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5215 questions

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Question 1301Question →

In a MacPherson strut front suspension assembly, which component is eliminated from the traditional double-wishbone design because the strut itself serves as the upper locating link?

Lower control arm

Upper control arm

Shock absorber

Steering knuckle

Show answer & explanation

Answer: Upper control arm

Answer

Upper control arm

A MacPherson strut incorporates the damper, spring, and upper pivot into one structural telescopic assembly attached directly to the body strut tower. Because this assembly handles the lateral and longitudinal forces at the top of the wheel assembly, the upper control arm and upper ball joint are not needed.

Step-by-Step Solution

Analyze the architecture of a traditional double-wishbone (short-and-long arm) suspension.

Identified that a double-wishbone system uses both an upper control arm and a lower control arm connected by ball joints to locate the steering knuckle.

Understanding the baseline multi-link architecture establishes which structural members guide the top and bottom of the wheel assembly.

Compare the double-wishbone layout to the MacPherson strut design.

The MacPherson strut combines the spring and damper into a rigid structural unit that bolts directly to the inner fender strut tower at the top.

This structural integration allows the top of the strut to serve as the upper locating pivot and support member.

Determine which component is made redundant by the structural strut.

The upper control arm and its associated upper ball joint are completely eliminated, simplifying the suspension and reducing packaging space.

The rigid strut replaces the mechanical locating role of the upper control arm.

Key Concept

MacPherson strut suspension architecture and component integration

Estimated Time:1m 0s

Question 1302Question →

A vehicle's brake pedal slowly sinks all the way to the floorboard while held at a stoplight, but an inspection reveals no external fluid leaks anywhere in the hydraulic system. What component is most likely failing internally?

Caliper slide pins

Master cylinder piston seals

Wheel cylinder piston cups

Drum brake retracting springs

Show answer & explanation

Answer: Master cylinder piston seals

Answer

Master cylinder piston seals

When a vehicle's brake pedal gradually sinks under constant foot pressure without any external loss of brake fluid, the master cylinder's internal rubber piston cups have degraded. This allows pressurized brake fluid to bleed backwards past the piston head into the unpressurized reservoir, collapsing line pressure without leaking outside the unit.

Step-by-Step Solution

Analyze the observed symptom

The brake pedal gradually drops to the floorboard under continuous holding force.

A sinking pedal indicates a progressive loss of hydraulic pressure within the closed braking circuit.

Evaluate the condition of no external fluid leakage

The fluid volume is retained inside the system rather than escaping via lines, calipers, or wheel cylinders.

External leaks produce visible drops, wet backplates, or a dropping fluid reservoir level; absence of these confirms the failure is entirely internal.

Identify the internal component capable of bypassing hydraulic pressure

The master cylinder piston rubber cup seals are allowing fluid to bleed past the piston head back into the unpressurized fluid reservoir chamber.

When internal piston seals deteriorate, fluid slips backwards under steady pedal load, dissipating line pressure while losing no total fluid volume.

Key Concept

Internal Hydraulic Pressure Bypass in Master Cylinders

Estimated Time:1m 0s

Question 1303Question →

A field technician is testing a direct-current (DC) sensor module. What is the current flowing through the circuit when a potential difference of $90\text{ V}$ is applied across a total circuit resistance of $15\text{ k}\Omega$ ?

Fill in the blanks below

Under an applied potential difference of

90\text{ V}

, a circuit having a resistance of

15\text{ k}\Omega

conducts a current of milliamperes.

Show answer & explanation

Answer

6 milliamperes (6 mA)

According to Ohm's law (

I = \frac{V}{R}

), dividing the

90\text{ V}

source by the

15,000\ \Omega

(

15\text{ k}\Omega

) total resistance yields

0.006\text{ A}

, which converts directly to

6\text{ mA}

Step-by-Step Solution

Identify known circuit quantities and convert resistance from kilohms to ohms.

V = 90\text{ V}

and

R = 15\text{ k}\Omega = 15 \times 1,000\ \Omega = 15,000\ \Omega

Standard electrical formulas require resistance in base units of ohms (

\Omega

) to calculate current in amperes (

\text{A}

Apply Ohm's law (

I = \frac{V}{R}

) to solve for the current in amperes.

I = \frac{90\text{ V}}{15,000\ \Omega} = 0.006\text{ A}

Current is directly proportional to voltage and inversely proportional to resistance.

Convert current from amperes to milliamperes as requested by the statement.

0.006\text{ A} \times 1,000\text{ mA/A} = 6\text{ mA}

One milliampere equals one thousandth of an ampere (

1\text{ mA} = 10^{-3}\text{ A}

Key Concept

Ohm's Law (

I = \frac{V}{R}

) defines the proportional relationship between potential difference, current, and resistance in a direct-current circuit.

Estimated Time:45s

Question 1304Question →

When a driver turns the steering wheel in a vehicle equipped with hydraulic power steering, fluid circulates through the system to provide assist. Arrange the steps of this hydraulic fluid flow path in the correct order from fluid intake to reservoir return.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence begins with the pump drawing fluid from the reservoir, followed by high-pressure delivery to the control valve, hydraulic assist acting inside the power cylinder, and low-pressure return of fluid back to the reservoir.

In a hydraulic power steering system, fluid circulation follows a continuous closed loop: the pump draws fluid from the reservoir, pressurizes it, and sends it through the high-pressure hose to the rotary control valve. The valve routes the fluid into the appropriate side of the power cylinder to provide turning assist, and displaced low-pressure fluid is routed through the return line back into the reservoir.

Step-by-Step Solution

Identify the fluid intake stage

The pump draws low-pressure fluid from the reservoir

The hydraulic circuit starts at the reservoir, which stores the fluid supply for the pump.

Trace the high-pressure delivery path

Pressurized fluid reaches the steering gear control valve

The mechanical pump pressurizes the fluid and delivers it to the rotary control valve via the high-pressure line.

Determine where hydraulic assist work occurs

Fluid is directed into the power cylinder chamber

The control valve channels high-pressure fluid to one side of the assist piston to push the steering mechanism.

Identify the return path to complete the loop

Low-pressure fluid returns through the return hose to the reservoir

Fluid displaced from the opposite chamber flows back through the low-pressure hose into the reservoir.

Key Concept

Hydraulic Power Steering Fluid Circuit and Flow Path

Estimated Time:45s

Question 1305Question →

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

Show answer & explanation

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 1306Question →

In a rear-wheel-drive vehicle, the driveshaft transmits engine torque lengthwise along the vehicle chassis. What is the primary purpose of the ring and pinion gear set located inside the rear axle assembly?

To allow the driveshaft to expand and contract in length during suspension travel

To change the rotational power direction by 90 degrees and provide gear reduction

To allow the left and right drive wheels to rotate at different speeds while turning

To absorb variations in driveshaft operating angles caused by rear axle movement

Show answer & explanation

Answer: To change the rotational power direction by 90 degrees and provide gear reduction

Answer

To change the rotational power direction by 90 degrees and provide gear reduction

In a longitudinal rear-wheel-drive powertrain, engine torque enters the rear axle parallel to the vehicle length. The drive pinion gear meshes with the ring gear at a right angle, converting this longitudinal rotation into transverse rotation to turn the left and right drive axles, while the gear ratio between them provides final torque multiplication.

Step-by-Step Solution

Identify the orientation of driveline torque entering the axle assembly.

Rotational torque travels longitudinally (front-to-back) along the driveshaft to the drive pinion gear.

The engine and transmission output shaft rotate along the vehicle's centerline.

Determine the mechanical interaction between the drive pinion and the ring gear.

The small drive pinion turns the larger ring gear, transferring rotational motion 90 degrees toward the transverse axle shafts while multiplying torque through gear reduction.

The drive wheels require transverse rotation and mechanical advantage to propel the vehicle.

Key Concept

Ring and Pinion Gear Function

Estimated Time:45s

Question 1307Question →

In electrical circuits, substances are categorized according to how their atomic structures facilitate, restrict, or modify the transfer of electric charge. Match each electrical material to its correct conduction property and circuit behavior.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Copper busbar pairs with the good conductor having abundant free electrons; Intrinsic silicon crystal pairs with the semiconductor having four valence electrons; Porcelain standoff pairs with the high-grade insulator having tightly bound electrons; Nichrome heating element pairs with the conductive alloy with relatively high resistance used for heat generation.

Each material corresponds to its specific atomic structure and functional role: Copper provides low-resistance conduction via free electrons; Silicon exhibits intermediate semiconductor behavior with four valence electrons; Porcelain serves as an insulator by immobilizing electrons; Nichrome operates as a specialized resistive conductor to produce heat.

Step-by-Step Solution

Classify the atomic and conduction properties of standard metals.

Copper is identified as a primary electrical conductor with low resistance and loosely held valence electrons.

Conductors must allow free electron mobility under an applied electromotive force.

Identify group IV elements and their electrical classification.

Silicon is identified as a semiconductor with four valence electrons and intermediate conductivity.

Semiconductors have conduction characteristics that sit between conductors and insulators.

Distinguish between non-conductive dielectric materials and resistive heating conductors.

Porcelain is classified as an insulator that blocks current, whereas Nichrome is a resistive conductor that generates heat.

Insulators prevent current flow entirely, while high-resistance alloys deliberately oppose current flow to produce thermal energy.

Key Concept

Classification of electrical materials into conductors, semiconductors, insulators, and resistive alloys based on valence electron structure and circuit function.

Estimated Time:1m 0s

Question 1308Question →

In a turbocharged gasoline engine, intake air travels through several induction components before entering the combustion chamber. What is the correct order of components that the air passes through from the outside atmosphere to the engine cylinder?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Air cleaner assembly → Turbocharger compressor → Charge air cooler (intercooler) → Throttle body assembly → Cylinder intake valve

In a turbocharged gasoline engine, ambient air is first drawn through the air cleaner assembly to filter out airborne contaminants. It then enters the turbocharger compressor housing where the impeller wheel pressurizes it. Because compression generates substantial heat, the hot air is directed through a charge air cooler (intercooler) to reduce temperature and increase charge density. The cooled air then passes through the throttle body assembly, which regulates intake air volume, before traveling through the intake manifold runners and past the cylinder intake valve into the combustion chamber.

Step-by-Step Solution

Identify the initial atmospheric entry point for induction air.

Air enters the air cleaner assembly.

Air must be filtered immediately upon entering the vehicle to prevent particulate damage to sensitive downstream induction components.

Trace the forced induction compression stage.

Filtered air enters the compressor housing of the turbocharger.

The exhaust-driven turbine spins the compressor wheel, drawing in clean air and compressing it above atmospheric pressure.

Account for charge air cooling.

Pressurized air routes through the charge air cooler (intercooler).

Air heats up during compression; passing through an intercooler removes heat, increasing air density and preventing engine knock.

Determine airflow regulation into the manifold.

Cooled charge air moves through the throttle body assembly.

The throttle valve controls engine speed and load by regulating the volume of boost air entering the intake manifold plenum.

Identify the final entry point into the combustion chamber.

Air passes the cylinder intake valve.

The mechanical intake valve opens during the piston's intake stroke to admit the fuel-air charge into the combustion chamber.

Key Concept

Turbocharged induction air path and charge cooling sequence

Estimated Time:1m 0s

Question 1309Question →

A motor pool inventory shows that $\frac{1}{4}$ of the vehicles are heavy trucks, $\frac{1}{3}$ are light tactical vehicles, and the remaining vehicles are armored personnel carriers. If a vehicle is randomly selected for an inspection, what is the probability that it is an armored personnel carrier?

\frac{7}{12}

\frac{2}{7}

\frac{5}{12}

\frac{5}{7}

Show answer & explanation

Answer: $\frac{5}{12}$

Answer

\frac{5}{12}

The total probability of all mutually exclusive events must sum to 1. First, find the combined probability of heavy trucks and light tactical vehicles by using a common denominator of 12:

\frac{1}{4}

becomes

\frac{3}{12}

and

\frac{1}{3}

becomes

\frac{4}{12}

. Their sum is

\frac{7}{12}

. Since the remaining vehicles are armored personnel carriers, subtract this sum from the total probability of 1 (or

\frac{12}{12}

). The result is

\frac{5}{12}

Step-by-Step Solution

Find the combined probability of selecting a heavy truck or a light tactical vehicle.

\frac{1}{4} + \frac{1}{3}

To determine the portion of the fleet that is not armored personnel carriers, the given fractions must be added.

Add the fractions by finding a common denominator.

\frac{3}{12} + \frac{4}{12} = \frac{7}{12}

Fractions must have the same denominator (12) before their numerators can be added.

Subtract the combined fraction from 1 (the total probability).

1 - \frac{7}{12} = \frac{12}{12} - \frac{7}{12} = \frac{5}{12}

The sum of probabilities for all possible events must equal exactly 1. The remainder represents the armored personnel carriers.

Key Concept

Single Event Probability from Complementary Events

Estimated Time:1m 0s

Question 1310Question →

During automotive brake inspections and road testing, technicians evaluate specific drivability symptoms to determine underlying mechanical or hydraulic faults. Match each brake system symptom with its most probable root cause.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Rhythmic brake pedal pulsation matches excessive rotor lateral runout or DTV; excessively hard pedal matches defective vacuum booster check valve or ruptured diaphragm; spongy pedal that firms up on pumping matches air entrained in hydraulic lines; vehicle pulling to the left matches seized right front caliper slide pin or grease-contaminated right pads.

Each symptom is paired directly with its underlying physical mechanism: rotor thickness variation creates cyclic feedback (pulsation); loss of vacuum multiplication increases pedal resistance (hard pedal); compressed gas bubbles create hydraulic softness (spongy pedal); and asymmetric friction coefficient or mechanical binding at one wheel assembly produces uneven directional torque (pulling).

Step-by-Step Solution

Analyze the pulsation symptom under dynamic braking

Pedal pulsation occurs when the disc rotor has uneven thickness or wobble, mechanically kicking back fluid pressure to the master cylinder and pedal.

Disc thickness variation (DTV) directly correlates with cyclic pedal pushback.

Analyze high pedal effort with reduced stopping assist

A hard pedal indicates a lack of power assist rather than hydraulic leakage or air.

The vacuum power booster multiplies driver pedal effort; when its vacuum supply fails or diaphragm tears, full manual force is required.

Analyze the spongy pedal symptom that recovers upon pumping

Unlike hydraulic fluid, air is compressible. Pumping the pedal compresses the pocket of air temporarily.

Entrained air in hydraulic circuits manifests uniquely as a spongy pedal that gains temporary firmness with rapid pumping.

Analyze directional brake pull during braking

A pull to the left indicates unequal braking torque across the front axle where the left brake functions normally but the right brake is restricted or inefficient.

A seized right caliper slide pin or oil/grease contamination on right friction pads reduces friction on the right side, causing an asymmetrical leftward pull.

Key Concept

Brake System Fault Isolation and Symptom Diagnosis

Estimated Time:1m 30s

Question 1311Question →

An electronics technician is preparing a piece of stranded 18 AWG copper hookup wire to attach an insulated crimp terminal. What is the correct chronological sequence of steps the technician should follow using standard wire preparation tools?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Cut wire with diagonal cutters → Strip insulation with calibrated wire stripper → Inspect and twist bare strands → Crimp terminal barrel with crimper → Perform mechanical pull test

The proper sequence follows standard electrical assembly protocol: cutting the conductor to length with diagonal cutters, removing insulation using a properly matched wire stripper notch, inspecting and twisting exposed stranded conductors, seating and compressing the terminal barrel with a crimper, and concluding with a physical pull test to verify mechanical integrity.

Step-by-Step Solution

Cut the conductor to length using diagonal cutters (dikes).

Clean cut at the desired total wire length.

Establishing the proper harness length must precede all end-preparation tasks.

Select the correct AWG stranded gauge notch on a wire stripper and remove the outer insulation jacket to the manufacturer's recommended strip length.

Cleanly exposed bare conductor strands without severed or nicked wire strands.

Using the matching gauge notch cuts through the insulation without cutting into the metal conductor.

Inspect the bare wire strands for any damage and twist the strands clockwise.

A compact, cohesive stranded bundle ready for insertion.

Twisting prevents stray strands from catching outside the barrel and creating a short circuit or high-resistance connection.

Seat the stripped conductor into the barrel of the crimp terminal and compress the joint using the appropriate color-coded or sized crimper jaw.

Cold-welded, mechanically deformed gas-tight crimp connection.

The crimping tool applies precise pressure to bond the terminal barrel to the conductor.

Gently pull the wire and terminal in opposing directions.

Verification that the wire does not slip out under tension.

A pull test confirms that the crimp has sufficient mechanical retention before placing the wire into service.

Key Concept

Standard wire preparation, stripping, and crimp termination procedure

Estimated Time:1m 0s

Question 1312Question →

Match each automotive cooling system component with its primary mechanical function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Radiator pressure cap matches maintaining system pressure; Heater core matches heating the passenger cabin; Cooling fan matches drawing airflow across the radiator at low speeds; Overflow reservoir matches collecting and returning expanding coolant.

Each component in a liquid cooling system performs a dedicated function: the radiator pressure cap increases system pressure to raise the coolant's boiling point; the heater core routes warm coolant under the dashboard to heat the cabin; the cooling fan ensures constant airflow across radiator tubes when vehicle motion is minimal; and the overflow reservoir accommodates the natural expansion and contraction of coolant across operating cycles.

Step-by-Step Solution

Identify the distinct function of each cooling system component

The pressure cap regulates pressure to increase boiling temperature, the heater core warms cabin air, the cooling fan supplies supplemental airflow, and the overflow tank manages fluid volume changes.

Understanding the individual purpose of each component enables accurate diagnosis and maintenance of the engine cooling system.

Pair each listed component to its unique operational description

Each component is correctly aligned with its corresponding mechanical role.

Differentiating between heat dissipation, passenger climate control, pressure management, and fluid recovery ensures correct component matching.

Key Concept

Automotive cooling system component roles and thermal regulation

Estimated Time:45s

Question 1313Question →

During the engine starting cycle in an automobile, several distinct electrical and mechanical events occur in a precise sequence when the ignition switch is turned to the START position. Arrange the following operational steps of the starting system in the correct chronological order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence begins with control current energizing the starter solenoid windings, followed by the plunger shifting the pinion gear into mesh with the flywheel ring gear, then the solenoid contact disc closing the high-current battery circuit, followed by the starter motor spinning to crank the engine, and finally the overrunning clutch freewheeling once the engine fires.

The starting system relies on a two-stage relay/solenoid operation. Turning the ignition key first energizes the low-current solenoid coils. The resulting electromagnetic pull shifts the pinion gear into the stationary flywheel ring gear. Only at the end of the plunger's travel does the contact disc bridge the large battery terminals, energizing the starter motor armature to spin the engine. Once the engine starts and rotates faster than the starter, the one-way overrunning clutch freewheels to protect the motor until the key is released.

Step-by-Step Solution

Identify the initial control circuit actuation.

Current flows from the battery through the ignition switch in START, passes through the neutral safety switch, and energizes the pull-in and hold-in coils of the starter solenoid.

The low-current control circuit must energize first to actuate the heavy-duty solenoid switch and mechanical shift lever.

Trace the mechanical motion of the solenoid plunger.

The solenoid magnetic field draws the plunger inward, pivoting the shift lever to advance the drive pinion along the armature shaft into engagement with the flywheel teeth.

The drive gear must mesh with the stationary ring gear before the motor begins high-speed rotation to avoid damaging gear teeth.

Identify the high-current circuit closure point.

At the end of the plunger travel, the internal copper contact plate bridges the main battery cable terminal and the starter motor feed terminal.

A heavy internal copper contact is required to handle the 150 to 300+ amperes needed by the starter motor.

Trace the electrical and mechanical power delivery in the starter motor.

Current energizes the armature and field windings, generating opposing magnetic fields that spin the armature shaft and deliver cranking torque through the flywheel.

High current through the starter motor windings creates the torque necessary to overcome engine compression and mechanical friction.

Determine the protective mechanism upon engine start.

When the engine starts running under its own power, the flywheel rotates faster than the starter pinion, causing the roller-type overrunning clutch to slip and freewheel.

Because of the gear reduction ratio (typically 15:1 to 20:1), an engine idling at 1,000 RPM would spin an uncoupled starter armature at 15,000 to 20,000 RPM, destroying it.

Key Concept

Starter solenoid dual electro-mechanical function and sequential cranking circuit operation

Estimated Time:1m 30s

Question 1314Question →

In a rear-wheel-drive vehicle, mechanical torque travels along a specific path from the transmission to the road. Arrange the following drivetrain components in the correct sequence of rotational power flow, starting from the transmission output and ending at the driving tires.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Driveshaft (propeller shaft) → Drive pinion gear → Ring gear and differential assembly → Axle shafts and drive wheels

Power flows sequentially down the chassis: torque exits the transmission into the driveshaft, turns the drive pinion gear at the rear axle housing, transfers into the ring gear and differential carrier assembly, and is finally distributed through the left and right axle shafts to the drive wheels.

Step-by-Step Solution

Identify the component directly connected to the transmission output.

The driveshaft (propeller shaft) is the primary link transmitting torque from the transmission along the length of the chassis.

Engine torque exits the transmission output shaft through the slip yoke and front universal joint into the driveshaft.

Determine the component at the differential input that is turned by the driveshaft.

The drive pinion gear is turned directly by the rear of the driveshaft.

The rear universal joint connects the driveshaft to the pinion shaft flange.

Identify the central gearing unit driven by the pinion gear.

The ring gear and differential assembly receive rotation from the pinion gear.

The drive pinion meshes with the larger ring gear to reduce gear ratio and redirect torque 90 degrees.

Identify the final components delivering torque to the ground.

The axle shafts and drive wheels receive torque from the differential side gears.

Axle shafts extend outward from the differential carrier to spin the wheels and tires.

Key Concept

Power flow sequence through a rear-wheel-drive driveline from the transmission output to the drive wheels.

Estimated Time:45s

Question 1315Question →

A $72\text{ V}$ direct-current (DC) power supply energizes an industrial sensor interface consisting of two resistors connected in parallel. The first resistor ( $R_1$ ) has a resistance of $24\ \Omega$ . If the power supply delivers a total current of $5\text{ A}$ to the circuit, what is the resistance of the second resistor ( $R_2$ )?

9.6\ \Omega

36\ \Omega

14.4\ \Omega

28.8\ \Omega

Show answer & explanation

Answer: $36\ \Omega$

Answer

36\ \Omega

In a parallel circuit, the voltage across every parallel branch is identical to the source voltage (

72\text{ V}

). The current through the known resistor is

I_1 = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

. Since the total current entering the parallel junction is

5\text{ A}

, the remaining current flowing through the second branch must be

I_2 = 5\text{ A} - 3\text{ A} = 2\text{ A}

. Using Ohm's law for the second branch yields

R_2 = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega

Step-by-Step Solution

Calculate the current flowing through the first branch resistor (

R_1

)

I_1 = \frac{V}{R_1} = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

In a parallel circuit, each branch experiences the full supply voltage of

72\text{ V}

Determine the current flowing through the second branch (

I_2

)

I_2 = I_{\text{total}} - I_1 = 5\text{ A} - 3\text{ A} = 2\text{ A}

By Kirchhoff's Current Law, total current in a parallel circuit equals the sum of the individual branch currents.

Calculate the resistance of the second resistor (

R_2

) using Ohm's law

R_2 = \frac{V}{I_2} = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega

Applying Ohm's law (

R = \frac{V}{I}

) with the branch voltage and branch current gives the branch resistance.

Key Concept

Parallel branch voltage equality, Kirchhoff's Current Law, and Ohm's law branch calculations

Estimated Time:1m 0s

Question 1316Question →

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 1317Question →

A technician is diagnosing a vehicle that will not crank and produces no clicking sound when the ignition is turned to the START position. Multimeter testing reveals a resting battery voltage of $12.6$ volts. During a cranking attempt, the voltage at the starter solenoid 'S' (ignition) terminal measures $12.4$ volts, the battery voltage remains steady at $12.6$ volts, and the voltage drop between the negative battery post and the starter casing is $0.1$ volts. Based on these test results, which of the following is the most likely cause of the failure?

A compromised ground connection between the engine and the chassis

Unacceptable electrical resistance in the ignition switch circuit

An open circuit within the starter solenoid windings

A severely discharged or internally sulfated battery

Show answer & explanation

Answer: An open circuit within the starter solenoid windings

Answer

An open circuit within the starter solenoid windings

The test results indicate that the battery is fully charged (12.6V) and the ignition switch is successfully delivering trigger voltage (12.4V) to the solenoid's 'S' terminal. The casing ground is also excellent (0.1V drop). Because the battery voltage does not dip when the key is turned, no electrical current is flowing. A solenoid that receives proper trigger voltage and ground but draws zero current and fails to click has an internal open circuit in its electromagnetic windings.

Step-by-Step Solution

Analyze the battery's state of charge using the resting voltage.

The resting voltage of 12.6V indicates a fully charged and healthy battery.

This establishes a baseline, ruling out a dead battery as the initial power source failure.

Analyze the circuit current flow during the starting attempt.

The battery voltage remains steady at 12.6V when the key is turned to START.

Because the voltage does not drop, it proves that 0 amps are flowing out of the battery. The starting circuit is completely open and not drawing a load.

Evaluate the ignition control circuit (the trigger wire).

The solenoid's 'S' terminal receives 12.4V.

This confirms the ignition switch, relay, and associated wiring are correctly delivering the trigger signal to the solenoid.

Evaluate the starter motor's ground path.

The casing shows a 0.1V drop during the start attempt.

A reading of 0.1V is well within specifications (typically under 0.2V for grounds), confirming the physical ground connection to the engine block is excellent.

Synthesize the findings to pinpoint the exact component failure.

The starter solenoid has an internal open circuit.

The solenoid receives the required voltage and has a proper ground, but does not click or draw any current, proving its internal electromagnetic windings are broken.

Key Concept

Voltage Drop Diagnostic Testing and Starting Circuit Logic

Question 1318Question →

In an automotive drum brake assembly, which component is forced outward against the rotating drum to generate friction and slow the vehicle?

Brake pads

Brake shoes

Brake rotors

Caliper pistons

Show answer & explanation

Answer: Brake shoes

Answer

Brake shoes are forced outward against the rotating drum to generate braking friction.

In a drum brake system, hydraulic pressure delivered to the wheel cylinder forces the curved brake shoes outward into direct contact with the inner friction surface of the rotating drum.

Step-by-Step Solution

Identify the specific brake configuration described in the stem.

The question specifies an automotive drum brake assembly.

Drum brake and disc brake configurations utilize distinctly shaped friction components.

Select the friction component designed to expand inside a brake drum.

Brake shoes are curved metal structures lined with friction material positioned inside the drum.

When hydraulic pressure enters the wheel cylinder, its pistons push the brake shoes outward against the drum.

Key Concept

Drum Brake Component Identification

Estimated Time:30s

Question 1319Question →

An automotive diagnostic technician connects an auxiliary relay coil to a $12\text{ V}$ direct-current (DC) power supply. An ammeter placed in series measures a current of $150\text{ mA}$ . What is the electrical resistance of the relay coil?

1.8\ \Omega

12.5\ \Omega

80\ \Omega

800\ \Omega

Show answer & explanation

Answer: $80\ \Omega$

Answer

The electrical resistance of the relay coil is

80\ \Omega

According to Ohm's law, electrical resistance is calculated by dividing voltage by current (

R = \frac{V}{I}

). Converting

150\text{ mA}

0.15\text{ A}

and dividing

12\text{ V}

0.15\text{ A}

gives

80\ \Omega

Step-by-Step Solution

Convert the current from milliamperes (

\text{mA}

) to base units of amperes (

\text{A}

I = 150\text{ mA} = 150 \times 10^{-3}\text{ A} = 0.15\text{ A}

Ohm's law requires values in standard SI base units (volts, amperes, and ohms) before performing calculations.

Rearrange Ohm's law (

V = I \times R

) to solve for resistance (

R

) and substitute the given values.

R = \frac{V}{I} = \frac{12\text{ V}}{0.15\text{ A}} = 80\ \Omega

Resistance is equal to the potential difference divided by the current flowing through the component.

Key Concept

Ohm's Law states that resistance equals voltage divided by current (

R = \frac{V}{I}

), requiring current to be expressed in amperes.

Estimated Time:45s

Question 1320Question →

During driveline inspection and diagnosis, a mechanic evaluates several critical powertrain components. Match each drivetrain component to its primary mechanical function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Slip yoke matches with sliding axially along splines to absorb driveshaft length changes; Differential spider gear matches with rotating on its shaft to permit drive wheels to turn at different speeds while cornering; Center support bearing matches with maintaining alignment and dampening vibrations at the midpoint of a two-piece driveshaft; Ring gear matches with transferring torque from the drive pinion to change the plane of rotation 90 degrees and reduce gear speed.

Each component serves a dedicated mechanical purpose in the vehicle drivetrain: the slip yoke provides linear slip to handle distance changes between transmission and rear axle; differential spider gears allow cornering speed differences between wheels; the center support bearing stabilizes multi-piece driveshafts against vibration; and the ring gear converts longitudinal driveshaft power into lateral axle shaft torque at a 90-degree angle with gear reduction.

Step-by-Step Solution

Identify the primary function of the slip yoke

Matches with sliding axially along splines to absorb driveshaft length changes during suspension travel

As the rear axle moves up and down over road bumps, the linear distance between the transmission and differential changes, requiring a telescoping slip joint.

Identify the primary function of the differential spider gear

Matches with rotating on its shaft to permit drive wheels to turn at different speeds while cornering

When cornering, the outer drive wheel must travel a longer radius; the differential spider gears rotate on their pin to drive the side gears at differential speeds.

Identify the primary function of the center support bearing

Matches with maintaining alignment and dampening vibrations at the midpoint of a two-piece driveshaft

Vehicles with longer wheelbases use split driveshafts; the center support bearing holds the intermediate joint firmly to the chassis to eliminate shaft whip and vibration.

Identify the primary function of the ring gear

Matches with transferring torque from the drive pinion to change the plane of rotation 90 degrees and reduce gear speed

The ring and pinion bevel gearset converts longitudinal driveshaft torque into lateral axle rotation while multiplying torque through final drive reduction.

Key Concept

Driveshaft and Differential Mechanical Functions

Estimated Time:1m 0s

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