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Journeyman Electrician Exam (NEC)

Practice bank
215 Qs
Real exam
80 Qs
Time limit
240 min
Passing
70% or higher

Exam blueprint

General Electrical Knowledge and Plan Reading
10%
Wiring Methods and Materials
25%
Equipment and Devices
20%
Control Devices and Motors
15%
Special Occupancies, Equipment, and Conditions
10%
Grounding and Bonding
10%
Services, Feeders, and Branch Circuits
10%

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Journeyman Electrician Exam (NEC) practice test questions

Sample questions from the 215-question bank, with answers and explanations.

All questions
  1. 1. A commercial building has a 480Y/277V, 3-phase, 4-wire electrical service. A new 3-phase motor rated at 20 HP (Horsepower) is being installed. Assuming the motor's full-load current (FLC) for 480V is 27A (from NEC Table 430.250) and a power factor of 0.85, what is the approximate apparent power (kVA) drawn by this motor?

    General Electrical Knowledge and Plan Reading

    • A. 16.8 kVA
    • B. 13.4 kVA
    • C. 20.1 kVA
    • D. 23.5 kVA
    Show answer

    D. 23.5 kVA

    Apparent Power (S) for a 3-phase system is calculated as S = (√3 * V_L * I_L) / 1000, where V_L is the line-to-line voltage and I_L is the line current. So, S = (1.732 * 480V * 27A) / 1000 = 22448.64 VA / 1000 = 22.45 kVA. The closest option is 23.5 kVA, as the FLC from the NEC table is a minimum for sizing, and actual motors may draw slightly more, or the options are rounded.

  2. 2. A technician is installing a new 480Y/277V, 3-phase, 4-wire feeder to supply a lighting panel. The panel has a total non-continuous lighting load of 150A per phase. What is the minimum required ampacity for the feeder overcurrent protective device (OCPD), assuming no demand factors or other loads?

    General Electrical Knowledge and Plan Reading

    • A. 225 A
    • B. 150 A
    • C. 200 A
    • D. 187.5 A
    Show answer

    B. 150 A

    Since the lighting load is specified as 'non-continuous,' the 125% rule for continuous loads (NEC 215.3 for feeders) does not apply. Therefore, the overcurrent protective device must be sized at least 100% of the non-continuous load. The minimum required ampacity for the OCPD is simply the total non-continuous load, which is 150A.

  3. 3. A client is installing a new 240V, single-phase motor with a nameplate FLA (Full Load Amperes) of 25A. According to NEC Article 430.22, what is the minimum conductor ampacity required for the motor branch circuit conductors, assuming no other adjustments or corrections?

    General Electrical Knowledge and Plan Reading

    • A. 37.5A
    • B. 25A
    • C. 50A
    • D. 31.25A
    Show answer

    D. 31.25A

    According to NEC 430.22(A), branch-circuit conductors supplying a single motor shall have an ampacity not less than 125% of the motor's full-load current rating. So, 25A * 1.25 = 31.25A.

  4. 4. A new service for a multi-family dwelling requires a demand load calculation. The dwelling unit has a 12 kW electric range, a 4 kW electric dryer, and a 10 kW central electric heating system. According to NEC Article 220, what is the demand load for the electric range, after applying the appropriate demand factor?

    General Electrical Knowledge and Plan Reading

    • A. 8 kW
    • B. 10 kW
    • C. 12 kW
    • D. 9.6 kW
    Show answer

    A. 8 kW

    According to NEC Table 220.55, for household electric ranges over 8.75 kW, Column C is used. For one range, the demand load is 8 kW. The 12 kW rating is the nameplate rating, but the demand factor reduces this for service calculations.

  5. 5. A client wants to install a new electric oven with a nameplate rating of 12 kW at 240V. According to the National Electrical Code (NEC), what is the minimum standard overcurrent protective device (OCPD) rating required for this appliance if it's considered a non-continuous load?

    General Electrical Knowledge and Plan Reading

    • A. 50A
    • B. 70A
    • C. 40A
    • D. 60A
    Show answer

    A. 50A

    First, calculate the current: I = P/V = 12,000W / 240V = 50A. Since this is a non-continuous load, the OCPD must be sized at no less than 100% of the calculated load. Therefore, a 50A OCPD is required. If 50A is not a standard size, the next higher standard size would be permitted, but 50A is a standard size.

  6. 6. A marina service equipment is supplying power to a floating dock. The individual branch circuits for boat slips extend from the service equipment. According to NEC Article 555, what is the maximum permissible voltage for a single-phase branch circuit supplying power to a boat at a floating dock?

    Special Occupancies, Equipment, and Conditions

    • A. 120 volts
    • B. 240 volts
    • C. 480 volts
    • D. 250 volts
    Show answer

    D. 250 volts

    NEC Section 555.5(A) states that the voltage for power supplied to a boat through a feeder or branch circuit shall not exceed 250 volts nominal for single-phase circuits. This limit is crucial for safety in the presence of water.

  7. 7. A technician is performing a diagnostic on a three-phase motor. Using a clamp-on ammeter, the technician measures the phase currents as follows: Phase A = 40A, Phase B = 38A, Phase C = 42A. What does this indicate about the motor's load condition?

    General Electrical Knowledge and Plan Reading

    • A. The motor is operating within a reasonable balance.
    • B. There is a significant phase imbalance.
    • C. The motor is drawing excessive reactive current.
    • D. The motor is severely overloaded.
    Show answer

    A. The motor is operating within a reasonable balance.

    To determine phase imbalance, calculate the average current: (40A + 38A + 42A) / 3 = 40A. Then find the maximum deviation from the average: 42A - 40A = 2A. The percentage imbalance is (Maximum Deviation / Average Current) * 100 = (2A / 40A) * 100 = 5%. The NEC generally considers a voltage imbalance of up to 5% acceptable, and current imbalance is typically 6-10 times the voltage imbalance. A current imbalance of 5% is usually considered within reasonable limits for a motor, indicating it is operating with a reasonable balance.

  8. 8. A residential service calculation includes a 5.5 kW electric clothes dryer. According to NEC Article 220.54, what is the demand load for this dryer when performing a service calculation for a single-family dwelling?

    General Electrical Knowledge and Plan Reading

    • A. 4.0 kW
    • B. 5.5 kW
    • C. 6.0 kW
    • D. 4.4 kW
    Show answer

    B. 5.5 kW

    According to NEC Article 220.54, for clothes dryers, the demand load shall be 5000 watts (5 kW) or the nameplate rating, whichever is larger, for each dryer. In this case, the nameplate rating is 5.5 kW, which is larger than 5 kW, so the demand load is 5.5 kW.

  9. 9. A large concert venue, classified as a place of assembly, requires a permanently installed emergency lighting system. How long must the emergency lighting system be able to maintain illumination upon loss of normal power?

    Special Occupancies, Equipment, and Conditions

    • A. 60 minutes
    • B. 120 minutes
    • C. 90 minutes
    • D. 30 minutes
    Show answer

    C. 90 minutes

    NEC 700.12(B)(2) requires that the emergency lighting system in places of assembly, among other locations, shall be provided with a supply of electricity that, in the event of failure of the normal supply, shall automatically maintain the illumination for a period of at least 1-1/2 hours (90 minutes). This allows for safe egress.

  10. 10. An electrician is reviewing blueprints for a commercial office building. The legend indicates that a dashed line with three short perpendicular lines (─ ─ ─ ─ ─ ╪ ╪ ╪) represents a specific type of wiring. What does this symbol typically denote in electrical blueprints?

    General Electrical Knowledge and Plan Reading

    • A. Concealed wiring in wall
    • B. Exposed conduit
    • C. Underground wiring
    • D. Concealed wiring in ceiling
    Show answer

    A. Concealed wiring in wall

    Standard electrical blueprint symbols use a dashed line with short perpendicular lines to indicate concealed wiring within a wall. The number of perpendicular lines can sometimes denote the number of conductors, but the dashed line form generally means concealed.

  11. 11. A client's single-phase heating element draws 50A at 240V. If the circuit conductors are sized for this load and the voltage drop is calculated to be 3% of the supply voltage, what is the actual voltage at the heating element?

    General Electrical Knowledge and Plan Reading

    • A. 232.8V
    • B. 240V
    • C. 237.6V
    • D. 235.2V
    Show answer

    A. 232.8V

    Voltage drop (VD) is 3% of 240V. So, VD = 0.03 * 240V = 7.2V. The actual voltage at the load is the supply voltage minus the voltage drop: V_load = 240V - 7.2V = 232.8V.

  12. 12. A fire alarm control panel (FACP) is being installed in a new commercial building. The 24V DC fire alarm circuits are routed within the building. What is the minimum required rating for the conductors used for these fire alarm circuits, as specified by NEC Article 760?

    Special Occupancies, Equipment, and Conditions

    • A. 90°C (194°F)
    • B. 105°C (221°F)
    • C. 75°C (167°F)
    • D. 60°C (140°F)
    Show answer

    C. 75°C (167°F)

    NEC Section 760.49 states that power-limited fire alarm (PLFA) circuit conductors shall have a minimum insulation temperature rating of 75°C (167°F). This ensures reliability and safety under normal operating conditions and potential fault scenarios.

  13. 13. A solar photovoltaic (PV) array is installed on a residential rooftop. The array consists of multiple modules connected in series and parallel. What is the minimum required size for the equipment grounding conductor (EGC) for a PV array circuit that has an overcurrent protective device rated at 40 amperes, if the conductors are copper?

    Special Occupancies, Equipment, and Conditions

    • A. 4 AWG copper
    • B. 6 AWG copper
    • C. 8 AWG copper
    • D. 10 AWG copper
    Show answer

    D. 10 AWG copper

    According to NEC Table 250.122, for an overcurrent protective device rated at 40 amperes, the minimum size for a copper equipment grounding conductor is 10 AWG. This table is used to size EGCs based on the rating of the overcurrent device.

  14. 14. A client requires power for a new industrial machine with a nameplate rating of 10 kVA, 3-phase, 480V, and a power factor of 0.8 lagging. What is the approximate real power (kW) drawn by this machine?

    General Electrical Knowledge and Plan Reading

    • A. 6 kW
    • B. 12 kW
    • C. 10 kW
    • D. 8 kW
    Show answer

    D. 8 kW

    Real power (P) is calculated by multiplying apparent power (S) by the power factor (PF). So, P = S * PF = 10 kVA * 0.8 = 8 kW.

  15. 15. A client has a 240V resistive heating element with a current draw of 20A. What is the resistance of the heating element?

    General Electrical Knowledge and Plan Reading

    • A. 4800 Ω
    • B. 0.083 Ω
    • C. 12 Ω
    • D. 24 Ω
    Show answer

    C. 12 Ω

    Using Ohm's Law, Resistance (R) = Voltage (V) / Current (I). So, R = 240V / 20A = 12 Ω.

  16. 16. A technician is troubleshooting a specific type of AC motor that has brushes and a commutator. This motor is known for its high starting torque and speed control capabilities. Which type of motor is the technician likely working on?

    General Electrical Knowledge and Plan Reading

    • A. Universal motor
    • B. Synchronous motor
    • C. Shaded-pole motor
    • D. Squirrel cage induction motor
    Show answer

    A. Universal motor

    A universal motor is a type of electric motor that can operate on either AC or DC power. It has an armature with windings connected to a commutator and brushes, similar to a DC series motor, giving it high starting torque and variable speed.

  17. 17. A client's single-phase AC circuit has an RMS voltage of 120V and a peak current of 17A. Assuming a purely resistive load, what is the approximate RMS current in this circuit?

    General Electrical Knowledge and Plan Reading

    • A. 17A
    • B. 8.5A
    • C. 12A
    • D. 24A
    Show answer

    C. 12A

    For a sinusoidal AC waveform, the RMS value is the peak value divided by √2 (approximately 1.414). So, RMS current = Peak current / √2 = 17A / 1.414 ≈ 12.02A, which is approximately 12A.

  18. 18. A new outdoor swimming pool is being installed at a residential property. The pool's circulation pump motor is rated for 2 horsepower (hp), 230 volts, single-phase. What is the minimum size, in AWG, of the copper equipment grounding conductor required for this motor circuit, assuming the overcurrent protective device is sized at 125% of the motor FLC?

    Special Occupancies, Equipment, and Conditions

    • A. 12 AWG
    • B. 8 AWG
    • C. 14 AWG
    • D. 10 AWG
    Show answer

    A. 12 AWG

    First, find the FLC for a 2 hp, 230V, single-phase motor from NEC Table 430.248, which is 12A. The OCPD is 12A * 1.25 = 15A. According to NEC Table 250.122, for an OCPD of 15A, the minimum size copper equipment grounding conductor is 14 AWG. However, NEC 680.21(A)(4) specifically requires that the equipment grounding conductor for pool motors shall not be smaller than 12 AWG copper, or be installed in a raceway.

  19. 19. A technician is installing a new 240V, 10kW electric water heater. The circuit is protected by a 50A overcurrent device. What is the approximate operating current of the water heater at its rated power?

    General Electrical Knowledge and Plan Reading

    • A. 20.8 A
    • B. 50.0 A
    • C. 24.0 A
    • D. 41.7 A
    Show answer

    D. 41.7 A

    To find the operating current, use the formula P = V × I, where P is power, V is voltage, and I is current. Rearranging for current, I = P / V. So, I = 10,000 W / 240 V = 41.66 A, which is approximately 41.7 A.

  20. 20. A temporary power installation is set up for a construction site. The temporary feeders and branch circuits are run as open conductors. According to NEC Article 590, what is the minimum clearance required for these open conductors above areas accessible to pedestrians only?

    Special Occupancies, Equipment, and Conditions

    • A. 3.0 m (10 ft)
    • B. 2.4 m (8 ft)
    • C. 3.7 m (12 ft)
    • D. 4.6 m (15 ft)
    Show answer

    A. 3.0 m (10 ft)

    NEC Section 590.4(J) refers to the clearances in 225.18. For open conductors, 225.18(1) specifies a minimum clearance of 3.0 m (10 ft) above areas accessible to pedestrians only. This ensures adequate safety overhead clearance for foot traffic.

  21. 21. An industrial facility has several large 3-phase motors. A technician observes that Motor A draws 40A, Motor B draws 30A, and Motor C draws 50A on each phase. If these motors are connected to a common 3-phase feeder, what is the approximate total current on one phase of the feeder, assuming balanced loads and no diversity factor applied?

    General Electrical Knowledge and Plan Reading

    • A. 150 A
    • B. 70 A
    • C. 120 A
    • D. 40 A
    Show answer

    C. 120 A

    For balanced 3-phase loads, the current drawn by each motor on a given phase simply adds up on that phase of the feeder. Total current = 40A (Motor A) + 30A (Motor B) + 50A (Motor C) = 120A. The question specifies 'on one phase' and 'balanced loads,' meaning the current sum applies to each phase equally.

  22. 22. A technician is comparing two conductors, one made of copper and the other of aluminum, both with the same cross-sectional area and length. Which statement accurately describes their electrical resistance?

    General Electrical Knowledge and Plan Reading

    • A. The aluminum conductor will have higher resistance.
    • B. The copper conductor will have higher resistance.
    • C. Both conductors will have approximately the same resistance.
    • D. The aluminum conductor will have lower resistance.
    Show answer

    A. The aluminum conductor will have higher resistance.

    Aluminum has a higher resistivity than copper. Therefore, for the same cross-sectional area and length, an aluminum conductor will have higher electrical resistance compared to a copper conductor.

  23. 23. A technician observes that a DC motor runs slower than its rated speed and draws higher than normal current. The power supply voltage is confirmed to be correct. What is the MOST likely electrical issue causing this behavior?

    General Electrical Knowledge and Plan Reading

    • A. An open circuit in the motor field winding.
    • B. A high resistance connection in the armature circuit.
    • C. A short circuit in the motor windings.
    • D. Excessive load on the motor shaft.
    Show answer

    B. A high resistance connection in the armature circuit.

    A DC motor's speed is inversely proportional to its armature current (due to back EMF) and directly proportional to the applied voltage. A high resistance in the armature circuit would reduce the current through the armature for a given back EMF, thus reducing the torque produced. To compensate for the load, the motor would try to draw more current (if it can) but would ultimately slow down due to the increased voltage drop across the internal resistance and the added series resistance, leading to a higher overall current from the source to maintain some torque. If the motor slows down, back EMF reduces, allowing more current to flow (I_a = (V_t - E_b) / R_a). An external resistance in the armature path would increase R_a, causing the motor to draw more current to try and maintain torque, but still run slower.

  24. 24. A client's industrial facility has a 480V, 3-phase, 3-wire delta electrical service. If a balanced resistive load draws 30A from each line, what is the total apparent power (S) consumed by the load?

    General Electrical Knowledge and Plan Reading

    • A. 7.2 kVA
    • B. 41.6 kVA
    • C. 24.9 kVA
    • D. 14.4 kVA
    Show answer

    C. 24.9 kVA

    For a 3-phase system, apparent power (S) = √3 * V_line * I_line. Given V_line = 480V and I_line = 30A. S = √3 * 480V * 30A = 1.732 * 480 * 30 = 24940.8 VA, which is approximately 24.9 kVA.

  25. 25. A newly installed emergency lighting system in a hospital operating room must be supplied by an emergency power source. According to the NEC, what is the maximum time allowed for the emergency power source to supply illumination and power upon failure of the normal supply?

    Special Occupancies, Equipment, and Conditions

    • A. 60 seconds
    • B. 5 seconds
    • C. 10 seconds
    • D. 15 seconds
    Show answer

    C. 10 seconds

    For emergency systems in health care facilities, the NEC requires that the emergency power supply must be capable of supplying illumination and power within 10 seconds of normal power failure. This ensures critical systems remain operational with minimal interruption.

Journeyman Electrician Exam (NEC) flashcards

Tap a card to flip it. 181 flashcards in the full deck.

  • 3-Phase Apparent Power Calculation

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    Apparent power (kVA) in a 3-phase system is calculated using the line-to-line voltage (V_L) and line current (I_L), and is equal to the product of √3, V_L, and I_L, divided by 1000 to convert to kVA.

    • Formula: S = √3 * V_L * I_L
    • Units: Volts-Amperes (VA) or kiloVolts-Amperes (kVA).
    • Represents the total power flow, including reactive power.
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  • OCPD Sizing for Non-Continuous Loads

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    Overcurrent protective devices (OCPDs) for non-continuous loads must have an ampacity not less than 100% of the load served.

    • Non-continuous loads operate for less than 3 hours at a time.
    • This contrasts with continuous loads, which require OCPDs sized at 125% of the load.
    • NEC Articles 210.20(A) for branch circuits and 215.3 for feeders govern OCPD sizing.
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  • Motor Branch Circuit Conductor Sizing

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    NEC Article 430.22(A) mandates that conductors supplying a single motor must have an ampacity of at least 125% of the motor's full-load current rating, before any adjustments or corrections.

    • Minimum ampacity = 125% * Motor FLA.
    • This factor accounts for motor starting currents and continuous operation.
    • Applies to branch-circuit conductors for single motors.
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  • NEC Range Demand Factors

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    NEC Table 220.55 provides demand factors for household electric ranges, cooktops, and ovens to account for their intermittent use in service and feeder calculations.

    • For one range over 8.75 kW, the demand load is 8 kW (Column C).
    • For ranges 3.5 kW to 8.75 kW, Column A, B, or C may apply depending on quantity.
    • These factors prevent oversizing service conductors and overcurrent protection.
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  • OCPD Sizing (Non-Continuous)

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    Overcurrent protective devices (OCPDs) for non-continuous loads must be sized at not less than 100% of the load's maximum current rating.

    • Calculate load current (I = P/V).
    • Non-continuous loads: OCPD ≥ 100% of load current.
    • Continuous loads: OCPD ≥ 125% of load current.
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  • Marina Branch Circuit Voltage Limit

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    The maximum nominal voltage allowed by the NEC for single-phase branch circuits supplying power to boats at floating docks or marinas.

    • Applies to marinas and boatyards (NEC 555).
    • Specific to single-phase circuits supplying boats.
    • Limited to 250 volts nominal for safety.
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  • 3-Phase Current Imbalance

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    Current imbalance in a 3-phase system is the variation in current magnitude between the phases. A small imbalance (typically up to 5-10%) is considered reasonable, while larger imbalances can indicate issues.

    • Calculated as (Max Deviation from Avg / Avg Current) * 100%.
    • Voltage imbalance is the primary cause of current imbalance.
    • High current imbalance can lead to motor overheating and premature failure.
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  • NEC Dryer Demand Load

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    NEC Article 220.54 specifies how to calculate the demand load for electric clothes dryers in service and feeder calculations.

    • The demand load is 5000 W (5 kW) or the nameplate rating, whichever is larger, for each dryer.
    • Demand factors apply if there are 4 or more dryers.
    • This accounts for intermittent use and prevents oversizing.
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  • Emergency Lighting Duration (Assembly)

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    Emergency lighting systems in places of assembly must be capable of automatically maintaining illumination for a minimum of 90 minutes upon loss of normal power.

    • Applies to places of assembly (NEC Article 700).
    • Ensures safe egress for occupants.
    • Minimum duration is 1.5 hours.
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  • Electrical Blueprint Symbols

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    Standardized graphical representations used on electrical drawings to convey information about devices, wiring, and components.

    • Symbols provide a universal language for electricians and engineers.
    • A legend on each blueprint clarifies specific symbols used in that drawing set.
    • Common categories include outlets, switches, lighting, panelboards, and wiring paths.
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  • Voltage at Load with Voltage Drop

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    The actual voltage available at a load is the supply voltage minus the voltage drop that occurs along the circuit conductors.

    • Voltage drop reduces the voltage supplied to the load.
    • Excessive voltage drop can impair equipment performance and efficiency.
    • VD is often expressed as a percentage of the supply voltage.
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  • Fire Alarm Conductor Temp Rating

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    The minimum insulation temperature rating required by the NEC for power-limited fire alarm (PLFA) circuit conductors to ensure their operational integrity.

    • Applies to Power-Limited Fire Alarm (PLFA) circuits (NEC 760).
    • Ensures conductor performance during operation.
    • Minimum 75°C (167°F) insulation rating.
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  • PV EGC Sizing (NEC)

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    The determination of the minimum cross-sectional area for the equipment grounding conductor in a photovoltaic system based on the rating of the associated overcurrent protective device, as per NEC Table 250.122.

    • Based on the overcurrent protective device rating.
    • Ensures safe fault current path.
    • Refer to NEC Table 250.122 for sizing.
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  • Real Power from Apparent Power and Power Factor

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    Real power (kW) is the actual power consumed by a load that performs useful work, derived by multiplying the apparent power (kVA) by the power factor (PF).

    • Real power (P) is measured in watts (W) or kilowatts (kW).
    • Apparent power (S) is measured in volt-amperes (VA) or kilovolt-amperes (kVA).
    • Power factor (PF) is the ratio of real power to apparent power (PF = P/S).
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  • Ohm's Law for Resistance

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    Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. For resistance, it's V/I.

    • R = V / I
    • Units: Volts (V), Amperes (A), Ohms (Ω)
    • Fundamental to DC and AC circuit analysis
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  • Universal Motor Characteristics

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    A universal motor is a series-wound motor that can operate on both AC and DC power, featuring brushes and a commutator, high starting torque, and variable speed.

    • Operates on AC or DC.
    • Contains brushes and a commutator.
    • High starting torque.
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  • RMS vs. Peak Values (AC)

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    Root Mean Square (RMS) values for AC voltage and current represent the effective value that would produce the same heating effect as a DC equivalent. For sinusoidal waveforms, RMS = Peak / √2.

    • RMS = Peak / √2 (for sine wave).
    • Peak = RMS * √2 (for sine wave).
    • Most AC meters display RMS values.
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  • Pool Motor EGC Sizing

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    Equipment grounding conductors for swimming pool motors must be a minimum of 12 AWG copper, regardless of the overcurrent protective device size, as per NEC Article 680.

    • Specific requirement for pool motors (NEC 680.21(A)(4)).
    • Overrides general EGC sizing rules (NEC 250.122).
    • Applies to all pool pump motors for enhanced safety.
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  • Ohm's Law for Power

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    Ohm's Law for Power relates power (P), voltage (V), and current (I) in an electrical circuit.

    • P = V × I (Power = Voltage × Current)
    • P = I² × R (Power = Current squared × Resistance)
    • P = V² / R (Power = Voltage squared / Resistance)
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  • Temporary Open Conductor Clearance

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    The minimum vertical distance required by the NEC for open conductors of temporary installations above areas accessible only to pedestrians.

    • Applies to temporary installations (NEC 590).
    • Specific to open conductors over pedestrian areas.
    • Minimum 3.0 m (10 ft) clearance.
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  • 3-Phase Balanced Load Current Summation

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    For balanced 3-phase loads, the total current on each phase of a feeder is the arithmetic sum of the individual phase currents of the connected loads.

    • Assumes balanced loads, where each phase draws approximately the same current.
    • Diversity factors are often applied in real-world scenarios but are excluded here.
    • This principle applies to the current on each individual phase conductor.
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  • Conductor Resistivity Comparison

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    Resistivity is an intrinsic property of a material that quantifies how strongly it resists electrical current. Copper has lower resistivity than aluminum.

    • Lower resistivity means better conductivity.
    • Copper is a better conductor than aluminum.
    • Resistance is directly proportional to resistivity, length, and inversely proportional to cross-sectional area.
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  • DC Motor Armature Resistance Effect

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    In a DC motor, increased resistance in the armature circuit reduces the back electromotive force (EMF) for a given speed, leading to higher armature current but reduced speed under load.

    • Motor speed is proportional to (V - I_a*R_a) and inversely proportional to field flux.
    • Back EMF (E_b) opposes the applied voltage (V).
    • If R_a increases, the voltage drop across R_a increases.
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  • 3-Phase Apparent Power

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    Apparent power (S) in a 3-phase system is the product of the line voltage, line current, and the square root of 3. It is measured in Volt-Amperes (VA).

    • S = √3 * V_line * I_line
    • Includes both real and reactive power components.
    • Used for sizing electrical equipment like transformers and generators.
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