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NEC Math – Practice

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Question 1 of 25

1. Question

What is the full-load current rating of a 15kW, 208 volt, single phase heat pump?

72

75

100

60

Correct

Incorrect

Hint

I=P/E

15,000/208
Question 2 of 25

2. Question

What is the full-load current rating of a 240 volt single phase 10kW commercial dishwasher?

41.6

50

52

38.6

Correct

Incorrect

Hint

I=P/E

10,000/240
Question 3 of 25

3. Question

Where K=12.9, what is the approximate voltage-drop on a circuit where an 80-amp, 240 volt, single phase load is supplied with size 3 awg copper conductors with THWN insulation, and placed 200 feet from a panelboard?

7.84

6.88

8.88

5.68

Correct

Incorrect

Hint

vd=2×12.9xIxD/Cmil

CH 9 Table 8
Question 4 of 25

4. Question

What is the total resistance of a 10-ohm, a 15-ohm, and a 20-ohm resistor connected in series?

45

15

25

35

Correct

Incorrect

Hint

R+R+R=TR
Question 5 of 25

5. Question

What is the minimum allowable trade size of IMC required to contain a total of 10 copper THW conductors in a 20′ run, 5 #1AWG, and 5 #3AWG?

2 1/2

2

3

Correct

Incorrect

Hint

CH 9 Table 5
5 THW #1AWG .1901
5 THW #3AWG .1134
Total: 1.5175

CH 9 Table 4 Article 342
Question 6 of 25

6. Question

Given the following information, what is the maximum approximate distance a single phase, 240 volt, 42 amp, load can be located from a panelboard? *copper conductors – K = 12.9 *size 8 AWG THWN *VD 3%

109'

50'

115'

125'

Correct

Incorrect

Hint

D=(CmilxVD)/(2xKxI)

CH 9 Table 8
#8AWG = CM 16510
vd= 240×3%=7.2
118,872/1,083.6 =
Question 7 of 25

7. Question

True or False I = P x E

True

False

Correct

Incorrect
Question 8 of 25

8. Question

True or False I = P / E

True

False

Correct

Incorrect
Question 9 of 25

9. Question

True or False E = I x R

True

False

Correct

Incorrect
Question 10 of 25

10. Question

True or False R = I x E

True

False

Correct

Incorrect
Question 11 of 25

11. Question

True or False P = I x E

True

False

Correct

Incorrect
Question 12 of 25

12. Question

True or False E = P / I

True

False

Correct

Incorrect
Question 13 of 25

13. Question

True or False I = E / R

True

False

Correct

Incorrect
Question 14 of 25

14. Question

True or False Voltage Drop for 3 Phase System VD = 1.732 x L x R x I / 1000

True

False

Correct

Incorrect
Question 15 of 25

15. Question

How many 20 amp circuits 120V are required for lighting for a 3000 sq. ft. house?

4

3

5

6

Correct

Incorrect

Hint

NEC 220.14(J)

3VAper sq.\’ 3,000 x 3 = 9,000VA

1 20A Ckt. = 120V x 20A = 2,400 VA

9,0000VA/2,400VA = 3.75 = 4
Question 16 of 25

16. Question

If a dwelling unit has a connected lighting load of 9,600 VA, what is the minimum number of 120-volt, 15-amp, general lighting branch circuits required?

6

5

3

4

Correct

Incorrect

Hint

9,600VA

120v x 15A = 1800va

9,600/1,800 = 5.3 = 6
Question 17 of 25

17. Question

What is the approximate diameter, on No. 2 RHW wire in inches?

0.472

0.582

0.412

0.000

Correct

Incorrect

Hint

CH. 9 Table 5
Question 18 of 25

18. Question

If a transformer is installed to supply 243 amp, 208-volt, 3-phase load, the kVA rating of the transformer must be a minimum of _______ kVA.

90 kva

85 kva

95 kva

100 kva

Correct

Incorrect

Hint

P=IxE

243 x (208×1.732)

243 x 360 = 87,542va

87.5kva
Question 19 of 25

19. Question

Size 12 AWG THWN copper conductors, for the purpose of determining box fill, are calculated at what volume per conductor?

2.25

2.00

2.50

1.75

Correct

Incorrect

Hint

NEC T314.16(B)
Question 20 of 25

20. Question

A nonmotor-generator arc welder has a primary current rating of 30A with a duty cycle of 40 percent. What’s the minimum size branch-circuit conductor permitted to be used for the welder?

12 AWG

10 AWG

8 AWG

6 AWG

Correct

Incorrect

Hint

Table 630.11(A)
30Ax.63 = 18.9A
T310.16 = 12AWG
Question 21 of 25

21. Question

What’s the general lighting load for a 200,000 square foot storage warehouse?

126.25 kVA

72.5 kVA

190.5 kVA

132.5 kVA

Correct

Incorrect

Hint

General Lighting Total =
200,000 sq ft Ã— 1.2 VA per sq ft
General Lighting Total = 240,000 VA
First 12,500 VA at 100% 12,500 VA Ã— 100% = 12,500 VA
Remainder VA at 50% 227,500 VA Ã— 50% = 113,750 VA
Total General Lighting Demand Load 126,250 VA
Question 22 of 25

22. Question

What’s the service load at 120/240V, single-phase for a 10,000 square office building containing the following loads? Outside lights – twenty 180W at 120V; Parking lots lights – five 200W at 120V; Receptacles – seventy-five at 120V; Receptacles continuous duty use – twenty at 120V; Water heater – two 4.50 kW at 240V; Computers – twenty 2A at 120V; Printers – two 12A at 120V; Miscellaneous load – 20 kVA at 120V; A/C – 30 kVA at 240V; Heating – 40 kW at 240V; Air handlers – two 3 hp at 240V.

634A

643A

653A

663A

Correct

Incorrect

Hint

General Lighting at 120V
10,000 sq ft x 3.5 VA = 35,000 VA at 125% = 43,750 VA
Outside Lights at 120V
20 lights x 180 VA = 3,600 VA at 125% = 4,500 VA
Parking Lot Lighting at 277V
5 lights at 200 VA = 1,000 VA at 125% = 1,250 VA
Receptacles at 120V
10,000 sq ft x 1 VA sq ft =
10,000 VA, omit smaller than actual
Receptacles at 120V
75 receptacles x 180 VA = 13,500 VA
First 10,000 VA at 100% = 10,000 VA
Next 3,500 VA at 50% = 1,750 VA
Receptacles at 120V Continuous
20 receptacles x 180 VA = 3,600 VA at 125% = 4,500 VA
Water Heater
4,500W x 2 units = 9,000W
Computer
20 computers x 2A x 120V = 4,800 VA
Printers
2 printers x 12A x 120V = 2,880 VA
Misc Loads 20,000 VA
Heating 40,000W
A/C
30,000 VA (Omit [220.61])
Air Handlers 3 hp
17A x 240 = 4,080 VA x 2 units = 8,160 VA
Sign at 120V 1,200 VA at 125% = 1,500 VA
Total Load for Service = 140,990 VA
Service Size in Amperes
I = P/E
I = 140,990 VA/240V
I = 634A
Question 23 of 25

23. Question

A 75-site manufactured home park is designed for homes that have a 14,000 VA load per site. The service demand load for the park is ________.

201 kVA

222 kVA

264 kVA

1,200 kVA

Correct

Incorrect

Hint

[550.31]
The calculation is based on a minimum of 16,000 VA.
Step 1: Determine the total connected load.
75 sites Ã— 16,000 VA = 1,200,000 VA
Step 2: Determine the demand factor [Table 550.31].
22%
Step 3: Determine the total calculated load.
1,200,000 VA Ã— 0.22 = 264,000 VA
264,000VA/1,000 = 264 kVA
Question 24 of 25

24. Question

Electricity costs 13 cents per kilowatt hour. Ten metal halide high bay lights draw 2A each at 240V for 12 hours a day, five days a week. How much would it cost to use these lights during a four-week period?

$105.44

$149.76

$210.86

$419.34

Correct

Incorrect

Hint

5 days per week Ã— 4 weeks Ã— 12 hours per day = 240 hours
10 luminaires Ã— 2A Ã— 240V/1,000 = 4.80 kWh
4.80 kWh Ã— 240 hours = 1,152 kWh
1,152 kWh Ã— $0.13 per kWh = $149.76
Question 25 of 25

25. Question

Using the standard method, what’s the service load for a 4,000 square foot restaurant containng the following loads?
SERVICE 120/240V, single-phase:
Lighting actual – 6 kVA;
Receptacles – sixty at 120V;
Gas furnaces – two 5A at 120V;
A/C – 50A at 240V;
Heat – 10 kW at 240V;
Sign – 1.2 kVA at 120V.

KITCHEN LOADS:
Mixer – 10A at 240V;
Gas ovens – two 250 W at 120V;
Gas steamer – 350 W at 120V;
Electric warmers – four 5 kW at 240V;
Electric steam table – 20 kW at 240V;
Appliances – 5,000 VA at 120V;
Electric deep fryer – 15,000 VA at 240V.

318A

415A

425A

435A

Correct

Incorrect

Hint

General Lighting: 4,000 sq ft x 2 VA x 125% = 10,000 VA
Actual Lighting = 6,000 VA (omit [220.60])
Sign, 1.20 kVA at 120V = 1,200 VA x 125% = 1,500 VA
[Receptacles 220.44 and Table 220.44]
60 receptacles Ã— 180 VA = 10,800 VA
First 10,000 VA at 100%: â€“10,000 VA x 100% = 10,000 VA
Remainder at 50%: 800 VA x 50% = 400 VA
Gas Furnaces, two 5A at 120V = 1,200 VA
A/C, 50A at 240V = 12,000 VA
Heat, 10 kW at 240V = 10,00 W (omit [220.60])
Kitchen Loads [220.56]:
Mixer, 10A at 240V = 2,400 VA
Gas Ovens, two 250W at 120V = 500 W
Gas Steamer, 350W at 120V = 350 W
Electric Warmers, four 5 kW at 240V = 20,000 W
Electric Steam Table, 20 kW at 240V = 20,000 W
Appliances, 5,000 VA at 120V = 5,000 VA
Electric Deep Fryer,
15,000 VA at 240V = 15,000 VA
Total Kitchen Loads 63,250 VA
Total Kitchen Loads = 63,250 VA x 65% = 41,113 VA
Total Service Load = 76,213 VA
Service Size in Amperes
I = P/E
I = 76,213 VA/240V
I = 318A

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NEC Math – Practice

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