Electrical and Charging Systems Lesson Test

Answer: B

Correct answer: The alternator. The battery stores energy, the starter spends it, and the belt-driven alternator replenishes it. While the engine runs, the alternator supplies the electrical loads and recharges the battery.

• A: The starter consumes battery energy to turn the engine; it is a motor, not a generator, and it works only during starting.
• B (correct): The belt-driven alternator generates electrical power whenever the engine runs, feeding loads and replacing the energy the battery used.
• C: The solenoid engages the starter pinion and switches starter current; it generates no electricity.
• D: The fuse box distributes and protects circuits but does not produce any electrical energy.

Answer: D

Correct answer: To open the circuit when current exceeds its rating, protecting the wiring. A fuse is a protective device. Its metal strip melts and opens the circuit when current goes above the rating, which keeps the wiring from overheating; a relay, by contrast, is a switch.

• A: That describes a relay, which uses a small coil current to switch a heavy load; a fuse is a protective device.
• B: A fuse adds no voltage at any time; it is just a calibrated conductor in the current path.
• C: Converting AC to DC is the job of the alternator's rectifier diodes.
• D (correct): A fuse's calibrated strip melts under excess current, stopping flow before the wire behind it overheats.

Answer: D

Correct answer: Six. Each lead-acid cell produces roughly 2.1 V. Six in series give about

the fully charged resting voltage of a 12-volt battery.

• A: This assumes each cell is 1 V; lead-acid cells are about 2.1 V, so twelve would give about 25 V.
• B: Four cells would give roughly 8.4 V, well short of a 12-volt battery.
• C: Two cells of about 2.1 V each would give only about 4.2 V.
• D (correct): Six cells of roughly 2.1 V each in series add up to about 12.6 V when fully charged.

Answer: A

Correct answer: The rectifier diodes. The stator produces AC because the spinning rotor field keeps reversing. A bridge of diodes, the rectifier, lets current flow only one way and turns that AC into DC.

• A (correct): The diodes let current pass in only one direction, steering both halves of the AC wave into DC.
• B: The belt and pulley supply mechanical rotation; they have no role in changing AC to DC.
• C: Brushes and slip rings carry the small field current to the spinning rotor; they do not rectify the output.
• D: The rotor is the spinning electromagnet that creates the moving field; it does not convert the output.

Answer: A

Correct answer: Pushes the pinion into the ring gear and closes heavy contacts that feed the motor. The solenoid is an electromagnet. Its plunger shoves the pinion forward to mesh with the flywheel ring gear and, at the same time, closes heavy internal contacts that send battery current to the starter motor.

• A (correct): The solenoid plunger both engages the pinion and connects battery current to the starter motor at the same moment.
• B: Those are the jobs of the alternator's rectifier and voltage regulator.
• C: Charging and running power come from the alternator once the engine runs; the solenoid only serves the starter.
• D: The solenoid does not touch the fuses or the alternator; it closes a circuit to the starter rather than opening one.

Answer: B

Correct answer: The door switch opens, breaking the loop so current stops. Current needs a complete path from the source, through the load and back. When the door switch opens, the loop is broken, so no current reaches the bulb even though every other part is good.

• A: Closing a door does not change battery voltage; the battery is fine in this case.
• B (correct): Current flows only around a closed loop, so an open switch anywhere in the loop stops current through the lamp.
• C: A switch does not change the bulb's resistance; it simply completes or breaks the path.
• D: Nothing is shorted here; the lamp goes dark because no current flows at all.

Answer: D

Correct answer: The current a new battery supplies for 30 seconds at 0 °F without dropping below 7.2 V. CCA describes a short, heavy demand: the current a new battery can deliver for 30 seconds at 0 °F without dropping below 7.2 V. Amp-hour capacity is a separate rating for slow delivery.

• A: That is the amp-hour (Ah) capacity rating, which describes slow, steady delivery over many hours.
• B: CCA rates the battery's own output, not the alternator's charging current after a start.
• C: Resting voltage reflects state of charge; it is not a current rating.
• D (correct): CCA measures the short, heavy current available for cranking a cold engine on a freezing morning.

Answer: D

Correct answer: 4 A. Rearrange Ohm's law to

• A: This subtracts the resistance from the voltage, which is not a valid relationship.
• B: This multiplies voltage by resistance instead of dividing.
• C: This divides resistance by voltage, which inverts the formula.
• D (correct): Current equals voltage divided by resistance: 14 ÷ 3.5 = 4 A.

Answer: B

Correct answer: The battery can release hydrogen gas, which is explosive. Charging and discharging can release hydrogen gas around the battery. Hydrogen is explosive, so flames, sparks and smoking are kept away and the area is ventilated.

• A: A flame does not reverse polarity; the real hazard is ignition of released gas.
• B (correct): Charging and discharging can give off hydrogen, which a spark or flame can ignite.
• C: Smoke does not conduct enough to short terminals; the concern is explosive gas.
• D: Sparks nearby do not change a battery's rating; the danger is an explosion.

Answer: C

Correct answer: 3 A. Rearrange

to

• A: This subtracts the voltage from the power, which is not how current is found.
• B: This divides voltage by power, which inverts the formula.
• C (correct): Current equals power divided by voltage: 36 ÷ 12 = 3 A.
• D: This multiplies power by voltage instead of dividing.

Answer: A

Correct answer: 1.5 A. Rearrange Ohm's law to

• A (correct): Ohm's law gives current as voltage divided by resistance: 13.5 V divided by 9 ohms is 1.5 A.
• B: This multiplies voltage by resistance instead of dividing, which gives a value far too large for a small heater.
• C: This divides resistance by voltage, turning Ohm's law upside down.
• D: This subtracts the resistance from the voltage, which is not an operation Ohm's law uses.

Answer: C

Correct answer: 84 W. Use

12 V times 7 A equals 84 W.

• A: This adds voltage and current, which does not produce a power value.
• B: This divides voltage by current, which gives resistance in ohms, not power.
• C (correct): Power equals voltage times current: 12 V times 7 A is 84 W.
• D: This divides current by voltage; power is the product of voltage and current.

Answer: D

Correct answer: Place an inductive clamp meter around one starter cable while cranking. Large currents such as starter draw are measured with an inductive clamp meter around a single conductor. An ordinary meter's current input is fused at about 10 A and must be in series, and an ohmmeter needs an unpowered circuit.

• A: A meter in current mode has very low resistance, so placing it across the battery creates a short circuit.
• B: A standard meter's fused current range is limited to about 10 A, far below starter draw, so the fuse or meter would be damaged.
• C: An ohmmeter must be used on an unpowered part, and resistance is not the same quantity as current draw.
• D (correct): A clamp meter senses the magnetic field around a single conductor, so it can read very large currents without breaking the circuit.

Answer: A

Correct answer: Surface charge may be raising the reading, so let the battery rest before judging its charge. Open-circuit voltage must be read after the battery rests, because recent charging leaves a surface charge that inflates the reading. Even a rested reading shows state of charge only, not battery health.

• A (correct): Surface charge from recent charging makes a battery read high; resting it or briefly running a light removes it before a true open-circuit reading.
• B: A voltage reading mostly reflects charge, not health; a load or conductance test is still needed to judge health.
• C: An engine-off reading cannot show the regulator's behavior, and a reading near 12.9 V right after driving is not abnormal.
• D: A slightly high reading just after driving is normal surface charge, not evidence of damaged plates or a ruined battery.

Answer: A

Correct answer: 8 V. In series the resistances add:

The drop across the 4 $\Omega$ part is

and the 4 V across the other part brings the total back to 12 V.

• A (correct): Total resistance is 6 $\Omega$, so current is 2 A, and 2 A × 4 $\Omega$ = 8 V.
• B: This is the drop across the 2 $\Omega$ resistance (2 A × 2 $\Omega$), not the 4 $\Omega$ one.
• C: Each series part gets only a share of the supply; the full 12 V would appear only across a parallel branch.
• D: This divides the supply by the 4 $\Omega$ value alone, ignoring that the resistances add in series.

Answer: B

Correct answer: 6 A. Each parallel branch sees the full 12 V. Branch currents are

A and

so the feed carries

• A: This is only the 6 $\Omega$ branch current; the 3 $\Omega$ branch adds another 4 A.
• B (correct): The 3 $\Omega$ branch draws 4 A and the 6 $\Omega$ branch draws 2 A, and parallel branch currents add to 6 A.
• C: This is only the 3 $\Omega$ branch current; the 6 $\Omega$ branch adds its own 2 A.
• D: This treats the branches as if they were in series (12 ÷ 9), but parallel branches each get the full voltage.

Answer: A

Correct answer: 20 h. Find stored energy first:

Wh. Then divide by the load power:

h. Equivalently, the load draws

h.

• A (correct): Energy is 12 × 90 = 1080 Wh, and 1080 Wh ÷ 54 W = 20 h.
• B: This divides amp-hours by watts directly, mixing charge with power.
• C: This is the load current in amperes (54 ÷ 12), not a run time.
• D: This divides capacity by the voltage instead of by the load current.

Answer: C

Correct answer: The charging system is not keeping up and should be inspected. When the alternator works, running voltage should rise clearly above the resting value. A reading that falls from 12.5 V to 12.4 V shows the charging system is not supplying the loads, so it deserves inspection, starting with simple items such as the belt and connections.

• A: Once the engine runs the starter disengages; this reading points to charging, not starting.
• B: A working charging system should raise the voltage noticeably above the engine-off value, typically into the 13s or 14s.
• C (correct): Running voltage at or below the engine-off value means the alternator is not taking over, so the belt, alternator and wiring need checking.
• D: A running reading says little about battery health; it mainly shows whether the charging system is supplying power.

Answer: C

Correct answer: 10.4 V. The cable drop is

Subtract it from the terminal voltage:

V reaches the starter.

• A: This uses a cable resistance ten times too large (0.05 $\Omega$) when finding the drop.
• B: This adds the cable drop to the battery voltage; resistance in the path removes voltage instead.
• C (correct): The cables drop 120 × 0.005 = 0.6 V, so the starter receives 11.0 − 0.6 = 10.4 V.
• D: This is the voltage lost in the cables, not the voltage left for the starter.

Answer: A

Correct answer: On fuel, spark and air supply. Cranking, starting and running are separate outcomes. A normal crank proves the starting circuit largely works; for the engine to fire it also needs fuel, ignition and air, so those systems are checked next.

• A (correct): A strong crank means the starting circuit largely works, so the engine is missing something else it needs to run.
• B: Bad cables usually cause slow cranking or clicking, not a normal-speed crank.
• C: A normal crank shows the starter and solenoid are doing their job.
• D: If the interlock were blocking the start request, the engine would not crank at all.

Answer: C

Correct answer: The button energizes a relay coil, and the relay's heavy contacts carry the horn current. A relay has two separate circuits. The light control side, a coil fed through the horn button, uses very little current; its magnetism closes heavy contacts on the load side that carry the horn's 20 A.

• A: A fuse does not reduce current; it only opens the circuit if current exceeds its rating.
• B: The ground is only the return path; the feed side still needs a switch heavy enough for 20 A, which the relay contacts provide.
• C (correct): A relay lets a small control current in its coil close separate heavy contacts that feed the large load.
• D: The vehicle's circuits run on DC; the horn's current is simply routed through relay contacts.

Answer: D

Correct answer: Find the fault drawing excess current, then use the specified fuse rating. A fuse's rating protects the wire behind it. Repeated blowing means too much current is flowing, so the fault must be found and fixed, and the replacement fuse must match the specified type and rating.

• A: A larger fuse lets the wire carry more current than it was designed for, which risks melting insulation or fire.
• B: Foil removes protection entirely and lets the wire overheat during the fault.
• C: Repeated blowing means something is drawing too much current; the fault must be found.
• D (correct): A fuse that keeps blowing signals a fault; diagnose it and replace the fuse only with the specified type and rating.

Answer: D

Correct answer: 700 W. Input power is

The losses are input minus output:

which heats the motor (efficiency 0.65).

• A: This adds output to input; losses are found by subtracting output from input.
• B: This is the useful mechanical output, not the loss.
• C: This is the total electrical input; part of it does become useful rotation.
• D (correct): Input is 10 × 200 = 2000 W; subtracting the 1300 W output leaves 700 W of heat.

Answer: C

Correct answer: The other lamp stays lit and the feed current drops to 4 A. Each parallel lamp has its own branch at full voltage and draws

When one opens, the other keeps its 4 A and stays lit, and the feed drops from 8 A to 4 A.

• A: The remaining lamp still sees 12 V and draws its own 4 A; it does not pick up the lost current.
• B: The burned-out lamp no longer draws its 4 A, so the shared feed current must fall.
• C (correct): Each lamp draws 48 ÷ 12 = 4 A on its own branch, so losing one branch removes only its 4 A.
• D: That would happen in a series circuit; parallel branches work independently.

Answer: B

Correct answer: 3 A. In series,

• A: This treats the parts as parallel branches and adds 12 A and 4 A, but a series circuit has only one path.
• B (correct): Series resistances add to 4 ohms, and 12 V divided by 4 ohms is 3 A, the same current through both parts.
• C: This divides the total resistance by the voltage, which inverts Ohm's law.
• D: This multiplies the voltage by the total resistance instead of dividing.

Answer: C

Correct answer: The battery may be fully charged but too worn to deliver cranking current. Resting voltage mostly shows state of charge. An aging battery can be fully charged yet sag badly under a cranking load because its plates are worn, so a load or conductance test is the next step.

• A: About 12.6 V on a rested battery suggests it is nearly full, so low charge is not indicated.
• B: Resting voltage reflects charge, not health; it cannot prove the battery can deliver cranking current.
• C (correct): State of charge and state of health differ; a load or conductance test is needed to judge an old battery's ability.
• D: Overcharging would not explain slow cranking with a full resting voltage.

Answer: D

Correct answer: The cable could still have high resistance; test its voltage drop while cranking. A continuity beep uses a tiny test current and proves only that a path exists. A cable with damaged strands or internal corrosion can still beep, so its voltage drop must be measured while it carries cranking current.

• A: Continuity only shows that some path exists; a few intact strands can beep while the cable cannot carry high current.
• B: Current mode must be placed in series, and a 10 A range could not handle cranking current anyway.
• C: The beep does not clear the cable, so blaming the battery is premature.
• D (correct): A voltage-drop test under load reveals resistance that a low-current continuity check cannot detect.

Answer: B

Correct answer: The voltage regulator, by adjusting rotor field current. Without control, alternator output would rise with speed. The voltage regulator adjusts the current fed to the rotor's electromagnet, strengthening or weakening the field so system voltage stays in its target range.

• A: A fuse responds to excess current, not to voltage, and does not regulate output.
• B (correct): The regulator raises or lowers field current to keep output in its target band at any engine speed.
• C: The battery helps smooth ripple, but it does not set the charging voltage.
• D: The diodes convert AC to DC; they do not adjust how much the alternator produces.

Answer: B

Correct answer: A short to ground. When a powered wire touches the body or another ground before the load, the load's resistance is bypassed. Current surges and the fuse opens, which is the signature of a short to ground.

• A: A short to power is contact with another live circuit, which tends to make devices run when they should be off.
• B (correct): A powered wire touching the body before its load lets resistance collapse and current surge, which blows the fuse.
• C: Extra ground resistance reduces current and dims the lamp rather than blowing the fuse.
• D: An open circuit stops current and would not cause a current surge that blows the fuse.

Answer: D

Correct answer: A 25 A fuse, as specified on the fuse panel. The blower's normal draw is

but fuse size is not picked from that arithmetic. The manufacturer chooses the rating to protect the circuit's wiring, so the 25 A fuse shown on the panel is installed.

• A: Choosing a fuse from the running current ignores the manufacturer's design for that wire size and parts.
• B: Fuse ratings are not averaged; only the specified rating matches the circuit design.
• C: A larger-than-specified fuse protects the wire less, not more.
• D (correct): The manufacturer sets the rating to protect that circuit's wiring and parts, so the labeled value is used.

Answer: A

Correct answer: 4 Ω. Add the reciprocals, then invert.

So

As a check, the branches draw 2 A and 1 A, and 12 V divided by 3 A is also 4 ohms.

• A (correct): The reciprocals add to 1/6 + 1/12 = 3/12, so the equivalent is 12/3 = 4 ohms.
• B: This stops at the sum of reciprocals, 0.25, and forgets to invert it to get resistance.
• C: This adds the resistances as if they were in series; parallel branches lower the total.
• D: This averages the two branches, but the parallel equivalent is always smaller than the smallest branch.

Answer: A

Correct answer: 9.9 V. Use

• A (correct): The internal drop is 180 A times 0.015 ohms, or 2.7 V, and 12.6 V minus 2.7 V leaves 9.9 V at the terminals.
• B: This subtracts the resistance value itself rather than the current times the resistance.
• C: This adds the internal voltage drop instead of subtracting it; a battery's terminal voltage sags under load.
• D: This is only the voltage lost inside the battery, not the voltage left at the terminals.

Answer: B

Correct answer: 45 W. Use

That heat in a single joint can melt insulation.

• A: This squares the resistance instead of the current.
• B (correct): Heat power is current squared times resistance: 900 times 0.05 is 45 W, concentrated in one small joint.
• C: This is current times resistance, which gives the voltage drop of 1.5 V, not the power.
• D: This divides current by resistance, which is not a valid power formula.

Answer: A

Correct answer: 1380 W. First find the electrical output:

The extra 552 W becomes heat in the alternator.

• A (correct): Output is 13.8 V times 60 A, or 828 W, and dividing by the 0.60 efficiency gives an input of 1380 W.
• B: This is only the electrical output; the belt must also supply the power lost as heat.
• C: This multiplies the output by the efficiency, which would make input smaller than output and create free energy.
• D: This divides the output by 0.40, the fraction lost, instead of by the efficiency of 0.60.

Answer: B

Correct answer: 15 h. Convert capacity to energy, then divide by the load.

In practice a starting battery should never be drained this far.

• A: This divides ampere-hours by watts, mixing charge and power without converting to energy.
• B (correct): Energy is 12 V times 75 Ah, or 900 Wh, and 900 Wh divided by 60 W gives 15 hours.
• C: This divides the load by the capacity, which inverts the runtime calculation.
• D: This divides the ampere-hour rating by the voltage, which does not give energy or time.

Answer: D

A click without cranking means the solenoid or relay responded but the motor did not turn. Dimming lights show system voltage sagging under load, so battery charge or health and cable or ground connections are the first areas to test.

• A: The fuel pump affects whether a cranking engine starts; it does not stop the starter from turning the engine.
• B: Spark matters only after the engine cranks, and a single symptom proves nothing by itself.
• C: The alternator does not power cranking, and an absolute claim like this ignores the battery and cables.
• D (correct): A click shows the solenoid responded, and sharp dimming shows voltage collapsing under load, which points to the battery or its cables.

Answer: C

An unwanted resistance forms a voltage divider with the load. With the bulb out no current flows, so the meter sees full voltage past the bad joint. Measuring voltage drops with the lamp lit shows where the voltage is lost.

• A: Fuse rating does not change voltage, and oversizing a fuse removes wire protection.
• B: An open circuit stops all current, so the lamp would be dark and the meter would not read full voltage at the socket.
• C (correct): With no load, a bad joint carries no current and drops no voltage; once the lamp draws current, the joint takes its share and the lamp starves.
• D: An unloaded reading cannot reveal a resistance that only shows up under current, so the wiring is not proven good.

Answer: C

Correct answer: 4 W. The connections are in series with the lamp, so the same 4 A flows through them. Their combined drop is

• A: This divides the 1.0 V combined drop by the current instead of multiplying.
• B: This is the total power from the battery, 12.4 V times 4 A, which includes the lamp.
• C (correct): The two drops add to 1.0 V, and 1.0 V times 4 A equals 4 W of heat in the connections.
• D: This is the power used by the lamp itself, 11.4 V times 4 A, not the power lost in the connections.

Answer: B

Computer modules stay awake for a while after the car is shut and locked, so early readings run high. Comparing the settled value of 0.03 A with the 0.05 A limit shows the draw is normal.

• A: A small parasitic draw for clocks, alarms and module memory is normal for a parked car.
• B (correct): Draw should be judged only after modules settle, and 0.03 A is below the 0.05 A limit.
• C: A high early reading identifies no guilty circuit, and the settled reading is within the limit.
• D: The first reading was taken while modules were still awake, so it should not be compared with the limit.

Answer: C

The new load draws

Parallel branch currents add, so the feed carries

above the 15 A fuse rating. The fix is a separate, correctly protected circuit, not a bigger fuse.

• A: Parallel loads do not share a fixed current; each added branch adds its own current to the feed.
• B: A larger fuse no longer protects the wire sized for that circuit and can let it overheat.
• C (correct): The new accessory draws 60 W divided by 12 V, or 5 A, so the total is 5 + 6 + 5 = 16 A, more than the fuse is meant to carry.
• D: This divides the total among the branches; in parallel, branch currents add on the shared feed.

Answer: D

Correct answer: It turns electrical energy from the battery into rotation that spins the engine. The starter is a motor: electrical energy in, mechanical rotation out, and only during starting. Its small pinion turns the flywheel ring gear to spin the crankshaft. The alternator does the reverse job, which is why the two are easy to mix up.

• A: That describes the alternator, a generator that works in the opposite direction from the starter.
• B: Storing energy is the battery's job, not the starter's.
• C: Holding voltage steady is the job of the voltage regulator in the charging system.
• D (correct): The starter is an electric motor that spends stored battery energy to crank the engine fast enough to fire.

Answer: C

Correct answer: Out of the positive terminal, through the lamp, and back to the negative terminal. By convention, diagrams show current leaving the battery's positive terminal, going through the load and returning to negative. Electrons really move the opposite way, but both descriptions give the same voltages, currents and powers as long as you stay consistent.

• A: That is the direction electrons actually drift in the wire, which is electron flow, not conventional current.
• B: Current flows around a single closed loop in one direction; it does not enter the load from both ends.
• C (correct): Conventional current is drawn leaving positive, passing through the load and returning to negative.
• D: That describes alternating current, but a battery circuit is DC and flows one way only.

Answer: D

Correct answer: 32.4 W. Series resistances add, so the total is

and the current is

The lamp gets

so its power is

The connector takes the other 1.2 V and turns 3.6 W into heat, which is why the lamp glows dimmer than it would with clean connections.

• A: This ignores the corroded connector and uses the full 12 V across 3.6 ohms.
• B: This is the power of the whole circuit (12 V x 3 A), which includes the power wasted in the connector.
• C: This is the heat made in the connector (3 squared x 0.4), not the power used by the lamp.
• D (correct): Current is 12 / 4.0 = 3 A, the lamp drops 3 x 3.6 = 10.8 V, and 10.8 V x 3 A = 32.4 W.

Answer: D

Correct answer: 32 h. First find the current:

Then runtime is capacity divided by current:

h. A real battery would give less, and a starting battery should not be run flat.

• A: This divides the 48 Ah directly by the 18 W, mixing up watts with amperes.
• B: This multiplies the capacity by the current instead of dividing by it.
• C: This divides the capacity by the 12 V system voltage, which is not the load current.
• D (correct): The light draws 18 / 12 = 1.5 A, and 48 Ah / 1.5 A = 32 h.

Answer: C

Correct answer: 4 h. Usable energy is half of

Wh, or 600 Wh. Running time is

h.

The same answer comes from current:

h.

• A: This uses the full 100 Ah and forgets the half-capacity limit.
• B: This is the charger's current in amperes, not a time in hours.
• C (correct): The charger draws 150 / 12 = 12.5 A, half the capacity is 50 Ah, and 50 / 12.5 = 4 h.
• D: This divides 50 Ah by 150 W, mixing up amperes and watts.

Answer: A

Correct answer: 0.016 Ω. Rearrange the model:

Only the voltage lost inside the battery is divided by the current.

• A (correct): The internal drop is 12.6 - 10.2 = 2.4 V, and 2.4 V / 150 A = 0.016 ohm.
• B: This divides the loaded terminal voltage by the current instead of the voltage lost inside the battery.
• C: This divides the no-load voltage by the current, which is not the internal drop.
• D: This has the right method but slips a decimal place; 2.4 / 150 is 0.016.

Answer: C

Correct answer: 60 percent. Find output power first:

So 60 percent becomes electrical output and the other 720 W becomes heat.

• A: This is the fraction lost as heat (720 / 1800), not the useful output fraction.
• B: This divides input by output; an efficiency above 100 percent would mean free energy.
• C (correct): Output is 14.4 x 75 = 1080 W, and 1080 / 1800 = 0.60.
• D: This divides the current (75 A) by the input power without first finding the output power in watts.

Answer: B

Correct answer: 0.5 V, meaning the post-to-clamp connection has resistance wasting voltage. A voltmeter shows

A clean clamp should lose almost nothing, so half a volt across one connection during cranking shows hidden resistance there, likely corrosion or a loose clamp.

• A: A voltmeter shows the difference between two points, not their sum, so no overcharge or regulator fault is shown.
• B (correct): The meter reads the difference between its probes, and a half-volt drop across one clamp under load points to a poor connection.
• C: A voltmeter across a connection does read the voltage lost there while current flows; that is the purpose of a voltage-drop test.
• D: The meter does not show the voltage of one point alone, and a large drop across a clamp is a fault sign.

Answer: D

Correct answer: 0.004 Ω. Rearrange Ohm's law:

Even this tiny resistance wastes 0.6 V that the starter never receives, which is why ground straps matter.

• A: This multiplies voltage by current, which gives power in watts, not resistance.
• B: This divides current by voltage, which is the inverse of Ohm's law for resistance.
• C: The method is right but a decimal place slipped; 0.6 / 150 is 0.004.
• D (correct): R = V / I = 0.6 / 150 = 0.004 ohm.

Answer: B

Correct answer: Shift to Park or Neutral, since the safety interlock blocks starting in gear. The start request passes through safety interlocks before reaching the starter relay or solenoid. On an automatic, the park or neutral switch blocks starting in gear. With no click and bright lights, the simplest likely cause is the interlock, so shift to Park or Neutral and try again before testing parts.

• A: A battery too weak to spin the starter would make the lights dim or cause clicking; bright lights argue against it.
• B (correct): Bright lights mean the battery is fine, and a lever left in Drive keeps the neutral safety switch from passing the start request.
• C: Fuel delivery matters only after the engine cranks; here it does not turn at all.
• D: The alternator plays no part in cranking; the starter and its control circuit do.

Answer: A

Correct answer: Lets current pass in one direction and blocks it in the other. A diode is a one-way valve for current. In an alternator, a bridge of diodes steers both halves of the AC wave the same way to make DC, and it also stops the battery from pushing current back through the stator when the engine is off.

• A (correct): A diode acts like a one-way check valve for current.
• B: That describes an ordinary wire; a diode blocks reverse current.
• C: That describes a fuse, a protective device.
• D: Storing energy is the battery's job.

Answer: C

Correct answer: The rotor loses its field current, so the alternator produces little or no output. The rotor is an electromagnet that gets its small field current through brushes and slip rings. Its spinning field induces AC in the stator, which the diodes rectify. With no brush contact there is no rotor field, so there is no induced voltage and the charging system stops charging.

• A: Output depends on the rotor's magnetic field; removing field current lowers output instead of raising it.
• B: The brushes are the path for rotor field current, so they are essential to charging.
• C (correct): Brushes carry field current to the spinning rotor; without a magnetic field, nothing is induced in the stator.
• D: The diodes are a separate part; the problem is that little or no AC is being made at all.

Answer: B

Correct answer: Ampere-hour (Ah) capacity. Amp-hour capacity measures how much charge a battery delivers slowly, often over a 20-hour test. Cold-cranking amps measure a different ability: a big current for 30 seconds in the cold. One rating does not replace the other.

• A: CCA describes a short, heavy current demand in the cold, not long, slow delivery.
• B (correct): Ah capacity describes how much charge the battery can supply at a low, steady rate.
• C: Resting voltage mostly reflects state of charge, not how much charge the battery holds.
• D: A 12 V battery always has six cells; that count does not tell its capacity.

Answer: C

Correct answer: About half charged. For a rested lead-acid battery, about 12.6 V or more suggests nearly full, around 12.4 V about three-quarters, and near 12.2 V about half. A reading tells charge, not health, so a load test is still needed to judge the battery's condition.

• A: Nearly full is about 12.6 V or more on a rested battery.
• B: Roughly three-quarters is around 12.4 V.
• C (correct): As a rough guide, a rested reading near 12.2 V suggests about 50 percent.
• D: 12.2 V is still about half charged; a fully discharged battery reads well below that.

Answer: D

Correct answer: 120 W. Each parallel branch sees the full 12 V. Branch currents are

A and

so the total is 10 A.

As a check, the equivalent resistance is 1.2 ohms, and

• A: This adds the resistances to 5 ohms as if the parts were in series, but parallel branches each get the full 12 V.
• B: This counts only the seat heater branch (12 V x 6 A) and leaves out the fan.
• C: This counts only the fan branch (12 V x 4 A) and leaves out the seat heater.
• D (correct): The branches draw 12/2 = 6 A and 12/3 = 4 A, so the feed carries 10 A and 12 V x 10 A = 120 W.

Answer: B

Correct answer: Through a ground wire or mounting into the metal body, then to battery negative. Most vehicles use negative ground: the battery negative cable bolts to the engine and body, so the metal structure acts as the return conductor. The headlamp's current flows through its ground wire or mount into the body and back to battery negative, and that return path is part of the circuit.

• A: Current must flow around a complete loop; a lamp converts energy, not charge.
• B (correct): The body and engine are bonded to battery negative, so the metal structure serves as the shared return path.
• C: One wire cannot carry current both out and back; the return needs its own path.
• D: A rubber drive belt is not an electrical path; the return goes through the body and ground cable to battery negative.

Answer: B

Correct answer: The metal could bridge a live terminal and ground and quickly heat red-hot. The battery and starter circuits can deliver hundreds of amperes with little or no fuse protection. A ring or watch band that touches positive and ground at the same time becomes a short, heating red-hot in seconds. That is why jewelry comes off and tools must never bridge the terminals.

• A: Vented hydrogen is a spark and flame hazard, but jewelry does not attract the gas.
• B (correct): High-current circuits can push hundreds of amperes through a ring or watch band, heating it enough to burn skin within seconds.
• C: The danger is a short circuit through the metal, not a meter reading error.
• D: Acid is a skin and eye hazard, but it does not dissolve jewelry almost instantly; the key danger from jewelry is the short circuit.

Answer: A

Correct answer: The charging system is not charging as expected, so the battery may be draining. The battery lamp is a charging-system lamp. With the engine running, it means the alternator system is not keeping up, perhaps from a slipping belt, a failed alternator or a wiring fault. Check the belt, then running voltage, then the alternator and its wiring before the battery runs down.

• A (correct): This lamp is really a charging-system warning; the car may be running on stored battery energy.
• B: The lamp does not measure battery health; the battery had enough energy to start the car, so immediate replacement is not indicated.
• C: The lamp warns of a charging problem; it is not a full-charge indicator.
• D: The lamp monitors charging, not the starter, and the engine started and runs normally.

Answer: B

Correct answer: This can be normal for smart charging, so compare with the maker's specification. On smart-charging vehicles, the engine computer sets the alternator's target and may let voltage fall while cruising and rise during braking. The right reading depends on the vehicle, so check the maker's specification and avoid answers that claim a reading 'proves' a fault.

• A: A single low reading does not prove failure, especially when the computer varies the target on purpose.
• B (correct): The engine computer may lower voltage to save fuel and raise it during braking, so a fixed rule of thumb can mislead.
• C: Raising voltage during braking is a typical smart-charging strategy, not proof that the regulator is overcharging.
• D: Charging voltage changes with temperature, load and battery condition on modern cars, so a change while driving does not mean the battery is bad.

Answer: D

Correct answer: The alternator belt is slipping under the extra electrical load. When heavy loads come on, the regulator raises field current and the alternator becomes harder to turn. A glazed, worn or loose belt then slips, squeals and lets output drop, so the lights dim. Check the belt and tensioner with the engine off.

• A: The starter works only during cranking, and a stuck starter would be noticed right after starting, not when loads are added.
• B: A blown fuse would stop the defroster but would not cause a belt squeal or dim the lights.
• C: CCA matters only for starting; with the engine running, the alternator supplies the loads.
• D (correct): Heavy loads make the alternator harder to turn, and a worn or loose belt then slips, squeals and cuts output.

Answer: A

Correct answer: In current mode the meter has very low resistance, so it shorts the battery. To measure current, the circuit is opened and the meter is placed in series so the current flows through it. Its current range is fused and usually limited to about 10 A. Placed across a battery like a voltmeter, it becomes a short. Large currents such as starter draw are measured with an inductive clamp meter.

• A (correct): An ammeter is meant to be inserted in series; across the terminals it becomes a near-zero-resistance short.
• B: The meter can read DC current as well; the danger is a short circuit, not a zero reading.
• C: No meter setting measures stored amp-hours by touching the terminals; the real danger is a short circuit.
• D: The current would be huge, not small, and would blow the meter's fuse or worse.

Answer: A

Correct answer: The meter supplies its own test current, which outside voltage or other paths would upset. An ohmmeter supplies its own small test current and calculates resistance from it. Any live voltage adds to that and gives false readings or damages the meter, and other parallel paths can let current bypass the part. Unpower and disconnect the part to get a true reading.

• A (correct): Live voltage adds to the meter's own test current, giving false readings or damage, and other paths can make a bad part look good.
• B: Resistance does not simply double with power applied; the problem is outside voltage interfering with the meter.
• C: Isolation, not removal from the vehicle, is what matters; unplugging the part is usually enough without unbolting it.
• D: Cooling lowers a winding's resistance only slightly; it never falls to zero.

Answer: C

Correct answer: A short section of smaller special wire that melts open during a severe overload. A fusible link is a short length of special wire, smaller than the circuit it protects, with insulation that will not burn. During a severe overload it melts open to protect the large feeds near the battery. Many newer vehicles use high-amp cartridge fuses for the same job, and either must be replaced only with the exact specified part after finding the fault.

• A: That describes a relay, which switches loads rather than protecting wire.
• B: That describes a ground strap, part of the return path.
• C (correct): A fusible link protects large feeds near the battery and has insulation that will not burn.
• D: That is a job of the alternator's rectifier diodes.

Answer: A

Correct answer: A short to power. When a wire touches another live circuit, it is a short to power: the device may run when switched off, or two circuits feed each other. A short to ground, by contrast, sends current straight to the body and usually blows the fuse. Naming the fault type guides the testing.

• A (correct): A wire touching another live circuit can feed a device so it runs when it should be off.
• B: A short to ground is a powered wire touching the body; that collapses resistance and usually blows the fuse instead of running the device.
• C: An open stops current, so the wiper would not run at all.
• D: Extra ground resistance would slow or weaken the wiper, not make it run with the switch off.

Answer: D

Correct answer: 8 W. First find the current:

That heat is concentrated in one small joint, which can melt insulation over time.

• A: This is I x R, which gives the 0.8 V dropped across the connector, not the power.
• B: This multiplies the load's 120 W by the connector resistance, which is not a valid power formula.
• C: This uses 12 V squared times the resistance; the connector does not have 12 V across it.
• D (correct): The defroster draws 120 / 12 = 10 A, and 10 squared x 0.08 = 8 W.

Answer: B

Correct answer: A failure in a shared part such as the fuse, relay, switch or common ground. In a parallel branch, each lamp has its own path, so a failed bulb darkens only that lamp. When both go out together, the fault is in something they share: the fuse, the relay contacts or coil, the dash switch, the common feed wire or the shared ground.

• A: Each bulb fails on its own; two failing at the exact same moment is very unlikely.
• B (correct): Only a part both lamps depend on can darken both at once.
• C: The lamps are wired in parallel, not in series, so one failed bulb darkens only its own lamp.
• D: A charging failure drains the whole system gradually; it would not suddenly darken only the fog lamps while everything else works.