RPM.

\frac{2,400\text{ RPM}}{2\text{ revolutions/cycle}} = 1,200\text{ power strokes per minute}

.

The cylinder fires once every two crankshaft revolutions.

3

Determine camshaft speed relative to crankshaft speed.

The camshaft rotates at a 1:2 ratio relative to the crankshaft:

\frac{2,400\text{ RPM}}{2} = 1,200\text{ RPM}

.

Each intake and exhaust valve opens only once per four-stroke cycle, requiring the camshaft to complete one full revolution for every two revolutions of the crankshaft.

Key Concept

Four-stroke engine cycle kinematics and camshaft-to-crankshaft 1:2 timing relationship

Estimated Time:1m 15s

Question 1046Question →

When inspecting a circuit board, a technician observes a four-band resistor with the color sequence Red, Red, Orange, and Gold. What is the rated resistance value and tolerance of this resistor?

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

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

220\ \Omega \pm 5\%

223\ \Omega \pm 5\%

Show answer & explanation

Answer: $22\text{ k}\Omega \pm 5\%$

Answer

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

According to standard electronic color code rules, the first band (Red = 2) and second band (Red = 2) give the two significant digits 22. The third band (Orange =

10^3

) multiplies this base number by

1{}000

, resulting in

22{}000\ \Omega

(

22\text{ k}\Omega

). The fourth band (Gold) indicates a tolerance rating of

\pm 5\%

.

Step-by-Step Solution

1

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

Band 1 (Red) = 2, Band 2 (Red) = 2. Combined digits = 22.

In standard four-band EIA color coding, the first two bands establish the base significant figures.

2

Determine the multiplier from the third band.

Band 3 (Orange) = multiplier of

10^3

(

1{}000

).

Orange corresponds to 3 zeros, or multiplying by

1{}000

.

3

Calculate nominal resistance and append the tolerance from the fourth band.

22 \times 1{}000 = 22{}000\ \Omega = 22\text{ k}\Omega

, with Band 4 (Gold) =

\pm 5\%

.

Combining the base value with the metric prefix gives

22\text{ k}\Omega

with a rated tolerance of

\pm 5\%

.

Key Concept

Four-Band Resistor Color Code Decoding

Estimated Time:35s

Question 1047Question →

A direct-current power supply in an automotive workshop delivers $24\text{ V}$ across two parallel branches. The first branch contains an $8\ \Omega$ inspection lamp, and the second branch contains a $24\ \Omega$ ventilation motor. What is the total current drawn by the circuit from the power supply?

0.75\text{ A}

4\text{ A}

3\text{ A}

6\text{ A}

Show answer & explanation

Answer: $4\text{ A}$

Answer

4\text{ A}

In a parallel circuit, the voltage across each branch is identical to the supply voltage (

24\text{ V}

). The current in each branch is calculated using Ohm's Law: the

8\ \Omega

load draws

3\text{ A}

(

24\text{ V} / 8\ \Omega

), and the

24\ \Omega

load draws

1\text{ A}

(

24\text{ V} / 24\ \Omega

). Summing these branch currents gives a total supply current of

4\text{ A}

.

Step-by-Step Solution

1

Calculate the current flowing through the first branch (

8\ \Omega

resistor).

I_1 = \frac{V}{R_1} = \frac{24\text{ V}}{8\ \Omega} = 3\text{ A}

In a parallel circuit, each branch receives the full source voltage of

24\text{ V}

.

2

Calculate the current flowing through the second branch (

24\ \Omega

resistor).

I_2 = \frac{V}{R_2} = \frac{24\text{ V}}{24\ \Omega} = 1\text{ A}

Ohm's Law applies independently to the second branch using the same

24\text{ V}

potential.

3

Sum the individual branch currents to find the total circuit current.

I_{\text{total}} = I_1 + I_2 = 3\text{ A} + 1\text{ A} = 4\text{ A}

According to Kirchhoff's Current Law, the total current supplied to a parallel network equals the sum of all branch currents.

Key Concept

In a parallel circuit, total current is the sum of the individual branch currents, each determined by dividing the constant source voltage by the branch resistance.

Alternative Method

Find the equivalent parallel resistance first:

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\ \Omega

. Then apply Ohm's Law to the whole circuit:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{24\text{ V}}{6\ \Omega} = 4\text{ A}

.

Estimated Time:1m 0s

Question 1048Question →

An electrician is joining two solid copper conductors to form a pigtail splice using standard electrical hand tools. What is the correct sequence of steps to complete this connection?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is: Strip the insulation with wire strippers, align the bare ends with lineworker pliers, twist the conductors clockwise into a tight spiral, trim the twisted tips flush with the plier cutting blades, and thread a twist-on wire connector over the splice.

The correct procedure begins with stripping the outer insulation to expose clean copper. The bare ends are then aligned and twisted clockwise with lineworker pliers to establish a secure mechanical bond. Trimming any uneven ends ensures the splice fits fully inside the twist-on wire connector, which is screwed on last to provide electrical insulation.

Step-by-Step Solution

1

Strip wire insulation

Clean bare copper conductors are exposed without nicking the metal.

Removing insulation is the mandatory first step to establish electrical conductivity.

2

Align and hold bare ends with lineworker pliers

The two conductors are positioned parallel in the plier jaws.

Even alignment ensures uniform tension and twisting across both wires.

3

Twist conductors clockwise

A tight, solid mechanical twist is formed.

Provides mechanical strength and prevents loose contacts under thermal cycling.

4

Trim the splice tip flush

The twisted conductors end at an equal length.

Prevents exposed bare copper from protruding past the connector housing.

5

Install the twist-on wire connector

The joint is mechanically locked and electrically insulated.

Encloses the bare connection to prevent short circuits and shock hazards.

Key Concept

Proper operational procedure for creating a mechanical pigtail wire splice using hand tools

Estimated Time:45s

Question 1049Question →

In a four-stroke internal combustion engine, completing all four strokes requires the crankshaft to complete two full rotations ( $720^\circ$ ). How many full rotations does the camshaft make during one complete four-stroke cycle?

Two full rotations, turning at the same speed as the crankshaft

One full rotation, turning at half the speed of the crankshaft

Four full rotations, turning at twice the speed of the crankshaft

One-half rotation, turning at one-fourth the speed of the crankshaft

Show answer & explanation

Answer: One full rotation, turning at half the speed of the crankshaft

Answer

The camshaft completes one full rotation, operating at half the rotational speed of the crankshaft.

In a four-stroke internal combustion engine, one full thermodynamic cycle spans four piston strokes, which equals two complete rotations (

720^\circ

) of the crankshaft. Because each valve needs to open only once during this entire sequence, the camshaft is geared to rotate at exactly half the speed of the crankshaft, making exactly one full rotation per cycle.

Step-by-Step Solution

1

Determine the crankshaft rotation required for one four-stroke cycle.

Two complete revolutions (

720^\circ

) of the crankshaft occur per cycle.

Each of the four strokes (intake, compression, power, exhaust) corresponds to

180^\circ

of crankshaft rotation (

4 \times 180^\circ = 720^\circ

).

2

Determine the required frequency of valve operation.

Each cylinder's intake and exhaust valves must open and close exactly once per four-stroke cycle.

A single cam lobe actuates its designated valve once per complete rotation of the camshaft.

3

Calculate the drive gear ratio between the crankshaft and camshaft.

The gear ratio is 2:1, meaning 2 crankshaft turns produce 1 camshaft turn.

To actuate valves once every two crankshaft revolutions, the camshaft must turn at half the engine speed (

1 \text{ camshaft rotation} / 2 \text{ crankshaft rotations}

).

Key Concept

Crankshaft to camshaft 2:1 speed and timing relationship in four-stroke engines

Estimated Time:1m 0s

Question 1050Question →

A technician evaluates four direct-current (DC) electrical components operating under steady-state conditions. Match each component and operating condition on the left with its corresponding dissipated electrical power on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The automotive lamp (

12\text{ V}, 750\text{ mA}

) matches

9\text{ W}

; the heating element (

50\ \Omega, 0.6\text{ A}

) matches

18\text{ W}

; the instrumentation sensor (

600\ \Omega, 200\text{ mA}

) matches

24\text{ W}

; and the relay coil (

100\ \Omega, 60\text{ V}

) matches

36\text{ W}

.

Each electrical component's power dissipation is determined by selecting the appropriate algebraic form of Watt's Law based on the known parameters:

P = V \times I

for the indicator lamp (

12\text{ V} \times 0.75\text{ A} = 9\text{ W}

),

P = I^2 \times R

for the heating element (

(0.6\text{ A})^2 \times 50\ \Omega = 18\text{ W}

) and sensor (

(0.2\text{ A})^2 \times 600\ \Omega = 24\text{ W}

), and

P = \frac{V^2}{R}

for the relay coil (

(60\text{ V})^2 / 100\ \Omega = 36\text{ W}

).

Step-by-Step Solution

1

Calculate power for the automotive indicator lamp using voltage and current

P = 12\text{ V} \times 0.75\text{ A} = 9\text{ W}

Convert milliamperes to amperes (

750\text{ mA} = 0.75\text{ A}

) and apply the fundamental Watt's Law equation

P = V \times I

.

2

Calculate power for the heating element using current and resistance

P = (0.6\text{ A})^2 \times 50\ \Omega = 0.36 \times 50 = 18\text{ W}

Apply Watt's Law in the form

P = I^2 \times R

when current and resistance are known.

3

Calculate power for the instrumentation sensor using current and resistance

P = (0.2\text{ A})^2 \times 600\ \Omega = 0.04 \times 600 = 24\text{ W}

Convert

200\text{ mA}

to

0.2\text{ A}

and calculate power using

P = I^2 \times R

.

4

Calculate power for the communications relay coil using voltage and resistance

P = \frac{(60\text{ V})^2}{100\ \Omega} = \frac{3600}{100} = 36\text{ W}

Apply Watt's Law in the form

P = \frac{V^2}{R}

when potential difference and resistance are known.

Key Concept

Watt's Law and Electrical Power Dissipation

Estimated Time:1m 15s

Question 1051Question →

A continuous-duty DC servomotor coil is connected to a $50\text{ V}$ power line and draws a steady current of $4\text{ A}$ . During a brownout condition, the source voltage drops to $40\text{ V}$ while the internal resistance of the coil remains constant. What is the electrical power dissipated by the servomotor coil under this reduced voltage condition?

40\text{ W}

128\text{ W}

160\text{ W}

200\text{ W}

Show answer & explanation

Answer: $128\text{ W}$

Answer

128\text{ W}

The coil's resistance is determined first using Ohm's Law:

R = \frac{50\text{ V}}{4\text{ A}} = 12.5\text{ }\Omega

. Under the reduced supply voltage of

40\text{ V}

, power dissipation is calculated using

P = \frac{V^2}{R} = \frac{40^2}{12.5} = \frac{1600}{12.5} = 128\text{ W}

. Alternatively, the new current is

I = \frac{40\text{ V}}{12.5\text{ }\Omega} = 3.2\text{ A}

, giving

P = 40\text{ V} \times 3.2\text{ A} = 128\text{ W}

.

Step-by-Step Solution

1

Calculate the constant internal resistance of the coil from initial operating values

R = \frac{V_1}{I_1} = \frac{50\text{ V}}{4\text{ A}} = 12.5\text{ }\Omega

The coil's physical resistance remains constant despite changes in external supply voltage.

2

Calculate the new current drawn by the coil at the reduced voltage of

40\text{ V}

I_2 = \frac{V_2}{R} = \frac{40\text{ V}}{12.5\text{ }\Omega} = 3.2\text{ A}

By Ohm's Law, reduced voltage across a fixed resistance causes a proportional reduction in current.

3

Calculate the new electrical power dissipated under the brownout condition

P_2 = V_2 \times I_2 = 40\text{ V} \times 3.2\text{ A} = 128\text{ W}

(or

P_2 = \frac{V_2^2}{R} = \frac{40^2}{12.5} = 128\text{ W}

)

Watt's Law defines power as the product of voltage and current or the square of voltage divided by resistance.

Key Concept

Electrical power in a resistive load varies with the square of voltage (

P = \frac{V^2}{R}

) when resistance is held constant.

Alternative Method

Proportional scaling using Watt's Law: Since

P \propto V^2

for constant resistance,

P_2 = P_1 \times \left(\frac{V_2}{V_1}\right)^2 = 200\text{ W} \times \left(\frac{40}{50}\right)^2 = 200 \times 0.64 = 128\text{ W}

.

Estimated Time:1m 30s

Question 1052Question →

Match each internal engine component to its primary mechanical function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Crankshaft matches with converting reciprocating motion to rotary motion; Camshaft matches with regulating valve timing; Connecting rod matches with transmitting force between piston and crankshaft; Piston matches with receiving combustion pressure and moving linearly.

In a standard internal combustion engine, the piston captures gas expansion, the connecting rod transmits that force to the crankshaft to produce rotary torque, and the camshaft opens and closes the valves in synchronization with engine cycle rotation.

Step-by-Step Solution

1

Differentiate between the engine's two primary rotating shafts

The crankshaft produces main engine output torque from reciprocating pistons, while the camshaft actuates cylinder valves via lobes.

Distinguishing the power output shaft from the valvetrain control shaft is essential for internal combustion understanding.

2

Identify the power transfer linkage inside the cylinder

The piston captures combustion pressure directly, and the connecting rod transfers that linear force directly to the crankshaft journal.

Following the direct mechanical load path confirms the roles of the cylinder-bound components.

Key Concept

Internal engine components work in a synchronized kinematic chain to convert thermal combustion pressure into usable mechanical rotary power.

Estimated Time:45s

Question 1053Question →

An avionics technician is servicing a step-down transformer in an aircraft subsystem. The transformer receives an input voltage of $115\text{ V AC}$ across its primary coil and delivers an output voltage of $23\text{ V AC}$ across its secondary coil. If the primary winding has $750\text{ turns}$ , what is the number of turns on the secondary winding?

Show answer & explanation

Answer: 150

Answer

150 turns

The relationship between primary and secondary voltage and turns in a transformer is

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

. Given a primary voltage of

115\text{ V AC}

and a secondary voltage of

23\text{ V AC}

, the transformer has a step-down ratio of

5:1

(

115 / 23 = 5

). Dividing the primary turns (

750

) by

5

gives

150\text{ turns}

for the secondary winding.

Step-by-Step Solution

1

State the fundamental transformer turns ratio formula relating voltage and winding turns.

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

The voltage induced across transformer windings is directly proportional to the number of turns on each winding.

2

Isolate the secondary turns variable (

N_s

).

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

Algebraic rearrangement allows direct calculation of the secondary coil turns.

3

Substitute the given values (

V_p = 115\text{ V AC}

,

V_s = 23\text{ V AC}

,

N_p = 750\text{ turns}

) and simplify.

N_s = 750 \times \frac{23}{115} = 750 \times 0.2 = 150\text{ turns}

The voltage reduction factor is

5:1

, meaning the secondary winding must have one-fifth the number of turns of the primary winding.

Key Concept

Transformer turns and voltage ratio relationship

Estimated Time:1m 0s

Question 1054Question →

Electronics technicians utilize specialized accessories and maintenance compounds to perform precision rework and protect delicate hardware. Match each soldering accessory to its primary operational function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Desoldering braid pairs with drawing solder via capillary action; thermal shunt clip pairs with dissipating heat away from semiconductor junctions; tip tinner and activator pairs with deoxidizing and restoring the iron tip coating; manual desoldering pump pairs with extracting molten solder using spring-loaded vacuum suction.

Each tool addresses a distinct physical or chemical requirement in electronics soldering: desoldering braid wicks solder via capillary action, thermal shunt clips provide thermal diversion to protect semiconductors, tip tinner/activator chemically deoxidizes and re-tins degraded iron tips, and desoldering pumps provide vacuum suction for clearing through-hole connections.

Step-by-Step Solution

1

Analyze the physical mechanism of desoldering braid (copper wick).

Identified as a fluxed copper mesh utilizing surface tension and capillary action to absorb solder from flat printed circuit board pads.

Capillary action pulls molten alloy into the braid when heated against the joint.

2

Analyze the thermal protection role of a thermal shunt clip.

Identified as a temporary metallic heat sink clamped onto solid leads.

Semiconductor junctions in diodes and transistors degrade or burn out when exposed to sustained soldering iron temperatures.

3

Analyze the chemical role of tip tinner and activator.

Identified as a restorative chemical compound that cleans oxidized plating and deposits fresh solder alloy on the iron face.

Damp sponges and brass wool cannot chemically dissolve heavy oxidation once a tip becomes detinned and non-wettable.

4

Analyze the pneumatic extraction mechanism of a manual desoldering pump.

Identified as a spring-loaded mechanical vacuum tool designed to clear through-hole barrels.

The sudden negative pressure created by releasing the compressed piston pulls fluid solder through the nozzle.

Key Concept

Identification and functional distinction of electronics soldering rework tools, protective accessories, and tip maintenance compounds.

Estimated Time:1m 30s

Question 1055Question →

An electronics technician is troubleshooting an intermittent circuit fault using an analog/digital oscilloscope. Match each oscilloscope control feature on the left with its correct functional operation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Trigger Level matches setting the voltage threshold to stabilize the sweep; AC Coupling matches inserting a series capacitor to block DC bias; Timebase (SEC/DIV) matches controlling horizontal sweep speed per division; Vertical Position matches shifting the trace vertically without changing signal amplitude.

Each control corresponds directly to its front-panel function: Trigger Level establishes the voltage threshold for horizontal sweep synchronization; AC Coupling blocks DC bias using an internal capacitor to view small AC variations; Timebase dictates horizontal sweep speed per division; and Vertical Position shifts the trace vertically for convenient graticule alignment.

Step-by-Step Solution

1

Identify the synchronization function of the oscilloscope.

Trigger Level sets the precise voltage threshold required to synchronize the horizontal sweep generator with the incoming signal.

Without proper trigger settings, periodic waveforms scroll or overlap asynchronously on the screen.

2

Identify the input filtering configuration.

AC Coupling engages an internal series capacitor to block steady DC levels while passing time-varying AC signals.

This allows small AC fluctuations (such as power supply ripple) to be viewed at high vertical sensitivity without driving the trace off-screen.

3

Identify the horizontal time-domain scaling control.

Timebase (SEC/DIV) calibrates the horizontal rate of trace movement across graticule divisions.

Adjusting SEC/DIV allows the user to expand or compress the horizontal waveform display to measure period and frequency.

4

Identify the trace alignment control.

Vertical Position relocates the waveform baseline up or down across the vertical graticule.

This provides visual alignment against graticule reference lines without modifying vertical gain (Volts/Div).

Key Concept

Oscilloscope Front-Panel Controls and Signal Conditioning

Estimated Time:1m 15s

Question 1056Question →

Complete the statement explaining the operational principle of a ground-fault circuit interrupter (GFCI) protective device.

Fill in the blanks below

A ground-fault circuit interrupter (GFCI) detects hazardous leakage currents by comparing the current flowing through the hot conductor with the current returning through the conductor, tripping open if any imbalance indicates current escaping to .

Show answer & explanation

Answer

Blank 1: neutral; Blank 2: ground

A GFCI functions by using an internal differential current transformer through which both the energized (hot) and grounded return (neutral) conductors pass. When the circuit is normal, the magnetic fields generated by the equal and opposing currents cancel out. If current leaks to an external ground path (such as through damp ground, plumbing, or a human body), the magnetic fields no longer cancel, creating an induced voltage in the sensing coil that trips the breaker within a fraction of a second.

Step-by-Step Solution

1

Identify the paired current-carrying conductors monitored by a GFCI in a single-phase AC circuit.

Under standard closed-circuit operation, normal electrical current supplied by the hot conductor must return completely through the neutral conductor.

Kirchhoff's Current Law dictates that the outgoing current in the energized wire must equal the returning current in the neutral line in an intact circuit.

2

Determine where leaked current flows during a ground-fault condition to cause a trip.

Any difference between hot and neutral currents signifies that current has diverted through an unintended path (such as a person or equipment chassis) into the ground.

The differential sensing coil in the GFCI detects the magnetic flux resulting from the current mismatch and trips the internal switch contacts to prevent electrical shock.

Key Concept

Differential Current Sensing in Ground-Fault Protective Devices

Estimated Time:1m 0s

Question 1057Question →

A cylindrical metal conductor with length $L$ and diameter $d$ has an electrical resistance of $16\ \Omega$ . If a replacement conductor made of the identical material has its length doubled ( $2L$ ) and its diameter doubled ( $2d$ ), what is the electrical resistance of the replacement conductor?

32\ \Omega

8\ \Omega

16\ \Omega

4\ \Omega

Show answer & explanation

Answer: $8\ \Omega$

Answer

The resistance of the replacement conductor is

8\ \Omega

.

The electrical resistance of a conductor is given by the formula

R = \rho \frac{L}{A}

, where

A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}

. When the length is doubled (

2L

), the resistance tends to double. However, when the diameter is doubled (

2d

), the cross-sectional area quadruples (

4A

), which divides the resistance by

4

. Combining both changes yields a net factor of

\frac{2}{4} = \frac{1}{2}

. Thus, the new resistance is

\frac{1}{2} \times 16\ \Omega = 8\ \Omega

.

Step-by-Step Solution

1

Express the relationship between resistance, length, and diameter for a cylindrical conductor.

R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4\rho L}{\pi d^2} = 16\ \Omega

Electrical resistance is directly proportional to conductor length and inversely proportional to its cross-sectional area.

2

Determine the scaling factor for the new conductor with length

2L

and diameter

2d

.

R_{new} = \frac{4\rho (2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{1}{2} \left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R

Doubling the diameter increases the cross-sectional area by a factor of

2^2 = 4

, while doubling length increases resistance by a factor of

2

.

3

Calculate the numerical resistance value of the replacement conductor.

R_{new} = \frac{1}{2} \times 16\ \Omega = 8\ \Omega

Multiplying the original resistance value by the net proportionality factor of

0.5

yields the final resistance.

Key Concept

The resistance of a uniform conductor is directly proportional to its length and inversely proportional to the square of its diameter (cross-sectional area).

Estimated Time:1m 0s

Question 1058Question →

An electronics technician is troubleshooting an inoperative single-phase ventilation blower that fails to operate when energized. In what sequence should the technician perform the following diagnostic and repair actions to follow standard, safe troubleshooting procedures from initial assessment to final verification?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct logical sequence is: verify supply voltage at the input terminals, measure the voltage drop across the closed control switch, de-energize and lock out the power source, isolate and measure motor winding resistance with an ohmmeter, and finally replace the defective motor and test operation.

Standard electrical troubleshooting begins with verifying the source power, followed by live circuit voltage checks across active control elements. Once the faulty branch or load is localized, the circuit must be safely de-energized before isolating components for resistance testing. The final step is component replacement followed by an operational verification test.

Step-by-Step Solution

1

Check the input power supply with a voltmeter.

Confirms whether electrical power is reaching the system.

Eliminates external power failure or tripped upstream breakers before diagnosing internal faults.

2

Check the closed control switch for voltage drop under live conditions.

Determines if the switch contacts are conducting correctly (

0\text{ V}

drop across good closed contacts; full source voltage across open contacts).

Allows quick live-circuit localization of series circuit faults.

3

Disconnect and lock out the circuit power.

De-energizes the system to establish an electrically safe work condition.

Essential for personnel safety and required before using an ohmmeter, which can be damaged by live voltage.

4

Isolate one winding terminal and measure resistance using an ohmmeter.

Provides an accurate internal resistance reading (infinite ohms indicating an open winding).

Isolating one lead prevents parallel circuit paths from yielding false resistance readings.

5

Replace the faulty motor and restore power to test the unit.

Restores system operation and validates that the root fault is fully resolved.

Every repair procedure concludes with an operational check under normal load conditions.

Key Concept

Logical Troubleshooting and Safety Sequence

Estimated Time:1m 30s

Question 1059Question →

A field communication station is powered by a $72\text{ V}$ direct-current supply connected across three parallel load branches. The resistance of the first branch is $18\ \Omega$ , the resistance of the second branch is $36\ \Omega$ , and the resistance of the third branch is $72\ \Omega$ . What is the total current, in amperes ( $\text{A}$ ), supplied to the circuit?

Show answer & explanation

Answer: 7

Answer

7 A

Each of the three branches is connected directly across the

72\text{ V}

source. Using Ohm's Law (

I = V/R

), the branch currents are

4\text{ A}

,

2\text{ A}

, and

1\text{ A}

. Adding these currents together yields the total current of

7\text{ A}

.

Step-by-Step Solution

1

Calculate the current in the first parallel branch using

I_1 = \frac{V}{R_1}

.

I_1 = \frac{72\text{ V}}{18\ \Omega} = 4\text{ A}

In a parallel DC circuit, the full supply voltage of

72\text{ V}

appears across every parallel branch.

2

Calculate the current in the second parallel branch using

I_2 = \frac{V}{R_2}

.

I_2 = \frac{72\text{ V}}{36\ \Omega} = 2\text{ A}

Applying Ohm's Law to the second branch with its specific resistance.

3

Calculate the current in the third parallel branch using

I_3 = \frac{V}{R_3}

.

I_3 = \frac{72\text{ V}}{72\ \Omega} = 1\text{ A}

Applying Ohm's Law to the third branch with its specific resistance.

4

Add all three individual branch currents together to determine the total current.

I_{\text{total}} = 4\text{ A} + 2\text{ A} + 1\text{ A} = 7\text{ A}

By Kirchhoff's Current Law, the total current leaving the power supply equals the sum of the currents through all parallel branches.

Key Concept

In a parallel circuit, the voltage across each parallel branch is identical to the source voltage, and the total circuit current is the sum of the individual branch currents (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

).

Alternative Method

Find the total equivalent resistance (

R_{\text{eq}}

) first:

\frac{1}{R_{\text{eq}}} = \frac{1}{18} + \frac{1}{36} + \frac{1}{72} = \frac{4}{72} + \frac{2}{72} + \frac{1}{72} = \frac{7}{72}\ \Omega^{-1}

, so

R_{\text{eq}} = \frac{72}{7}\ \Omega

. Then find total current using

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{72}{72/7} = 7\text{ A}

.

Estimated Time:1m 0s

Question 1060Question →

An electronics technician is bench-testing a de-energized two-position rotary selector switch using a digital multimeter set to the low-resistance continuity function. With the switch set to Position 1, the meter connected across the common terminal and output terminal 1 displays $0.2\ \Omega$ with an audible tone. When the switch is turned to Position 2, the meter across the common terminal and output terminal 1 continues to display $0.2\ \Omega$ , while testing across the common terminal and output terminal 2 displays $\text{O.L.}$ (over limit). What specific switch fault do these multimeter readings indicate?

The common terminal insulator has broken down into a conductor, creating an open circuit on all switch positions.

The internal contacts for Position 1 are welded or shorted closed, while the contacts for Position 2 have failed open.

The meter is reading residual circuit voltage, which forces the resistance display to read zero ohms regardless of contact position.

The switch is functioning normally because a reading of

\text{O.L.}

indicates zero resistance and continuous current flow.

Show answer & explanation

Answer: The internal contacts for Position 1 are welded or shorted closed, while the contacts for Position 2 have failed open.

Answer

The internal contacts for Position 1 are welded or shorted closed, while the contacts for Position 2 have failed open.

A reading of

0.2\ \Omega

with an audible tone confirms a closed, continuous electrical path. When the rotary switch is moved to Position 2, output terminal 1 should disconnect and display

\text{O.L.}

; because it remains at

0.2\ \Omega

, the Position 1 contacts are welded or shorted closed. Simultaneously, output terminal 2 should establish continuity in Position 2; showing

\text{O.L.}

indicates an open circuit where contacts failed to close.

Step-by-Step Solution

1

Analyze the baseline continuity measurement in Position 1.

A reading of

0.2\ \Omega

with an audible tone indicates a normal, closed, continuous connection between the common terminal and output terminal 1.

Digital multimeters indicate continuity with near-zero resistance and an audible beeper.

2

Evaluate the resistance between common and terminal 1 after switching to Position 2.

The meter continues to read

0.2\ \Omega

, meaning the contacts remained closed instead of opening.

When rotated away from Position 1, the contacts should open and show infinite resistance (

\text{O.L.}

); failing to do so indicates contacts welded or mechanically shorted closed.

3

Evaluate the resistance between common and terminal 2 in Position 2.

The meter displays

\text{O.L.}

, indicating an open circuit.

In Position 2, terminal 2 should connect to common with near-zero ohms; an over-limit reading proves the contacts failed to make contact (open circuit defect).

Key Concept

Interpreting multimeter resistance and continuity values (

0\ \Omega

vs.

\text{O.L.}

) to diagnose internal short-circuit and open-circuit defects in multi-position switches.

Estimated Time:1m 15s

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