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Revised Date: June 23, 2010

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2 Mike Holt’s Illustrated Guide to NEC Exam Preparation

Answer Key Unit 1— Practice Questions

Chapter 1 eLeCtrICaL theOrY

Unit 1—electrician’s Math and Basic electrical Formulas

Unit 1—Practice Questions1. (a) 0.50

2. (a) 0.20

3. (b) left

4. (b) 0.75

5. (b) 2.25

6. (c) 3

7. (d) multiplier

8. (b) 20A

The overcurrent protection device must be sized 1.25 times larger than the load.16A x 1.25 = 20A

9. (b) 80A

The continuous load must be limited to 80 percent of the rating of the protection device. 100 x 0.80 = 80A

10. (b) 9.60 kW

Step 1: Change the % to its decimal multiplier, 20% increase = 1.20

Step 2: Multiply the number by the multiplier, 8 kW x 1.20 = 9.60 kW

11. (a) 0.80

Reciprocal of 1.25 = 1/1.25Reciprocal of 1.25 = 0.80

12. (b) 80A

The continuous load must be limited to 80 percent of the rating of the protection device.100A x 0.80 = 80A

13. (a) True

14. (a) 50W

P = I2 x RP = 16A2 x 0.20 ohmsP = (16A x 16A) x 0.20 ohmsP = 51.20W

15. (c) 3 sq in.

Area = Pie x r2

Pie = 3.14r = radius (½ of the diameter)Area = 3.14 x (½ x 2)2 Area = 3.14 sq in.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (888.632.2633) 3

Unit 1— Practice Questions Answer Key

16. (c) 1642 = 4 x 4 = 16

17. (c) 144122 = 12 x 12 = 144

18. (c) 100 ft

D = (Cmil x VD)/(2 x K x I)D = (4,110 Cmil x 10V)/(2 wires x 12.90 ohms x 16A)D = 41,100/4,128D = 99 ft

19. (b) 50A

I = VA/(E x √3)I = 18,000W/(208V x 1.732) Current = 18,000W/360 Current = 50A

20. (b) False

21. (b) 32Enter the number on your calculator, then push the square root key (√).

22. (b) 1.732Enter the number on your calculator, then push the square root key (√).

23. (a) cubic inches

24. (b) 24 cu in.

Volume = 4 in. x 4 in. x 1.50 in.Volume = 24 cu in.

25. (a) 0.075 kW

kW = W/1000kW = 75W/1000kW = 0.075 kW

26. (b) 252 + 7 + 8 + 9 = 26, the multiple choice selections are rounded to the nearest “fives.”

27. (c) 110W

The input must be greater than the output.Input = Output/EfficiencyInput = 100W/0.90Input = 111W

28. (d) all of these

29. (b) negative, positive

30. (b) False

31. (a) True

32. (b) False

33. (a) True



34. (a) True

35. (b) False

36. (d) silver, copper, gold, aluminum


38. (a) True

39. (d) directly, inversely


41. (b) False


43. (c) 6.40V

EVD = I x REVD = 16A x 0.40 ohmsEVD = 6.40V

44. (a) 0.14 ohms

R = E/IR = 7.20V/50AR = 0.14 ohms

45. (a) 175W

P= I x EP = 24A x 7.20VP = 172.80W

46. (a) 8.20 kW

The power consumed by this resistor will be 10,000W if connected to a 230V source. But, because the applied voltage (208V) is less than the equipment voltage rating (230V), the actual power consumed will be less.

Step 1: Determine the resistance rating of a 10 kW, 230V load.R = E2/PR = 230V2/10,000WR = 52,900V/10,000WR = 5.29 ohms

Step 2: Determine the power consumed for a 5.29 ohm load connected to a 208V source.P = E2/RP = 208V2/5.29 ohmsP = (208V x 208V)/5.29 ohmsP = 43,264/5.29P = 8,178W or 8.20 kW

47. (d) a and b

48. (a) True

49. (d) power loss



50. (b) 100W

P = I2 x RP = 16A2 x 0.40 ohmsP = (16A x 16A) x 0.40 ohmsP = 102.40W

51. (a) 43W

P = I x EP = 12A x (120V x 3%)P = 12A x 3.60VP = 43.20W

52. (c) $72

Formula: Cost per Year = Power for the Year in kWh x $0.08

Step 1: Determine the power loss per hour.P = I2 x RP = 16A2 x 0.40 ohmsP = (16A x 16A) x 0.40 ohmsP = 102.40W per hour

Step 2: Determine the power loss in kWh for the year.Power for the Year in kWh = (Power per hour x 24 hours per day x 365 days)/1,000Power for the Year in kWh = (102.40W x 24 hours x 365 days)/1,000Power for the Year in kWh = 897 kWh

Step 3: Determine the cost per year for the conductor power losses.Formula: Cost per Year = kWh per Year x Cost per kWhCost per Year = 897 kWh x $0.08Cost per Year = $71.76

53. (b) False

54. (a) 2.50 kW

The power consumed by this resistor will be 10,000W if connected to a 230V source. But, because the applied voltage (115V) is less than the equipment voltage rating (230V), the actual power consumed will be less.

Step 1: Determine the resistance rating of a 10 kW, 230V load.R = E2/PR = 230V2/10,000WR = (230V x 230V)/10,000WR = 52,900/10,000R = 5.29 ohms

Step 2: Determine the power consumed for a 5.29 ohm load connected to a 115V source.P = E2/RP = 115V2/5.29 ohmsP = (115V x 115V)/5.29 ohmsP = 13,225/5.29 ohmsP = 2,500W or 2.50 kW

Note: Power changes with the square of the voltage. If the voltage is reduced to 50%, then the power consumed will be equal to the new voltage percent2 or 50%2, or 10,000W x (0.50 x 0.50 = 0.25 = 25%) = 2,500W.


Answer Key Unit 1— Challenge Questions

Unit 1—Challenge Questions1. (d) 1,000 VA

2. (d) Salt water

3. (a) increase

4. (b) reduced by half

According to Ohm’s Law, current is inversely proportional to resistance. This means that if the resistance goes down, assuming voltage remains the same, the current will increase. It also works in the opposite direction; if the resistance increases, again assuming the voltage remains the same, the current will decrease.

Example: What is the current of a 120V circuit if the resistance is 5 ohms, 10 ohms, or 20 ohms? Formula: I =E/R

Answer: At 5 ohms the current is equal to 24A, at 10 ohms the current is equal to 12A, and at 20 ohms, the current is only equal to 6A

I = 120V/5 ohms = 24AI = 120V/10 ohms = 12AI = 120V/20 ohms = 6A

5. (c) a shorted coil

If the reading is less than 30 ohms, this indicates that the length of the coil’s conductor must be shorted.

6. (d) any of these pairs of variables

7. (d) 1,440W

The formula I2 x R in the question has nothing to do with the actual calculation. If we know the voltage of the circuit and the resistance in ohms of the resistor, the formula we need to use is:

P = E2/RP = 120V2/10 ohmsP = (120V x 120V)/10 ohmsP = 1,440W

8. (c) P = I2 x R

9. (b) less

If current remains the same and resistance increases, then energy consumed will increase. Example:

P = I2 x RP = 10A2 x 5 ohms P = 500WP = 10A2 x 10 ohms P = 1,000W


Unit 1— Challenge Questions Answer Key

10. (c) 4.50AThe power consumed by this resistor will be 500W if connected to a 115V source. But, because the applied voltage (120V) is greater than the equipment voltage rating (115V), the actual power consumed will be greater than 500W.

Step 1: Determine the resistance rating of a 500W, 115V load.R = E2/PR = 115V2/500WR = 13,225/500R = 26.45 ohms

Step 2: Determine the current of a 26.45 ohm load connected to a 120V source.I = E/RP = 120/26.45 ohmsP = 4.54A

11. (b) less

When the resistance is not changed, the power will decrease with decreasing voltage. For example a 100 ohm resistor will consume 144W of power at 120V, but only 132W of power at 115V.

P = E2/RP = 120V2/100 ohmsP = 144WP = 115V2/100 ohmsP = 132W

12. (c) 400W

The power consumed by this resistor will be 100W if connected to a 115V source. But, because the applied voltage (230V) is greater than the equipment voltage rating (115V), the actual power consumed will be greater than 100W.

Step 1: Determine the resistance rating of a 100W, 115V lamp.R = E2/PR = 115V2/100WR = (115V x 115V)/100WR = 13,225/100R = 132.25 ohms

Step 2: Determine the power consumed for a 132.25 ohm load connected to a 230V source.P = E2/RP = 230V2/132.25 ohmsP = (230V x 230V)/132.25 ohmsP = 52,900/132.25 ohmsP = 400W



13. (a) 1,225W

The power consumed by this resistor will be 1,500W if connected to a 230V source. But, because the applied voltage (208V) is less than the equipment voltage rating (230V), the actual power consumed will be less than 1,500W.

Step 1: Determine the resistance rating of a 1,500W, 230V load.R = E2/PR = 230V2/1,500WR = (230V x 230V)/1,500WR = 52,900/1,500R = 35.27 ohms

Step 2: Determine the power consumed for a 35.27 ohm load connected to a 208V source.P = E2/RP = 208V2/35.27 ohmsP = (208V x 208V)/35.27 ohmsP = 43,264/35.27 ohmsP = 1,227W

14. (b) 29W

P = I2 x RP = 12A2 x 0.20 ohmsP = 29W

Unit 2— electrical Circuits

Unit 2—Practice Questions1. (a) True

2. (c) a and b

3. (b) False

4. (a) True

5. (a) True

6. (a) True

7. (a) True

8. (d) the same

9. (b) False

10. (a) True

11. (a) True

12. (a) True

13. (a) True

14. (b) False

15. (b) False



16. (a) True

17. (b) False

18. (a) True

19. (a) True

20. (b) False

21. (d) any of these

22. (b) False

23. (b) two

24. (a) True

25. (a) True

26. (c) series-parallel circuit

27. (a) True

28. (a) True

29. (a) True

30. (a) True

31. (b) neutral point

32. (a) True

33. (b) neutral point

34. (a) True

35. (d) 100 percent

36. (a) 0 percent

37. (a) True

38. (a) True

39. (a) True

40. (a) 0A

41. (a) True

42. (a) True

43. (b) False

44. (b) False

45. (b) False

46. (a) True



47. (b) 6.40V

EVD = I x R I = 16A R = 2 ohms/1,000 ft = 0.002 ohms per ftResistance of Conductors = 0.002 ohms x 100 ft x 2 conductorsResistance of Two Conductors = 0.40 ohmsVoltage Drop of Conductors: EVD = I x REVD = 16A x 0.40 ohmsEVD = 6.40V

48. (a) 3.20V

EVD = I x R I = 16A R = 2 ohms/1,000 ft = 0.002 ohms per ftResistance of Conductors = 0.002 ohms x 100 ftResistance of One Conductor = 0.20 ohmsVoltage Drop of Conductors: EVD = I x REVD = 16A x 0.20 ohmsEVD = 3.20V

49. (d) b and c

50. (a) 2-wire

51. (b) grounded

Unit 2—Challenge Questions1. (a) 1A

The current through any resistor of a series circuit is equal to the current of the circuit.

I = E/RE = 12VR = 12 ohmsI = 12V/12 ohmsI = 1A

2. (d) all of these

Voltage in a series circuit is distributed among all equal value resistors equally according to Kirchoff’s Law on voltage. This means that since there are four equal value resistors in series, each resistor will have one-quarter of the source voltage. 120V/4 resistors = 30V each resistor.

3. (d) all of these



4. (d) 10V

Step 1: Determine the current of the circuit.I = E/RTE = 30VRT = 22.50 ohmsI = 30V/22.50 ohmsI = 1.33A

Step 2: Determine the voltage of Resistor 2.E2 = I x R2I = 1.33AR2 = 7.50 ohmsE2 = 1.33A x 7.50 ohms E2 = 9.98V

5. (b) 6V

This is tricky. By placing the voltage meter across the switch, the circuit conductors and the load are used as part of the voltage meter leads.

6. (c) parallel

7. (b) parallel

8. (d) b and c

9. (c) in parallel

Series Example: When connected in series, each resistor will operate at 30V (one-quarter of the voltage source).

P = E2/RP = 30V2/10 ohmsP = 90W for each resistor in seriesP = 90W x 4P = 360W

Parallel Example: If the four resistors are connected in parallel, each resistor will operate at 120V.

P = E2/RP = 120V2/10 ohmsP = 1,440W for each resistor in parallelP = 1,440W x 4 P = 5,760W

10. (b) 1 ohm

Rule: The total resistance of a parallel circuit is always less than the smallest resistor. Formula:

RT = 1/(1/R1 + 1/R2 + 1/R3)RT = 1/(½ + 1/3 + 1/5) RT = 1/(0.50 + 0.33 + 0.20)RT = 1/1RT = 1 ohm

Note: Figure 2-53 applies to the next three questions.

11. (c) A3



12. (d) 48 ohms

R = E2/PE = 24VP = 12WR = 24V2/12WR = 576/12R = 48 ohms

13. (a) 22 ohms

R1 (Bell 1) = E1/ITE1 = 30VIT = 0.75AR1 = 30V/0.75AR1 = 40 ohmsR2 (Bell 2) = 48 ohmsRT = Product/SumRT = (R1 x R2)/(R1 + R2)RT = (40 x 48)/(40 + 48) RT = 1,920/88RT = 21.82 ohms

Note: Figure 2-54 applies to the next three questions.


15. (a) 1.50V

Step 1: Determine the Resistance of the two parallel resistors R3 and R4: RT = Product/SumRT = (R3 x R4)/(R3 + R4)RT = (4 ohms x 4 ohms)/(4 ohms + 4 ohms) RT = 16 ohms/8 ohms RT = 2 ohms

Note: R3 and R4 can now be thought of as one 2 ohm resistor.

Step 2: Determine the current of the circuit (current flowing through R3,4).IT = ES/RTES = 6VRT = 2 ohms + 2 ohms + 2 ohms + 2 ohms RT = 8 ohmsIT = 6V/8 ohmsIT = 0.75A

Step 3: Calculate the voltage across R3,4E3,4 = I x R3,4E3,4 = 0.75A x 2 ohmsE3,4 = 1.50V



16. (b) 3V

IT = 0.75A (last answer)R3,4,5 = R3,4 + R5R3,4,5 = 2 ohms + 2 ohmsR3,4,5 = 4 ohmsE4 = IT x R3,4,5E3,4,5 = 0.75A x 4 ohms E3,4,5 = 3V

Note: Figure 2-55 applies to the next two questions.

17. (a) 10 ohms

Calculate the parallel resistance and then add the resistance of resistor R1.

Resistance of one resistor/number of resistors = 15 ohms/3 resistors = 5 ohms

Note: The 3 parallel resistors can be thought of as a single 5 ohm resistor in series with resistor R1.

RT = R1 + (R2,3,4)RT = 5 ohms + 5 ohmsRT = 10 ohms

18. (c) 60V

Voltage drop across R1 is determined by: E1 = IT x R1

IT = ES/RTIT = 120V/10 ohmsIT = 12AR1 = 5 ohmsE1 = IT x R1E1 = 12A x 5 ohms E1 = 60V

19. (b) 0.58A

If the neutral is opened, the multiwire branch circuit becomes one 240V series circuit.

IT = ES/RT ES = 240VRT = R1 + R2

Use the rated voltage to determine the resistance of each resistor.

R1 = E2/PR1 = 130V2/75WR1 225 ohmsR2 = 120V2/75WR2 = 192 ohmsRT = 225 ohms + 192 ohmsRT = 417 ohms

IT = ES/RTIT = 240V/417 ohmsIT = 0.575A



Unit 3—Understanding alternating Current


2. (a) True

3. (a) True

4. (a) True

5. (a) True

6. (d) all of these

7. (b) False

8. (a) True

9. (c) sine

10. (d) Hertz

11. (a) True

12. (a) in-phase

13. (d) 360º

14. (b) 120º

15. (d) degrees

16. (c) voltage

17. (a) lead

18. (b) lags

19. (d) “Instantaneous”

20. (b) times 1.41

21. (a) “Peak”

22. (a) times 0.707

23. (a) True

24. (b) “Root-mean-square”

25. (a) “Capacitance”

26. (b) False

27. (b) charged

28. (d) shorted

29. (a) conductive

30. (a) True




32. (a) True

33. (a) lead the applied voltage by 90º

34. (a) XC

35. (a) True

36. (c) induced

37. (c) magnetic

38. (c) self-induced


40. (c) 180°

41. (a) True

42. (a) True

43. (c) resistance

44. (b) False


46. (b) False

47. (d) skin effect

48. (a) True

49. (a) True

50. (b) applied voltage

51. (a) XL

52. (c) 180º

53. (a) 90º


55 (a) True

56. (c) apparent power

57. (a) True

58. (b) False

59. (b) False

60. (a) True

61. (c) resistive

62. (b) $300

10 luminaires x 150W = 1,500WPower for the Year in kWh = (1,500W x 6 hours x 365 days)/1,000Power for the Year in kWh = 3,285 kWhCost per Year = 3,285 kWh x $0.09Cost per Year = $295.65



63. (b) False

64. (a) True

65. (c) 1,200W

Power (Watts) = Volts x Amperes x Power FactorW = 120V x 10A x 1.00 PFW = 1,200W

66. (b) 25 kVA

Transformer kVA = (Volts x Amperes)/1,000Transformer kVA = (240V x 100A)/1,000Transformer kVA = 24 kVA

67. (c) 37.50 kVA

Load kW = (Volts x Amperes)/1,000Load kW = (240V x 100A)/1,000Load kW = 24 kWTransformers are sized to the VA of the load, not the kW. VA = Watts/Power FactorVA = 24,000 W/0.85VA = 28,235 VAThe first choice large enough to handle this load is 37.50 kVA

68. (d) 6 circuits

VA per Circuit = Volts x AmperesVA per Circuit = 120V x 20AVA per Circuit = 2,400 VALights per Circuit = 2,400 VA/300WLights per Circuit = 8Circuits = 42 luminaires/8 per circuitCircuits = 6

69. (c) 7 circuits

VA per Circuit = Volts x AmperesVA per Circuit = 120V x 20AVA per Circuit = 2,400 VAVA per Luminaire = Watts/Power FactorVA per Luminaire = 300W/0.85 PFVA per Luminaire = 353 VALights per Circuit = 2,400 VA/353 VA = 6.8Lights per Circuit = 6Circuits = 42 luminaires/6 per circuitCircuits = 7

70. (a) True

71. (a) True

72. (b) 73%

Efficiency = Output/InputEfficiency = 1,320W/1,800WEfficiency = 0.7333 or 73.33%



73. (a) 66%

Efficiency = Output/InputEfficiency = 746W/1,128WEfficiency = 0.6613 or 66.13%

74. (b) 15A

Input = Output/EfficiencyInput = 1,600W/0.88 EffInput = 1,818WInput Amperes = Watts/VoltsInput Amperes = 1,818W/120VInput Amperes = 15.15A

75. (a) 970W

Output = Input x EfficiencyOutput = 1,000W x 0.97 EffOutput = 970W

Unit 3—Challenge Questions1. (a) True

2. (a) reduced voltage dropWhen the system voltage is increased, the current decreases and voltage drop is determined by E = I x R.

3. (b) ease of voltage variation

4. (d) all of these

5. (b) 1 second

6. (a) 1/120 secondIt takes 1/120 of a second for 60 Hz ac to travel through 180 degrees.

7. (a) the same

8. (c) 35A

Effective (RMS) Current = Peak Current x 0.707RMS Current = 50A x 0.707 RMS Current = 35A

9. (b) False

The peak value of ac is equal to:EPeak = Effective (RMS) x 1.41EPeak = 120V x 1.41EPeak = 169V

10. (c) effective

11. (c) dielectric

12. (b) Connect all three in parallel.

Connecting capacitors in parallel has the same effect as increasing the plate area of one capacitor. Three capacitors of 20, 30, and 60 microFarads connected in parallel have a capacitance of: CT = C1 + C2 + C3



13. (b) lags the current by 90º

14. (c) XL = XC

15. (a) ohms

16. (c) Lenz’s

17. (c) XL

18. (a) True

19. (c) ohmsInductance is measured in Henrys, but inductive reactance is measured in ohms.

20. (a) remain constant regardless of the current and voltage changesInductive reactance changes with frequency, not with voltage or current.

21. (a) impedance

22. (c) alternating-currentImpedance is the opposition to current flow because of resistance, capacitive reactance (XC), and inductive reactance (XL).

23. (a) higher than

24. (c) apparent power

25. (c) 2.30 kVA

Apparent Power = Volts x AmperesVolts = 120Amperes = 19.20Apparent Power = 120V x 19.20AApparent Power = 2,304 VA or 2.304 kVA

26. (c) series-parallelSeries to measure the current and parallel to measure the voltage.

27. (d) b and c

28. (c) I2 x R

29. (a) 24,367W

True Power (Watts) = Apparent Power x Power FactorApparent Power = Volts x Amperes x 1.732Volts = 208VAmperes = 76AApparent Power = 208V x 76A x 1.732 Apparent Power = 27,379 VATrue Power = Apparent Power x Power FactorApparent Power = 27,379 VAPower Factor = 89% or 0.89True Power = 27,379 VA x 0.89 PF True Power = 24,367W or 24.367 kW



30. (c) 1.91 kW

True Power = Apparent Power x Power FactorApparent Power = 2,100 VAPower Factor = 91% or 0.91True Power = 2,100 VA x 0.91 PF True Power = 1,911W = 1.911 kW

31. (d) resistive loadsPower factor is unity (100%) for resistive loads.

32. (c) 65A

Current = Watts/(Volts x 1.732 x Power Factor)Watts = 24,000WVolts = 230VPower Factor = 92% or 0.92Current = 24,000W/(230V x 1.732 x 0.92)Current = 65.49A

33. (c) 87%

PF = W/VAPF = 68W/(0.65A x 120V)PF = 68W/(78VA)PF = 0.8718 or 87%

34. (a) 76 VA

VA = W/PFVA = 68W/0.90 PFVA = 75.55 VA

35. (b) 1,150W

Watts = VA x PFWatts = (120V x 12A) x 0.80 PFWatts = 1,152W

36. (a) True

37. (d) $105

Step 1: Power per hour = E2/RE = 120VR = 10 ohmsP = E2/RP = (120V x 120V)/10 ohmsP = 1,440W

Step 2: Power consumed per day:1,440W x 24 hours = 34,560 Wh or 34.56 kWh

Step 3: Power consumed in 30 days:34.56 kWh x 30 days = 1,036.80 kWh

Step 4: Cost of power at $0.10 per kWh:1,036.80 kWh x $0.10 = $103.68

38. (a) True



39. (c) 6 kVA

kVA = (50 Fixtures x 100W)/1,000kVA = 5,000W/1,000kVA = 5 kVA Apparent Power = kW/PFApparent Power = 5 kW/0.90 PFApparent Power = 5.56 kVA

40. (a) 3 circuits

Circuits are loaded according to VA, not watts!VA of each luminaire equals: VA = Watts/PFVA = 100W/0.90VA = 111 VAEach circuit has a capacity of: 120V x 20A = 2,400 VAEach circuit can have: 2,400 VA/111 VA = 21 luminaires The number of circuits required is: 50 luminaires/21 luminaires per circuit = 3 circuits

41. (c) hp x 746W/W Input Motor efficiency is equal to motor output watts (hp x 746W) divided by motor input watts.

42. (c) 90Input = 4,000 VAOutput = 3,600 VAEfficiency = Output/InputEfficiency = 3,600 VA/4,000VAEfficiency = 0.90 or 90%

Note: Efficiency is never 100% or greater.

Unit 4—Motors and transformers


2. (b) parallel, series

3. (a) 230V, 460V

4. (c) 746W

5. (c) 40 hp

hp = Output Watts/746Whp = 30,000W/746Whp = 40 hp

6. (a) 11 kW

Output Watts = hp x 746WOutput Watts = 15 hp x 746WOutput kW = 11,190W/1,000Output kW = 11.19 kW



7. (a) 3.75 kW

Output Watts = hp x 746WOutput Watts = 5 hp x 746WOutput Watts = 3,730WOutput kW = 3,730W/1,000Output kW = 3.73 kW

8. (d) voltage

9. (a) True

10. (b) False

11. (b) 20A

FLA = hp x 746W/(Volts x PF x Eff)FLA = 5 x 746W/(230V x 0.93 PF x 0.87)FLA = 3,730/186FLA = 20A

12. (b) 58A

FLA = hp x 746W/(Volts x 1.732 x PF x Eff)FLA = 20 x 746W/(208V x 1.732 x 0.90 PF x 0.80)FLA = 14,920/259FLA = 57.60A

13. (d) 6

14. (a) True

15. (c) LRC

16. (a) FLA

17. (d) commutator

18. (b) False

19. (b) False

20. (d) a or b

21. (a) True

22. (c) synchronous

23. (b) Universal

24. (b) two

25. (d) transformer

26. (b) primary, secondary

27. (a) True

28. (a) True

29. (a) True

30. (a) True

31. (a) primary, secondary



32. (c) 2:1

33. (b) False

34. (c) eddy currents

35. (c) Eddy currents

36. (d) hysteresis loss

37. (a) True

38 (d) kVA

39. (a) True

40. (a) True

Unit 4—Challenge Questions

1. (c) 6,100 VA

VAInput = Volts x Motor Amperes x 1.732VAInput = 230V x 15.20A x 1.732 VAInput = 6,055 VA

2. (b) 1,840 VA

VAInput = Volts x Motor AmperesVAInput = 115V x 16A VAInput = 1,840 VA

3. (a) dc series

4. (b) dc

5. (c) run the same as beforeThe field or armature leads must be reversed, not the line (supply) leads.

6. (b) FalseThe field or armature leads must be reversed, not the line (supply) leads.

7. (d) wound-rotor

8. (d) b or c

9. (a) higher thanA ratio of 2:1 means that the secondary voltage is less, and the power remains the same. The current on the secondary will therefore be greater, I = P/E.

10. (a) 0.50AThe primary voltage is 10 times greater, and the power remains the same. Therefore, the current on the primary will be 10 times lower (5A x 0.10 = 0.50A).

11. (a) PrimaryA current transformer (CT) is often used to measure current in large conductors. The phase conductors act as the primary winding and the CT serves as the secondary winding. Naturally the current flowing through the CT (secondary winding) is much lower than the primary winding. Typically, CTs have a current ratio of 100:1; that is, for every 100A on the primary (phase conductor) the secondary (CT) will measure 1A.

12. (b) Secondary



13. (b) SecondaryThe secondary has lower voltage and the power remains the same; therefore, the current will be greater than the primary.

14. (b) 2A

If the primary voltage is 5 times more than the secondary, then the primary current will be 5 times less than the secondary.

IPrimary = ISecondary x (Secondary Winding Turns/Primary Winding Turns)

IPrimary = 10A x (1/5)

IPrimary = 10A x 0.20

IPrimary = 2A

15. (b) 17A

I = P/EP = 200WE = 12VI = 200W/12VI = 16.67A

Efficiency does not change the load on the secondary; it impacts the primary current.

16. (a) 6.80A

I = VA/EPrimary VA = Secondary VA/EfficiencyPrimary VA = 1,500W/0.92 EffPrimary VA = 1,630 VAE = 240VPrimary Current = Primary VA/Primary VoltsPrimary Current = 1,632 VA/240VI = 6.79A

17. (c) 42 kVA

Secondary VA = Volts x Amperes x 1.732Secondary VA = 208V x 100A x 1.732Secondary VA = 36,000 VAPrimary VA = Secondary VA/EfficiencyPrimary VA = 36,000 VA/0.86Primary VA = 41,860 VA or 41.86 kVA

18. (c) 50A

Primary Current = Primary VA/(Volts x 1.732)Primary Current = 41,860 VA/(480V x 1.732)Primary Current = 41,860 VA/831VPrimary Current = 50A



19. (a) 6V

The turns (voltage) ratio is 200/10 or 20/1, which means that the secondary voltage will be 20 times less than the primary.

Secondary Volts = Primary Volts/Voltage RatioSecondary Volts = 120V/(200/10)Secondary Volts = 120V/20Secondary Volts = 6V

20. (c) 526W

Primary Power = Secondary Power/EfficiencyPrimary Power = 500W/0.95Primary Power = 526W

21. (c) 4.38A

I = P/EI = 526W/120VI = 4.38A

22. (a) 200 VAThe load (output) is given as two 100W lamps; the efficiency only affects the input, not the output.

23. (a) 120 VA

Efficiency only affects the input, not the output.Secondary VA = Secondary Volts x Secondary AmperesVA = 24V x 5A VA = 120 VA

24. (c) 217 VA

Primary W = Secondary W/EfficiencyPrimary W = 200W/0.92Primary W = 217W or 217 VA (since there is no power factor)

25. (c) 31 kW

Input = Output/EfficiencyInput = 20 kW/0.65Input = 30.80 kW

26. (b) saturated

27. (a) 0.06 kVA

Primary VA = Secondary VASecondary VA = Volt x AmperesVolts = 12VAmperes = 5ASecondary VA = 12V x 5A Secondary VA = 60 VA or 0.06 kVA



Chapter 2—NEC CaLCULatIOns

Unit 5—raceway and Box Calculations

Unit 5—Practice Questions

1. (c) Annex CChapter 9, Table 1, Note 1

2. (a) TrueChapter 9, Table 1, Note 3

3. (d) 60Chapter 9, Table 1, Note 4

4. (b) 29Annex C, Table C.1

5. (a) 7Annex C, Table C.2

6. (b) 7Annex C, Table C.3

7. (d) 11Annex C, Table C.4

8. (c) 350 kcmilAnnex C, Table C.8(A)

9. (b) 40Chapter 9, Table 1

10. (b) 0.0243Chapter 9, Table 5

11. (b) 0.0209Chapter 9, Table 5

12. (c) 0.0211Chapter 9, Table 5

13. (d) 0.0353Chapter 9, Table 5

14. (a) 0.013

Chapter 9, Table 88 AWG Solid = 0.013 sq in.8 AWG Stranded = 0.017 sq in.




16. (a) Trade size 2

Step 1: Area of the conductors [Chapter 9, Table 5]3—3/0 THHN: 0.2679 sq in. x 3 =0.8037 sq in.1—2 THHN: 0.1158 sq in. x 1 = 0.1158 sq in.1—6 THHN: 0.0507 sq in. x 1 = 0.0507 sq in.

Step 2: Total sq in. area of the conductors: 0.9702 sq in.

Step 3: Permitted conductor fill at 40% fill [Chapter 9, Tables 1 and 4]Trade size 2 Schedule 80 RNC area = 1.15 sq in.

17. (a) Trade size 1½

Step 1: Area of the conductors [Chapter 9, Table 5]3—4/0 THHN: 0.3237 sq in. x 3 = 0.9711 sq in.1—1/0 THHN: 0.1855 sq in. x 1 = 0.1855 sq in.1—4 THHN: 0.0824 sq in. x 1 = 0.0824 sq in.

Step 2: Total sq in. area of the conductors: 1.239 sq in.

Step 3: Size the conduit at 60% fill [Chapter 9, Table 4, Note 3 use 60% Column]Trade size 1½: 1.243 sq in.

18. (d) 11

Step 1: Area of conductor fill permitted for a trade size ¾ nipple is 0.329 sq in. [Chapter 9, Table 4]

Step 2: Determine the sq in. area of the existing conductors– [Chapter 9, Table 5]4—10 THHN: 0.0211 sq in. x 4 = 0.0844 sq in.1—10 AWG bare stranded [Chapter 9, Table 8]: 0.0110 sq in. x 1 = 0.0110 sq in.Total area of existing conductors: 0.0954 sq in.

Step 3: Subtract the area of the existing conductors from the area of permitted conductor fillSpare Space Area:Permitted Area Fill less Existing Conductors AreaSpare Space Area = 0.3290 sq in. - 0.0954 Spare Space Area = 0.2336 sq in.

Step 4: Determine the number of 10 THHN conductors that can be added to the available spare space.Number of conductors permitted: Spare Space Area/Area of 10 THHNNumber of 10 THHN conductors permitted: 0.2336 sq in./0.0211 sq in.Number of conductors permitted = 11 conductors

19. (b) 16 sq in.Cross sectional area is found by multiplying height by depth: 4 in. x 4 in. = 16 sq in.

20. (b) 3.20 sq in.16 sq in. x 0.20 = 3.20 sq in. [376.22]

21. (d) 12

36 sq in. x 0.20 = 7.20 [376.22(A)]400 kcmil THHN = 0.5863 sq in. [Chapter 9, Table 5]Maximum Allowable Area/Area per Conductor = Number of Conductors7.20 sq in. /0.5863 sq in. = 12.2812 conductors could be installed.

Note: ampacity adjustment for bundling not required for under 30 conductors [376.22(B)]



22. (a) 4 x 1¼ squareInsulation is not a factor [Table 314.16(A)], 9 - 14 AWG.

23. (a) 8 conductors [Table 314.16(A)]

24. (d) all of these[Table 314.16(A)]

25. (d) all of these [314.16(B)]

26. (d) all of these [314.16(B)]

27. (d) b and c[314.16(B)(1) Ex]

28. (a) Yes

314.16(B)(1) Ex permits us to omit fixture wires that enter the outlet box from a luminaire canopy.

Step 1: Determine the number and size of conductors14/2 NM cable [314.16(B)(1) 2—14 AWGGround wire 1—14 AWGOne Cable Clamp [314.16(B)(2)] 1—14 AWG

Step 2: Volume of the conductors [Table 314.16(B)]:4 conductors x 2 cu in. = 8 cu in.

29. (c) 4 x 21⁄8 square

Step 1: Determine the number and size of conductors [314.16(B)]12/2 NM cable 2—12 AWG conductors12/3 NM cable 3—12 AWG conductorsCable clamps 1—12 AWG conductorsSwitch 2—12 AWG conductorsReceptacles 2—12 AWG conductorsGround wire 1—12 AWG conductorsTotal Number 11—12 AWG conductors

Step 2: Determine the volume (cubic inches) of the above conductors [Table 314.16(B)]11 conductors x 2.25 = 24.75 cu in.

Step 3: Select the outlet box from Table 314.16(A)4 x 21⁄8 in. square = 30.30 cu in.



30. (b) two

Step 1: Determine the number and size of the existing conductors [314.16(B)]Two Receptacles 4—14 AWG conductorsFive 14 AWG 5—14 AWG conductorsTwo Ground Wires 1—14 AWG conductorTotal Conductors 10—14 AWG conductors

Step 2: Determine the volume of the existing conductors [Table 314.16(B)]10 conductors x 2 cu in. = 20 cu in.

Step 3: Determine the spare space.A. 4 x 1½ square box = 21 cu in. + 3.60 (ring) = 24.60 cu in.B. Spare Space: 24.60 cu in. - 20 cu in. = 4.60 cu in.

Step 4: 14 THHN conductors permitted in spare space: Spare Space/Conductor Volume4.60 cu in./2 cu in. = 2 conductors

31. (d) all of these[314.28(A)(1) and (2)]

32. (d) none of these[314.28(A)(2) Ex]

33. (b) 20 in.

[314.28(A)(1)]Left wall to the right wall angle pull: (6 x 2.50 in.) + 2.50 = 17.50 in.Left wall to the right wall straight pull: 8 x 2.50 in. = 20 in.Right wall to the left wall angle pull: No calculationRight wall to the left wall straight pull: 8 x 2.50 in. = 20 in.

34. (a) 15 in.

[314.28(A)]Bottom wall to the top wall angle pull: 6 x 2.5 in. = 15 in.Bottom wall to the top wall straight pull: No calculationTop wall to the bottom wall angle pull: No calculationTop wall to the bottom wall straight pull: No calculation

35. (a) 15 in.

[314.28(A)(2)]6 x 2.50 in. = 15 in.

36. (a) 14 in.

[314.28(A)]Left wall to the right wall angle pull: (6 x 2 in.) + 2 in. = 14 in.Left wall to the right wall straight pull: No calculationRight wall to the left wall angle pull: No calculationRight wall to the left wall straight pull: No calculation



37. (a) 14 in.

[314.28(A)]Bottom wall to the top wall angle pull: No calculationBottom wall to the top wall straight pull: No calculationTop wall to the bottom wall angle pull: (6 x 2 in.) + 2 in. = 14 in.Top wall to the bottom wall straight pull: No calculation

38. (a) 12 in.

[314.28(A)(2)] 2 in. x 6 = 12 in.

Unit 5—Challenge Questions1. (a) 11

Area conductor fill for a trade size 3, Schedule 40 PVC (RNC) raceway: 2.907 sq in. [Chapter 9, Table 4, 40% column]Area 1 RHW without cover: 0.1901 sq in. [Chapter 9, Table 5]Area of seven 1 RHW conductors: 0.1901 x 7 conductors = 1.3307 sq in.Spare space: 2.907 sq in. - 1.3307 sq in. = 1.5763 sq in.Quantity of 2 THW permitted in spare space: 1.5763 sq in./0.1333 = 11.8 conductors

Note: The answer is 11 conductors. We only round up to the next size when all of the conductors are the same size (total cross-sectional area including insulation) and the calculation results in a decimal of 0.80 or greater. See Chapter 9, Table 1, Note 7.

2. (b) 6 in. x 6 in.

Conductor area [Chapter 9, Table 5]:400 kcmil THHN = 0.5863 sq in. x 3 = 1.7589 sq in.250 kcmil THHN = 0.3970 sq in.4/0 THHN = 0.3237 sq in. x 4 = 0.1295 sq in.8 THHN = 0.0366 sq in. x 3 = 0.1098 sq in.Total Conductor Area = 3.5607 sq in.The wireway must not be filled over 20%, or 1⁄5 [376.22(A)]Conductor Area x 5 = Required Wireway Minimum Area3.5607 sq in. x 5 = 17.8035 sq in. A 6 in. x 6 in. wireway has a cross-sectional area of 36 sq in. and would be large enough.

3. (d) 22 cu in.

2—10 AWG passing through 2—10 AWG1—1 yoke (receptacle) 2—12 AWG4—14 AWG spliced in the box 4—14 AWG2—12 AWG for terminating 2—12 AWG1—12 AWG bonding jumper 0—12 AWG

Total - two 10 AWG conductors, four 12 AWG conductors, and four 14 AWG conductors (note: insulation type does not matter for box fill calculations)

Volume of the conductors: [Table 314.16(B)]10 AWG: 2.50 cu in. x 2 conductors 5.0 cu in.12 AWG: 2.25 cu in. x 4 conductors 9.0 cu in.14 AWG: 2 cu in. x 4 conductors 8.0 cu in.Total 22.0 cu in.



4. (b) 7 conductors[314.16(B)]

Volume of the conductors: 2 - 18 AWG Not counted according to 314.16(B)(1) Ex1—14/3 3—14 AWG conductors1—Ground 1—14 AWG conductor1—switch 2—14 AWG conductors2—clamps 1—14 AWG conductorTotal Count 7—14 AWG conductors

5. (d) 24 in. [314.28(A)]

Straight: Left to Right: 8 x 3 in. = 24 in.Right to Left: 8 x 3 in. = 24 in.Angle: Left to Right: (6 x 3 in.) + 2.50 in. + 2 = 22.50 in.Right to Left: (6 x 3 in.) + 2.50 in. = 20.50 in.

6. (a) 16 in. [314.28(A)]

Straight: None possibleTop to Bottom: Angle; None possibleBottom to Top: Angle; (6 x 2 in.) + 2 in. + 2 in. = 16 in.

7. (a) 12 in. [314.28(A)(2)]

6 x 2 in. = 12 in.

8. (b) 4½ in.

Unit 6—Conductor sizing and protection Calculations

Unit 6—Practice Questions1. (d) a and c

[310.8 and Table 310.13]

2. (d) 4/0 AWG[110.6 and Table 310.16]

3. (d) a and b[310.5]

4. (a) True[110.14(C)(1)(a)]

5. (c) 4 AWGConductors must be selected according to the lowest temperature rating of the equipment (60°C). A THHN/THWN conductor can be used, but it must be sized according to the 60°C terminal rating of the equipment. This requires a 4 AWG conductor [110.14(C)(1)(a)].

6. (b) 8 AWGThis requires an 8 AWG conductor, rated 50A at 75°C in accordance with Table 310.16.



7. (a) True[110.14(C)(1)(b)]

8. (c) 10 AWG[110.14(C)(1)(a)(3) and Table 310.16]

9. (a) 1/0 AWG[110.14(C)(1)(b) and Table 310.16]

10. (a) True[110.14(C)(1)(a)(2) and 110.14(C)(1)(b)(2)]

11. (b) 3/0 AWG[110.14(C)(1)(b) and Table 310.16]

12. (b) 1/0[310.4]

13. (a) True[310.4]

14. (a) True[310.4 and 250.122(F)]

15. (b) False[310.4(E)]

16. (d) 1/0 AWG[310.4 and 250.122(F)]

17. (c) 500 kcmil

750A/2 raceways = 375A [Table 310.16] = 500 kcmil rated 380A each.380A x 2 = 760A conductors, next size up protection (800A) permitted [240.4(B)].

18. (c) 1/0 AWG

Ampere per Parallel Conductor = Total Amperes/Number of Parallel ConductorsAmpere per Parallel Conductor = 250A/2 conductorsAmpere per Parallel Conductor = 125AThe conductor required to carry the load is only 1 AWG according to Table 310.16, however the smallest size conductor permitted to be paralleled is 1/0 AWG [310.4].

19. (a) True[240.1 FPN]

20. (b) 5,000A, 10,000A

Circuit Breakers, 5,000A [240.83(C)]Fuses, 10,000A [240.60(C)]

21. (d) any of these[240.6(A)]

22. (c) 125 percent[210.20(A) and 215.3]

23. (a) True[240.4(D)]



24. (b) False[240.4(B)(1)]

25. (c) 4 AWG [210.20(A) and Table 310.16]

26. (b) 125A [215.3 and 240.6(A)]

27. (d) all of these[240.21(B)(2)]

28. (c) a and b [240.21(C)(4)]

29. (d) all of these[310.10 FPN]

30. (b) FalseTable 310.16 heading, 30°C and 3 current-carrying conductors.

31. (d) 57A

Table 310.16, 55A x 1.04 (Temperature Correction Factor) = 57.2A.

Note: Ampacity increases if the ambient temperature is less than 78°F.

32. (b) 12 AWG

The conductor must have an ampacity of 16A after applying the ambient temperature adjustment factor and it must be protected by a 20A protection device. 240.4(D) requires the conductor to be no smaller than 12 AWG.

Method 1: Ampacity = Table Amperes x Temperature Adjustment14 THHN/THWN—Not permitted to be protected by a 20A protection device.12 THHN/THWN 30A x 0.91 = 27A

Method 2: Conductor Ampacity = Load/Correction factorConductor Ampacity = 16A/0.91 = 17.60A, Table 310.16 14 AWG, this conductor size is not permitted to be protected by a 20A protection device [240.4(D)].

Note: Since there is no mention of the terminations (terminals) being rated for a higher temperature, 110.14C(1)(a) requires us to use the 60°C ampacity column of Table 310.16.

33. (b) 24 in.[310.15(B)(2)(a)]

34. (b) 136A

1/0 THHN at 90°C = 170A [Table 310.16]Ampacity = Table Ampacity x Adjustment Factor of 80%Ampacity = 170A x 0.80 = 136A

35. (c) 10 THHN

The conductors must have an ampacity of 21A and must be protected by the 30A circuit protection device. Table 310.16 (note on the bottom of the table) requires a 10 AWG wire for a 30A overcurrent protection device [240.4(D)]. The adjustment factor is 70% [310.15(B)(2)(a)].Ampacity = Table Ampacity x Adjustment Factor10 THHN 40A x 0.70 = 28A



36. (b) 25A

Ampacity = Ampacity x Temperature Correction x Bundle Adjustment[Table 310.15(B)(2)(a) and Table 310.16]Temperature Correction = 0.91Bundle Adjustment = 0.70Ampacity = 40A x 0.91 x 0.70 = 25.48A

37. (b) False[310.15(B)(4)(a)]

38. (b) False[310.15(B)(4)(c)]

39. (b) False[310.15(B)(4)(b)]

40. (a) True[310.15(A)(2) Ex]

41. (a) True[110.14(C)(1)]


1. (c) 35A

Overcurrent devices must be sized not less than 125% of the continuous load [210.20(A), 215.3, and 240.6(A)]. The overcurrent protection device must be sized not less than 27A x 1.25 = 33.75A. We may round up to the next standard size overcurrent protection device listed in 240.6(A) [240.4(B)].

2. (c) 60A

The overcurrent protection device must be sized not less than 125% of the continuous load [210.20(A), 215.3, and 240.6(A)].

45A x 1.25 = 56.25A. We may round up to the next standard size overcurrent protection device listed in 240.6(A) [240.4(B)].

3. (d) 90A

Overcurrent devices must be sized not less than 125% of the continuous load [210.20(A), 215.3, and 240.6(A)].


4. (c) 150A

Overcurrent devices must be sized not less than 125% of the continuous load [240.6(A), 215.3, and 230.42(A)].


5. (b) 0.82

Table 310.16. Bottom, 60°C wire at 102°F. Be sure to use a straightedge when using a table!



6. (b) 90°C

No correction factors are listed for the 60°C or 75°C columns of Table 310.16.

7. (b) 144A

[310.15(B)(2)(a) and Table 310.16]Ampacity = Table Ampacity x Bundle AdjustmentTable 310.16 Ampacity: 4/0 XHHW aluminum in a wet location = 180A* Six current-carrying conductors factor = 0.80Ampacity = 180A x 0.80 = 144A

*Note: Table 310.13 requires that when XHHW is used in a wet location, the 75°C ampacity column of Table 310.16 must be used. See the definition of “Location, Wet.”

8. (b) 16A[310.15(B)(2)(a) and Table 310.16]

Ampacity = Amperes x Temperature x BundleTable 310.16 Ampacity: 10 RHW aluminum = 30A at 75°CTemperature Adjustment: 75°C wire at 75°F = 1.0515 current-carrying conductors factor = 0.50Ampacity = 30A x 1.05 x 0.50 Ampacity = 15.75A

9. (b) 12[310.15(B)(2)(a) and Table 310.16]

Ampacity = Ampere x Temperature x BundleTable ampacity:14 THHN—25A at 90°C12 THHN—30A at 90°C10 THHN—40A at 90°C 8 THHN—55A at 90°C

Ambient Temperature Correction: 90°C wire at 75°F = 1.049 current-carrying conductors adjustment = 0.7014 THHN Ampacity = 25A x 1.04 x 0.70 = 18.20A (too small)12 THHN Ampacity = 30A x 1.04 x 0.70 = 21.84A

10. (c) 35A

Bundle factors do not apply to raceways that are 24 in. in length or less (nipples) [310.15(B)(2)(a) Ex 3]. Table 310.16 Ampacity: 10 THW at 75°C = 35A

11. (b) 68A

Bundle factors do not apply to raceways that are 24 in. in length or less (nipples) [310.15(B)(2)(a) Ex 3 and Table 310.16].

Ampacity = Amperes x Temperature x BundleTable Ampacity 310.16: 6 THHW = 75A at 90°C

Note: THHW is listed in both the 75°C as well as 90°C column of Table 310.16. Use the 90°C column for dry locations or if the question does not specify a wet location.

Temperature Adjustment 90°C wire at 39°C = 0.91Bundle Adjustment = Does not apply to nipplesAmpacity = 75A x 0.91 = 68.25A



12. (c) 9 conductors[310.15(B)(4)(c) and 310.15(B)(5)]

Current-Carrying Conductors4 wire incandescent luminaires 3 conductors4 wire fluorescent luminaires 4 conductors2 wire receptacles 2 conductors1 ground wire 0 conductors Total current-carrying 9 conductors

Note: electric discharge lighting such as fluorescent is a nonlinear load and produces harmonic currents, so the neutral is counted as current carrying

13. (b) 7 conductors[310.15(B)(4)(c) and 310.15(B)(5)]

Current-Carrying Conductors4 wire incandescent luminaires 3 conductors4 wire fluorescent luminaires 4 conductors1 ground wire 0 conductorsTotal current-carrying 7 conductors

Note: electric discharge lighting such as fluorescent is a nonlinear load and produces harmonic currents, so the neutral is counted as current carrying

Unit 7—Motor and air-Conditioning Calculations


[430.32(A) and 430.52]

2. (c) a and b

Overcurrent is any current in excess of the equipment rating, and can be caused by overload, short circuit, or ground fault [Article 100 definition]. Motor overload is in Article 430, Part III, short-circuit and ground-fault protection is in Article 430, Part IV.

3. (b) full-load current[430.17]

4. (d) 3 hp, 120V, single-phase

25 hp, 460V, three-phase synchronous = 26A [Table 430.250] (be sure to use the synchronous motor section of the table)20 hp, 460V, three-phase = 27A [Table 430.250] 15 hp, 460V, three-phase = 21A [Table 430.250] 3 hp, 120V single-phase = 34A [Table 430.248]

5. (c) 10 AWG

The FLC for a 5 hp, 230V, single-phase motor is 28A [Table 430.248] 28A x 1.25 = 35A, [430.22(A)].10 AWG is rated 35A at 75°C [Table 310.16].

6. (d) a and b[430.24]



7. (d) all of these[430.31]

8. (b) False[430.6(A)(2), 430.32(A)(1), and 430.32(B)(1)]

9. (a) True[430.32(A)(1), 430.32(B)(1), and 430.32(C)]

10. (c) 125[430.32(A)(1)]

11. (c) 125[430.32(A)(1)]

12. (b) 115[430.32(A)(1)]

13. (a) 20A[430.6(A)(2), 430.32(A)(1), and 430.55]

Overloads are sized according to the nameplate current rating, not the motor FLC, 16A x 1.25 = 20A.

14. (d) 40A[430.32(A)(1) and 240.6(A)]

Since no nameplate current is given, the motor FLC listed in Table 430.250 must be used (Be sure to look in the synchronous motor section of the table).FLC = 32A, 32A x 1.25 = 40A.

15. (d) a and b[430.51]

16. (d) all of these[430.52 and Table 430.52]

17. (b) larger[430.52(C)(1) Ex 1]

18. (d) all of these[Table 430.52]

19. (d) 225430.52(C)(1) Ex 2(b)

20. (d) 125, 115, 250[430.22(A), 430.32(A)(1), 430.32(B)(1), and Table 430.52]

There is no relationship between the branch-circuit conductor ampacity and the short-circuit, ground-fault protection device!


10 hp, 208V, three-phase FLC = 30.80A [Table 430.250].

Conductor: 30.80A x 1.25 = 38.50A [430.22(A)]8 AWG is rated 40A at 60°C [Table 310.16]Overload protection [430.32(A)(1)]: 29A (nameplate) x 1.15 = 33A

Short-circuit and ground-fault protection :30.80A (FLC) x 2.50 = 77A [430.52], next size up circuit breaker = 80A [430.52(C)(1) Ex 1]



22. (d) a and b[430.62(A)]

23. (d) b and c

VA three-phase = Motor Voltage rating x Motor Ampere rating x 1.732FLC of 5 hp, 460V, three phase motor = 7.60A FLC of 5 hp, 230V, three phase motor = 15.20A [Table 430.250]VA of 5 hp 460V = 460V x 7.60A x 1.732 = 6,055 VAVA of 5 hp, 230V = 230V x 15.20A x 1.732 = 6,055 VA

24. (a) 3,890 VA

FLC of 3 hp, 208V, single-phase motor = 18.70A [Table 430.248]VA of 3 hp, 208V, single-phase, 208V x 18.70A = 3,890 VA

25 (a) 40A

24A x 1.75 = 42A, next size down = 40A

If the 40A overcurrent device isn’t capable of carrying the starting current, then the overcurrent device can be sized up to 225 percent of the equipment load current rating. 24A x 2.25 = 54A, next size down 50A [240.6(A) and 440.22(A)].

26. (a) 30A

18A x 1.75 = 31.50, next size down = 30A

If the 30A overcurrent device isn’t capable of carrying the starting current, then the overcurrent device can be sized up to 225 percent of the equipment load current rating. 18A x 2.25 = 40.50A, next size down 40A [240.6(A) and 440.22(A)].

27. (c) 10AWG[Table 310.16]

24A x 1.25 = 30A, 10 AWG, rated 30A at 75°C

28. (b) 12 AWG18A x 1.25 = 22.50A, 12 AWG, rated 25A at 75°C


1. (a) 21A[430.22(E)]

Table 430.22(E) Intermittent and 5-minute rated motor: Branch-circuit conductor ampacity must not be less than 85% of the motor nameplate amperes. 25A x 0.85 = 21.25A.



2. (c) 4/0 AWG[430.24(a) and Table 310.16]

The feeder conductor must be sized not less than 125% of the largest motor FLC, plus 100% of the FLCs of all other motors on the same line. Motor FLCs—Tables 430.248 and 430.250.15 hp, 208V, three-phase FLC = 46.20A [Table 430.250]3 hp, 208V, single-phase FLC = 18.70A [Table 430.248]1 hp, 120V, single-phase FLC = 16A [Table 430.248] L1 L2 L33–15 hp, three-phase 46.20 46.20 46.20 46.20 46.20 46.20 46.20 46.20 46.203–3 hp, 208V 18.70 18.70 18.70 18.70 18.70 18.703–1 hp, 120V 16.00 16.00 16.00Total 192.0 192.0 192.0

Feeder conductors must be sized not be less than:(46.20A x 1.25) + 46.20A + 46.20A + 18.70A + 18.70A + 16A = 203.55A4/0 AWG is rated 230A at 75°C.

Note: 3/0 AWG is only rated for 200A at 75°C.

3. (b) 24.70A[430.32(A)(1)]

The motor overload for this motor must be sized no more than 115% of the motor nameplate current rating: 21.5A x 1.15 = 24.73A

4. (d) 60A

The branch-circuit protection device must not be greater than 250% of the motor FLC [430.52(C)(1) and Table 430.52].2 hp, 120V, single phase motor FLC = 24A [Table 430.248]24A x 2.50 = 60A [240.6(A)]

5. (d) 125A

The motor branch-circuit protection device must not be greater than 250% of the motor FLC [430.52(C)(1) and Table 430.52].10 hp, 230V, single-phase motor FLC = 50A [Table 430.248]50A x 2.50 = 125A [240.6(A)]

6. (c) 700A

The branch-circuit protection device must not be greater than 150% of the motor FLC [Table 430.52]125 hp, 240V, dc motor FLC = 425A [Table 430.247]425A x 1.50 = 637.5A, but 430.52(C)(1) Ex 1 permits the next size up, 700A. [240.6(A)]



7. (d) 90A[430.62(A)]

The feeder protection device must not be greater than the largest branch-circuit protection device plus the sum of the motor FLCs on the same line. Largest branch-circuit protection device [430.62(A)]:5 hp, 230V, single phase motor FLC = 28A [Table 430.248]

The branch-circuit protection device must not exceed 250% of the motor FLC: 28A x 2.50 = 70A [430.52(C)(1) and Table 430.52].

Feeder protection not to exceed: 70A device + 20A = 90A [240.6(A)]

Note: The service factor is not used to size short-circuit and ground-fault protection for motor branch-circuits or feeders.

8. (b) 175A[240.6(A), 430.62(A), and Table 430.52]

Feeder protection must be sized not greater than the largest branch-circuit protection device plus the FLCs of all other motors on the same line. Branch-circuit protection devices must not be greater than the values of Table 430.52, except as permitted by 430.52(C)(1) Ex 1.

Motor 1 = 40 hp, 52A x 250% = 130A, Next size up, 150A (Largest Motor Protection)Motor 2 = 20 hp, 27AMotor 3 = 10 hp, 14AMotor 4 = 5 hp, 7.60A

The feeder protection device must not be greater than: 150A branch overcurrent device + 27A + 14A + 7.60A = 199A; the next size down is 175A

9. (b) 225A

The feeder short-circuit protection device must not be greater than the largest branch-circuit protection device, plus the sum of the FLCs of all other motors on the same line [430.62(A)].3 hp, 120V, single-phase motor FLC = 34A [Table 430.248]25 hp, 208V, three-phase motor FLC = 74.80A [Table 430.250]Branch-circuit protection for a 25 hp, 208V, three-phase motor must be sized not more than 250% of the motor full-load current [430.52(C)(1) and Table 430.52].74.80A x 2.50 = 187A, next size up, 200A [240.6(A) and 430.52(C)(1) Ex 1]The feeder protection must not be greater than: 200A + 34A = 234A, next standard size down, 225A [240.6(A) and 430.62(A)].




Step 1: FLC [Table 430.250]30 hp, 460V, three-phase synchronous motor = 32A (be sure to look in the synchronous motor section of the table)10 hp, 460V, three-phase motor = 14A

Step 2: Branch-circuit conductors [430.22(a) and Table 310.16]30 hp: 32A x 1.25 = 40A, 8 AWG, Table 310.16 rated 50A10 hp: 14A x 1.25 = 18A, 14 AWG, rated 20A

Step 3: Branch-circuit protection size [240.6(A), 430.52(C)(1) Ex 1, and Table 430.52]:30 hp: 32A x 2.50 = 80A10 hp: 14A x 2.50A = 35A

Step 4: Feeder conductor [430.24]: (32A x 1.25) + 14A = 54A, 6 THHN conductor

Step 5: Feeder protection [430.62]: The feeder protection device must not be greater than the largest branch-circuit device plus the sum of the FLCs on the same line, 80A + 14A = 94A, Next size down = 90A.

Unit 8—Voltage-Drop Calculations

Unit 8—Practice Questions1. (a) greater

2. (d) all of these

3. (a) Silver

4. (c) Aluminum

5. (c) circular mils[110.6]

6. (a) True

7. (a) positive

8. (d) 8, 9

9. (a) 0.20 ohms

6 AWG resistance = 0.491 ohms/1,000 ft [Chapter 9, Table 8](0.491 ohms/1,000 ft) x 400 ft = 0.1964 ohms

10. (a) 0.0308 ohms


11. (c) 0.403 ohms[Chapter 9, Table 8]

12. (b) 0.16 ohms

1/0 AWG aluminum resistance = 0.201 ohms/1,000 ft [Chapter 9, Table 8](0.201 ohms/1,000 ft) x 800 ft = 0.1608 ohms



13. (a) 0.0154 ohms

1 AWG resistance = 0.154 ohms/1,000 ft [Chapter 9, Table 8](0.154 ohms/1000 ft) x 100 ft = 0.0154 ohms

14. (d) 0.2015 ohms

3 AWG aluminum resistance = 0.403 ohms/1,000 ft [Chapter 9, Table 8](0.403 ohms/1,000 ft) x 500 ft = 0.2015 ohms

15. (a) 0.014 ohms

300 kcmil resistance = 0.0429 ohms/1,000 ft [Chapter 9, Table 8]RTotal = Resistance of One Conductor/Number of Parallel ConductorsRTotal = 0.0429 ohms/3 conductors = 0.0143 ohms

16. (a) True

17. (a) True

18. (c) Eddy

19. (b) skin


21. (a) True

22. (a) 0.03 ohms

2/0 AWG resistance = 0.10 ohms/1,000 ft [Chapter 9, Table 9](0.10 ohms/1,000 ft) x 300 ft = 0.03 ohms

23. (d) 0.032 ohms[Chapter 9, Table 9]

24. (b) 0.025 ohms


25. (a) 0.0108 ohms

500 kcmil resistance = 0.027 ohms/1,000 ft [Chapter 9, Table 9](0.027 ohms/1,000 ft) x 400 ft = 0.0108 ohms

26. (a) 0.05 ohms

1 AWG aluminum resistance = 0.25 ohms/1,000 ft [Chapter 9, Table 9](0.25 ohms/1,000 ft) x 200 ft = 0.05 ohms

Note: You must double the distance to include the resistance in both conductors.

27. (b) 2 AWG[Chapter 9, Table 9]

1/0 AWG aluminum in a steel raceway has a resistance of 0.20 ohms; a 2 AWG copper conductor can be used because its ohms-to-neutral resistance is also 0.20 ohms. The ampacity of 2 AWG copper is 115A at 75°C, Table 310.16

28. (a) 110V

Voltage Source = VD/0.03Voltage Source = 3.30V/0.03Voltage Source = 110V



29. (a) 0.067 ohms

1/0 AWG aluminum = 0.20 ohms/1,000 ft [Chapter 9, Table 9]Assume a steel raceway.R = Resistance of One Conductor/Number of Parallel ConductorsR = 0.20 ohms/3 conductors = 0.0666 ohms, round to 0.067

30. (a) True

31. (b) suggests[90.5(C), 210.19(A)(1) FPN No. 4, and 215.2(A)(3) FPN No. 2]

32. (a) Inductive

33. (b) Electronic

34. (a) True


36. (b) 6.24V

VD = 208V x 0.03 = 6.24V

37. (d) 7.20VVD = 240V x 0.03 = 7.20V

38. (c) 4.76V

The continuous load does not affect voltage drop calculations.

VD = I x RI = 12AR/kft = 1.98 ohms/kft for 12 AWG copper [Chapter 9, Table 8]R = (1.98 ohms/1,000 ft) x 200 ftR = 0.396 ohmsVD = 12A x 0.396 ohmsVD = 4.75V

39. (d) 9.50V

VD = (2 x K x I x D)/CmilK = 12.90 ohms, copperI = 24AD = 160 ftCmil = 10,380VD = (2 wires x 12.90 ohms x 24A x 160 ft)/10,380 cmilVD = 9.50V

40. (c) 4.40V

VD = (1.732 x K x I x D)/CmilK = 21.20 ohms, aluminumI = 100AD = 100 ftCmil = 83,690VD = (1.732 x 21.20 ohms x 100A x 100 ft)/83,690 cmilVD = 4.39V



41. (d) 3 AWG

Cmil = (2 x K x I x D)/VDK = 12.90 ohms, copperI = 52A, not FLCD = 110 ftVD = 115V x 0.03 = 3.45V [210.19(A)(1) FPN No. 4]Cmil = (2 x 12.90 ohms x 52A x 110 ft)/3.45VCmil = 42,776 cmil[Chapter 9, Table 8 = 3 AWG]

42. (b) 8 AWG

Cmil = (2 x K x I x D)/VDK = 12.90 ohms, copperI = 26A at 230VD = 110 ftVD = 6.90V (230V x 0.03) [210.19(A)(1) FPN No. 4]Cmil = (2 x 12.90 ohms x 26A x 110ft)/6.90VCmil = 10,694[Chapter 9 Table 8 = 8 AWG]

43. (a) 10 AWG

Cmil = (1.732 x K x I x D)/VDK = 12.90 ohms, copperI = 18AD = 300 ftVD = 480V x 0.03VD = 14.40VCmil = (1.732 x 12.90 ohms x 18A x 300 ft)/14.40VCmil = 8,379[Chapter 9, Table 8 = 10 AWG]

44. (c) 145 ft

D = (Cmil x VD)/(2 x K x Q x I)Cmil = 16,510VD = 7.20V (240V x 0.03)K = 12.90 ohmsQ = Less than 2/0 AWG, does not applyI = 31A D = (16,510 cmil x 7.20V)/(2 x 12.90 ohms x 31A)D = 118,872/806D = 148.60 or approximately 145 ft

Note: Do not confuse distance (D) with length (L). This formula gives the distance between two points, not the length of conductors in the circuit.



45. (c) 375 ft

D = (Cmil x VD)/(1.732 x K x I)Cmil = 26,240VD = 490V x 0.03VD = 14.40VK = 12.90 ohms, copperI = 45AD = (26,240 cmil x 14.40V)/(1.732 x 12.90 ohms x 45A)D = 375.80 ft

46. (d) 147A

I = (Cmil x VD)/(2 x K x D)Cmil = 1/0 AWG = 105,600 cmilVD = 240V x 0.03VD = 7.20VK = 12.90 ohms, copperD = 200 ftI = (105,500 cmil x 7.20V)/(2 x 12.90 ohms x 200 ft)I = 147A

Note: The maximum load permitted on 1/0 AWG is 150A at 75°C [110.14(C)(1)(b) and Table 310.16].

47. (c) 74A

I = (Cmil x VD)/( 1.732 x K x D)Cmil = 2 AWG, 66,360 cmilVD = 480V x 0.03VD = 14.40VK = 21.20 ohms, aluminumD = 350 ftI = (66,360 cmil x 14.40V)/(1.732 x 21.20 ohms x 350 ft)I = 74A

Note: The maximum load permitted on 2 AWG aluminum is 75A at 60°C [110.14(C)(1)(a) and Table 310.16].

48. (b) 125 [695.6(C)(1)]

49. (c) 15[695.7]

50. (a) 5[695.7]

Unit 8—Challenge Questions1. (b) temperature coefficient

2. (a) impedance



3. (c) 0.05 ohms[Chapter 9, Table 9]

Assume a steel raceway.The ac resistance of 4 AWG aluminum is 0.51 ohms in steel conduit for 1,000 ftR = (Resistance/1000 ft) x Conductor LengthR = (0.51 ohms/1000 ft) x (2 wires x 50 ft)R = 0.051 ohms

4. (b) 6 AWG

No calculations are required for this problem. From Chapter 9, Table 9, the ac resistance of 4 AWG aluminum in a steel raceway (assumed) = 0.51ohms. 6 AWG copper in a steel raceway has an ac resistance of 0.49 ohms [Chapter 9, Table 9].

5. (c) 232.80V

The NEC recommends a maximum 3% voltage drop on the branch-circuit conductors, which calculates out to be: 240V x 0.03 = 7.20V. The minimum voltage at the load is 240V less 7.20V = 232.80V, or 240V x 0.97 = 232.80V.

6. (b) 5.31V

VD = (2 x K x I x D)/CmilK = 21.20 ohms, aluminumI = 55A (use the nameplate, not the FLC)D = 95 ftCmil = 4 AWG, 41,740 cmil, [Chapter 9, Table 8]VD = (2 wires x 21.20 ohms x 55A x 95 ft)/41,740 cmil = 5.31V

7. (b) 3 AWG

Cmil = (2 x K x I x D)/VDK = 12.90 ohms, copperI = 40AD = 150 ftVD = 240V – 236V = 4V already dropped.VD = 7.20V – 4V = 3.20VCmil = (2 x 12.90 ohms x 40A x 150 ft)/3.20VCmil = 48,375 cmil[Chapter 9, Table 8 = 3 AWG]

Note: Table 310.16, 110.14(C)(1)(a) 3 AWG rated 85A at 60°C, okay for 40A load.



8. (d) 4 AWG

Step 1: Determine the voltage drop of the existing conductors:VD = (2 x K x I x D)/CmilK = 21.20 ohms, aluminumI = 50AD = 65 ftCmil = 41,740 [Chapter 9, Table 8]VD = (2 x 21.20 ohms x 50A x 65 ft)/41,740 cmilVD = 3.30V

Step 2: Determine the voltage drop permitted for the extension by subtracting the voltage drop of the existing conductors from the permitted voltage drop. The voltage drop recommended for the extension would be:Recommended VD = 208V x 0.03 = 6.24V6.24V – 3.30 = 2.94V

Step 3: Determine the extended conductor size:Cmil = (2 x K x I x D)/VDK = 12.90 ohms, copperI = 50AD = 85 ftVD = 2.94VCmil = (2 x 12.90 ohms x 50A x 85 ft)/2.94VCmil = 37,296 cmil [Chapter 9, Table 8 = 4 AWG]

Note: 4 AWG is rated for 70A at 60°C [110.14(C)(1)(a) and Table 310.16], which is sufficient to carry the load.

9. (c) 325 ft

D = (Cmil x VD Allowable)/(1.732 x K x I)Cmil = 52,620 cmil [Chapter 9, Table 8]VD = 230V x 3% = 6.90VK = 12.90 ohms, copperI = 50AD = (52,620 cmil x 6.90V)/(1.732 x 12.90 ohms x 50A)D = 325 ft

10. (c) 10A

I = (Cmil x VD Allowable)/(2 x K x D)Cmil = 16,510 cmil [Chapter 9, Table 8]VD = 120V x 3%VD = 3.60VK = 12.90 ohms, copperD = 225 ftI = (16,510 cmil x 3.60V)/(2 x 12.90 ohms x 225 ft)I = 10A



Unit 9—Dwelling Unit Calculations

Unit 9—Practice Questions1. (d) any of these

[220.5(A)]

2. (c) 0.50[220.5(B)]

3. (b) two[210.11(C)(1) and 210.52(B)(1)]

4. (c) 3,000 VA[220.52(A)]

5. (c) 33A

8 kW [Table 220.55, Column C]I = VA/EI = 8,000W/240VI = 33A

6. (b) 37A[Table 220.55, Note 4]

The Column C value (8 kW) must be increased 5% for each kW or major fraction of a kW (0.50 kW or larger) over 12 kW [Table 220.55, Note 1]. 13.50 kW – 12 kW = 1.50 kW (round up to 2)2 x 5% = 10%, an increase of 10% of the Column C value results in 110%, or 1.10 multiplier

Calculated Load = Column C Value x MultiplierCalculated Load = 8 kW x 1.10Calculated Load = 8.80 kW

I = VA/EI = 8,800W/240VI = 37A

7. (b) 25A[Table 220.55, Note 4]

Nameplate = 6,000WI = VA/EI = 6,000W/240VI = 25A



8. (b) 20A[Table 220.55, Note 4]

Nameplate = 4,800WI = VA/EI = 4,800W/240VI = 20A

9. (c) 33A[Table 220.55, Note 4]

Step 1: Total connected load: 6 kW + 3 kW = 9 kW

Step 2: Calculated load for one range: 8 kW, [Table 220.55 Column C]I = VA/EI = 8,000W/240VI = 33A

10. (d) 37A[Table 220.55, Note 4]

Step 1: Total connected load:6 kW + 4 kW + 4 kW = 14 kW

Step 2: Calculated load for one range: 8 kW, [Table 220.55 Column C]

Step 3: Increase Column C value (8 kW) 5% for each kW, or major fraction (0.50 kW), that the range exceeds 12 kW: 14 kW – 12 kW = 2 kW2 x 5% = 10%, resulting in 110% or a 1.10 multiplier8 kW x 1.10 = 8.80 kWI = 8,800W/240VI = 37A

11. (a) True[210.11(C)(2)]

12. (a) 1,500 VA[220.52(B)]

13. (d) all of these[220.12]

14. (a) True[220.12 and 220.14(J)]

15. (c) 6,300 VA(2,100 sq ft x 3 VA) No additional load is required for general-use receptacles and lighting outlets [Table 220.12, Note a].

16. (a) True[210.11(A)]



17. (c) 4 circuits[210.11(A), See Annex D, Example D1(a)]

Step 1: General Lighting VA: 2,340 sq ft x 3 VA = 7,020 VA

Step 2: General Lighting Amperes: I = VA/EI = 7,020 VA/120VI = 58.50A

Step 3: Determine the number of circuits:Circuits = General Lighting Amperes/Circuit AmperesCircuits = 59A/15ACircuits = 3.93Circuits = 4 circuits

Note: Use 120 or 120/240V, single-phase unless specified otherwise [220.5].

18. (b) 3 circuits[210.11(A), See Annex D, Example D1(a)]

Step 1: General Lighting VA: 2,100 sq ft x 3 VA = 6,300 VA

Step 2: General Lighting Amperes: I = VA/EI = 6,300 VA/120VI = 53A



19. (a) 2 circuits[210.11(A), See Annex D, Example D1(a)]

Step 1: Determine the General Lighting VA load: Square footage x 3 VA1,500 sq ft x 3 VA = 4,500 VA

Step 2: Determine the General Lighting Amperes: I = VA/EI = 4,500 VA/120VI = 37.50AI = 38A



20. (a) demand factor[Article 100, 220.42, and Table 220.42]

21. (a) True[220.60]



22. (d) four[220.53]

23. (d) the greater of a or b[220.54]

24. (b) False220.54 requires a dryer load only if the dryer is installed in the dwelling unit.

25. (b) FalseTable 220.55 applies to cooking appliances (household) over 1¾ kW.

26. (a) 3-wire, single-phase, 120/240V systems up to 400A[Table 310.15(B)(6)]

27. (c) 8,100 VA[220.12]

2,700 sq ft x 3 VA = 8,100 VA

28. (d) 24,120 VA[220.12, 220.52]

General Lighting and Receptacles

6,540 sq ft x 3 VA 19,620 VASmall-Appliance Circuits 1,500 VA x 2 3,000 VALaundry Circuit 1,500 VA x 1 1,500 VATotal Connected Load 24,120 VA

29. (d) 9,000 VA

A/C: 5 hp, 230V FLC = 28A [Table 430.248]A/C VA = (V x A) A/C VA = 230V x 28AA/C VA = 6,440 VA* (omit)Heat: [220.51] 3,00 x 3 = 9,000 VA*Omit the smaller of the two loads [220.60].

30. (c) 10,000 VA

A/C: 5 hp, 230V FLC = 28A [Table 430.248] A/C VA = (V x A)A/C VA = (230V x 28A) A/C VA = 6,440 VA* (omit)Heat: 10,000 VA*Omit the smaller of the two loads [220.60].

31. (c) 6,690 VA

Disposal 940 VADishwasher 1,250 VAWater Heater + 4,500 VA

6,690 VA



32. (a) 5,843 VA

Disposal 940 VADishwasher 1,250 VATrash Compactor 1,100 VAWater Heater + 4,500 VAConnected Load 7,790 VA

Calculated Load = Connected Load x Demand Factor [220.53]Calculated Load = 7,790 VA x 0.75 [220.53]Calculated Load = 5,843 VA

33. (c) 5 kW

The dryer load must not be less than 5,000 VA or the nameplate rating if greater than 5 kW (for standard calculations) [220.14(B) and 220.54]. This does not apply to optional calculations [220.82].

34. (d) 5.50 kW

The dryer load must not be less than 5,000 VA or the nameplate rating if greater than 5 kW (for standard calculations) [220.14(B) and 220.54]. This does not apply to optional calculations [220.82].

35. (c) 4.50 kW

[Table 220.55 Column A]: 3 kW x 2 units = 6 kW

Calculated Load = Total Nameplate Rating x Demand Factor [Table 220.55]Calculated Load = 6 kW x 0.75 [Table 220.55 Column A]Calculated Load = 4.50 kW

36. (c) 4.80 kW[Table 220.55 Column B]

Calculated Load = Total Nameplate Rating x Demand Factor [Table 220.55]Calculated Load = 6 kW x 0.80 [Table 220.55 B]Calculated Load = 4.80 kW

37. (d) 9.30 kW[Table 220.55 and Table 220.55 Note 3]

Column B: 6 kW x 0.80 4.80 kWColumn A: 3 kW x 2 units 6 kW x 0.75 + 4.50 kWCalculated Load 9.30 kW

38. (b) 8 kW[Table 220.55 Column C]: 8 kW

39. (c) 8.80 kW

The Column C value (8 kW) must be increased 5% for each kW or major fraction of a kW (0.50 kW or larger) over 12 kW [220.55 Note 1]. 13.60 kW – 12 kW = 1.60 kW (round up to 2)2 x 5% = 10%, an increase of 10% of the Column C value results in 110%, or 1.10 multiplier


40. (b) 2/0 AWG[Table 310.15(B)(6)]



41. (c) 5,100 VA[220.52(A), 220.52(B), Table 220.12, and Table 220.42]

General lighting1,500 sq ft x 3 VA 4,500 VA2 Small-Appliance Circuits (1,500 x 2) 3,000 VALaundry +1,500 VA 9,000 VA -3,000 VA at 100% = 3,000 VA 6,000 VA at 35% = + 2,100 VATotal Calculated Load 5,100 VA

Note: 15A and 20A receptacles are considered part of the general lighting load (3 VA), Table 220.12.

42. (d) 4,140 VAA/C VA Load = 18A x 230VA/C VA Load = 4,140 VAHeat (4 kW) omitted because it’s smaller than air-conditioning [220.60].

43. (c) 5.50 kVA[220.53]

Dishwasher 1,500 VAWater Heater + 4,000 VACalculated Load at 100% 5,500 VA /1,000 = 5.50 kVA

44. (b) 5 kW[220.54]

45. (c) 8.80 kW[Table 220.55, Note 1]

Column C (8 kW) is increased 5% for each kW or major fraction of a kW over 12 kW [220.55 Note 1].14 kW – 12 kW = 2 kW 2 x 5% = 10%, an increase of 10% of the Column C value results in a 110%, or 1.10 multiplier


46. (b) 2 AWG[Table 310.15(B)(6)]I = VA/EI = 30,000 VA/240VI = 125AService Conductor: 2 AWG copper, rated 125A [Table 310.15(B)(6)]

47. (a) 8 AWG[250.102(C) and Table 250.66]

48. (a) 8 AWG[250.66 and Table 250.66]



49. (d) all of these[220.82]

50. (a) 1 AWG AL[220.82]

Step 1: Total connected loadGeneral Lighting 1,500 sq ft x 3 VA 4,500 VASmall-Appliance Circuits 1,500 VA x 2 circuits 3,000 VALaundry Circuit + 1,500 VA 9,000 VAAppliances (nameplate ratings)Disposal 1,000 VADishwasher 1,500 VAWater Heater 5,000 VADryer 5,500 VACooktop 6,000 VAOvens 3,000 VA x 2 + 6,000 VAConnected Load 34,000 VA

Step 2: Determine the calculated load.Connected Load 34,000 VAFirst 10,000 at 100% 10,000 VA at 100% = 10,000 VARemainder at 40% 24,000 VA at 40% = +9,600 VACalculated Load 19,600 VA

Step 3: A/C versus HeatAir-conditioning or heat-pump compressors at 100% versus 65% of one, two, or three separately controlled electric space-heating units [220.82(C)].

A/C: 3 hp, 230V FLC = 17A [Table 430.248]A/C VA = V x AA/C VA = 230V x 17A A/C VA= 3,910 VA, (omit) [220.60]Heat [220.82(C)(4)]: 10,000 VA x 0.65 = 6,500 VA

Step 4: Determine the total calculated load.From Step 2 19,600 VAFrom Step 3 + 6,500 VATotal Calculated Load 26,100 VA

I = VA/EI = 26,100 VA/240VI = 109A

Feeder/Service Conductors: 1 AWG AL rated 110A [Table 310.15(B)(6)].



51. (b) 1/0 AWG AL[220.82]

Step 1: Total connected load.General lighting (and receptacles) 2,330 sq ft x 3 VA 6,990 VASmall-Appliance 1,500 x 2 3,000 VALaundry Circuit 1,500 VA

Appliance (nameplate ratings)Disposal 1,000 VADishwasher 1,500 VATrash Compactor 1,500 VAWater Heater 6,000WRange 14,000 VADryer + 4,500 VAConnected Load 39,990 VA

The dryer load is 4,500 VA, not 5,000 VA! Be careful - 220.54 does not apply to 220.82.

Note: The porch and carport are not counted in the dwelling living space [220.12].

Step 2: Determine the calculated load.Connected Load 39,990 VAFirst - 10,000 VA at 100% = 10,000 VARemainder 29,990 VA at 40% = 11,996 VACalculated Load 21,996 VA

Step 3: A/C versus HeatAir-conditioning or heat-pump compressors at 100% versus 40% of four or more separately controlled electric space-heating units [220.82(C)].A/C: 5 hp, 230V FLC = 28A [Table 430.248]A/C VA = V x AA/C VA = 230V x 28A A/C VA = 6,440 VAHeat [220.82(C)(5)]: 10,000 x 0.40 = 4,000 VA, (omit) .

Step 4: Determine the total calculated load.From Step 2 21,996 VAFrom Step 3 + 6,440 VATotal Calculated Load 28,436 VA

Step 5: Feeder and service conductors, Table 310.15(B)(6).I = VA/EI = 28,436/240VI = 118AFeeder/Service conductors: 1/0 AWG AL rated 125A [310.15(B)(6)].

52. (a) True[220.61]

53. (b) Two 750 kcmil and one 3/0 AWG[110.14(C), 220.61 and Table 310.16]

The ungrounded (hot) conductor is sized at 475A with 75°C terminals - 750 kcmil . The neutral conductor is sized at the maximum unbalanced load, 475A - 275A = 200A; 3/0 AWG.



54. (c) 70[220.61(B)(1)]

55. (b) 5.60 kW

The neutral load is based on the feeder calculated load derived from Table 220.55 Column C, which is 8 kW for one unit. The neutral calculated load is based on 70% of the feeder calculated load of 8 kW [220.61(B)(1)].

Range Neutral Load = Range Load x 70% [220.61(B)(1)]Range Neutral Load = 8 kW x 0.70Range Neutral Load = 5.60 kW

56. (c) 70[220.61(B)(1)]

57. (a) 4.20 kW[220.61(B)(1)]

The neutral load is based on the feeder calculated load derived from Table 220.54, which is the nameplate for one unit. The neutral calculated load is based on 70% of the feeder calculated load of the nameplate [220.61(B)(1)].

Dryer Neutral Load = Dryer Nameplate x 0.70Dryer Neutral Load = 6 kW x 0.70 Dryer Neutral Load = 4.20 kW

Unit 9—Challenge Questions1. (b) 10.40 kW

Table 220.55, Note 4 permits Table 220.55 to be used.Column C (8 kW) is increased 5% for each kW or major fraction of a kW over 12 kW [220.55 Note 1].18 kW – 12 kW = 6 kW 6 x 5% = 30%, an increase of 30% of the Column C value results in a 130%, or 1.30 multiplier


2. (b) 8.80 kW

Table 220.55, Note 4 specifies that the branch-circuit load for one cooktop, and not more than two wall-mounted ovens, can be calculated by combining the units, treating them as one range, and applying the demand factors according to Table 220.55.

Step 1: Combine the ratings together: 9 kW + 5.30 kW = 14.30 kW

Step 2: Determine the Column C demand factor for one unit = 8 kW

Step 3: Increase the Column C demand factor 5% for each kW or major fraction of a kW that the range exceeds 12 kW.14.30 kW - 12 kW = 2.30 kW2 x 5% = 10% increase of Column C, resulting in 110%, or a 1.10 multiplier.




3. (b) 3,882 VA[220.12, 220.42, and 220.52]

Table 220.42 permits a demand factor for the general lighting load (3 VA per sq ft of living space), the two small-appliance circuits, and the laundry circuit.

Note: The laundry circuit load is not required [210.52(F) Ex 1].

Step 1: Total LoadGeneral Lighting 840 sq ft x 3 VA 2,520 VA2 Small-Appliance Circuits 3,000 VALaundry Circuit + 0 VATotal Connected Load 5,520 VA

Step 2: Calculated LoadConnected Load 5,520 VAFirst 3,000 VA at 100% 3,000 VA at 100% = 3,000 VARemainder at 35% 2,520 VA at 35% = 882 VATotal Calculated Load 3,882 VA

4. (c) 5,415 VA[220.12, 220.42, and 220.52]

Table 220.42 demand factors are permitted to be applied to the small-appliance circuit and the laundry circuit. The 3 VA per sq ft includes the general-use receptacles and all general lighting.

Step 1: Total LoadGeneral Lighting 1,800 sq ft (omit porch and carport) 5,400 VASmall-Appliance Circuits 3,000 VALaundry Circuit 1,500 VA x 1 circuit) + 1,500 VATotal Connected Load 9,900 VA

Step 2: Calculated Load 9,900 VAFirst 3,000 VA at 100% 3,000 VA at 100% = 3,000 VARemainder at 35% 6,900 VA at 35% = 2,415 VAGeneral Lighting Calculated Load 5,415 VA

5. (c) 3 circuits

See Annex D, Example D1(a).

Step 1: General Lighting load, 3 VA per sq ft [Table 220.11].

Step 2: Total general lighting load in volt-amperes: 1,800 sq ft x 3 VA = 5,400 VA.

Step 3: Total general lighting load in amperes: 5,400 VA/120V = 45A.

Step 4: Number of general lighting load circuits: 45A/15A = 3 circuits.

6. (b) 4 circuits

See Annex D, Example D1(a).

Step 1: General Lighting load, 3 VA per sq ft [Table 220.11].

Step 2: Total general lighting load in volt-amperes: 2,800 sq ft x 3 VA = 8,400 VA.

Step 3: Total general lighting load in amperes: 8,400 VA/120V = 70A.

Step 4: Number of general lighting load circuits: 70A/20A = 3.50 = 4 circuits.



7. (b) 6.70 kVA

220.53 permits a 75% demand factor for 4 or more appliances. Less than 4 are calculated at their nameplate rating.

Dishwasher 1,200 VATrash Compactor 1,500 VAWater Heater + 4,000 VATotal Connected Load 6,700 VA

8. (b) 5.10 kVA

220.53 permits a 75% demand factor for 4 or more appliances. Less than 4 are calculated at their nameplate rating.

Dishwasher ½ hp (115V x 9.80A) 1,127 VAWater heater +4,000 VATotal Connected Load 5,127 VA

9. (d) 5.50 kW[220.54] Use a minimum of 5 kW or the nameplate rating, whichever is larger.

10. (d) 21 kW

The demand factors of Table 220.55 do not apply to cooking appliances of 1¾ kW and less.1.75 kW x 12 units = 21 kW

11. (a) 10 AWG[210.19(A)(3) Ex 2 and Table 220.55, Note 3]

12. (a) 38.40 kW[Table 220.55]

The word “maximum” is asking for the larger of Column B or C.Table 220.55, Note 3 permits Column B demand factors to be used.

Step 1: Determine the total connected load: 15 units x 8 kW = 120 kW

Step 2: Determine the Column B demand factor for 15 units: 32%

Step 3: Apply the Column B demand factor to the total connected load, Maximum Calculated Load = 120 kW x 0.32 Maximum Calculated Load = 38.40 kW (maximum). Column C for 15 units = 30 kW minimum calculated load.



13. (d) 33 kW[Table 220.55, Note 2]

Step 1: Determine the total connected load.Five ranges at 10 kW = 5 x 12 kW* 60 kWFive ranges at 14 kW = 5 x 14 kW 70 kWFive ranges at 16 kW = 5 x 16 kW +80 kWTotal Connected Load 210 kW* Use 12 kW for any range less than 12 kW

Step 2: Determine average range kW:Average range = 210 kW/15 unitsAverage range = 14 kW

Step 3: Determine the calculated load using Table 220.55 Column C, 15 units = 30 kW.

Step 4: Column C (8 kW) is increased 5% for each kW or major fraction of a kW the average is over 12 kW [220.55 Note 2].14 kW – 12 kW = 2 kW 2 x 5% = 10%, an increase of 10% of the Column C value results in 110%, or 1.10 multiplier

Calculated Load = Column C Value x MultiplierCalculated Load = 30 kW x 1.10Calculated Load = 33 kW

14. (c) 7.80 kW[Table 220.55, Note 3]

Step 1: Determine the total connected load: 6 kW + 6 kW = 12 kW.

Step 2: Column B demand factor for 2 units: 65%.

Step 3: Apply the Column B demand factor to the total connected load. 12 kW x 0.65 = 7.80 kW

Note: Column C for two units = 11 kW.

15. (b) 8 kW

The word “maximum” in a range question is asking for the larger of Column B or C. Table 220.55, Note 3 permits Column C demand factors to be used.

Step 1: Determine the total connected load. 8 kW

Step 2: Determine the Column B demand factor for 1 unit. 80%

Step 3: Apply the Column B demand factor to the total connected load, 8 kW x 0.80 = 6.40 kW.Column C for one unit = 8 kW (maximum).

16. (c) 19.50 kW

Table 220.55, Note 3 permits Column B demand factors to be used.

Step 1: Determine the total connected load.5 units rated 5 kW 25 kW2 units rated 4 kW 8 kW4 units rated 7 kW + 28 kWTotal Connected Load 61 kW

Step 2: Column B demand factor for 11 units: 32% [Table 220.55].

Step 3: Apply the Column B demand factor to the total connected load, 61 kW x 0.32 = 19.52 kW.

Note: Column C for 11 units = 26 kW.



17. (b) 9.30 kWTable 220.55, Note 3 permits Column A and Column B demand factors to be used.

Step 1: Determine the total connected load:Column A = 3 kW + 3 kW = 6 kWColumn B = 6 kW

Step 2: Determine the Column A and Column B demand factors:Column A (2 units) = 75%Column B (1 unit) = 80%

Step 3: Column A demand factor: 6 kW x 0.75 = 4.50 kWColumn B demand factor: 6 kW x 0.80 = 4.80 kWTotal Calculated Load = 4.50 kW + 4.80 kW = 9.30 kW

18. (c) 3/0 AWG[Table 310.16]

The use of Table 310.15(B)(6) is allowed only for the feeder/service conductors to individual dwelling units.

19. (c) 110A[Table 310.15(B)(6)]

I = VA/EI = 24,221 VA/240VI = 101A

20. (b) 100A[220.82]

I = VA/EI = 21,560 VA/240VI = 90A

Note: 220.82 requires a minimum 100A service!

21. (b) 4,000 VA[220.82(C)]

The larger of:Air-conditioning at 100% [220.82(C)(1)] = 4,000 VAHeat at 65% [220.82(C)(4)]: 6,000W x 0.65 = 3,900 VA, (omit)



22. (b) 22 kVA[220.82]

Step 1: Apply demand factors:Total Connected Load 25,000 VAFirst 10 kW at 100% - 10,000 VA at 100% = 10,000 VARemaining at 40% 15,000 VA at 40% = 6,000VACalculated Load 16,000 VA

Step 2: Larger of Air-Conditioning versus HeatA/C at 100% [220.82(C)(1)]: 6,000 VAHeat at 40% [220.82(C)(5)]: 10,000 x 0.40 = 4,000 VA (omit)Total Calculated Load (Step 1 and Step 2) 22,000 VA

Step 3: Total Calculated Load in Amperes:I = VA/EI = 22,000 VA/240VI = 92A

Note: The minimum size service using the optional method is 100A [220.82(A)].

Step 4: Feeder/Service Conductor: 4 AWG rated 100A [Table 310.15(B)(6)]

23. (c) 110A[220.82]

Note: Living space is 1,200 sq. ft + 600 sq. ft = 1,800 sq. ft. The 200 sq. ft for the porch does not count [220.12].

Step 1: General Lighting 1,800 sq ft x 3 VA 5,400 VA

Step 2: Small-Appliance and Laundry Circuits2 Small-Appliance Circuits 1,500 VA x 2 3,000 VA1 Laundry Circuit 1,500 VA

Step 3: Appliances (nameplate rating)Dishwasher 115V x 9.80A 1,127 VAWater Heater 4,000 VA[Table 430.248]Dryer (nameplate, not 5,000W) 4,000 VAOven 6,000 VACooktop 6,000 VA

Step 4: MotorsPool Pump 115V x 13.80A 1,587 VA[Table 430.248]

Step 5: Total Connected Load 32,614 VA

Step 6: Calculated LoadTotal Connected Load 32,614 VAFirst 10,000 at 100% -10,000 VA at 100% = 10,000 VARemainder at 40% ( x 0.40) 22,614 VA at 40% = 9,046 VA 19,046 VA

continued...



Step 7: Larger of Air-Conditioning versus HeatA/C 100%:VA =Volts x AmperesVA = 230V x 28AVA = 6,440 VA + 6,440 VAHeat at 65%: 6,000W x 0.65 = 3,900W [220.82(4)] (omit)

Step 8: Total Calculated Load in VA (Step 6 + Step 7) 25,486 VA


Step 10: Feeder/Service Conductors: 3 AWG rated 110A [Table 310.15(B)(6)]

24. (b) 110A[220.82]


Step 2: Small-Appliance and Laundry CircuitsSmall-Appliance Circuits 1,500 VA x 2 3,000 VALaundry Circuit 1,500 VA 1,500 VA

Step 3: Appliances (nameplate rating)Dishwasher 1,500 VAWater Heater 4,000 VADryer 4,500 VAOvens 2 units x 3,000 VA 6,000 VARange 6,000 VA

Step 4: Totals 31,900 VA

Step 5: Demand FactorsTotal Connected load 31,900 VAFirst 10,000 VA at 100% -10,000 VA at 100% = 10,000 VARemainder at 40% 21,900 VA at 40% = 8,760 VA

Step 6: Larger of A/C versus HeatA/C 100%: 6,000 VA 6,000 VAHeat 40%: 10,000W x 0.40 = 4,000 VA [220.82(5)] (omit)






25. (b) 110A service with 3 AWG[220.82]


Step 2: Small-Appliance and Laundry CircuitsSmall-Appliance Circuits 1,500 VA x 2 3,000 VALaundry Circuit 1,500 VA 1,500 VA

Step 3: Appliances (nameplate rating)Dishwasher 1,200 VAWater Heater 4,000 VADryer (nameplate) 4,000 VARange 13,900 VA

Step 4: MotorsPool Pump (115V x 13.80A or 230 x 6.90) 1,587 VA

Step 5: Totals 34,587 VA

Step 6: Demand FactorsTotal Connected Load 34,587 VAFirst 10,000 VA at 100% -10,000 VA at 100% = 10,000 VARemainder at 40% 24,587 VA at 40% = 9,835 VA

Step 7: Larger of A/C versus HeatA/C 100%: 230V x 28A = 6,440 VA 6,440 VAHeat 65%: 6,000W x 0.65 = 3,900 VA [220.82(4)] (omit)






Chapter 3—aDVanCeD NEC CaLCULatIOns

Unit 10—Multifamily Dwelling Calculations

Unit 10—Practice Questions1. (b) 40,590 VA

[Tables 220.12, 220.42, and 220.52]

General Lighting 840 sq ft x 3 VA 2,520 VASmall Appliance 1,500 VA x 2 3,000 VALaundry Circuit 1,500 VA x 0* + 0 VATotal Connected Load 5,520 VA x 20 units = 110,400 VADemand Factor, Table 220.42Total Connected Load 110,400 VAFirst 3,000 VA at 100% - 3,000 VA x 1.00 = 3,000 VANext 117,000 VA at 35% 107,400 VA x 0.35 = + 37,590 VATotal Calculated Load 40,590 VA*Laundry facilities provided.

2. (a) 51,300 VA[220.52, Table 220.12, and Table 220.42]

General Lighting 990 sq ft x 3 VA 2,970 VASmall Appliance 1,500 VA x 2 3,000 VALaundry Circuit 1,500 VA x 1 + 1,500 VATotal Connected Load 7,470 VA x 20 = 149,400 VADemand Factor, Table 220.42Total Connected Load 149,400 VAFirst 3,000 VA at 100% = - 3,000 VA at 100% = 3,000 VA 146,400 VANext 117,000 VA at 35% = - 117,000 VA at 35% = 40,950 VARemaining VA at 25% = 29,400 VA at 25% = + 7,350 VATotal Calculated Load 51,300 VA

3. (d) 240 kVA

A/C VA Load = 17A x 230V x 40 unitsA/C VA Load = 156,400 VA (156.4 kVA), (omit) [220.60]

Dwelling Heat Load = VA x Number of Heaters in Dwelling Dwelling Heat Load = 3,000 VA x 2 HeatersDwelling Heat Load = 6,000 VA

Heat Load = Dwelling Heat Load x Number of Units [220.51]Heat Load = 6,000 VA x 40 unitsHeat Load = 240,000 VA (240 kVA)



4. (c) 125 kVA

A/C VA Load = 17A x 230V x 24 unitsA/C VA Load = 93,840 VA (93.84 kVA), (omit) [220.60]

Heat Connected Load = VA x Number of Units [220.51] Heat Connected Load = 5,000 VA x 25 UnitsHeat Connected Load = 125,000 VA (125 kVA)

5. (b) 80 kVA[220.53]Waste Disposal 940 VADishwasher 1,250 VAWater Heater + 4,500 VAConnected Load 6,690 VA

Total Connected Load = Connected Load x Number of UnitsTotal Connected Load = 6,690 VA x 16 unitsTotal Connected Load = 107,040 VATotal Calculated Load = Connected Load x Demand Factor [220.53]Total Calculated Load = 107,040 VA x 0.75 [220.53]Total Calculated Load = 80,280 VA

6. (c) 149 kVA[220.53]Waste Disposal 900 VADishwasher 1,200 VAWater Heater + 5,000 VAConnected Load 7,100 VA

Total Connected Load = Connected Load x Number of UnitsTotal Connected Load = 7,100 VA x 28 unitsTotal Connected Load = 198,800 VATotal Calculated Load = Connected Load x Demand Factor [220.53]Total Calculated Load = 198,800 VA x 0.75 [220.53]Total Calculated Load = 149,100 VA

7. (b) 53 kW[220.54]

Dryer Demand Factor % = 35 – [0.50 x (Number of Dryers Exceeding 23)]Dryer Demand Factor % = 35 – [0.50 x (40 dryers – 23)]Dryer Demand Factor % = 35 – (0.50 x 17)Dryer Demand Factor % = 35 – 8.50Dryer Demand Factor % = 35 – 8.50 = 26.50%, change the percent to a decimalDryer Calculated Load = 5 kW* x 40 units x 0.265Dryer Calculated Load = 53 kW*The minimum load is 5 kW for standard calculations

8. (a) 26 kW5.25 kW x 10 units = 52.5 kW x 0.50 = 26.25 kW

9. (c) 18 kW[Table 220.55, Note 3]

Column A: 3.25 kW x 12 units x 0.45 = 17.55 kW



10. (c) 20 kW[Table 220.55, Note 3]

Column B: 7 kW x 8 units x 0.36 = 20.16 kW

11. (b) 20 kW[Table 220.55, Column C]

Note: 12.40 kW is over 12 kW by 0.40 kW, so technically, Table 220.55 Note 1 does apply. But in this case, 0.40 kW is not a major fraction so there is no increase in the Column C calculated load.

12. (c) 17 kW[Table 220.55, Note 1]

Step 1: Column C calculated load, 3 units = 14 kW

Step 2: The 15.50 kW range exceeds 12 kW by 3.50 kW. 0.50 is a major fraction, so the Column C value must be increased by 4 x 5% = 20%. An increase of 20% of the Column C value results in 120%, or a 1.20 multiplier.

Calculated Load = Column C Value x MultiplierCalculated Load = 14 kW x 1.20 Calculated Load = 16.80 kW

13. (a) 22 kW

Step 1: Determine the total connected load:11 kW (use minimum 12 kW): 12 kW x 3 units 36 kW14 kW x 3 units 42 kWTotal Connected Load 78 kW

Step 2: Determine the average range rating:Average = Total Connected/Number of Units78 kW/6 units = 13 kW average

Step 3: Demand load, Table 220.55 Column C: 6 ranges = 21 kW

Step 4: The average range (13 kW) exceeds 12 kW by 1 kW.

Increase the calculated load from Column C (21 kW) by 5%. An increase of 5% of the Column C value results in 105%, or a 1.05 multiplier.


14. (c) 700 kcmil AL[Table 310.16]

I = VA/EI = 90,000 VA/240VI = 375AService Conductors: 700 kcmil aluminum is rated 375A at 75°C [110.14(C) and Table 310.16]

Note: Table 310.15(B)(6) can only be used for individual dwelling units, not for multifamily dwelling feeder/service conductors.



15. (a) 500 kcmil copper[Table 310.16]

I = VA/EI = 90,000 VA/240VI = 375A

Service Conductors: 500 kcmil copper is rated 380A at 75°C [110.14(C) and Table 310.16]

Note: Table 310.15(B)(6) can only be used for individual dwelling units, not for multifamily dwelling feeder/service conductors.

16. (c) Two—500 kcmil

I = VA/(E x 1.732)I = 260,000 VA/(208V x 1.732)I = 260,000 VA/360VI = 722A

Each set must have an ampacity of 722A/2 parallel sets = 361A per parallel set. 500 kcmil is rated 380A at 75°C [110.14(C)(1)(b) and Table 310.16]Two sets (parallel) of 500 kcmil have an ampacity of 760A at 75°C. These conductors (760A) are permitted to be protected by an 800A protection device [240.4(B)].

17. (d) Two—700 kcmil AL


Each set must have an ampacity of 722A/2 parallel sets = 361A per parallel set. 700 kcmil AL is rated 375A at 75°C [110.14(C)(1)(b) and Table 310.16]Two sets (parallel) of 700 kcmil AL have an ampacity of 750A at 75°C. These conductors (750A) are permitted to be protected by an 800A protection device [240.4(B)].

18. (a) 40 kVA[220.52, Table 220.12, and Table 220.42]

(a) General Lighting 1,500 sq ft x 3 VA 4,500 VA(b) 2 Small-Appliance Circuits 3,000 VA(c) Laundry Circuit + 1,500 VATotal Connected Load 9,000 VA x 12 units = 108,000 VADemand Factor, Table 220.11Total Connected load 108,000 VAFirst 3,000 VA at 100% - 3,000 VA x 1.00 = 3,000 VARemaining VA at 35% 105,000 VA x 0.35 = + 36,750 VACalculated Load 39,750 VA

19. (c) 60 kVA[220.51, 220.60, and Table 430.28]

A/C VA Load = 18A x 240V x 12 unitsA/C VA Load = 49,680 VA (49.68 kVA (omit) [220.60]

Heat Connected Load = VA x Number of Units [220.51] Heat Connected Load = 5,000 VA x 12 UnitsHeat Connected Load = 60,000 VA (60 kVA)



20. (c) 50 kVA[220.53]Dishwasher 1,500 VAWater heater +4,000 VATotal Calculated Load 5,500 VA x 12 units x 0.75 = 49,500 VA

Note: The total number of appliances on the feeder is 24, so the 75% demand factor applies.

21. (a) 27.60 kW[220.54]

Dryer Demand Factor % = 47 – (Number of Dryers Exceeding 11)Dryer Demand Factor % = 47 – (12 dryers – 11)Dryer Demand Factor % = 47 – 1Dryer Demand Factor % = 46%Dryer Calculated Load = 5 kW* x 12 units x 0.46Dryer Calculated Load = 27.60 kW*The minimum load is 5 kW for standard calculations

22. (b) 30 kW[Table 220.55, Note 1]

Column C = 27 kWColumn C (27 kW) increased by 5% for each kW or major fraction of a kW that the range (14.45 kW) is over 12 kW. 14.45 kW – 12 kW = 2.452 kW x 5% = 10% increase, resulting in 110%, or a 1.10 multiplier27 kW x 1.10 = 29.70 kW

23. (c) 1,000A[240.4(C)]

I = VA/EI = 206,000 VA/240VI = 858A

Note: The 858A service/feeder calculated load requires a 1,000A service [240.6(A)]. Since this is over 800A, the next size up rule [240.4(B)] does not apply. For devices rated over 800A [240.4(C)], the ampacity of the conductors must be equal to, or greater than, the rating of the 1,000A overcurrent device.

24. (b) 1/0 AWG

The equipment bonding conductor on the supply side of the service, for parallel service conductors, is sized based of the conductor size in each raceway and on Table 250.66 [250.102(C)].

25. (c) 3/0 AWG

The grounding electrode conductor is sized on the largest ungrounded conductor, or equivalent area where run in parallel. 400 kcmil x 3 raceways = 1,200 kcmil, Table 250.66, use 3/0 AWG.

26. (b) The larger of the air-conditioning load or the space-heating load[220.84(C)]

100% A/C versus 100% Heat



27. (b) 130 kVA[Table 220.84](a) General Lighting 1,500 sq ft x 3 VA = 4,500 VA [Table 220.12, 220.84(C)(1)](b) 2 Small-Appliance Circuits 3,000 VA [ 220.84(C)(2)](c) Laundry + 1,500 VA [ 220.84(C)(2)] 9,000 VA

Total Calculated Load = VA x Number of Units x Demand Factor [Table 220.84]Total Calculated Load = 9,000 VA x 60 Units x 0.24 [Table 220.84]Total Calculated Load = 540,000 VA x 0.24 Total Calculated Load = 129,600 VA (130 kVA)

28. (d) 72 kVA[220.84, 220.60, and Table 430.248]

A/C: 3 hp, 230V FLC = 17AA/C VA = V x AA/C VA = 230V x 17AA/C VA = 3,910 VAA/C Fan: 1⁄8 hp* 230V FLC = 1.45A

Fan VA = V x AFan VA = 230V x 1.45AFan VA = 334 VA* 1⁄8 use ½ of the FLC for ¼ hp from Table 430.248.

Total A/C and Fan VA = 3,910 VA + 334 VATotal A/C and Fan VA = 4,244 VA

Total Calculated A/C Load = VA x Number of Units x Demand Factor [Table 220.84]Total Calculated A/C Load = 4,244 VA x 60 units x 0.24 [Table 220.84]Total Calculated A/C Load = 61,114 VA, (omit)

Heat VA = VA x Number of Units x Demand Factor [Table 220.84]Heat VA = 5,000 VA x 60 units x 0.24 = 72,000 VA (72 kVA)

29. (d) 80 kVA[220.84(C)(3)]Dishwasher 1,500 VAWater heater + 4,000 VAUnit Connected Load 5,500 VA

Total Calculated Load = Unit Connected Load x Number of Units x Demand Factor [Table 220.84]Total Calculated Load = 5,500 VA x 60 Units x 0.24 [Table 220.84]Total Calculated Load = 79,200 VA (79.20 kVA)

30. (a) 58 kW[220.84(C)(3)]

Dryer Calculated Load = Nameplate Rating x Number of Units x Demand FactorDryer Calculated Load = 4 kW x 60 Units x 0.24 [220.84(C)(3)]Dryer Calculated Load = 57.6 kW

Note: When using the optional method, all loads are at the nameplate rating.



31. (c) 65 kW[220.84(C)(3)]

Dryer Calculated Load = Nameplate Rating x Number of Units x Demand FactorDryer Calculated Load = 4.50 kW x 60 Units x 0.24 [220.84(C)(3)]Dryer Calculated Load = 64.80 kW

32. (d) 200 kW[220.84(C)(3)]

Range Calculated Load = Nameplate Rating x Number of Units x Demand FactorRange Calculated Load = 14 kW x 60 Units x 0.24 [220.84(C)(3)]Range Calculated Load = 201.60 kW

33. (b) 800A[240.6(A)]


34. (b) 250 kVA[220.82]

Step 1: Determine the total connected load.General Lighting 900 sq ft x 3 VA 2,700 VATwo Small-Appliance Circuits (required) 3,000 VALaundry Circuit (required) 1,500 VAWater Heater 5,000 VARange 14,000 VAAir-Conditioning (230V x 28A) 6,440 VAHeat 5 kW, (omit) [220.60] + 0 VATotal Connected Load 32,640 VA x 20 units = 652,800 VA

Step 2: Calculated load = 652,800 VA x 0.38 = 248,064 VANote: Service size

I = VA/EI = 248,064 VA/240VI = 1,034A

35. (c) 1/0 AWG

The equipment bonding conductor on the service side of the service, for parallel service conductors, is sized based on the conductor size in each raceway and Table 250.66 [250.102(C)].

36. (b) 2/0 AWG

The grounding electrode conductor is sized base on the largest ungrounded conductor, or equivalent area where run in parallel. 500 kcmil x 2 raceways = 1,000 kcmil, Table 250.66, use 2/0 AWG.



Unit 10—Challenge Questions1. (c) 54 kVA

[Table 220.12]

900 sq ft x 3 VA x 20 units = 54,000 VA

2. (b) 41 kVA

Tables 220.12, 220.42, and 220.52 permit a demand factor for the general lighting load (3 VA per sq ft of living space), the two small-appliance circuits, and the laundry circuit.

Note: Laundry circuit load not required [210.52(F) Ex 1].

Step 1: Total Connected Load.General Lighting and Receptacles (840 sq ft x 3 VA) 2,520 VASmall-Appliance Circuits (1,500 VA x 2 circuits) 3,000 VALaundry Circuit + 0 VAUnit Load 5,520 VATotal Connected Load 5,520 VA x 20 units = 110,400 VA

Step 2: Demand Factors, Table 220.42First 3,000 VA at 100%: - 3,000 VA x 1.00 = 3,000 VANext 117,000 VA at 35%: 107,400 VA x 0.35 = 37,590 VATotal Calculated Load 40,590 VA

3. (b) 106,500 VA[220.53]Waste Disposal 900 VADishwasher 1,200 VAWater Heater + 5,000 VATotal Unit Connected Load 7,100 VA

Total Calculated Load = Unit Connected Load x Number of Units x Demand Factor [220.53]Total Calculated Load = 20 Units x 0.75Total Calculated Load = 106,500 VA

4. (b) 25 kW

220.54 requires a minimum of 5 kW for each dryer, and the demand factor for 10 units is 50% [Table 220.54],

Calculated Load = VA x Number of Units x Demand FactorCalculated Load = 5 kW x 10 Units x 0.50 [Table 220.54]Calculated Load = 25 kW.

5. (d) 54 kW[Table 220.55, Note 1]

15.80 exceeds 12 kW by 3.80, 0.80 is a major fraction so the Column C value must be increased by 4 x 5% = 20%. An increase of 20% of the Column C value results in 120%, or a 1.20 multiplier

Column C: 30 units = 15 kW + 1 kW for each range in the calculation, 15 kW + 30 kW = 45 kW

Calculated Load = Column C Value x MultiplierCalculated Load = 45 kW x 1.20 Calculated Load = 54 kW



6. (c) 33 kW[Table 220.55, Note 2]

Step 1: Determine the total connected load (using 12 kW as the minimum*).*Five ranges at 10 kW = 5 units x 12 kW 60 kWFive ranges at 14 kW = 5 units x 14 kW 70 kWFive ranges at 16 kW = 5 units x 16 kW + 80 kWTotal Connected Load 210 kW

Step 2: Determine the average range kW;Average range = Total kW/Number of Ranges210 kW/15 units = 14 kW.


Step 4: Increase the Column C demand 5% for each kW or major fraction of a kW that the average range exceeds 12 kW. The average range (14 kW) exceeds 12 kW by 2 kW; therefore, Column C must be increased by 2 x 5% = 10%.An increase of 10% of the Column C value results in 110%, or a 1.10 multiplier

Calculated Load = Column C Value x MultiplierCalculated Load = 30 kW x 1.10 Calculated Load = 33 kW.

7. (d) 36.75 kW[Table 220.55, Note 2]

Step 1: Determine the total connected load (using 12 kW as minimum*).Ten ranges at 12 kW (10 units x 12 kW) 120 kWEight ranges at 14 kW (8 units x 14 kW) 112 kW*Two ranges at 9 kW (2 units x 12 kW) + 24 kWTotal Connected Load 256 kW

Step 2: Determine the average range kW; Average Range = Total kW/Number of Ranges256 kW/20 units = 12.80 kW.


Step 4: Increase the Column C demand 5% for each kW or major fraction of a kW that the average range exceeds 12 kW. The average range (12.8 kW) exceeds 12 kW by 1 kW; therefore, Column C must be increased by 5%, which results in 105%, or a 1.05 multiplier.


Note 3 to Table 220.55 Calculations

8. (d) 19.50 kW[Table 220.55, Note 3]

Step 1: Determine the total connected load.5 units rated 5 kW 25 kW2 units rated 4 kW 8 kW4 units rated 7 kW + 28 kWTotal Connected Load 61 kW

Step 2: Apply the Column B demand factor (11 units) to the total connected load, 61 kW x 0.32 = 19.52 kW.

Note: The Column C Value for eleven units = 26 kW.



9. (b) 38.40 kW[Table 220.55, Note 3]

The word “maximum” in a range question is asking for the larger of Columns B or C.Table 220.55, Note 3 permits the Column B demand factors to be used.

Step 1: Determine the total connected load = 15 units x 8 kW = 120 kW

Step 2: Apply the Column B demand factor to the total connected load, 120 kW x 0.32 = 38.40 kW.Column C for fifteen units = 30 kW minimum calculated load.

10. (c) 70[220.61(B)(1)]

11. (a) 21 kW[220.61(B)(1) and Table 220.55]

Column C, Fifteen units: 30 kW

Neutral Calculated Load = Range Calculated Load x 0.70 [220.61(B)(1)]Neutral Calculated Load = 21 kW

12. (a) 17.50 kW[220.54 and 220.61(B)(1)]

Dryer Calculated Load = Nameplate x Number of Units x Demand Factor [Table 220.54]Dryer Calculated Load = 5 kW x 10 Units x 0.50Dryer Calculated Load = 25 kW

Neutral Calculated Load = Dryer Calculated Load x Demand Factor [220.61(B)(1)]Neutral Calculated Load = 25 kW x 0.70 Neutral Calculated Load = 17.50 kW

13. (a) 18.10 kW[220.54 and 220.61(B)(1)]

Dryer Calculated Load = Nameplate x Number of Units x Demand Factor [Table 220.54]Dryer Calculated Load = 5 kW* x 11 Units x 0.47Dryer Calculated Load = 25.85 kW

Neutral Calculated Load = Dryer Calculated Load x Demand Factor [220.61(B)(1)]Neutral Calculated Load = 25.85 kW x 0.70 Neutral Calculated Load = 18.10 kW*The minimum is 5 kW for the standard calculation method.

14. (d) 17.50 kW[220.61(B)(1) and Table 220.55]

Range Calculated Load = Column C, ten units = 25 kW [Table 220.55]

Neutral Calculated Load = Range Calculated Load x Demand Factor [220.61(B)(1)]Neutral Calculated Load = 25 kW x 0.70 Neutral Calculated Load = 17.50 kW



Unit 11— Commercial Calculations

Unit 11—Practice Questions1. (b) ampacity

[Article 100 definition]

2. (d) overcurrent protection[Article 100 definition of overcurrent]

3. (a) True[240.6(A)]

4. (b) 0.49[220.5(B)]

5. (b) 15 kVA[Tables 220.12 and 220.42]

24 motel rooms x 685 sq ft = 16,440 sq ft16,440 sq ft x 2 VA 32,880 VAFirst 20,000 VA at 50% - 20,000 VA x 0.50 10,000 VANext 80,000 VA at 40% 12,880 VA x 0.40 + 5,152 VATotal Calculated Load 15,152 VA

6. (b) 56 kVA[Tables 220.12 and 220.42]General lighting (250,000 sq ft x 0.25 VA) 62,500 VAReceptacles (200 x 180 VA)* + 36,000 VATotal Connected Load 98,500 VAFirst 12,500 VA at 100% - 12,500 VA x 1.00 12,500 VARemainder at 50% 86,000 VA x 0.50 + 43,000 VATotal Calculated Load 55,500 VA

*The receptacle load (180 VA for each receptacle) is permitted to be included with the general lighting for storage warehouses, see 220.44 of the NEC for details.

7. (c) 8,000 VA[215.2(A), 215.3, 230.42(A), and Table 220.12]

General Lighting Load (3,200 sq ft x 2 VA x 1.25) = 8,000 VA

8. (d) 338 kVA[215.2(A), 215.3, 230.42(A), and Table 220.12]

General Lighting (90,000 sq ft x 3 VA x 1.25) = 337,500 VA

9. (b) 33 kVA[220.43(A), 210.19(A)(1), and 210.20(A)]

Total VA = 130 ft x 200 VA per footTotal VA = 26,000 VA

Total VA for Continuous Load = VA x 1.25Total VA for Continuous Load = 26,000 VA x 1.25 Total VA for Continuous Load = 32,500 VA.

32,500 VA/1,000 = 32.50 kVASee Annex D Example D3.



10. (a) 3,600 VA

Multioutlet Receptacle Assembly [220.14(H)]VA per 5 ft = 180 VA when unlikely to be used simultaneously:

Number of 180 VA Sections = 50 ft/5 ft perNumber of 180 VA Sections = 10

VA Load = Number of sections x 180 VAVA Load = 10 x 180 VAVA Load = 1,800 VA

VA per 1 ft = 180 VA when likely to be used simultaneously:Number or 180 VA Sections = 10 ft/1 ft perNumber of 180 VA Sections = 10 sections

VA Load = Number of sections x 180 VAVA Load = 10 x 180 VAVA Load= 1,800 VA

Total multioutlet receptacle assembly load = 3,600 VA

11. (b) 13[220.14(I)]

Circuit VA = Circuit Volts x Circuit AmperesCircuit VA = 120V x 20ACircuit VA = 2,400 VA

Receptacles per Circuit = Circuit VA/180VAReceptacles per Circuit = 2,400 VA/180 VAReceptacles per Circuit = 13 receptacles.

Note: Receptacles are not considered a continuous load.

12. (c) 15 kVA[220.44 and Table 220.44]110 receptacles x 180 VA 19,800 VAFirst 10,000 VA at 100%: -10,000 VA x 1.00 = 10,000 VARemainder at 50%: 9,800 VA x 0.50 = + 4,900 VATotal Receptacle Calculated Load 14,900 VA

13. (c) 162 kVA[220.12, 220.14(K)(2), 220.44, and 215.3]

General Lighting (30,000 sq ft x 3.5 VA x 1.25) = 131,250 VAReceptacle load, Table 220.12 Note and 220.14(K)1 VA per sq ft 30,000 VATotal General Lighting and Receptacle Calculated Load 161,250 VA 151,250 VA



14. (d) 56 kVA[220.12, 220.14(K), 215.2(A), 230.42(A), and 220.44]General Lighting (10,000 sq ft x 3.5 VA x 1.25) 43,750 VAReceptacle Demand: Larger of 220.14(K)(1) and (2)220.14(K)(1): 75 receptacles x 180 VA 13,500 VA220.14(K)(2): load at 1 VA per sq ft: 10,000 VA (omit)Apply 220.44 13,500 VA1st 10,000 VA at 100% -10,000 VA x 1.00 = 10,000 VARemainder at 50% 3,500 VA x 0.50 = + 1,750 VAReceptacle Calculated Load 11,750 VA

Total General Lighting and Receptacle Calculated Load = 43,750 VA + 11,750 VATotal General Lighting and Receptacle Calculated Load = 55,500 VA

15. (b) 1,500 VA[220.14(F), 600.5, 215.2(A), and 230.42(A)]

The sign calculated conductor load:1,200 VA x 1.25 = 1,500 VA

16. (b) 550A555.12 and Table 555.12 permit a demand factor according to the number of receptacles. 54 receptacles = 40% demand factor.24 receptacles x 20A 480A30 receptacles x 30A 900A 1,380A x 0.40 (demand factor) = 552A

17. (c) 300 kcmil

570A/2 raceways = 260A[Table 310.16] 300 kcmil conductors have a rating of 285A each x 2 conductors = 570A combined. Because the load is less than 800A, the conductor must [240.4(B)]:(1) have an ampacity of at least 570A,(2) be protected by a 600A protection device, and(3) must be sized according to Table 310.16, 300 kcmil, using the 75°C terminal rating [110.14(C)(1)(b)].

Note: 250 kcmil THHN conductor has an ampacity of 290A each, but we cannot use the 90°C ampacity rating for sizing conductors [110.14(C)(1)(b)].

18. (c) 644A[Table 550.31]

The feeder and service for mobile home parks is sized using the demand factors of Table 550.31.Be careful. The demand factors of Table 550.31 apply to the larger of:(1) 16,000 VA for each mobile home lot [550.13(I)] or(2) The calculated load for the largest typical mobile home each lot will accept [550.31(2)].In this question, 16,000 VA must be used for the calculation because we don’t know the calculated load.16,000 VA per site x 42 sites x 0.23 (demand factor) = 154,560 VA

I = VA/EI = 154,560 VA/240VI = 644A



19. (b) 400A[551.73(A)]

Recreational vehicle parks are calculated using the demand factors of Table 551.73.

Step 1: Each 20A receptacle site dedicated to tents is rated a minimum of 600 VA, each 20A or 30A site is rated 3,600 VA, and each 50A site is rated 9,600 VA [551.73(A)].

Step 2: The total connected load is:Total Load = (17 sites x 600 VA) + (35 sites x 3,600 VA) + (10 sites x 9,600 VA)Total Load = 10,200 VA + 126,000 VA + 96,000 VATotal Load = 232,200 VA

Step 3: The demand factor for 62 sites listed in Table 551.73 is 0.41.

Step 4: Total calculated load: 232,200 VA x 0.41 = 95,202 VA

Step 5: The total ampere rating is calculated by:I = VA/EI = 95,202 VA/240VI = 396.70A

20. (a) 210 kVA

Calculated Load = 50% x (amount over 325 kVA) + 172.50 kVA [Table 220.88]Calculated Load = 0.50 x (400 kVA – 325 kVA) + 172.50 kVACalculated Load = 0.50 x 75 kVA + 172.50 kVACalculated Load = 37.50 kVA + 172.50 kVACalculated Load = 210 kVA

21. (b) 296 kVA

Calculated Load = 45% x (amount over 325 kVA) + 262.50 kVA [Table 220.88]Calculated Load = 0.45 x (400 kVA – 325 kVA) + 262.50 kVACalculated Load = 0.45 x 75 kVA + 262.50 kVACalculated Load = 33.75 kVA + 262.50 kVACalculated Load = 296.25 kVA

Unit 11—Challenge Questions1. (a) 23 kVA

[Table 220.12 and Table 220.42]

Hotel:12 x 15 ft = 180 sq ft x 2 VA x 100 Rooms 36,000 VAFirst 20,000 VA at 50% -20,000 VA x 0.50 10,000 VARemainder at 40% 16,000 VA x 0.40 6,400 VA

Office:60 x 20 ft = 1,200 sq ft x 3.5 VA x 1.25 (continuous load) 5,250 VAOffice Receptacles: 1,200 sq ft x 1 VA* 1,200 VAHalls: 120 sq ft x 0.5 VA x 1.25 (continuous load) + 75 VA 22,925 VA /1,000 = 22.93 kVA

* [220.12(K)(2) and Table 220.12 Note b] add 1 VA per sq ft for unknown receptacle count



2. (c) 14,000 VA[220.44]

Table 220.44; Nondwelling Unit Receptacles at 180 VA each100 Receptacles x 180 VA 18,000 VAFirst 10,000 VA at 100%: 10,000 VA x 1.00 10,000 VARemainder at 50%: 8,000 VA x 0.50 + 4,000 VATotal Calculated Load 14,000 VA

3. (b) 20,000 VA[220.14(K)(2) and Table 220.12]

General-Use Receptacles:20,000 sq ft x 1 VA per sq ft* 20,000 VA

* Allow 1 VA per sq ft for receptacles when the number of receptacles is unknown in banks and office buildings. The demand factors of 220.42 apply when the receptacle calculation is based on 220.14(I) but not when based on 220.14(K) [220.44]

4. (c) 161 kVA[220.14(K)(2), Table 220.12, and Table 220.44]

General Lighting (30,000 sq ft x 3.5 VA x 1.25) 131,250 VAGeneral-Use Receptacles:30,000 sq ft x 1 VA per sq ft* 30,000 VATotal Lighting and Receptacles 161,250 VA161,250/1,000 = 161.25 kVA

* Allow 1 VA per sq ft for receptacles when the number of receptacles is unknown in banks and office buildings. The demand factors of 220.42 apply when the receptacle calculation is based on 220.14(I) but not when based on 220.14(K) [220.44]

5. (b) 630A[555.12]

Step 1: Determine the total connected receptacle ratings:20 receptacles rated 20A (20 x 20A) 400A17 receptacles rated 30A (17 x 30A) 510A7 receptacles rated 50A (7 x 50A) + 350ATotal Connected Receptacle Rating 1,260A

Step 2: Determine the demand factor from Table 555.12, which is 50% for 44 receptacles

Step 3: Determine the calculated load, 1,260A x 0.50 = 630A

6. (c) 264 kVA[550.31]

The calculation is based on a minimum of 16,000 VA

Step 1: Determine the total connected load: 75 sites x 16,000 VA = 1,200,000 VA

Step 2: Determine the demand factor from Table 550.31, 22%

Step 3: Determine the total calculated load:1,200,000 VA x 0.22 = 264,000 VA264,000VA/1,000 = 264 kVA



7. (a) 83 kVA[551.71 and 551.73(A)]

Note: A minimum of 70% of the sites must have a 30A or 20A facility (3,600 VA per site) and a minimum of 20% of the sites must have a 50A facility (9,600 VA per site). Check: 42 sites x 0.70 = 29 minimum 3,600 VA sites42 sites x 0.20 = 8.40 or 9 minimum 9,600 VA sites.

Step 1: Determine the total connected load:9 sites at 50A (9 sites x 9,600 VA) 86,400 VA30 sites at 30A (30 sites x 3,600 VA) 108,000 VA3 sites at 20A (3 sites x 2,400 VA) + 7,200 VATotal Connected Load 201,600 VA

Step 2: Determine the demand factor for 42 sites [Table 551.73]: 41%

Step 3: Determine the calculated load:Calculated Load = 201,600 VA x 0.41Calculated Load = 82,656 VACalculated Load in kVA = 82,656 VA/1,000 = 82.66 kVA

8. (d) 64 kVA[220.88]

Step 1: Determine the connected load:General Lighting 30 kVADishwasher 5 kWCoffee Makers (2 kW x 2 units) 4 kWKitchen Appliances (2 kW x 5 units) 10 kVASmall-Appliances Circuits (1.50 kVA x 10 units) + 15 kVATotal Connected Load 64 kVA

Step 2: Apply the Table 220.88 demand factor for not all electric: 64 kVA at 100% = 64 kVA calculated load.

Unit 12—transformer Calculations

Unit 12—Practice Questions

1. (a) series

2. (a) Delta

3. (b) line

4. (b) Line

5. (a) True

6. (d) 208V

High Leg Voltage = Voltage to Ground x 1.732High Leg Voltage = 120V x 1.732 High Leg Voltage = 208V

7. (d) orange [110.15]



8. (b) Wye

9. (b) False

10. (b) 2:1, 4:1

11. (a) 4:1480 primary phase volts:120 secondary phase volts

12. (d) 2:1480 primary phase volts to 240 secondary phase volts

13. (a) True

14. (b) False

15. (b) 54A

ILine = VALine/(ELine x 1.732)I = 45,000 VA/(480V x 1.732)I = 45,000 VA/831VI = 54A

16. (c) 108A

ILine = VALine/(ELine x 1.732)I = 45,000/(240V x 1.732)I = 45,000/416VI = 108A

17. (a) True

18. (b) 26A

ILine = VALine/(ELinee x 1.732)I = 22,000 VA/(480V x 1.732)I = 22,000 VA/831VI = 26A

19. (d) 61A

ILine = VALine/(ELine x 1.732)I = 22,000 VA/(208V x 1.732)I = 22,000 VA/360VI = 61A

20. (a) 125[Table 450.3(B)]

21. (b) 125[Table 450.3(B)]

22. (b) 4 AWG4 AWG is rated 85A at 75°C [Table 310.16]

23. (c) 2 AWG2 AWG is rated 115A at 75°C [Table 310.16]

24. (a) 2/0AWG2/0 is rated 175A at 75°C [Table 310.16]



25. (c) 350 kcmil350 kcmil is rated 310A at 75°C [Table 310.16]. A conductor with the full 300A ampacity must be used to terminate into the 300A overcurrent device, as this is a transformer secondary tap [240.21(C)].

26. (c) at any single point[250.30(A)(1)]

27. (a) True[250.30(A)(3)]

28. (d) a and b[250.30(A)(6)]

29. (a) 250.66 [250.30(A)(3)]

30. (d) 4 AWG[250.66(B)]

31. (a) True[110.9]

Unit 12—Challenge Questions1. (a) 45A

IPrimary = VAPrimary/(EPrimary x 1.732)IPrimary = (35.70 kVA x 1000)/(480V x 1.732)IPrimary = (35.70 kVA x 1000)/831VIPrimary = 45.10A

2. (c) 113A

There are two ways to calculate this problem.

(1) Calculate the primary line current, then use the ratio to determine the secondary line current:

Primary Line Current = Primary Phase Power/(Primary Phase Volts x 1.732)Primary Line Current = 45,000 VA/(460V x 1.732)Primary Line Current = 45,000 VA/797VPrimary Line Current = 56.50A

Next, use the ratio to determine the secondary voltage, and then calculate the secondary line current.The ratio is 2:1, therefore the secondary voltage is ½ of the primary voltage and the current will be twice the primary line current.

Secondary Line Current = 56.50A x 2Secondary Line Current =113A

(2) The voltage ratio is 2:1. If the primary voltage is 460V, the secondary is 230VSecondary line current = 45,000 VA/(230V x 1.732)Secondary line current = 113A



3. (a) 17A

Step 1: Determine the primary line power.A. The primary line power is the same as the secondary line power, assuming 100% efficiency.B. The secondary line power is calculated as:

VA = Secondary Line Volts x Secondary Line Amperes x 1.732Secondary Line Power = 240V x 300A x 1.732Secondary Line Power = 124,704 VA

Step 2: Determine the primary line current.

Primary Line Current = Primary Line Power/(Primary Line Voltage x 1.732)Primary Line Current = 124,704 VA/(4,160V x 1.732) Primary Line Current = 17A

4. (d) 87A

Assuming 100% efficiency, the primary VA is equal to the secondary VA.

Step 1: Determine the transformer VA.Secondary VA = E x I x 1.732Secondary VA = 208V x 200A x 1.732Secondary VA = 72,051 VA

Step 2: Determine the primary line current.ILine = VALine/(ELinee x 1.732)ILine = 72,051 VA/(480 x 1.732)ILine = 72,051 VA/831VILine = 86.70A

5. (c) 12A

Line current for wye system = Line Power/(Line Volts x 1.732)If the phase voltage is 277V, the line voltage is 277V x 1.732 = 480V.

Line Current = 10,000 VA/(480V x 1.732)Line Current = 10,000 VA/831VLine Current = 12A

6. (b) (E x I)/1.732

Example: 15 hp, 208V, three-phase motor.Table 430.250 Amperes = 46.20A

Line Power = (E x I)/1.732Line Power = (208V x 46.20A)/1.732Line Power = 5,548 VA

This can also be calculated by:Line Power = Volts x Amperes x 1.732Line Power = 208V x 46.20A x 1.732Line Power = 16,644 VA/3 phasesLine Power = 5,548 VA per line

7. (d) 60A

[Table 450.3(B)] not more than 125% of primary current, 45.10A x 1.25 = 56.40A,[Table 450.3(b) Note 1] next size up, [240.6(A)], 60A device

8. (c) 70A



45,000VA/(480V x 1.732 = 54A54A x 1.25 = 67.50A, next size up is 70A[240.6(A), Table 450.3(B), Note 1]

9. (c) 125A

75,000VA/(480V x 1.732 = 90A90A x 1.25 = 112.50A, next size up is 125A[240.6(A), Table 450.3(B), Note 1]

10. (d) 175A

112,500VA/(480V x 1.732 = 135A135A x 1.25 = 169A, next size up is 175A[240.6(A), Table 450.3(B), Note 1]


Chapter 4—Practice Quizzes Answer Key

Chapter 4— neC praCtICe QUIzzes, ChaLLenge QUIzzes, anD FInaL neC exaMs

practice Quizzes

UnIt 1 praCtICe QUestIOns In straIght OrDer—90.1 thrOUgh 225.6

1. (d) 90.1(B)2. (d) 90.2(A)3. (c) 90.2(B)(4)4. (d) 90.2(B)(5) FPN5. (a) 90.36. (d) 90.4, 100 Special Permission7. (c) 90.48. (c) 90.79. (a) 100 Accessible (as applied to equipment)10. (b) 100 Accessible (as applied to wiring methods)11. (c) 100 Accessible, Readily12. (d) 100 Attachment Plug (Plug Cap) (Plug)13. (b) 100 Branch Circuit14. (d) 100 Cabinet15. (d) 100 Circuit Breaker16. (b) 100 Circuit Breaker, Inverse Time17. (d) 100 Communications Equipment18. (c) 100 Disconnecting Means19. (a) 100 Electric Sign20. (d) 100 Equipment21. (a) 100 Garage22. (b) 100 Ground23. (c) 100 Grounded24. (c) 100 Ground-Fault Circuit Interrupter (FPN)25. (c) 100 Grounding Electrode26. (b) 100 Handhole Enclosure27. (d) 100 In Sight from (Within Sight)28. (a) 100 Location, Dry29. (c) 100 Location, Wet30. (c) 100 Outlet31. (d) 100 Receptacle Outlet32. (b) 100 Remote-Control Circuit33. (c) 100 Service34. (c) 110.535. (c) 110.836. (d) 110.12(A)37. (c) 110.14 FPN38. (b) 110.14(C)(1)(a)39. (c) 110.14(C)(1)(a)(3)40. (c) 110.14(C)(1)(b)41. (d) 110.1542. (d) 110.1643. (d) 110.20 FPN

44. (b) 110.26(A)(1) and Table 110.26(A)(1), Condition 245. (c) 110.26(A)(1) and Table 110.26(A)(1), Condition 346. (c) 110.26(A)(1)47. (b) 110.26(A)(2)48. (b) 110.26(A)(3)49. (b) 110.26(B)50. (c) 110.26(C)(2)51. (d) 110.26(E)52. (b) 110.26(F)(1)(a)53. (b) 110.27(B)54. (b) 200.6(A)55. (c) 200.6(B)56. (b) 200.10(B)(1)57. (c) 210.358. (d) 210.4(B)59. (a) 210.7(B)60. (b) 210.8(A)(2)61. (b) 210.11(C)(1)62. (b) 210.11(C)(2)63. (b) 210.19(A)(1) FPN 464. (a) 210.21(B)(2) and Table 210.21(B)(2)65. (b) 210.21(B)(3) and Table 210.21(B)(3)66. (a) 210.23(A)(2)67. (d) 210.25(B)68. (a) 210.52(A)(1)69. (a) 210.52(A)(2)(1)70. (c) 210.52(A)(2)(2)71. (d) 210.52(C)(1)72. (b) 210.52(C)(3)73. (c) 210.52(C)(5)74. (a) 210.52(D)75. (b) 210.52(E)(3)76. (d) 210.52(G)(1)77. (b) 210.6278. (d) 210.6379. (d) 210.70(B) Ex 180. (c) 210.70(C)81. (c) 215.682. (b) 215.883. (d) 215.12(C)84. (b) 220.5(B)85. (a) 220.14(H)(1)86. (b) 220.14(I)


Answer Key Chapter 4—Practice Quizzes

26. (a) 100 Interrupting Rating27. (c) 210.52(C)(3)28. (c) 100 Bonded29. (c) 100 Conduit Body30. (b) 100 Identified (as applied to equipment)31. (d) 210.8(A)(7)32. (a) 210.70(A)(1) Ex 133. (c) 100 Intersystem Bonding Termination34. (d) 100 Kitchen35. (b) 100 Labeled36. (c) 110.1437. (b) 100 Demand Factor38. (b) 215.10 Ex 239. (d) 110.1140. (b) 100 Voltage of a Circuit41. (d) 210.6(A)42. (a) 100 Bonding Jumper, Main43. (d) 100 Grounding Electrode Conductor44. (a) 210.70(A)(1) Ex 245. (b) 90.446. (d) 110.3(A)47. (d) 200.2(B)48. (b) 210.52(B)(3)49. (b) 210.52(E)(1)50. (b) 210.60(B)

1. (b) 100 Grounded, Solidly2. (b) 200.7(C)(1)3. (a) 200.10(C)4. (a) 200.115. (a) 100 Ground-Fault Protection of Equipment6. (b) 100 Neutral Conductor7. (c) 100 Grounding Conductor, Equipment8. (a) 100 Guest Room9. (b) 100 Guest Suite10. (d) 210.8(A)(1) and (B)(1)11. (b) 100 Clothes Closet12. (a) 210.8(A)(2)13. (d) 200.114. (c) 100 Approved15. (d) 210.8(B)(1), (2), and (3)16. (b) 110.617. (b) 100 Concealed18. (d) 100 Outline Lighting19. (c) 110.26(A)(1) and Table 110.26(A)(1), Condition 220. (a) 100 Branch Circuit, Individual21. (d) 100 Qualified Person22. (c) 90.2(B)(5)(a)23. (a) 110.1224. (a) 210.8(A)(6)25. (d) 100 Premises Wiring

UnIt 1 praCtICe QUestIOns In ranDOM OrDer—artICLe 100 thrOUgh 215.10

87. (b) 220.43(A)88. (b) 220.43(B)89. (b) 220.5190. (c) 220.52(A)91. (c) 220.5492. (c) 220.54 and Table 220.5493. (a) 220.55 and Table 220.55

94. (d) 220.55, Table Column C95. (d) 220.5696. (b) 220.61(B)(1)97. (c) 220.61(C)(2)98. (a) 220.82(B)99. (a) 220.84 and Table 220.84100. (a) 225.6(A)(1)



51. (c) 250.2—Ground-Fault Current Path52. (d) 250.2—Ground-Fault Current Path FPN53. (c) 250.4(A)(3)54. (d) 250.4(A)(4)55. (d) 250.4(A)(5)56. (c) 250.4(A)(5)57. (b) 250.4(B)(1)58. (c) 250.20(D) FPN No. 159. (d) 250.21(B)60. (c) 250.24(A)(4)61. (b) 250.24(B)62. (a) 250.24(C)(2) and 310.463. (d) 250.28(B)64. (c) 250.28(D)(1)65. (c) 250.30(A)(1) 66. (c) 250.30(A)(4)(a)67. (a) 250.30(A)(4)(b)68. (a) 250.32(E)69. (a) 250.34(A)70. (a) 250.34(B)71. (c) 250.50 Ex72. (b) 250.52(A)(1)73. (c) 250.52(A)(1)74. (c) 250.52(A)(3)75. (c) 250.52(A)(3)76. (b) 250.52(A)(5) and 250.53(G)77. (a) 250.52(A)(5)(b)78. (c) 250.53(A)79. (b) 250.53(D)(1)80. (b) 250.53(D)(2) and (E)81. (c) 250.53(F)82. (b) 250.53(G)83. (d) 250.53(G)84. (b) 250.53(G)85. (a) 250.5486. (d) 250.64(A)87. (c) 250.64(B)88. (c) 250.64(B)89. (d) 250.64(C)90. (b) 250.66, Table91. (c) 250.66, Table92. (d) 250.66, Table93. (d) 250.66(B)94. (d) 250.92(B)95. (b) 250.92(B)(2)96. (d) 250.9497. (d) 250.96(A)98. (c) 250.102(C), Table 250.6699. (a) 250.102(C), Table 250.66100. (c) 250.102(D), Table 250.122

1. (b) 225.16(A)2. (b) 225.19(A)3. (b) 225.19(A) Ex 34. (b) 225.19(A) Ex 45. (a) 225.19(D)(2)6. (b) 225.227. (a) 225.268. (b) 225.329. (a) 225.3510. (d) 225.38(A)11. (b) 230.2(D)12. (b) 230.313. (a) 230.9(A)14. (a) 230.9(B)15. (b) 230.23(B)16. (b) 230.24(A)17. (b) 230.24(A) Ex 318. (b) 230.24(A) Ex 419. (b) 230.24(D), 680.8(A), and Table 680.820. (b) 230.2621. (a) 230.2822. (c) 230.31(A)23. (d) 230.31(B)24. (c) 230.31(B) Ex25. (c) 230.42(A)(1)26. (b) 230.51(A)27. (a) 230.54(C)28. (d) 230.54(F)29. (b) 230.70(A)(1)30. (c) 230.70(B)31. (c) 230.72(C) Ex32. (d) 230.76(2)33. (a) 230.79(A)34. (d) 230.79(B)35. (a) 230.9536. (a) 240.4(A)37. (a) 240.4(C)38. (d) 240.4(D)39. (c) 240.5(B)(3)40. (c) 240.6(A)41. (d) 240.6(A)42. (a) 240.1343. (c) 240.24(A)44. (d) 240.24(A)45. (b) 240.24(F)46. (b) 240.51(A)47. (d) 240.53(A) and 240.54(A)48. (c) 240.54(C)49. (c) 240.8050. (a) 240.86(A)

UnIt 2 praCtICe QUestIOns In straIght OrDer—artICLe 225.16 thrOUgh 250.102



26. (b) 250.102(C)27. (a) 225.19(D)(3)28. (a) 230.9(C)29. (c) 250.102(C)30. (d) 110.1031. (c) 230.9032. (c) 90.1(B) FPN33. (d) 250.1234. (d) 100 Fitting35. (d) 210.5236. (d) 250.52(A)(8)37. (d) 100 Connector, Pressure (Solderless)38. (c) 100 Device39. (b) 230.54(B)40. (d) 210.70(A)(2) Ex41. (d) 230.4342. (d) 230.50(B)(1)43. (b) 100 Plenum44. (c) 100 Voltage, Nominal45. (a) 100 Separately Derived System46. (b) 100 Feeder47. (a) 230.72(C)48. (b) 100 Service Point49. (d) 250.104(A)(2)50. (a) 100 Service Equipment

1. (c) 230.72(B)2. (b) 250.4(B)(4)3. (d) 250.4(A)(1)4. (d) 250.70 and 250.85. (d) 250.8(A)6. (c) 230.54(B) Ex7. (d) 225.32 Ex 18. (c) 250.10(2)9. (c) 250.30(A)(3)10. (b) 250.92(B)(3)11. (c) 240.5(B)(1)12. (d) 220.14(I)13. (a) 225.3014. (c) 100 Dwelling Unit15. (d) 100 Dwelling Multifamily16. (d) 100 Continuous Load17. (c) 100 Exposed (as applied to wiring methods)18. (a) 240.15(B)19. (a) 225.38(D)20. (d) 220.1221. (d) 90.2(A)(1)22. (b) 250.5623. (c) 100 Lighting Outlet24. (c) 100 Cutout Box25. (b) 110.26(D)

UnIt 2 praCtICe QUestIOns In ranDOM OrDer—artICLe 90.2 thrOUgh 250.104



51. (d) 310.4(B)52. (a) 310.5, Table53. (c) 310.10 FPN54. (c) 310.13(A), Table55. (b) 310.15(A)(1) FPN No. 156. (a) 310.15(B)(2) Ex57. (c) 310.15(B)(2)(a), Table58. (b) 310.15(B)(2)(a) Ex 359. (d) 310.15(B)(2)(a) Ex 560. (d) 310.15(B)(2)(c), Table61. (a) 310.15(B)(4)(a)62. (b) 312.263. (d) 312.364. (c) 312.465. (a) 312.5(A)66. (b) 312.5(C) Ex67. (c) 312.868. (d) 314.369. (d) 314.16(A), Table 314.16(B), and 314.16(B)(1)–(4)70. (b) 314.16(A),Table71. (b) 314.17(C) Ex72. (b) 314.2073. (c) 314.2074. (d) 314.23(B)(1)75. (a) 314.23(B)(2)76. (d) 314.23(D)(1)77. (b) 314.23(F)78. (d) 314.27(A)79. (b) 314.27(A) Ex80. (d) 314.27(B)81. (b) 314.27(C)82. (a) 314.27(D)83. (a) 314.27(E) Ex84. (b) 314.28(B)85. (d) 314.29 Ex86. (d) 314.3087. (c) 314.30(B)88. (c) 314.30(D)89. (c) 314.30(D)90. (d) 320.23(A)91. (d) 320.30(B)92. (c) 320.30(C)93. (d) 320.30(D)(3)94. (c) 320.4095. (d) 330.1296. (b) 330.24(B)97. (b) 330.30(B)98. (c) 330.30(B)99. (b) 330.40100. (c) 334.2

1. (b) 250.118(5)2. (d) 250.1193. (c) 250.1194. (b) 250.122(B)5. (c) 250.122(D)(1)6. (c) 250.142(B) Ex 27. (a) 250.1468. (c) 250.146(B)9. (a) 250.146(D)10. (d) 250.148(C)11. (d) 285.1212. (d) 300.3(C)(1)13. (a) 300.4(A)(1)14. (a) 300.4(A)(2)15. (c) 300.4(B)(1)16. (d) 300.4(B)(1)17. (a) 300.4(B)(2)18. (c) 300.4(E)19. (a) 300.5, Table Column 220. (c) 300.5, Table Column 321. (b) 300.5, Table Column 422. (a) 300.5, Table Column 523. (c) 300.5(D)(1)24. (c) 300.5(D)(3)25. (d) 300.5(F)26. (d) 300.5(I) and 300.3(B)27. (d) 300.5(K)28. (d) 300.6(A)29. (d) 300.6(B)30. (d) 300.6(C)(1)31. (d) 300.6(C)(2)32. (a) 300.6(D)33. (b) 300.834. (c) 300.1035. (d) 300.11(A)(1)36. (c) 300.1237. (a) 300.12 Ex 238. (a) 300.13(A)39. (b) 300.1440. (a) 300.18(A)41. (b) 300.18(A) Ex42. (b) 300.19(A), Table43. (c) 300.2144. (d) 300.22(A)45. (a) 300.22(B)46. (b) 300.22(C) FPN47. (d) 300.22(C)(1)48. (b) 310.349. (d) 310.4(A) 50. (d) 310.4(B)




26. (a) 250.10627. (a) 312.5(C)28. (c) 240.6129. (c) 240.1030. (b) 110.26(C)(3)31. (d) 110.732. (d) 240.2—Current Limiting33. (a) 100 Short-Circuit Current Rating34. (c) 100 Ampacity35. (c) 300.5(J)36. (c) 210.50(C)37. (d) 90.2(A)38. (a) 314.16(B)(3)39. (a) 314.16(B)(5)40. (b) 314.28(A)(1)41. (a) 314.28(A)(2)42. (b) 220.6043. (d) 110.14(A)44. (d) 250.6245. (b) 314.22 Ex46. (d) 300.5(H)47. (d) 314.548. (d) 320.3049. (a) 110.26(B)50. (a) 240.85

1. (a) 300.4(D)2. (d) 250.1263. (b) 300.5, Table Note (1)4. (a) 250.146(A)5. (b) 250.146(A)6. (b) 300.11(A)7. (b) 314.16(B)(1) 8. (d) 314.16(B)(1) Ex9. (b) 314.23(E)10. (d) 310.15(B)(6)11. (d) 300.11(C)12. (d) 300.1713. (b) 250.118(5)(d) and (6)(e)14. (d) 300.3(B), see 300.5(I)15. (c) 300.7(A)16. (c) 310.13(A), Table17. (a) 314.23(A)18. (d) 300.5(G)19. (d) 250.104(C)20. (d) 230.721. (d) 250.92(A)(1) and (2)22. (b) 100 Service Drop23. (c) 210.25(A)24. (d) 90.5(A)25. (c) 90.5(B)




51. (b) 348.2452. (c) 348.2653. (b) 348.2854. (a) 348.30(A) 55. (d) 348.30(A) Ex 156. (b) 348.4257. (d) 350.10(3)58. (a) 350.30(A) Ex 159. (b) 350.4260. (c) 350.6061. (d) 352.10 FPN62. (d) 352.12(A), (B) and (C)63. (c) 352.2664. (d) 352.30(A)65. (d) 353.266. (d) 353.10(1)67. (a) 353.12(4)68. (d) 353.2469. (b) 353.4870. (d) 353.48 FPN71. (b) 353.6072. (c) 354.273. (a) 354.674. (d) 354.10(1), (2), and (3)75. (d) 354.12(1), (2), and (3)76. (a) 354.2477. (c) 354.2678. (c) 354.2879. (c) 354.4880. (d) 355.281. (a) 356.282. (a) 356.2283. (a) 356.2484. (c) 356.2685. (b) 358.286. (c) 358.10(C)87. (d) 358.12(1), (2) and (5)88. (b) 362.289. (c) 362.10(2) Ex90. (c) 362.10(6)91. (b) 362.12(9)92. (a) 362.2293. (a) 362.2894. (a) 362.30(A) Ex 395. (d) 376.10(1), (2), and (3)96. (a) 376.10(4)97. (a) 376.23(A)98. (d) 376.56(B)(3) and (4)99. (c) 378.2100. (d) 378.10(1) and (3)

1. (c) 100 Panelboard2. (d) 100 Special Permission3. (a) 100 Supplementary Overcurrent Protective Device4. (b) 110.14(C)5. (c) 110.22(A)6. (a) 210.4(C)7. (d) 210.70(A)(3)8. (d) 230.23(A)9. (b) 240.83(D)10. (b) 240.8511. (a) 250.28(D)(1)12. (d) 250.30(A)13. (a) 250.30(A)(1) 14. (d) 250.30(A)(6)15. (a) 250.102(C)16. (a) 250.104(A)(1)17. (a) 250.104(A)(3)18. (a) 300.4(C)19. (d) 300.18(B)20. (a) 300.2121. (d) 314.23(C)22. (a) 314.30(C)23. (d) 334.624. (d) 334.12(A)(9)25. (b) 334.15(B)26. (a) 334.15(C)27. (d) 334.1728. (d) 320.3029. (b) 334.8030. (c) 334.11231. (d) 336.1032. (c) 338.233. (c) 338.10(B)(1)34. (b) 338.10(B)(2)35. (b) 340.236. (d) 340.12(4), (5) and (6)37. (d) 340.12(7), (8) and (9)38. (a) 340.8039. (b) 342.1440. (d) 342.2641. (b) 342.30(B)(1)42. (a) 342.4643. (d) 344.10(B)(1)44. (d) 344.1445. (b) 344.24 and Chapter 9, Table 246. (d) 344.2647. (b) 344.30(C)48. (c) 344.42(B)49. (a) 344.4650. (d) 348.12(1), (6) and (7)

UnIt 4 praCtICe QUestIOns In straIght OrDer—artICLe 100 thrOUgh 378.10



26. (c) 110.22(A)27. (a) 320.80(A)28. (b) 210.12(B) Ex 129. (c) 352.4830. (b) 250.122(C)31. (d) 250.3632. (c) 240.21(B)(5)33. (d) 240.4(B)34. (d) 215.1035. (d) 250.24(A)(1)36. (b) 225.34(A)37. (b) 230.72(A)38. (c) 250.30(A)(2)39. (c) 240.21(B)(5)40. (c) 220.5641. (c) 110.2142. (d) 310.4(A) Ex 143. (b) 344.2844. (d) 250.97 Ex (1), (2), (3), and (4)45. (d) 250.92(B)(1) through (4)46. (d) 354.20(B)47. (b) 230.848. (c) 100 Overcurrent49. (a) 90.1(C)50. (a) 90.2(B)(1)

1. (b) 240.852. (b) 348.423. (b) 350.424. (d) 250.30(A)(6)5. (d) 358.12(1), (2) and (5)6. (a) 300.217. (d) 230.23(A)8. (a) 314.30(C)9. (d) 352.10 FPN10. (a) 100 Supplementary Overcurrent Protective Device11. (d) 250.30(A)12. (a) 210.4(C)13. (a) 300.4(C)14. (d) 300.18(B)15. (b) 240.83(D)16. (c) 100 Panelboard17. (a) 250.30(A)(1) 18. (a) 250.28(D)(1)19. (a) 250.102(C)20. (a) 250.104(A)(1)21. (a) 250.104(A)(3)22. (d) 314.23(C)23. (b) 110.14(C)24. (d) 100 Special Permission25. (d) 210.70(A)(3)




51. (b) 406.2(B)52. (c) 406.3(B)53. (b) 406.3(D)(1)54. (c) 406.3(D)(2)55. (d) 406.3(D)(3)56. (b) 406.4(A)57. (a) 406.4(B)58. (b) 406.4(E)59. (c) 406.4(G)60. (a) 406.8(A)61. (b) 406.8(A)62. (d) 406.8(A)63. (d) 406.8(E)64. (a) 406.9(E)65. (d) 406.1166. (b) 408.3(C)67. (b) 408.3(D)68. (b) 408.3(E)69. (a) 408.570. (b) 408.771. (b) 408.3672. (a) 408.36(B)73. (a) 408.4174. (a) 408.5475. (a) 408.56, Table76. (d) 410.177. (a) 410.278. (c) 410.279. (b) 410.10(A)80. (b) 410.10(C)(1)81. (c) 410.10(D)82. (c) 410.10(D)83. (d) 410.10(E)84. (d) 410.16(A)85. (a) 410.1886. (d) 410.24(A)87. (d) 410.30(A)88. (d) 410.30(B)(1) Ex 289. (c) 410.30(B)(5)90. (d) 410.36(B)91. (d) 410.36(B)92. (a) 410.42(A)93. (c) 410.115(C)94. (b) 410.116(A)(1)95. (d) 410.116(A)(2)96. (b) 410.116(B)97. (a) 410.117(C)98. (c) 410.117(C)99. (c) 410.136(B)100. (b) 410.151(C)(8)

1. (a) 378.10(4)2. (a) 378.223. (b) 378.444. (b) 378.565. (a) 380.2(A)6. (d) 380.2(B)(2), (4) and (5)7. (c) 384.12(1) and (2)8. (d) 384.30(A)9. (b) 384.5610. (b) 386.2211. (a) 386.3012. (d) 386.5613. (c) 386.6014. (a) 388.10(1)15. (b) 388.2216. (d) 388.5617. (d) 392.318. (c) 392.319. (d) 392.3(A)20. (d) 392.421. (d) 392.6(C)22. (d) 392.6(H)23. (d) 392.6(J)24. (a) 392.6(J)25. (a) 392.7(B)(1)26. (c) 392.8(A)27. (c) 392.8(C)28. (b) 392.8(D)29. (a) 400.4 Table30. (a) 400.5(A)31. (d) 400.8(2), (3), and (4)32. (d) 400.10 FPN33. (c) 400.1434. (b) 400.1435. (a) 400.2236. (a) 402.5, Table37. (c) 402.638. (a) 402.739. (b) 402.1040. (d) 404.741. (b) 404.8(A) Ex 242. (c) 404.8(B)43. (c) 404.8(C)44. (c) 404.9(B) Ex45. (b) 404.1246. (d) 404.14(A)47. (b) 404.14(C)48. (a) 404.14(E)49. (d) 404.15(A)50. (b) 404.15(B)




26. (a) 250.6(A)27. (a) 250.24(D)28. (a) 250.30(A)(7)29. (b) 250.136(A)30. (b) 285.1131. (a) 314.16(C)(2)32. (a) 320.1733. (a) 352.2834. (a) 406.8(B)(2)(a)35. (b) 100 Luminaire36. (a) 230.70(A)(3)37. (a) 250.6(C)38. (a) 250.32(A)39. (a) 250.138(A)40. (a) 340.12(10)41. (b) 376.2142. (a) 404.8(A)43. (a) 406.8(C)44. (a) 410.24(B)45. (a) 110.3(B)46. (a) 210.20(A)47. (b) 250.32(A) Ex48. (a) 250.104(B)49. (a) 250.142(A)50. (b) 285.24

1. (d) 406.8(B)(1) Ex2. (c) 406.8(A)3. (d) 386.704. (d) 388.705. (d) 410.506. (a) 392.27. (d) 400.8(3) and (5)8. (b) 400.5(A)9. (b) 400.8(1)10. (d) 386.22(1), (2) and (3)11. (d) 384.22(1), (2), and (3)12. (c) 410.30(B)(1) and (3)13. (a) 210.52(B)(2) Ex 114. (a) 285.615. (a) 320.1516. (a) 342.42(B)17. (b) 404.6(A)18. (a) 406.2(D)(2)19. (a) 410.9020. (a) 90.4, 100 Special Permission21. (b) 210.19(A)(1) Ex 222. (a) 210.52(B)(2) Ex 223. (a) 210.60(B)24. (a) 230.2(E)25. (a) 230.70(A)(2)




50. (c) 440.22(A)51. (b) 440.3252. (a) 440.3353. (b) 440.62(C)54. (d) 440.6455. (c) 445.1156. (b) 445.1357. (a) 450.3(B)58. (b) 450.3(B), Table59. (b) 450.960. (c) 450.1361. (c) 450.21(B)62. (b) 450.43(C)63. (d) 450.45(A)64. (d) 460.2(B)65. (d) 460.8(A)66. (a) 460.8(B)67. (d) 460.8(C)68. (a) 500.269. (b) 500.270. (d) 500.371. (d) 500.5(A)72. (c) 500.5(B)(1)73. (a) 500.5(C)74. (b) 500.5(C)(1)75. (b) 500.5(D)76. (a) 500.5(D)(1)77. (b) 500.5(D)(2)78. (d) 500.779. (d) 500.8(A)(1)80. (d) 500.8(B)(1)81. (d) 500.8(C)82. (c) 500.8(E)83. (b) 500.8(E) Ex84. (d) 501.185. (d) 501.15(A)(4)86. (a) 501.15(A)(4) Ex 287. (c) 501.15(B)(2)88. (b) 501.15(B)(2) Ex 189. (d) 501.15(C)(3)90. (a) 501.15(C)(6)91. (c) 501.105(B)(1) Ex92. (a) 501.120(A)93. (b) 501.125(A)94. (a) 501.130(A)(1)95. (a) 501.130(A)(3)96. (a) 501.130(B)(3)97. (d) 501.130(B)(4)98. (a) 501.13599. (d) 501.140(B)100. (c) 502.1

1. (d) 410.151(C)(9) and 410.10(D)2. (c) 411.23. (c) 411.4(B)4. (d) 411.5(D)5. (b) 422.10(A)6. (c) 422.10(A)7. (c) 422.11(A)8. (b) 422.11(E)(2)9. (c) 422.11(E)(3)10. (b) 422.1211. (d) 422.1312. (b) 422.16(B)(1)(2) and (3)13. (b) 422.16(B)(2)(2)14. (c) 422.16(B)(3)15. (d) 422.33(B)16. (b) 422.3417. (d) 422.5118. (a) 422.5219. (c) 424.3(B)20. (d) 424.1921. (c) 424.19(C)(3)22. (d) 424.44(G)23. (b) 424.6524. (c) 430.6(A)(1)25. (b) 430.6(A)(1)26. (b) 430.14(A)27. (b) 430.1728. (d) 430.28(2)29. (d) 430.3130. (c) 430.3131. (d) 430.32(A)(1)32. (c) 430.32(A)(2)33. (c) 430.37 and Table 430.37 and Table 430.39 Ex34. (c) 430.52, Table35. (d) 430.5536. (a) 430.62(A)37. (d) 430.72(B)(2) and Table 430.72(B), Column B,

Note 238. (c) 430.72(B)(2) and Table 430.72(B), Column C,

Note 339. (a) 430.83(A)(1)40. (c) 430.102(B)(1)41. (d) 430.102(B)(1) Ex42. (b) 430.10343. (a) 430.10444. (d) 430.10945. (a) 430.109(A)(1)46. (d) 430.11147. (d) 440.248. (d) 440.3(B)49. (d) 440.3(C)




26. (a) 210.8(A)(4)27. (a) 210.70(A)(1)28. (b) 220.5629. (a) 250.53(B) and 250.5830. (a) 250.64(E)31. (a) 300.6(D) FPN32. (a) 310.15(B)(2)(a)33. (a) 384.234. (b) 410.235. (b) 424.936. (a) 90.5(C)37. (a) 200.7(C)(2)38. (a) 250.20(A)(1)39. (a) 300.1540. (a) 300.2341. (a) 314.2242. (a) 334.15(A)43. (a) 352.4644. (a) 362.12(1)45. (a) 388.246. (a) 404.9(B)47. (a) 460.8(B) Ex48. (b) 90.2(B)(2)49. (a) 210.52(C)(4)50. (a) 250.20(B)(2)

1. (c) 445.182. (c) 440.633. (d) 500.5(B)(2) FPN No. 14. (d) 501.15(C)(1), (3), and (4)5. (d) 501.10(A) (1), (2), and (3)6. (a) 502.10(A)(1)7. (b) 430.848. (a) 501.10(B)(4)9. (a) 411.3(A) and (B)10. (a) 430.8711. (b) 411.4(A)(1)12. (b) 501.125(A)(1)13. (b) 320.23(B)14. (b) 334.10(1)15. (a) 100 Neutral Point16. (a) 100 Service Conductors17. (a) 210.21(B)(1)18. (a) 250.32(B)19. (a) 300.3(A)20. (b) 334.12(A)(2)21. (a) 348.12(7) and 348.1022. (a) 362.10(5) Ex23. (a) 380.324. (a) 430.10325. (a) 501.125(A)




51. (d) 513.3(D)52. (c) 513.4(A)53. (d) 513.7(A)54. (d) 513.7(B)55. (c) 513.7(C)56. (a) 513.8(A)57. (d) 513.1258. (d) 514.159. (b) 514.3(A)60. (d) 514.8 Ex 261. (d) 514.9(A)62. (d) 514.11(B)63. (c) 514.11(C)64. (b) 514.1665. (a) 517.166. (d) 517.2 Health Care Facilities67. (c) 517.2 Limited Care Facility68. (a) 517.2 Patient Bed Location69. (d) 517.2 Patient Care Area70. (a) 517.2 FPN Patient Care Area71. (d) 517.13(B)72. (d) 517.18(B)73. (a) 517.18(C)74. (a) 517.13(B) Ex 275. (d) 518.2(A)76. (d) 518.3(B)77. (d) 518.4(A)78. (b) 518.4(A)79. (b) 518.4(B)80. (b) 525.181. (a) 525.2, Operator82. (d) 525.5(A) and 225.1883. (d) 525.5(B)(1) Overhead Conductor Clearances84. (b) 525.685. (a) 525.10(A)86. (a) 525.20(E)87. (b) 525.21(A)88. (a) 525.22(A)89. (b) 525.23(C)90. (b) 525.3091. (a) 525.3192. (d) 547.1 and (A)93. (d) 547.1(B)94. (d) 547.295. (a) 547.5(B)96. (c) 547.5(C)(2)97. (b) 547.5(F)98. (d) 547.5(G)99. (b) 547.10(B)100. (a) 550.2

1. (b) 502.10(A)(1)(4)2. (d) 502.10(A)(2)3. (d) 502.10(B)(1)4. (a) 502.10(B)(1)(2)5. (d) 502.10(B)(2)6. (d) 502.157. (a) 502.30(A)8. (b) 502.115(A)(1)9. (a) 502.115(B)10. (d) 502.120(A)11. (d) 502.125(A)12. (d) 502.130(A)(3)13. (b) 502.130(B)(2)14. (b) 502.140(1)15. (b) 502.145(A)16. (b) 502.150(B)(1)17. (b) 503.118. (d) 503.10(A) and 503.10(B)19. (c) 503.10(A)(1) and 503.10(B)20. (c) 503.11521. (d) 503.12022. (c) 503.130(B)23. (a) 503.130(C)24. (a) 503.130(D)25. (d) 503.140(1), (2), and (3)26. (c) 503.14527. (d) 504.128. (a) 504.229. (b) 504.10(A)30. (a) 504.10(B)31. (d) 504.30(A)(1)32. (a) 504.30(A)(2)(1)33. (a) 504.30(A)(3)34. (b) 504.60(A)35. (c) 504.80(B)36. (b) 504.80(C)37. (c) 511.138. (d) 511.2 Major repair garage39. (c) 511.3(C)(1)(b)40. (c) 511.3(C)(2)41. (a) 511.3(C)(3)(b)42. (b) 511.3(D)(3)(b)43. (d) 511.3(E)(2)44. (c) 511.4(B)(2)45. (c) 511.7(A)(2)46. (b) 511.7(B)(1)(b)47. (b) 513.148. (a) 513.3(A)49. (c) 513.3(B)50. (c) 513.3(C)(1)




26. (a) 250.20(B)(3)27. (a) 250.102(A)28. (a) 300.5(E)29. (a) 300.15(G)30. (a) 314.1531. (a) 410.130(G)(1)32. (a) 430.52(C)(1) Ex 133. (b) 460.8(C) Ex34. (a) 501.15(D)(2) Ex 235. (a) 501.130(A)(4)36. (a) 547.10(A)(2)37. (a) 210.8(A)(8)38. (a) 210.52(C)(5) Ex39. (b) 210.70(A)(2)(b)40. (b) 230.1041. (a) 300.15(L)42. (a) 342.2843. (a) 376.56(B)(1)44. (a) 388.2145. (a) 410.36(G)46. (a) 460.947. (a) 501.15(E)(1) Ex48. (a) 90.2(B)(5) FPN49. (a) 100 Signaling Circuit50. (b) 210.70(A)(2)(c)

1. (a) 514.11(A)2. (b) 525.20(A)3. (a) 547.5(C)(1)4. (a) 517.805. (a) 525.20(F)6. (a) 514.137. (a) 547.5(C)(2)8. (b) 502.130(B)(4)9. (a) 511.3(A)10. (a) 525.21(B)11. (a) 525.23(A)(1)12. (a) 300.7(B)13. (a) 300.15(C)14. (a) 310.2(A)15. (b) 362.12(5)16. (a) 406.4(C)17. (a) 430.10718. (b) 525.23(B)19. (a) 547.10(A)(1) and (2)20. (a) 90.9(D)21. (b) 100 Overload22. (a) 110.923. (a) 210.70(A)(2)(b)24. (a) 230.50(A)25. (a) 250.4(A)(2)




52. (d) 604.453. (c) 604.7 and 330.3054. (a) 620.23(C)55. (d) 620.24(A)56. (a) 620.24(B)57. (a) 620.24(C)58. (c) 620.5159. (b) 620.51(A)60. (b) 620.51(C)61. (a) 620.51(D)62. (c) 620.8563. (b) 620.8564. (d) 630.12(A)65. (c) 630.1366. (c) 630.31(A)(2)67. (d) 630.3368. (d) 640.469. (d) 640.6(B)70. (c) 640.6(D)(1)71. (c) 640.10(A)72. (a) 640.2573. (a) 645.274. (d) 645.4(1), (2), and (5)75. (c) 645.5(A)76. (c) 645.5(D)(3)77. (d) 645.5(D)(6)78. (d) 645.5(D)(6)(c)79. (a) 645.5(F)80. (d) 645.1081. (a) 645.1082. (d) 645.1583. (b) 647.184. (c) 647.385. (a) 647.4(D)86. (c) 647.4(D)87. (b) 680.2, Permanently Installed Swimming, Wading,

Immersion, and Therapeutic Pools88. (d) 680.7(A) and (B)89. (d) 680.8 Table90. (d) 680.8(B)91. (d) 680.8(C)92. (c) 680.993. (a) 680.1094. (d) 680.10 and Table 680.10 95. (d) 680.1296. (c) 680.22(A)(1)97. (b) 680.22(A)(1)98. (d) 680.22(A)(4)99. (b) 680.22(B)100. (b) 680.22(C)(1)

1. (d) 550.13(B)2. (b) 550.13(E)3. (c) 550.25(B)4. (c) 550.32(A)5. (c) 550.32(C)6. (b) 550.32(F)7. (c) 550.33(A)8. (c) 550.33(B)9. (d) 551.7110. (c) 551.7111. (b) 551.73(A)12. (a) 555.113. (c) 555.2, Marine Power Outlet14. (b) 555.2, Electrical Datum Plane15. (c) 555.516. (a) 555.717. (d) 555.918. (b) 555.12, Note 119. (d) 555.17(A)20. (b) 555.17(B)21. (c) 555.17(B)22. (c) 555.19(A)(4)23. (b) 555.19(B)(1)24. (d) 555.21(A)25. (c) 590.2(B)26. (d) 590.3(B)27. (d) 590.3(C)28. (b) 590.3(D)29. (a) 590.4(D)30. (b) 590.4(F)31. (c) 590.4(G)32. (d) 590.4(H)33. (d) 590.4(J)34. (c) 590.4(J) and Ex35. (c) 590.636. (d) 590.6(A)37. (d) 590.6(A)38. (d) 590.6(B)39. (d) 600.140. (a) 600.2, Section Sign41. (d) 600.342. (a) 600.3(B)43. (b) 600.5(A)44. (c) 600.6(A)(1)45. (b) 600.9(A)46. (b) 600.9(B)47. (c) 600.9(C)48. (d) 600.10(C)(2)49. (c) 600.21(D)50. (d) 600.21(F)51. (c) 604.2




26. (a) 210.8(B)(4)27. (a) 250.30(A)(3)28. (b) 250.68(A) Ex 129. (a) 250.122(A)30. (a) 250.148(B)31. (a) 300.11(A)32. (a) 310.15(B)(4)(c)33. (a) 330.17 and 300.434. (a) 338.10(B)(4)(a)35. (a) 354.4636. (a) 358.28(B)37. (a) 384.6038. (a) 392.11(A)(1)39. (a) 410.42(B)40. (b) 410.151(B)41. (a) 440.142. (a) 501.30(B)43. (b) 503.30(A)44. (b) 511.7(A)(1)45. (a) 525.3246. (a) 550.2 Mobile Home47. (a) 630.11(A)48. (a) 680.22(A)(3)49. (a) 100 Structure50. (a) 110.26(E) Ex

1. (d) 647.4(B)2. (a) 555.15(B)3. (a) 640.14. (a) 645.5(D)(4)5. (a) 640.2 Abandoned Audio Distribution Cable6. (a) 620.37(A)7. (a) 645.5(D)(6)(c)8. (a) 555.19(A)(3) FPN9. (a) 640.6(C)10. (a) 680.1111. (b) 640.21(E)12. (a) 590.2(A)13. (a) 640.2214. (a) 220.14(J)15. (a) 240.1016. (a) 250.68(A)17. (a) 250.14818. (a) 406.5(B)19. (b) 408.420. (b) 410.151(A)21. (a) 501.30(A)22. (a) 502.40 Ex23. (b) 640.23(A)24. (a) 90.2(B)(4)(5) FPN25. (a) 110.11 FPN No. 2




51. (b) 695.6(C)(1)52. (a) 695.6(H)53. (c) 695.754. (a) 695.755. (d) 700.1 FPN No. 356. (b) 700.1 FPN No. 357. (a) 700.258. (a) 700.4(D)59. (a) 700.5(A)60. (c) 700.5(B)61. (d) 700.6(A)62. (d) 700.6(C)63. (a) 700.6(D)64. (d) 700.765. (d) 700.9(B)(1), (2), and (3)66. (a) 700.9(B)(5)67. (d) 700.9(C)68. (b) 700.1269. (b) 700.12(A)70. (a) 700.12(B)(2) and 701.11(B)(2)71. (b) 700.2572. (a) 700.2773. (b) 701.274. (b) 701.5(A)75. (c) 701.5(C)76. (a) 701.677. (a) 701.678. (d) 701.7(A)79. (b) 701.7(C)80. (b) 701.11(E)81. (b) 701.1582. (b) 701.18 Coordination83. (c) 702.184. (a) 702.2 FPN85. (b) 702.5(B)86. (d) 702.6 Ex87. (b) 725.288. (a) 725.2489. (d) 725.2590. (a) 725.31(A)91. (d) 725.121(A)(1), (2), and (3)92. (a) 725.12493. (b) 725.136(A)94. (c) 725.136(B)95. (d) 725.139(F)96. (d) 725.154(B)97. (a) 725.179(A)98. (b) 725.179(B)99. (c) 760.1100. (d) 760.1 FPN No. 1

1. (c) 680.23(A)(3)2. (c) 680.23(A)(5)3. (d) 680.23(B)(2)(b)4. (d) 680.23(F)(1)5. (a) 680.23(F)(1)6. (a) 680.23(F)(2)7. (a) 680.24(A)(1)8. (a) 680.24(A)(2)(a)9. (c) 680.24(A)(2)(b)10. (b) 680.24(A)(2)(b)11. (d) 680.24(B)(1)12. (a) 680.24(D)13. (a) 680.24(F)14. (b) 680.25(A)15. (a) 680.26(B)(1)16. (d) 680.26(B)(3) through (6)17. (b) 680.26(B)(7) Ex 1 and Ex 218. (d) 680.27(B)(1)19. (a) 680.27(B)(2)20. (d) 680.3221. (b) 680.3222. (b) 680.3423. (b) 680.4124. (c) 680.42(A)(1) and (2)25. (a) 680.43 Ex26. (c) 680.43(A)27. (a) 680.43(A)(2)28. (d) 680.43(B)(1)(a)29. (c) 680.43(B)(1)(b)30. (a) 680.43(D)(2)31. (b) 680.43(D)(3)32. (d) 680.43(E)(1), (2), and (3)33. (d) 680.4434. (b) 680.51(A)35. (d) 680.51(C)36. (c) 680.51(F)37. (a) 680.5338. (a) 680.55(B)39. (b) 680.56(A)40. (d) 680.56(B)41. (a) 680.57(A) and (B)42. (a) 680.57(C)(2)43. (d) 680.5844. (d) 680.7445. (d) 680.7446. (c) 695.4(A)47. (a) 695.4(B)(1)48. (b) 695.5(A)49. (d) 695.5(B)50. (d) 695.6(B)




26. (a) 500.2 FPN27. (b) 501.40 Ex28. (a) 725.229. (a) 100 Qualified Person FPN30. (a) 110.12 FPN31. (a) 110.22(C)32. (a) 110.26(F)(1)(a)33. (a) 210.52(F) Ex 134. (a) 240.15(B)(2) and (3)35. (b) 250.4(A)(5)36. (a) 250.5437. (a) 250.68(B)38. (b) 285.139. (a) 300.11(A)(1)40. (a) 310.4(C)41. (a) 314.16(B)(2)42. (a) 342.30(B)(4)43. (a) 352.10(H)44. (a) 410.62(B)45. (a) 430.75(A)46. (a) 501.100(A)(1)47. (b) 501.140(A)48. (a) 504.50(A)49. (a) 517.13(A)50. (a) 90.3

1. (a) 700.162. (a) 725.243. (a) 700.4(E)4. (a) 725.255. (a) 760.26. (b) 700.267. (a) 701.178. (b) 701.5(B)9. (a) 702.610. (a) 700.8(A)11. (a) 701.5(D)12. (b) 701.5(E)13. (a) 702.914. (a) 725.14315. (a) 725.216. (a) 210.8(C)17. (a) 230.82(3)18. (a) 250.30(A)(4)19. (a) 250.68(A) Ex 220. (a) 310.15(B)(5)21. (a) 352.10(F)22. (a) 386.223. (a) 402.1124. (b) 410.1625. (a) 430.9(B)




51. (d) 800.9352. (d) 800.100(A)(4)53. (a) 800.100(A)(4) Ex54. (a) 800.100(A)(4) FPN55. (d) 800.100(B)(2)56. (c) 800.100(D)57. (a) 800.11058. (b) 800.113 Ex59. (d) 800.133(A)(2)60. (c) 800.154(C)(4) and (5)61. (d) 800.179(A)62. (b) 800.179(B)63. (a) 810.1264. (a) 810.18(A)65. (a) 810.18(B)66. (a) 810.18(C)67. (a) 810.20(A)68. (b) 810.21(D)69. (c) 810.21(J)70. (c) 810.5471. (d) 820.2, Coaxial Cable72. (a) 820.2173. (a) 820.2474. (d) 820.2675. (d) 820.44(F)(3)76. (c) 820.4877. (d) 820.100(A)(1), and (3)78. (d) 820.100(A)(4)79. (d) 820.100(A)(4) Ex80. (a) 820.100(A)(4) FPN81. (c) 820.100(A)(6)82. (c) 820.100(D)83. (b) 820.133(A)(1)(b) Ex 184. (c) 820.154(B)85. (b) Chapter 9, Table 186. (b) Chapter 9, Table 187. (b) Chapter 9, Table 1, Note 388. (d) Chapter 9, Table 1, Note 789. (a) Chapter 9, Table 1, Note 990. (d) Chapter 9, Table 491. (b) Chapter 9, Table 492. (c) Chapter 9, Table 493. (a) Chapter 9, Table 594. (b) Chapter 9, Table 895. (b) Chapter 9, Table 896. (b) Chapter 9, Table 897. (b) Chapter 9, Table 898. (b) Chapter 9, Table 999. (b) Chapter 9, Table 9100. (c) Chapter 9, Table 9

1. (a) 760.242. (a) 760.253. (c) 760.254. (b) 760.305. (c) 760.41(A)6. (d) 760.41(B)7. (a) 760.438. (a) 760.48(B)9. (b) 760.51(A)10. (b) 760.121(B)11. (b) 760.12412. (d) 760.130(B)(1)13. (b) 760.136(A)14. (b) 760.139(D)15. (b) 760.14316. (c) 760.17917. (b) 760.17918. (b) 760.179(C)19. (c) 770.2 Optical Fiber Cable20. (c) 770.221. (d) 770.222. (a) 770.223. (a) 770.224. (a) 770.2125. (d) 770.24 FPN26. (a) 770.2427. (a) 770.2428. (a) 770.2529. (d) 770.2630. (c) 770.93(A)31. (a) 770.11032. (b) 770.133(A)33. (a) 770.133(A)34. (d) 770.133(C)35. (c) 770.154(B)(2)36. (a) 770.179(A)37. (b) 770.179(B)38. (c) 770.179(C)39. (d) 770.179(D)40. (a) 800.241. (a) 800.2, Exposed to Accidental Contact42. (b) 800.2, Point of Entrance43. (b) 800.3(A)44. (a) 800.1845. (a) 800.2446. (a) 800.2547. (d) 800.2648. (d) 800.44(A)(1) and (2), and 800.44(B)49. (c) 800.4850. (a) 800.53

UnIt 10 praCtICe QUestIOns In straIght OrDer—artICLe 760.24 thrOUgh Chapter 9



51. (a) 404.2(A)52. (b) 250.24(C)53. (c) 100 Building54. (c) 502.30(B)55. (c) 503.30(B)56. (b) 100 Service Lateral57. (b) 430.52(B)58. (d) 110.2659. (a) 240.3360. (b) 680.7161. (d) 240.6162. (d) 240.60(B)63. (d) 110.14(C)(2)64. (d) 388.10(2)65. (d) 344.30(B)(4)66. (d) 386.10(4)67. (c) 502.130(A)(2)68. (c) 502.130(B)(3)69. (c) 501.130(A)(2)70. (c) 501.130(B)(2)71. (b) 504.272. (c) 344.10(D)73. (c) 342.10(D)74. (d) 511.7(B)(1)75. (a) 300.5(B)76. (a) 300.977. (d) 547.5(C)(3)78. (a) 725.48(B)(1)79. (d) 384.10(1), (5), and (7)80. (d) 240.24(C),(D), and (E)81. (d) 336.1282. (c) 410.151(C)(1) and (2)83. (b) 240.51(B)84. (d) 353.10(2) and (4)85. (d) 386.12(1), (3), and (4)86. (b) 680.23(B)(6)87. (c) 210.60(A)88. (d) 210.52(C)(1)89. (d) 250.28(A)90. (c) 200.991. (d) 800.2192. (c) 440.14 Ex 193. (b) 410.62(C)(1)94. (d) 250.110(1), (2), and (3)95. (c) 600.6(A)(2)96. (b) 408.3(F)97. (a) 210.4(A) FPN98. (b) 314.2199. (a) Chapter 9, Table 8100. (a) 240.81

1. (a) 820.154(A)2. (a) 820.243. (a) 830.24. (a) 820.255. (b) 810.20(B)6. (a) 820.93(A)7. (a) 810.21(E)8. (a) 810.21(F)(1)9. (a) 810.21(H)10. (a) 820.211. (a) 820.11312. (a) 100 Raceway13. (a) 225.1714. (a) 240.2115. (a) 250.52(A)(3)16. (a) 285.1 FPN 217. (a) 340.10(3)18. (a) 362.4619. (a) 404.2(A) Ex20. (b) 410.16(B)21. (a) 501.100(A)(2)22. (a) 555.10(A)23. (a) 590.4(B)24. (a) 605.625. (a) 725.3(B)26. (a) Chapter 9, Table 1, Note 427. (a) Annex C28. (b) 450.43(A)29. (c) 230.71(A)30. (c) 517.2 Hospital31. (c) 517.2 Nursing Home32. (c) 225.33(A)33. (d) 100 Branch Circuit, Multiwire34. (d) 334.15(C)35. (d) 702.8(A)36. (d) 701.9(A)37. (a) 410.130(F)(5)38. (c) 604.6(A)(1)39. (c) 725.154(A)40. (b) 100 Surge Protective devices (SPDs)41. (d) 348.242. (b) 240.83(D)43. (d) 230.644. (c) 590.4(E)45. (b) 338.12(B)(1)46. (d) 250.52(B)(1) and (2)47. (c) 210.8(A)(4) and (5)48. (a) 600.649. (b) 300.13(B)50. (b) 400.23

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Chapter 4—Challenge Quizzes Answer Key

26. (d) 314.71(B)(1)27. (d) 314.71(B)(2)28. (b) 110.14(A)29. (c) 540.1330. (d) 408.18(A)31. (d) 410.140(B)32. (b) 424.3533. (c) 440.12(A)(1)34. (c) 322.10435. (d) 340.10436. (a) 620.12(A)(1)37. (c) 810.52 Table38. (c) 460.6(A)39. (d) 430.72(C)(4)40. (a) 225.39(A)41. (d) 225.39(B)42. (b) 324.10(B)(2)43. (a) 440.55(B)44. (b) 660.945. (b) 408.5246. (d) 720.447. (a) 330.10448. (a) 727.649. (d) 550.13(D)50. (a) 344.10(C)

1. (a) 670.3(B)2. (c) 326.23. (c) 366.100(E)4. (a) 430.12(C) and Table 430.12(C)(1)5. (c) 332.24(1)6. (b) 390.3(B)7. (c) 450.21(C)8. (a) 370.4(C)9. (b) 374.410. (a) 430.81(A)11. (c) 353.12012. (d) 480.6(B)13. (a) 810.16(A) Table14. (b) 810.16(A) Table15. (c) 830.100(D)16. (b) 368.3017. (a) 460.2(A)18. (d) 430.32(C)(1)19. (b) 430.110(B)20. (a) 450.4621. (a) 240.83(B)22. (b) 520.53(G)23. (b) 720.624. (d) 427.1225. (d) 314.71(A)

ChaLLenge QUIz 1—artICLe 90 thrOUgh Chapter 9


Answer Key Chapter 4—Challenge Quizzes

26. (c) 422.11(C)27. (c) 550.10(A) Ex 128. (d) 810.11 Ex29. (d) 540.230. (c) 324.4131. (b) 727.932. (c) 398.15(C)33. (c) 330.24(A)(2)34. (a) 338.2435. (d) 830.100(A)(4)36. (a) 830.100(A)(4) Ex and 830.100(B)(3)(2)37. (c) 366.12(2)38. (c) 660.6(A)39. (a) 660.540. (d) 366.23(A)41. (d) 366.23(A)42. (a) 368.10(C)(2)(a)43. (a) 830.44(I)(3)44. (b) 382.30(A)45. (a) 398.30(A)(1)46. (c) 408.5, Table47. (d) 424.3448. (c) 310.13(A), Table49. (d) 310.15(B)(3)50. (c) 324.10(B)(1)

1. (b) 394.172. (d) 470.33. (a) 250.52(A)(7)4. (a) 620.12(A)(2)5. (c) 392.7(B), Table Note b6. (a) 322.56(B)7. (c) 110.27(A)(2)8. (b) 410.54(B), See 402.69. (d) 551.45(B)10. (c) 424.22(B)11. (d) 424.22(B)12. (b) 430.23213. (c) 250.112(K)14. (c) 310.17, Table15. (c) 650.816. (d) 332.3017. (c) 344.12018. (c) 410.5 Ex19. (c) 410.1220. (c) 701.11(G)21. (c) 324.2 Definition22. (d) 394.30(A)23. (c) 360.20(B)24. (a) 727.125. (d) 408.36(A)




26. (b) 547.9(D)27. (d) 240.60(C)(1) through (5)28. (c) 470.18(B)29. (c) 625.2230. (d) 520.5(A)31. (c) 692.632. (a) 410.56(C)33. (d) 625.2634. (d) 344.13035. (d) 690.14(C)36. (d) 830.44(A) and (F)37. (d) 324.40(A)38. (d) 427.2 FPN39. (a) 370.340. (d) 314.2941. (a) 620.22(A)42. (a) 382.10(C)43. (a) 408.3(A)(2)44. (d) 332.40(B)45. (d) 426.20(E)46. (b) 404.2(B) Ex47. (d) 404.13(A)48. (d) 100 Exposed (as applied to live parts)49. (d) 547.2, Distribution Point FPN No. 150. (d) 100 Automatic

1. (a) 450.62. (c) 328.23. (b) 480.24. (a) 430.9(C)5. (b) 390.56. (a) 404.14(B)(2)7. (b) 410.115(A)8. (c) 455.6(A)(1)9. (a) 424.1110. (c) 520.4211. (b) 230.50(B)(2)12. (a) 400.4, Table Note 413. (b) 424.3914. (c) 430.110(A)15. (c) 230.95(A)16. (c) 424.3617. (c) 517.61(A)(5)18. (d) 230.95(C)19. (d) 517.35(A)20. (a) 320.221. (b) 330.222. (b) 410.74(A)23. (c) 410.130(E)(1)24. (d) 545.6 Ex25. (d) 332.2




26. (d) 100 Utilization Equipment27. (c) 430.4428. (b) 322.40(D)29. (b) 620.91(C)30. (c) 517.3231. (a) 830.2432. (a) 404.10(B)33. (d) 553.7(B)34. (d) 324.56(B)35. (d) 392.5(F)36. (a) 410.151(D)37. (c) 320.2438. (d) 352.10039. (d) 340.11640. (c) 330.11641. (b) 410.82(A)42. (c) 324.6043. (b) 620.61(B)(2)44. (d) 100 Branch Circuit, Appliance45. (c) 250.2—Effective Ground-Fault Current Path46. (a) 250.2—Bonding Jumper, System47. (c) 314.448. (c) 426.50(A)49. (b) 250.2—Ground Fault50. (d) 100 Enclosed

1. (c) 372.22. (b) 515.3 Table3. (b) 516.3(F)4. (a) 374.105. (a) 366.26. (d) 430.7(A)(6)7. (d) 440.41(A)8. (d) 334.116(A) and (B)9. (c) 426.3010. (b) 480.9(A)11. (a) 324.40(C)(1)12. (b) 332.11213. (c) 240.15(A)14. (d) 310.11(C)15. (a) 332.2416. (d) 702.7(1) and (2)17. (d) 701.8(A),(B), and (C)18. (d) 382.4019. (d) 372.1320. (c) 430.111(B)(3)21. (a) 408.2022. (a) 100 Dusttight23. (d) 240.5224. (a) 250.9025. (d) 503.155(A)




26. (d) 520.68(A)(1)27. (d) 372.628. (d) 110.27(A)(1), (3), and (4)29. (b) 700.2130. (d) 545.231. (d) 408.432. (d) 406.633. (d) 110.13(A)34. (a) 250.64(A)35. (d) 390.736. (d) 366.100(A)37. (b) 240.54(E)38. (d) 353.10039. (d) 310.8(C)40. (c) 530.1441. (d) 396.10(A)42. (c) 372.943. (b) 336.244. (c) 427.445. (c) 100 Conductor, Covered46. (b) 430.35(B)47. (b) 324.10(E)48. (d) 100 Switch, Isolating49. (b) 408.3(A)(1)50. (c) 240.50(C)

1. (c) 100 Sealable Equipment2. (d) 240.41(B)3. (d) 100 Hoistway4. (d) 230.70(C)5. (a) 424.38(B)(4)6. (b) 500.6(A)(3) FPN7. (a) 500.6(A) FPN No. 28. (c) 372.59. (d) 360.10 and 360.1210. (d) 392.5(C), (D), and (E)11. (c) 410.12012. (d) 445.12(A)13. (b) 702.1114. (d) 240.54(D)15. (d) 100 Duty, Varying16. (a) 675.217. (a) 100 Isolated (as applied to location)18. (a) 332.1019. (b) 332.3120. (a) 372.721. (d) 322.10(1)22. (b) 430.75(B)23. (d) 450.2424. (a) 250.96(B) and 300.10 Ex 225. (d) 408.19




26. (b) 410.56(E)27. (b) 230.95(C) FPN No. 228. (d) 550.2 Definition, Appliance, Portable and FPN29. (a) 424.1330. (d) 324.1231. (d) 440.1332. (d) 368.17(C) and 404.8(A) Ex 133. (d) 338.234. (a) 398.1735. (c) 370.236. (d) 100 Branch-circuit overcurrent device37. (c) 547.238. (b) 390.639. (c) 680.27(C)(3)40. (b) 374.641. (d) 430.4342. (d) 332.24(2)43. (c) 398.30(D)44. (b) 727.145. (d) 100 Solar Photovoltaic System46. (a) 332.10447. (d) 760.49(C)48. (c) 408.1749. (c) 322.12(4)50. (c) 396.30

1. (d) 324.1202. (a) 430.14(B) Ex3. (b) 324.184. (c) 460.28(A)5. (a) 230.776. (b) 396.27. (d) 430.408. (d) 517.64(A)(1), (2), and (3)9. (d) 400.5(B)10. (c) 100 Explosionproof Apparatus11. (b) 422.60(A)12. (d) 625.2 Electric Vehicle13. (c) 702.10(A)14. (a) 370.215. (b) 660.216. (d) 692.217. (d) 830.133(A)(2)18. (b) 250.8(B)19. (d) 230.8120. (b) 480.10(A)21. (b) 501.1522. (b) 455.8(A)23. (b) 424.38(A)24. (d) 100 Rainproof25. (c) 324.42(A)




27. (a) 322.228. (b) 324.10(A)29. (a) 332.10830. (a) 370.7(2)31. (a) 424.6632. (a) 427.2233. (b) 427.4734. (a) 430.109(B)35. (a) 500.6(A)(1)36. (a) 500.6(B)(1)37. (a) 500.6(B)(2)38. (a) 500.6(B)(3)39. (a) 501.115(B)(2)40. (b) 502.1541. (a) 505.21 Ex42. (a) 517.30(E)43. (a) 517.41(E)44. (a) 520.2445. (a) 520.7346. (a) 540.11(B)47. (a) 545.148. (b) 550.1549. (b) 605.250. (a) 625.2 Electric Vehicle Connector

1. (c) 430.12(A)2. (a) 324.10(D)3. (a) 100 Switchboard4. (a) 310.13(A), Table5. (d) 410.526. (d) 398.27. (a) 520.2, Performance Area8. (c) 760.459. (c) 725.45(A)10. (b) 390.811. (c) 392.6(B)12. (d) 727.413. (d) 230.4614. (b) 300.6(A)(3)15. (a) 394.30(B)16. (a) 250.114(3)(a) and (b)17. (d) 240.83(A)18. (a) 100 Raintight19. (a) 100 Thermal Protector FPN20. (a) 110.14(B)21. (b) 250.6(D)22. (a) 250.8023. (a) 300.16(B)24. (a) 300.18(A)25. (b) 314.16(C)(1)26. (a) 314.44



Answer Key Chapter 4—Final Exams

51. (a) 230.9552. (b) 408.36(D)53. (c) 100 Grounded Conductor54. (c) 90.1(A)55. (a) 344.42(A) and 314.1556. (d) 501.10(B)(1)(7)57. (a) 230.258. (a) 800.15659. (b) 450.4160. (d) 300.20(A)61. (d) 440.1462. (d) 100 Controller63. (c) 501.14564. (c) 250.5865. (b) 314.16(B)(4)66. (a) 314.23(H)(1)67. (d) 600.4(A)68. (c) 250.4(B)(2)69. (d) 503.12570. (d) 502.125(B)71. (c) 100 Switch, General-Use Snap72. (b) 90.373. (a) 100 Authority Having Jurisdiction FPN74. (b) 110.26(F)(1)(d)75. (a) 210.11(C)(3) Ex76. (a) 210.52(G)(2)77. (a) 250.52(A)(3)78. (b) 250.8679. (a) 250.122(F)80. (b) 285.481. (b) 300.11(A)(2)82. (a) 310.1083. (b) 340.12(1)84. (a) 342.42(A)85. (b) 392.386. (a) 410.6487. (a) 501.105(A)88. (b) 511.10(A)89. (a) 514.11(A)90. (a) 517.13(B) Ex 191. (a) 525.10(B)92. (a) 547.293. (a) 555.10(B)94. (a) 590.4(B)95. (a) 680.2, Spa or Hot Tub96. (b) 680.26(B)97. (a) 701.1098. (a) 725.2199. (a) 760.139(A)100. (c) 312.2

1. (a) 240.812. (b) 220.52(B)3. (b) 225.18(2)4. (b) 230.24(B)(2)5. (b) 215.2(A)(2)6. (d) 250.102(E)7. (b) 550.308. (b) 430.22(A)9. (c) 400.5(A), Table Note10. (b) 406.2(C)11. (c) 406.8(B)(1) 12. (a) 358.30(C)13. (b) 110.26(A)(1) and Table 110.26(A)(1), Condition 114. (b) 422.31(B)15. (c) 770.48(A)16. (a) 250.24(C)(1)17. (c) 330.30(C)18. (b) 680.22(D)19. (b) 410.16(C)(2)20. (b) 386.12(2)21. (b) 410.15422. (a) 378.30(A)23. (c) 210.52(A)(3)24. (a) 820.1525. (c) 300.4(G)26. (d) 630.32(A)27. (b) 645.5(C)28. (d) 250.30(A)(4)(c)29. (d) 620.23(A)30. (c) 362.4831. (d) 240.21(B)(2)32. (a) 348.6033. (c) 100 Receptacle34. (d) 100 Listed35. (b) 210.4(D)36. (c) 430.83(A)(2) and (3)37. (d) 110.12(B)38. (a) 352.2239. (c) 725.179(C)40. (b) 500.4(A)41. (d) 820.133(B)42. (c) 300.643. (c) 334.3044. (c) 230.54(G)45. (b) 700.4(A)46. (a) 356.647. (d) 504.7048. (c) 680.7149. (a) 800.2450. (b) 353.28

FInaL exaM nO. 1 praCtICe QUestIOns In ranDOM OrDer—artICLe 90 thrOUgh annex C


Chapter 4—Final Exams Answer Key

51. (b) 410.16052. (a) 502.130(A)(1)53. (c) 250.30(A)(4)(c)(2)54. (c) 358.4255. (c) 440.1456. (c) 517.1657. (a) 240.1 FPN58. (a) 230.5659. (b) 110.13(B)60. (c) 100 Coordination, Selective61. (b) 342.42(A)62. (d) 344.42(A)63. (b) 110.26(G)64. (b) 680.7365. (c) 376.266. (b) 314.16(B)(4)67. (d) 503.130(A)68. (c) 700.4(C)69. (c) 300.20(B) FPN70. (d) 504.80(A)71. (b) 230.2(B)(2)72. (d) 250.52(A)(4)73. (b) 680.2, Wet-Niche Luminaire74. (b) 90.475. (a) 100 Ungrounded76. (a) 210.12(B)77. (b) 210.52(B)(2)78. (a) 215.2(A)(3)79. (b) 250.6080. (a) 250.122(G)81. (a) 285.582. (a) 300.11(B)(2)83. (a) 334.3084. (b) 340.12(2)85. (a) 386.2186. (a) 404.487. (a) 500.5(A) FPN88. (a) 501.10(B)(2)(2)89. (a) 501.115(B)(1)90. (b) 547.5(A)91. (a) 590.4(C)92. (a) 645.5(D)(2)93. (c) 220.5394. (b) 450.4295. (a) 404.496. (d) 320.1097. (d) 400.7(A)(2), (3) and (6)98. (b) 700.12(F)99. (d) 680.43(B)(1)(c)(2)100. (b) 100 Bathroom

1. (b) 110.26(A)(1) and Table 110.26(A)(1), Condition 12. (b) 410.683. (d) 225.18(4)4. (d) 230.24(B)(4)5. (c) 210.52(H)6. (b) 406.8(B)(1) 7. (c) 250.24(C)(1)8. (d) 240.21(B)(4)(2)9. (b) 410.16(C)(3)10. (b) 550.31 Table 16,000 VA x 6 = 96,000 VA x 0.29 =

27,840 VA11. (b) 680.43(C)12. (b) 555.12 Table13. (a) 240.83(C)14. (a) 378.30(B)15. (d) 525.1116. (d) 300.5, Table Column 117. (d) 680.26(B)18. (d) 430.83(C)(2)19. (d) 630.32(B)20. (c) 430.2421. (b) 422.33(A)22. (c) 501.150(A)23. (b) 310.10 FPN24. (a) 406.2(D)25. (b) 514.11(A)26. (d) 680.23(A)(2)27. (a) 250.4(B)(3)28. (c) 353.4629. (b) 352.2430. (b) 100 Receptacle31. (c) 250.86 Ex 232. (d) 314.2533. (d) 356.10(3) and (4)34. (b) 620.23(B)35. (b) 100 Location, Damp36. (c) 100 Ground-Fault Circuit Interrupter37. (d) 511.1238. (a) 700.1539. (d) 330.30(D)40. (a) 312.241. (c) 350.242. (c) 90.1(B)43. (b) 210.5(C)44. (c) 408.4045. (d) 250.104(A)(1)46. (d) 701.11(B)(1), (2), and (3)47. (d) 320.12(1), (2), and (4)48. (c) 392.349. (d) 600.4(C)50. (d) 300.6(A)

FInaL exaM nO. 2 praCtICe QUestIOns In ranDOM OrDer—artICLe 90 thrOUgh annex C


Answer Key Notes

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