Electrical Safety and Troubleshooting Lesson Test

Answer: A

Correct answer: Remove it from service until it is properly repaired and tested. Insulation is the barrier that keeps your hand away from the live conductors, so a break in it means a shock or short can happen with an ordinary grip. A cord that still powers its load proves only that the conductors carry current, not that they are safely enclosed. Tape is tempting, but it wears, peels and lacks the cord jacket's rating. Tagging the cord out of service keeps others from picking it up in the meantime.

• A (correct): Visible damage can expose users to shock, so the cord should not be used until it has been properly repaired and tested.
• B: Continued operation is not evidence that damaged insulation is safe.
• C: Improvised tape does not restore the original insulation rating or protection against crushing and abrasion.
• D: Altering a grounding pin defeats its protective purpose.

Answer: C

Correct answer: An arc-fault circuit interrupter (AFCI). Match the device to the hazard. Repeated sparking at a loose connection is arcing, and the AFCI is the device built to recognize arcing patterns in the current and open the circuit before a fire starts.

• A: A fuse responds only to current above its rating, and this arcing happens at roughly normal current.
• B: A GFCI looks for current leaking away to ground, not for the waveform pattern of sparking.
• C (correct): An AFCI watches the current waveform for the patterns of dangerous arcing, mainly to prevent fires.
• D: The grounding conductor gives fault current a path; it does not sense arcing.

Answer: B

Correct answer: An imbalance between monitored outgoing and returning currents. A GFCI passes the hot and neutral conductors through a sensing coil. Normally the current out equals the current back, their magnetic effects cancel, and the coil senses nothing. If some current leaks to ground, for example through a person, the return is short, and when the difference reaches about 5 mA the device trips. It does not measure voltage or temperature, and it is not an overcurrent device: a breaker or fuse still handles overloads and shorts.

• A: Current imbalance sensing does not directly measure insulation temperature.
• B (correct): Leakage outside the expected return path can produce a differential current.
• C: A GFCI is not a precision voltage regulator or voltage meter.
• D: A line-to-line fault may not cause the differential imbalance a GFCI detects.

Answer: B

Correct answer: Ohmmeter. The name gives it away: ohms are the unit of resistance, just as amperes are measured by an ammeter and watts by a wattmeter. An ohmmeter works by sending a small current from its own battery through the part and measuring the resulting voltage, then applying

Because it supplies its own source, the circuit under test must be de-energized, or outside voltage will give false readings and can damage the meter.

• A: A frequency counter measures repetition rate rather than resistance.
• B (correct): An ohmmeter is intended to determine resistance in ohms under suitable de-energized conditions.
• C: A wattmeter is for power, not direct resistance measurement.
• D: An ammeter measures electric current in amperes.

Answer: A

Correct answer: In series with the lamp. An ammeter has very low internal resistance and measures the current that flows through it, so it is placed in series, in line with the lamp. A voltmeter, by contrast, goes in parallel across the two points being compared.

• A (correct): Current must pass through the meter to be measured, so the ammeter goes in series with the lamp.
• B: An ammeter across a source creates a near short circuit, which is the most dangerous mistake.
• C: Across the lamp, the meter's very low resistance would act as a near short around it.
• D: With no source in the circuit there is no current to measure at all.

Answer: B

Correct answer: Qualified personnel using suitable equipment and procedures. Testing live or possibly live equipment exposes the tester to shock and arc flash, so safety depends on three things together: training to recognize hazards, equipment rated for the circuit, and a procedure such as verifying the meter before and after use. A qualified person is someone trained in those hazards and practices. Theory knowledge or simply owning a meter covers only one piece, which is why those answers are tempting but incomplete.

• A: Hearing a symptom provides no qualification or safe isolation.
• B (correct): OSHA restricts testing work on electrical circuits or equipment to qualified persons.
• C: Owning a meter does not establish qualification or show that the device is suitable for the circuit.
• D: A formula and drawings do not provide the training, ratings, or precautions needed for real work.

Answer: B

Correct answer: Provide a protective fault-current path in the intended design. If a hot wire touches a metal case, the grounding conductor gives that fault current a low-resistance path back to the source. The resulting surge of current quickly trips the breaker or blows the fuse, removing voltage from the case before someone touches it. Under normal conditions it carries no current; that job belongs to the neutral. A useful way to remember it: the neutral carries working current, while the ground carries only trouble current.

• A: Grounding is one layer and cannot guarantee freedom from all hazards.
• B (correct): This connection helps suitable protection respond when exposed metal becomes energized under fault.
• C: A ground conductor does not act as a voltage amplifier.
• D: Normal load current uses intended circuit conductors, not the protective path.

Answer: A

Correct answer: Avoid contact, call emergency services, and stay clear until the scene is safe. While the person is still in contact, their body is part of the live circuit, and anyone who touches them can become part of it too. The instinct to grab and pull is exactly what creates a second victim. Getting the power shut off, by trained responders or at a disconnect you can reach safely, comes first; only then can first aid such as CPR begin.

• A (correct): Approaching or touching the person before the power is controlled can create a second casualty.
• B: Water can worsen conductive exposure and other hazards.
• C: Direct contact can put the rescuer in the same current path and expose them to dangerous current.
• D: A conductive tool can carry current into the rescuer.

Answer: C

Correct answer: Stop use and seek qualified evaluation before re-energization. A breaker is sized to protect the wiring, so a trip means current exceeded what that wiring can carry safely, from an overload, a short or a ground fault. One trip can be a fluke; repeated trips are a symptom of a real problem. Resetting, upsizing or bypassing all treat the warning instead of the cause, and they allow the wires to overheat. Finding the cause first is what actually makes the circuit safe.

• A: Bypassing a protective device removes a safety function.
• B: Repeated reclosing without finding the cause can reintroduce hazards; OSHA bars manual re-energization until the circuit is determined safe.
• C (correct): A recurring protective trip can indicate an unresolved overload or fault.
• D: Increased overcurrent rating can let the wiring overheat before the breaker opens, defeating its protection.

Answer: B

Correct answer: Capacitor. A capacitor holds separated charge on two plates, and that charge stays put after the supply is disconnected unless a discharge path drains it. Large capacitors in power supplies, microwave ovens and motor starters can hold enough energy to cause a serious shock long after unplugging. That is why technicians discharge and verify capacitors before working. Ideal wires and open switches carry or block current but store no energy themselves.

• A: An ideal wire does not function as a capacitor in the stated simple model.
• B (correct): A charged capacitor stores energy in an electric field.
• C: An ideal open switch does not itself act as an energy-storage element.
• D: A label has no electrical-energy storage function.

Answer: D

Correct answer: Overload. An overload happens when properly working loads, connected as intended, together draw more current than the circuit wiring is rated for. Each heater is normal; the problem is the total. The current still flows through the intended path, which is what separates it from a short circuit, where current bypasses the load through an accidental low-resistance contact, and from a ground fault, where current escapes to grounded parts. Overloads build heat more slowly, so breakers typically trip on them after a delay.

• A: An open interrupts current; the described loads are operating and drawing too much.
• B: The stem specifically excludes a direct low-impedance fault path.
• C: A ground fault is current leaking to grounded parts; the stem describes normal loads drawing too much current.
• D (correct): Too many normal loads together can draw more current than the circuit is designed to carry.

Answer: B

Correct answer: Ground fault. The word fault signals that current is going somewhere it was never meant to, and ground tells you where: to a grounded part such as a metal enclosure or a person. Normal current leaves on the hot and returns on the neutral. In a ground fault, some or all of it returns through the grounding conductor or earth instead. A ground fault is the condition a GFCI is designed to detect, and the grounding conductor helps the breaker clear it.

• A: An accidental enclosure connection is a fault, not designed consumption.
• B (correct): Current travels outside the normal intended circuit to a grounded part.
• C: The metal enclosure is not an intended normal load-current return path.
• D: The hazard is an unintended conducting path, not an interrupted path.

Answer: A

Correct answer: It reduces some shock hazards from leakage current but does not remove every risk. A GFCI trips when about 5 mA of current goes missing between hot and neutral, which typically means it is leaking to ground, often through a person. It is very effective for that common case. But a person touching both hot and neutral, with no path to ground, creates no imbalance, and a GFCI does not replace overcurrent protection or good insulation. Words like always and guarantees are a strong sign of an overstatement in safety questions.

• A (correct): A differential current threshold can help interrupt certain hazardous ground faults.
• B: Actual response depends on fault path and device characteristics.
• C: Physical insulation and safe use remain essential.
• D: A balanced supply-and-return shock path, such as hot to neutral through a person, may not trip the differential detector.

Answer: C

Correct answer: Circuit breaker. A breaker opens its contacts with a mechanism, either a heated bimetal strip for slow overloads or an electromagnet for sudden high currents. Because nothing is destroyed, the contacts can be closed again with the handle once the cause is fixed. A fuse protects by melting its metal link, so it is used up and must be replaced. Grounding conductors and resistors have no interrupting function at all.

• A: A conventional blown fuse is replaced according to its design, not reset.
• B: A grounding conductor is not a resettable overcurrent interrupter.
• C (correct): A circuit breaker is a protective switching device designed for defined overcurrents and appropriate subsequent restoration.
• D: A fixed resistor is not a normal resettable protective switching device.

Answer: A

Correct answer: That is an unsafe and inappropriate resistance measurement. An ohmmeter measures resistance by pushing a small current from its own battery and reading the result, so it expects a dead circuit. Connected to a live outlet, the outlet's 120 V overwhelms that tiny test source, producing meaningless readings and possibly damaging the meter or causing an arc. To check a live outlet, a technician switches to the voltage function and uses rated leads; to measure resistance, the power must be off first.

• A (correct): An ohmmeter uses its own test source and is not connected to energized circuits for resistance testing.
• B: A screen indication alone cannot validate safety or the selected mode.
• C: Internal batteries do not remove an external circuit hazard.
• D: Voltage measurement is a different function and requires qualified practices.

Answer: A

Correct answer: 12 V. An open stops current everywhere in a series loop. With zero current, there is no IR drop across the working lamps, so by Kirchhoff's voltage law the whole 12 V appears across the gap. Zero current does not mean zero voltage.

• A (correct): With no current, the good lamps drop no voltage, so the entire 12 V appears across the open filament.
• B: Current is zero, but voltage can still be present; with no drops across the other lamps, the gap takes the full source voltage.
• C: Four volts is each lamp's share while current flows; once the loop is open, no current flows to divide the voltage.
• D: There is no reason for the voltage to split in half; the two good lamps have no drop with zero current.

Answer: D

Correct answer: 4 A. Each parallel branch has the full 18 V. Before the fault:

With the 9 $\Omega$ branch open, only the 3 A and 1 A branches remain, so the source supplies 4 A.

• A: That would be true of a series loop; in parallel, the other branches keep their full voltage and current.
• B: Two amperes is the current the open branch used to carry, not what is still flowing.
• C: Six amperes is the total before the break; it ignores the 2 A that the open branch no longer carries.
• D (correct): The remaining branches still carry 3 A and 1 A, so the total falls from 6 A to 4 A.

Answer: C

Correct answer: 36 W. The load sets the current, so use

That heat is concentrated in a tiny spot, and the breaker may never trip because the current is normal.

• A: This divides current by resistance, which is not a power formula.
• B: This multiplies current by resistance, which gives the 3 V drop, not the power.
• C (correct): Heat at the joint is I squared R: 12 squared times 0.25 equals 36 W.
• D: This squares the current but never multiplies by the 0.25 ohm resistance.

Answer: A

Correct answer: 960 W. For a fixed resistance,

so power scales with the square of the voltage ratio.

• A (correct): 96 V is 0.8 of 120 V, and 0.8 squared is 0.64, so the power is 1,500 times 0.64 = 960 W.
• B: This reduces power in proportion to voltage, but power depends on voltage squared.
• C: This is how much the power drops, not the power the heater still produces.
• D: The heater's resistance is fixed, so its power must fall when the voltage falls.

Answer: A

Correct answer: The designed overcurrent interruption is defeated. A fuse is meant to be the weakest link, melting before the wiring overheats. A plain wire has a much higher melting current, so in an overload or short the wiring insulation can char and start a fire before anything opens. The circuit may seem to work normally, which is why the practice is tempting and dangerous. A blown fuse should be replaced only with one of the same type and rating, after finding why it blew.

• A (correct): A wire bridge can permit fault current beyond the intended protective limit.
• B: The bridge creates a path and may permit dangerous current.
• C: A wire bridge does not detect ground leakage.
• D: Bypassing a fuse does not regulate voltage.

Answer: C

Correct answer: About 23.3 A, which overloads the circuit. Add the loads and convert with

That exceeds the 20 A rating, so the circuit is overloaded.

• A: This divides by 240 V instead of the 120 V the circuit actually supplies.
• B: This leaves out the kettle; every load on the branch adds to the current.
• C (correct): Total power is 2,800 W, and 2,800 divided by 120 is about 23.3 A, above the 20 A rating.
• D: This is the total power in watts, which still needs to be divided by the voltage to give amperes.

Answer: D

Correct answer: The meter and leads must be rated and suitable for the circuit and environment. Meters carry voltage ratings and measurement categories, such as CAT II or CAT III, that tell you which circuits they can safely handle, including brief voltage spikes. A meter used beyond its rating can fail violently. The leads are part of the system too: cracked insulation can expose a live tip to your hand. Accurate readings only show that the meter works, not that it is safe for the job.

• A: Handheld design is not evidence of safe voltage or transient suitability.
• B: Display technology does not replace electrical ratings and work practice.
• C: Damaged insulation or connectors can create a hazard irrespective of accuracy.
• D (correct): Appropriate ratings, condition, and qualification are required for electrical testing.

Answer: B

Correct answer: A low-resistance fault path that lets the breaker or fuse trip quickly. When a hot wire touches a grounded case, the grounding conductor completes a path back to the source with very little resistance. Current jumps far above normal, which trips the breaker or blows the fuse and removes the voltage. Without that path, the case could stay energized silently until someone touched it. In normal operation, the grounding conductor carries no current; the neutral does that work.

• A: Protective grounding is not the intended normal load-current path.
• B (correct): Bonding the case to ground lets fault current rise high enough for the overcurrent device to open.
• C: Grounding does not exist to increase ordinary load current.
• D: The case can still carry some voltage briefly or through a poor connection, so grounding reduces risk rather than eliminating it.

Answer: B

Correct answer: 5 J. Stored energy in a capacitor is E = ½CV². The ½ appears because charging builds voltage gradually from zero, so the average voltage during charging is half the final value. Square the voltage first: 50² = 2,500. Then

and half of that is 5 J. The most common slips are forgetting to square V or dropping the ½. Even a few joules from a charged capacitor can give a painful shock, so disconnected equipment is not automatically safe.

• A: This multiplies C and V without the squared-voltage relation.
• B (correct): E = 1/2 CV² = 0.5 × 0.004 × 2500 = 5 J.
• C: This omits the one-half factor in stored energy.
• D: This uses the displayed voltage as though it were energy.

Answer: D

Correct answer: A crushed cord with exposed inner wiring. The question is which sign shows that protection has failed. A cord's insulation is what separates you from live conductors, so exposed wiring means an ordinary touch can complete a circuit through your body, and the bare conductors can also short and start a fire. The other choices describe normal features or normal operation. When inspecting tools, look for damaged insulation, missing ground pins, scorch marks, and loose plugs.

• A: Normal label information is not evidence of a defect.
• B: Normal control operation does not establish a fault.
• C: The condition itself is not a warning sign.
• D (correct): Damaged insulation and exposed wires create a foreseeable injury hazard.

Answer: D

Correct answer: It carries up to 10 A normally and may be used on circuits up to 125 V. Voltage and current ratings describe different limits. The 10 A rating tells how much current the fuse carries before it is designed to open, and the 125 V rating tells the highest circuit voltage it can safely interrupt. Neither rating substitutes for the other.

• A: The 125 is a voltage rating, not a current rating; the fuse is rated to carry 10 A.
• B: A current rating says nothing about voltage; above 125 V the fuse may not interrupt safely.
• C: Fuses respond to sustained excess current, not to voltage.
• D (correct): The current rating sets what it carries without opening, and the voltage rating sets the highest voltage it can safely interrupt.

Answer: C

Correct answer: The neutral, or grounded, conductor. A branch circuit has three conductors with different roles. The hot carries current to the load, the neutral carries the normal return current, and the equipment grounding conductor stays idle unless a fault occurs.

• A: The grounding conductor normally carries essentially no current; it is a fault path only.
• B: The hot conductor carries current out to the load, not back from it.
• C (correct): The neutral is part of the working circuit and carries the normal return current.
• D: The case should carry no current at all; current on it would mean a fault.

Answer: C

Correct answer: A protective event occurred; more evidence is needed to find the cause. A breaker trips on any of several conditions: an overload, a short circuit, or, for some breakers, a ground fault, and occasionally from wear in the breaker itself. The trip tells you something happened, not which thing. Jumping to a specific cause is the tempting error in an inference question. Troubleshooting means gathering evidence, such as what was running and whether it trips immediately or after a delay.

• A: Many causes can operate protection, including an ordinary overload from too many loads.
• B: Repeated restoration without evaluating safety is hazardous.
• C (correct): An opened breaker provides a clue without uniquely identifying its cause.
• D: The breaker position alone cannot determine this specific failure.

Answer: A

Correct answer: The lock physically holds the switch off, while a tag can only warn. A tag is a warning sign. A lockout device physically holds the energy-isolating device in the off position, usually with a padlock the worker controls, so no one can re-energize the equipment while it is being serviced. A full program also isolates every source, releases stored energy and verifies the result.

• A (correct): A tag communicates but cannot stop anyone from restoring power; a lock physically prevents it.
• B: Releasing stored energy is a separate step; a padlock on a switch does nothing to capacitors.
• C: Verification is a separate step done by qualified people with rated test equipment.
• D: A lock changes nothing electrically; it only keeps the isolating device in the off position.

Answer: C

Correct answer: The shorted lamp goes dark, and the other lamp glows brighter. A short across a load gives current an easier path around it, so that lamp goes dark. Total circuit resistance drops to one lamp's resistance, so current rises and the remaining lamp now has the full battery voltage and glows brighter.

• A: The other lamp is still in a complete loop with the battery, so current still flows through it.
• B: This reverses the effect; the bypassed lamp receives almost no current.
• C (correct): Current takes the near-zero path around the shorted lamp, and with less total resistance, more current flows through the remaining lamp.
• D: The shorted lamp is bypassed, so almost no current flows through it.

Answer: C

Correct answer: 80 mA, which is far too small to trip the 15 A breaker. Use Ohm's law.

A few tens of milliamperes through the body can be dangerous, yet 0.08 A is nowhere near 15 A, so the breaker would not trip. Shock protection comes from a GFCI and sound insulation, not from the breaker.

• A: This is a decimal slip; the current is 80 mA, and tens of milliamperes can be dangerous.
• B: This divides 1,500 by 120, treating the resistance as if it were a power in watts.
• C (correct): I = 120/1,500 = 0.08 A, which can be dangerous to a person but is far below the breaker's trip level.
• D: Eighty milliamperes is 0.08 A, a tiny fraction of 15 A, so the breaker sees no overcurrent.

Answer: D

Correct answer: An open in the L2 branch. Eliminate causes that conflict with the evidence. The battery and L1 must work because current reaches L3. A short across L2 would also short L3, which is still lit. Only an open in the L2 branch, such as a broken filament or lead, explains all three observations.

• A: An open in the main line would stop all current, so L1 and L3 could not be lit.
• B: A short across L2 would also bypass L3, which sits across the same two points, so L3 would go dark too.
• C: A dead battery would leave every lamp dark, but two lamps are lit.
• D (correct): An open confined to the L2 branch stops only that lamp, while L1 and L3 still have a complete path.

Answer: D

Correct answer: 6.25 times. With R fixed, heat is proportional to $I^2$.

So the wire produces 6.25 times as much heat.

• A: This subtracts the currents, but heat depends on the ratio of currents squared, not the difference.
• B: This squares only the new current and forgets to compare it with the old current squared, which is 16.
• C: This uses the current ratio directly; heating grows with the square of the current.
• D (correct): Heat follows I squared R, so the factor is (10/4) squared = 2.5 squared = 6.25.

Answer: A

Correct answer: 0.03 A (30 mA). A differential detector, like the sensing coil in a GFCI, responds only to the difference between outgoing and returning current, because matched currents cancel each other's magnetic effect. Here

and multiplying by 1,000 converts it to 30 mA. That missing current must be leaving through some other path, such as ground. Reporting the full 4 A load or adding the two currents misses the point: the detector ignores the size of the load and looks only for the mismatch.

• A (correct): The imbalance magnitude is 4.00 − 3.97 = 0.03 A, which equals 30 mA.
• B: The subtraction gives three hundredths, not three thousandths.
• C: This reports supply current instead of the mismatch.
• D: This adds opposite-path magnitudes instead of finding the discrepancy.

Answer: C

Correct answer: The second stores nine times as much. Because energy is E = ½CV², energy grows with the square of voltage. With C the same for both, only the voltage changes, so the energy ratio is the voltage ratio squared:

The tempting answer is three times, which assumes energy grows in direct proportion to voltage. A quick check: doubling the voltage would quadruple the energy. This square law is why high-voltage capacitors are especially dangerous.

• A: Equal capacitance does not imply equal energy at different voltages.
• B: Increasing voltage raises energy in this model.
• C (correct): E = 1/2 CV² and equal C mean energy ratio is the voltage ratio squared: 3² = 9.
• D: This ignores the square on voltage.

Answer: B

Correct answer: 0 V across the 4 ohm resistor and 24 V across the gap. Before the break, current is

A and the 4 $\Omega$ resistor drops 8 V. After the break, current is 0 A, so

across every resistor. Kirchhoff's voltage law then puts the whole 24 V across the open gap.

• A: Zero current gives zero drop across the resistor, but the source voltage still appears across the gap.
• B (correct): With no current, every resistor drops 0 V, so the full 24 V appears across the open gap.
• C: Eight volts is the drop only while 2 A flows; after the break there is no current, so the drop vanishes.
• D: This keeps the old drop and assumes an open has no voltage, but both are wrong once current stops.

Answer: D

Correct answer: 18,000 J. A watt is one joule per second, so energy equals power times time measured in seconds. Convert first: 5 minutes × 60 s per minute = 300 s. Then 60 J/s × 300 s = 18,000 J. The key step people miss is the unit conversion; multiplying 60 by 5 minutes gives 300, a number with the wrong units. For larger amounts, utilities use kilowatt-hours instead: 1 kWh equals 3,600,000 J.

• A: This treats the time alone as energy and omits 60 watts.
• B: This divides 60 by five without converting time to seconds.
• C: This assumes one minute rather than five minutes.
• D (correct): Five minutes is 300 seconds, so E = P t = 60 J/s × 300 s = 18,000 J.

Answer: D

Correct answer: It rises from 1.5 A to 4.5 A. An ideal source holds its voltage steady, so each parallel branch sees the full 12 V and draws current independently. The 8 Ω branch keeps drawing

and the new 4 Ω branch adds

for a total of 4.5 A. You can check with equivalent resistance: 8 × 4 ÷ (8 + 4) ≈ 2.67 Ω, and 12 ÷ 2.67 ≈ 4.5 A. Adding parallel branches always lowers total resistance, which is the common trap.

• A: The additional branch draws its own current at the same ideal source voltage, so the total must rise.
• B: Adding a parallel branch reduces equivalent resistance and increases source current.
• C: No branch draws 12 A given the stated resistances and voltage; the total is 1.5 A plus 3 A.
• D (correct): Initially I = 12/8 = 1.5 A; the new 4 Ω branch draws 3 A, making total 4.5 A.

Answer: B

Correct answer: 37.6 J. Convert first: 470 $\mu$F = 0.00047 F. Then use

The capacitor holds this energy even with no supply connected.

• A: This forgets to square the voltage; stored energy grows with V squared.
• B (correct): With C = 0.00047 F, one-half times 0.00047 times 400 squared gives 37.6 J.
• C: This leaves out the factor of one-half in the energy formula.
• D: This uses 470 as if it were in farads; microfarads must be converted to 0.00047 F first.

Answer: D

Correct answer: The unit may still hold energy; qualified procedures must confirm it is de-energized. Pulling the plug removes only one energy source. A backup battery is a second source that can keep circuits live, and a capacitor stores charge that may remain for minutes or longer. That is why lockout/tagout procedures require isolating every source, discharging stored energy, and then testing with a verified meter before touching anything. Assuming unplugged means safe is one of the most common causes of shock during repair work.

• A: Capacitor discharge depends on the circuitry; many capacitors hold charge for a long time.
• B: The battery and charged capacitor contradict this assertion.
• C: A battery is another source of electrical energy, not a safety guarantee.
• D (correct): Unplugging one input does not remove all stored and alternative electrical sources.

Answer: C

Correct answer: In parallel, across the two ends of the resistor. A voltmeter measures the potential difference between two points, so it is connected in parallel, across the component. Its very high resistance lets it do this while drawing almost no current.

• A: Series connection is for an ammeter measuring current; a high-resistance voltmeter in series would nearly stop the current.
• B: A voltmeter needs two points to compare; one lead alone measures no potential difference.
• C (correct): Voltage is a difference between two points, so the meter's leads go on either side of the resistor.
• D: This is again a series connection, which suits current measurement, not voltage across a part.

Answer: B

Correct answer: Live parts are enclosed by two independent layers of insulation. Double-insulated equipment surrounds live parts with two independent layers of insulation (or one reinforced layer), so if one layer fails the other still protects the user. That is why it can use a two-prong plug, but damage still requires removing the tool from service.

• A: Removing a grounding pin defeats a fault path; a double-insulated tool was designed without one.
• B (correct): The square-in-square mark identifies a double-insulated design, which does not rely on a grounding conductor.
• C: The mark refers to insulation design, not to a device that compares outgoing and returning current.
• D: Physical damage can defeat both layers of insulation, so a cracked tool should be removed from service.

Answer: D

Correct answer: About 600 A. Ohm's law gives

Even a low-voltage battery can drive a huge short-circuit current, which can heat a tool or ring red-hot and cause severe burns.

• A: 2 A is the lamp's normal current; the wrench is a far lower resistance than the lamp.
• B: This multiplies voltage by resistance instead of dividing.
• C: This computes $V^2/R$, which is power in watts, not current.
• D (correct): $I=V/R=12/0.02=600$ A, hundreds of times the lamp's normal 2 A.

Answer: B

Correct answer: The current returns on the neutral, so outgoing and return currents stay equal. A GFCI compares current leaving on the hot conductor with current returning on the neutral. Current that flows from hot through a person to neutral still returns on the neutral, so there is no imbalance to detect. That is a key limit of GFCI protection.

• A: A GFCI does not trip on total current; it reacts to a few milliamperes of imbalance.
• B (correct): A GFCI only senses an imbalance; current through a person from hot to neutral still comes back on the neutral.
• C: Detecting arcing is the job of an AFCI, not a GFCI.
• D: The mat blocks a path to ground, but the hand-to-hand path between hot and neutral is still complete.

Answer: D

Correct answer: Stay far away, keep others back, and call 911 and the utility. A downed line must be treated as energized even when it is quiet. Current can spread through wet ground, fences and puddles, so keep a wide distance, keep others away, and let emergency services and the utility make it safe.

• A: Approaching or touching the line exposes the person to hazardous voltage in the line and the ground around it.
• B: Lack of sparks or noise does not mean a line is dead; getting close is dangerous.
• C: Touching or approaching the line to check it is never a safe test.
• D (correct): A silent line may still be energized, and current can spread through wet ground near it.

Answer: A

Correct answer: 18 V. Total resistance is

kΩ, so the current is 24 V ÷ 8 kΩ = 3 mA. The drop across the 6 kΩ resistor is 3 mA × 6 kΩ = 18 V, and the 2 kΩ resistor gets the other 6 V, adding up to 24 V.

• A (correct): The current is 24/8000 = 3 mA, and 3 mA × 6 kΩ = 18 V.
• B: This is the drop across the 2 kΩ resistor; the larger resistor takes the larger share.
• C: Series drops split in proportion to resistance, not equally between the two parts.
• D: The full source voltage appears across one part only if the loop is open; here both resistors share it.

Answer: C

Correct answer: A high-resistance connection at the outlet. Normal current and a working load rule out an open and a short, and 8 A on a 20 A circuit rules out overload. A poor contact adds resistance in a tiny spot, and

concentrates heat there without raising current enough to trip the breaker.

• A: A short would send a very large current and trip the breaker, and the heater would not work normally.
• B: An open would stop the current, so the heater would not run at all.
• C (correct): A loose or corroded contact heats at that spot while the circuit current stays normal, so the breaker does not trip.
• D: 8 A is well under the 20 A rating, so the circuit is not overloaded.

Answer: A

Correct answer: Capacitor B, 11.25 J. Use

with capacitance in farads.

B has one quarter of the capacitance but nine times the $V^2$, so it stores more.

• A (correct): $0.5\times 250\times 10^{-6}\times 300^2=11.25$ J, more than A's 5 J, because voltage is squared.
• B: A stores 5 J, but its larger capacitance does not outweigh B's three-times-higher voltage, which is squared.
• C: This leaves out the factor of one half in $E=\frac{1}{2}C{V}^2$.
• D: This leaves out the one half and also misses that B's higher voltage gives it more energy.

Answer: C

Correct answer: 3,600 J. Find the heating power with

Energy is power times time in seconds:

• A: This uses I rather than $I^2$, so the power comes out as 3 W instead of 60 W.
• B: 60 is the power in watts; energy also requires multiplying by the 60 s of operation.
• C (correct): Power is $20^2\times 0.15=60$ W, and 60 W for 60 s is 3,600 J.
• D: This multiplies 60 W by 3,600 s, treating one minute as one hour.

Answer: A

Correct answer: An open in L2. L3 lights, so the battery is fine and the problem is in the L1–L2 branch. Both lamps are dark, so that branch carries no current, which points to an open. With no current, the good lamp drops 0 V and the full 12 V appears across the break, so the open is at L2.

• A (correct): With L2 open the series branch carries no current, L1 drops 0 V, and the full 12 V appears across the gap at L2.
• B: If L1 were open, the full 12 V would appear across L1 and 0 V across L2, the reverse of the readings.
• C: A short across L1 would leave L2 lit, and even brighter, with 12 V across it, but L2 is dark.
• D: A dead battery could not keep L3 lit, and it could not put 12 V across L2.