Journeyman Electrician (NEC) — All Questions

A heating element of 30 ohms is connected across 240 volts. What current flows?

• a.8 amperes✓
• b.0.125 ampere
• c.7,200 amperes
• d.12 amperes

Ohm's law gives I = E / R = 240 / 30 = 8 amperes. The 0.125-ampere answer inverts the division, 7,200 multiplies voltage by resistance instead of dividing, and 12 amperes would require a 20-ohm element.

What power is consumed by a 120-volt load drawing 12 amperes at unity power factor?

• a.1,440 watts✓
• b.10 watts
• c.132 watts
• d.1,200 watts

Power in a resistive single-phase circuit is P = E x I = 120 x 12 = 1,440 watts. The 10-watt answer divides instead of multiplying, 132 watts adds the two values, and 1,200 watts does not follow from the given numbers.

How much power is dissipated when 5 amperes flows through a 20-ohm resistor?

• a.100 watts
• b.500 watts✓
• c.4 watts
• d.2,000 watts

Using P = I squared x R, the calculation is 5 x 5 x 20 = 500 watts. The 100-watt answer multiplies current by resistance without squaring the current, 4 watts divides the wrong way, and 2,000 watts squares the resistance instead of the current.

Three resistors of 10, 15 and 25 ohms are connected in series across 120 volts. What is the circuit current?

• a.24 amperes
• b.2.4 amperes✓
• c.4.8 amperes
• d.0.42 ampere

Series resistances add: 10 + 15 + 25 = 50 ohms, so I = 120 / 50 = 2.4 amperes. The 24-ampere answer misplaces the decimal, 4.8 amperes would result from a 25-ohm total, and 0.42 ampere inverts the division.

A 20-ohm resistor and a 30-ohm resistor are connected in parallel. What is the total resistance?

• a.50 ohms
• b.12 ohms✓
• c.25 ohms
• d.10 ohms

For two resistors in parallel, R total = (R1 x R2) / (R1 + R2) = (20 x 30) / 50 = 600 / 50 = 12 ohms. The 50-ohm answer adds them as if they were in series, and 25 ohms averages them, but total parallel resistance is always less than the smallest branch resistance.

A balanced three-phase load draws 40 amperes at 480 volts with a power factor of 0.85. What is the true power?

• a.Approximately 33.3 kilowatts
• b.Approximately 28.3 kilowatts✓
• c.Approximately 19.2 kilowatts
• d.Approximately 16.3 kilowatts

True power for a three-phase load is P = 1.732 x E x I x PF = 1.732 x 480 x 40 x 0.85, or about 28,270 watts. The 33.3-kilowatt answer omits the power factor and gives apparent power in kVA instead, 19.2 kilowatts leaves out the 1.732 factor, and 16.3 kilowatts omits both.

On a 208Y/120-volt wye system, what is the relationship between line voltage and phase voltage?

• a.Line voltage equals phase voltage divided by 1.732
• b.Line voltage equals phase voltage
• c.Line voltage equals phase voltage multiplied by 3
• d.Line voltage equals phase voltage multiplied by 1.732✓

In a wye configuration the line voltage is the square root of three times the phase voltage: 120 x 1.732 = 208 volts, while line current equals phase current. In a delta configuration the relationship reverses, with line voltage equal to phase voltage and line current equal to 1.732 times phase current.

A balanced delta-connected load has a phase current of 20 amperes. What is the line current?

• a.20 amperes
• b.Approximately 34.6 amperes✓
• c.Approximately 11.5 amperes
• d.60 amperes

In a delta connection each line conductor feeds two phase windings, so line current equals 1.732 times phase current: 20 x 1.732 = 34.6 amperes. The 20-ampere answer describes a wye connection, 11.5 amperes divides instead of multiplying, and 60 amperes triples the value rather than applying the square root of three.

A single-phase transformer has a 480-volt primary and a 120-volt secondary. If the secondary carries 100 amperes, what is the approximate primary current?

• a.50 amperes
• b.100 amperes
• c.400 amperes
• d.25 amperes✓

Voltage and current transform inversely, so the 4 to 1 voltage ratio produces a 1 to 4 current ratio: 100 / 4 = 25 amperes on the primary. Neglecting losses, the volt-amperes match on both sides, 480 x 25 = 120 x 100 = 12,000 VA. The 400-ampere answer inverts the ratio and would represent an impossible power gain.

What is the secondary full-load current of a 25 kVA single-phase transformer with a 240-volt secondary?

• a.Approximately 60 amperes
• b.Approximately 104 amperes✓
• c.Approximately 208 amperes
• d.Approximately 52 amperes

For a single-phase transformer, I = VA / E = 25,000 / 240, or about 104 amperes. The 60-ampere answer divides by 480 volts, 208 amperes divides by 120 volts, and 52 amperes is half the correct value.

A 480-volt panelboard is mounted on a wall with a grounded concrete wall directly behind the working space. What is the minimum depth of working space?

• a.2 and 1/2 feet
• b.3 feet
• c.4 feet
• d.3 and 1/2 feet✓

For nominal voltages of 151 to 600 volts to ground, Condition 2 exists when exposed live parts face a grounded surface such as a concrete or masonry wall, and the required depth is 3 and 1/2 feet. Condition 1, with no live or grounded parts opposite, requires 3 feet, and Condition 3, with exposed live parts on both sides, requires 4 feet.2023 NEC §110.26(A)(1)

What is the minimum width of working space in front of electrical equipment?

• a.30 inches in all cases regardless of equipment width
• b.30 inches, or the width of the equipment, whichever is greater✓
• c.36 inches, or the width of the equipment, whichever is greater
• d.24 inches, measured from the centerline of the equipment

The working space must be at least 30 inches wide or as wide as the equipment, whichever is greater, and it must allow all doors and hinged panels to open at least 90 degrees. The space need not be directly centered on the equipment, but it must be continuous across the full width required.2023 NEC §110.26(A)(2)

Entrances at each end of the working space are generally required for equipment rated at least what value and over 6 feet wide?

• a.1,000 amperes
• b.800 amperes
• c.1,200 amperes✓
• d.600 amperes

Equipment rated 1,200 amperes or more and over 6 feet wide containing overcurrent, switching or control devices needs an entrance at each end of the working space so a worker is never trapped behind an arcing fault. The 1,000-ampere threshold belongs to service ground-fault protection, and 800 amperes is the ceiling for the next-standard-size overcurrent rule.2023 NEC §110.26(C)(2)

The dedicated equipment space above a panelboard extends from the top of the equipment to what height?

• a.It extends all the way up to the structural ceiling in every case, with no allowance to stop at any lower height
• b.It extends only 3 feet above the top of the equipment in all cases, regardless of the height of the structural ceiling
• c.6 feet above the equipment or to the structural ceiling, whichever is lower✓
• d.6 and 1/2 feet above the finished floor

The dedicated space is the footprint of the equipment extended from the floor to a height of 6 feet above it or to the structural ceiling, whichever is lower, and no foreign piping, ducts or equipment may be installed in it. Suspended ceilings with removable panels are not considered structural ceilings, and the 6 and 1/2-foot dimension is the minimum headroom of working space, a separate requirement.2023 NEC §110.26(E)(1)

Which sequence correctly describes verifying that equipment is de-energized before beginning work?

• a.Test the meter on a known live source, test the de-energized conductors, then retest the meter on the known live source✓
• b.Test the de-energized conductors, then test the meter on a known live source
• c.Rely on the open position of the disconnect and the applied lock
• d.Test the de-energized conductors only, since the meter was checked at the start of the shift

The live-dead-live method proves the instrument is working both before and after the absence-of-voltage test, so a meter that failed during the check cannot lead a worker to treat energized conductors as safe. Locking and tagging the disconnect is required, but it does not by itself confirm that the conductors ahead of the work are actually de-energized, since backfeeds and mislabeled circuits are common.

As defined in Article 100, ampacity is best described as which of the following?

• a.The maximum current a conductor can carry continuously under the conditions of use without exceeding its temperature rating✓
• b.The current that will instantly melt the conductor
• c.The rated current of the overcurrent device protecting the conductor, taken directly from the marking on the breaker or fuse without regard to installation conditions
• d.The current at which voltage drop reaches 3 percent

Article 100 defines ampacity as the maximum current, in amperes, that a conductor can carry continuously under the conditions of use without exceeding its temperature rating. It is a property of the conductor and its environment, not the rating of the protective device ahead of it.2023 NEC Art. 100

Under the Article 100 definitions, what does it mean for metal parts to be bonded?

• a.Connected to the earth through a ground rod
• b.Insulated from all other metal parts
• c.Connected together to establish electrical continuity and conductivity✓
• d.Painted to prevent corrosion

Bonding means connecting parts together to establish electrical continuity and conductivity, which is what allows fault current to return to the source and open the overcurrent device. Connecting a part to the earth is grounding, a related but distinct concept.2023 NEC Art. 100

Three 30-ohm resistors are connected in parallel. What is the total resistance?

• a.90 ohms
• b.10 ohms✓
• c.30 ohms
• d.15 ohms

For equal resistors in parallel, total resistance equals the value of one resistor divided by the number of resistors: 30 / 3 = 10 ohms. The 90-ohm answer adds them as if in series, and total parallel resistance is always less than the smallest branch.

How much energy does a 1,500-watt electric heater consume when operated continuously for 4 hours?

• a.375 watt-hours
• b.1.5 kilowatt-hours
• c.600 watt-hours
• d.6 kilowatt-hours✓

Energy equals power multiplied by time: 1,500 watts x 4 hours = 6,000 watt-hours, or 6 kilowatt-hours. The 1.5-kilowatt-hour answer reports only the power rating, and 375 watt-hours divides instead of multiplying.

A 40-ohm resistor and a 20-ohm resistor are connected in series across a 120-volt source. What is the voltage drop across the 40-ohm resistor?

• a.80 volts✓
• b.40 volts
• c.60 volts
• d.48 volts

In series the total resistance is 40 + 20 = 60 ohms, so the current is 120 / 60 = 2 amperes. The drop across the 40-ohm resistor is 2 x 40 = 80 volts, and the remaining 40 volts appears across the 20-ohm resistor, confirming the two drops sum to the source voltage.

Article 100 defines a qualified person as one who has which of the following?

• a.A journeyman license issued in any state
• b.Skills and knowledge related to the construction and operation of the electrical equipment and installations, plus safety training to recognize and avoid the hazards involved✓
• c.At least eight thousand hours of documented field experience working under the direct supervision of a licensed master electrician, regardless of whether any formal safety or hazard-recognition training has been completed
• d.A current first-aid and CPR certification

A qualified person is defined by demonstrated skills and knowledge related to the construction and operation of the equipment and installations, together with safety training to recognize and avoid the hazards involved. The definition is competency-based and does not hinge on a specific license, hour count or certificate.2023 NEC Art. 100

A heating element of 10 ohms is connected across 120 volts. What current flows?

• a.1,200 amperes
• b.10 amperes
• c.12 amperes✓
• d.0.083 ampere

Ohm's law gives I = E / R = 120 / 10 = 12 amperes. Inverting the division gives the 0.083-ampere trap, and multiplying gives 1,200.

What power is consumed by a 240-volt resistive load drawing 15 amperes?

• a.16 watts
• b.3,000 watts
• c.255 watts
• d.3,600 watts✓

Power in a resistive single-phase circuit is P = E x I = 240 x 15 = 3,600 watts. Dividing gives 16 and adding gives 255.

How much power is dissipated when 6 amperes flows through a 25-ohm resistor?

• a.150 watts
• b.6.25 watts
• c.3,600 watts
• d.900 watts✓

Using P = I squared x R = 6 x 6 x 25 = 900 watts. Multiplying current by resistance without squaring gives 150 watts.

Three resistors of 5, 10 and 20 ohms are connected in series across 105 volts. What is the circuit current?

• a.0.33 ampere
• b.6 amperes
• c.30 amperes
• d.3 amperes✓

Series resistances add: 5 + 10 + 20 = 35 ohms, so I = 105 / 35 = 3 amperes.

A 10-ohm resistor and a 40-ohm resistor are connected in parallel. What is the total resistance?

• a.4 ohms
• b.8 ohms✓
• c.25 ohms
• d.50 ohms

For two resistors in parallel, R = (R1 x R2) / (R1 + R2) = (10 x 40) / 50 = 400 / 50 = 8 ohms. Total parallel resistance is always less than the smallest branch.

Three 60-ohm resistors are connected in parallel. What is the total resistance?

• a.30 ohms
• b.180 ohms
• c.20 ohms✓
• d.60 ohms

For equal resistors in parallel, total resistance equals one value divided by the number of resistors: 60 / 3 = 20 ohms. Adding them gives the 180-ohm series answer.

A balanced three-phase load draws 30 amperes at 208 volts with a power factor of 0.90. What is the true power?

• a.Approximately 6.2 kilowatts
• b.Approximately 10.8 kilowatts
• c.Approximately 5.6 kilowatts
• d.Approximately 9.7 kilowatts✓

True power = 1.732 x E x I x PF = 1.732 x 208 x 30 x 0.90, about 9,730 watts. Omitting the power factor gives about 10.8 kVA of apparent power.

On a 208Y/120-volt wye system, what is the relationship between line current and phase current?

• a.Line current equals phase current divided by 1.732
• b.Line current equals 3 times phase current
• c.Line current equals 1.732 times phase current
• d.Line current equals phase current✓

In a wye configuration the line current equals the phase current, while the line voltage is 1.732 times the phase voltage. In a delta configuration those relationships reverse.

A single-phase transformer has a 240-volt primary and a 48-volt secondary. If the secondary carries 50 amperes, what is the approximate primary current?

• a.5 amperes
• b.10 amperes✓
• c.50 amperes
• d.250 amperes

Voltage and current transform inversely and the volt-amperes match: 240 x I = 48 x 50, so I = 2,400 / 240 = 10 amperes. Inverting the ratio gives the 250-ampere trap.

What is the secondary full-load current of a 15 kVA single-phase transformer with a 240-volt secondary?

• a.62.5 amperes✓
• b.125 amperes
• c.31 amperes
• d.104 amperes

For a single-phase transformer, I = VA / E = 15,000 / 240 = 62.5 amperes.

How much energy does a 2,000-watt load consume when operated continuously for 3 hours?

• a.666 watt-hours
• b.2 kilowatt-hours
• c.6 kilowatt-hours✓
• d.0.67 kilowatt-hour

Energy equals power multiplied by time: 2,000 watts x 3 hours = 6,000 watt-hours, or 6 kilowatt-hours.

A 30-ohm resistor and a 60-ohm resistor are connected in series across a 90-volt source. What is the voltage drop across the 60-ohm resistor?

• a.60 volts✓
• b.30 volts
• c.90 volts
• d.45 volts

Total resistance is 30 + 60 = 90 ohms, so the current is 90 / 90 = 1 ampere. The drop across the 60-ohm resistor is 1 x 60 = 60 volts, and 30 volts appears across the 30-ohm resistor.

A 480-volt panelboard is mounted so there are no live or grounded parts opposite the working space. What is the minimum required depth of working space?

• a.3 feet✓
• b.3 and 1/2 feet
• c.2 and 1/2 feet
• d.4 feet

For 151 to 600 volts to ground, Condition 1 (no live or grounded parts opposite) requires 3 feet of depth. Condition 2 requires 3 and 1/2 feet and Condition 3 requires 4 feet. See NEC 110.26(A)(1).

A 480-volt panelboard has exposed energized parts on both sides of the working space. What is the minimum required depth of working space?

• a.3 feet
• b.5 feet
• c.4 feet✓
• d.3 and 1/2 feet

For 151 to 600 volts to ground, Condition 3, with exposed live parts on both sides of the working space, requires a depth of 4 feet. See NEC 110.26(A)(1).

What is the minimum headroom of the working space about electrical equipment such as panelboards?

• a.6 and 1/2 feet✓
• b.3 feet
• c.7 feet
• d.6 feet

The minimum headroom of working space is 6 and 1/2 feet, or the height of the equipment if taller. See NEC 110.26(A)(3).

Working spaces about service equipment, switchboards, panelboards or motor control centers installed indoors must be provided with what?

• a.A dedicated telephone
• b.Illumination✓
• c.A portable fire extinguisher
• d.A rubber grounding mat

Indoor working spaces about service equipment, switchboards, panelboards and motor control centers must have illumination. See NEC 110.26(D).

Under the Article 100 definitions, what does it mean to connect equipment to the earth?

• a.Grounding✓
• b.Insulating
• c.Bonding
• d.Derating

Grounding is the connection to earth or to a conductive body that extends the earth connection, while bonding is the connection of parts together for electrical continuity. See NEC Article 100.

Article 100 defines a continuous load as one where the maximum current is expected to continue for how long?

• a.More than 24 hours
• b.Any duration over 100 amperes
• c.Only at night
• d.3 hours or more✓

A continuous load is one where the maximum current is expected to continue for 3 hours or more, which is why such loads are sized at 125 percent. See NEC Article 100.

What is the primary purpose of a lockout/tagout procedure?

• a.To improve the power factor of the circuit
• b.To increase the available fault current
• c.To reduce conductor voltage drop
• d.To prevent the unexpected energization or startup of equipment during servicing✓

Lockout/tagout keeps a disconnecting means in the open position so equipment cannot be unexpectedly energized or started while a worker is servicing it. This reflects OSHA 1910.147 and NFPA 70E safe-work practice.

A Class A ground-fault circuit interrupter is designed to trip when the difference between the ungrounded and grounded conductor currents reaches approximately what value?

• a.100 milliamperes
• b.20 amperes
• c.4 to 6 milliamperes✓
• d.1 ampere

A Class A GFCI trips when the ground-fault (imbalance) current reaches 4 to 6 milliamperes, the level intended to protect people from electrocution, not to protect the circuit conductors.