• Author / Uploaded
• Omar Al-hamid

Citation preview

Mike Holt’s Illustrated Guide to

ELECTRICAL

EXAM PREPARATION Includes Mike’s NEC6 Exam Practice Questions Workbook

ANSWER KEY While great care was used in creating the questions and answers, mistakes do sometimes slip by. Any errors found in the textbooks or Answer Keys after printing are listed on our Website. If you find an error, first check to see if it has already been corrected. Go to www.MikeHolt.com, click on the "Books” link, and then the "Corrections” link (www.MikeHolt.com/bookcorrections.htm). If you don’t find the error listed on the website, contact us by e-mailing [email protected]. Be sure to include the book title, page number, and any other pertinent information. The Answer Keys are updated regularly and the most recently revised are listed on our website. Instructors that have registered to receive a log-in password can access these updated Answer Keys and download them from the website. Go to www.MikeHolt.com, click on the “INSTRUCTORS," link on the left panel, then the “Answer Keys" link, and then follow the instructions listed on this page to download the Answer Keys.

Date: September 29,2014

Mike Holt Enterprises, Inc. ^ IM

888.NEC.CODE (632.2633) ‘ W W W .M ikeHoit.com

Unit 1— Electrician’s Math and Basic Electrical Formulas I Answer Key

CHAPTER 1 ELECTRICAL THEORY

14. (a) 50W P = I2 x R

Unit 1— Electrician’s Math and Basic Electrical Formulas Unit 1— Review Questions 1.

(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

15.

(c) 3 sq in. Area = Pie x r2 Pie - 3.14 r = radius (Vz of the diameter) Area = 3 .1 4x(J6 x2)2 Area = 3.14 sq in.

16.

(c)

17.

(c)

16 42 = 4 x 4 = 16 144 122 = 12 x 12 = 144

The overcurrent protection device must be sized 1.25 times larger than the load.

18.

(c)

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

16Ax 1.25 = 20A 9.

P = 16A2 x0.20 ohms P = (16Ax 16A) x 0.20 ohms P = 51.20W

(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 percent to its decimal multiplier: 20% increase = 1.20 Step 2: Multiply the number by the multiplier: 8 kW x 1.20 = 9.60 kW

D = 41,100/4,128 D —99 ft 19. (b) 50A I - VA/(E x V3) I = 18.000W/(208V x 1.732) Current = 18,000W/360 Current = 50A 20. (b) False 21. (b) 32 Enter the number on your calculator then push the square

11. (a) 0.80

root key (V).

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

22. (b) 1.732 Enter the number on your calculator then push the square

12. (b) 80A

root key (V).

The continuous load must be limited to 80 percent of the rating of the protection device.

23. (a) cubic inches

100Ax 0.80 = 80A

24. (b) 24 cu in.

13. (a) True

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

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (632.2633)

3

Answer Key I Unit 1— Electrician's Math and Basic Electrical Formulas

25. (a) 0.075 kW

45. (a) 175W

kW = W/1000 kW = 75W/1000 kW = 0.075 kW 26. (b) 25 2 + 7 + 8 + 9 — 26, the multiple choice selections are rounded to the nearest “fives.”

P- Ex I P = 7.20V x 24A P = 172.80W 46. (d) a and b 47. (a) True 48. (d) power loss

27. (c) 110W The input must be greater than the output.

49. (b) 100W P — I2 x R

Input = Output/Efficiency Input = 100W/0.90 Input = 111W 28. (d) all of these

P = 16A2 x0.40 ohms P - (16A x 16A) x 0.40 ohms P = 102.40W 50. (a) 43W

29. (b) negative, positive

P= PP= P=

30. (b) False 31. (a) True 32. (b) False 33. (a) True

Ix E 12A x (120V x 3%) 12Ax 3.60V 43.20W

51. (b) False

34. (a) True 35. (b) False 36. (d) silver, copper, gold, aluminum

Unit 1— Challenge Questions 1.

(d) 1,000 VA

2.

(d) Salt water

39. (d) directly, inversely

3.

(a) increase

40. (d) all of these

4.

(b) reduced by half

37. (d) all of these 38. (a) True

41. (b) False 42. (d) all of these 43. (c) 6.40V EVD - I x R Evd = 16Ax 0.40 ohms Evd = 6.40V 44. (a) 0.14 ohms R = E/I R = 7.20W50A R = 0.14 ohms

According to Ohm’s Law, current is inversely proportional to resistance, This means that if the resistance decreases, assuming voltage remains the same, the current will increase. It also works in the opposite direction; if the resis­ tance increases, again assuming the voltage remains the same, the current will decrease. Example: What's the current of a 120V circuit if the resis­ tance 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 = 24A ! - 120V/10 ohms = 12A I = 120V/20 ohms = 6A

Mike Holt's Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Unit 2— Electrical Circuits I Answer Key

5.

Step 2: Determine the power consumed by a 132.25 ohm load connected to a 230V source.

(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 hasnothing to do with the actual calculation. If we know thevoltage of thecircuit and the resistance in ohms of the resistor, the formula we need

PPP= P=

12. (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.

to use is: P - E2/R P = 120V2/1 0 ohms

Step 1: Determine the resistance rating of a 1,500W, 230V

P = {120V x120V)/10 ohms P = 1,440W 8.

(c) P - !2 x R

9.

(b) less

E2/R 230V2/1 32.25 ohms {230V x230V)/132.25 ohms 52,900/132.25 P -4 0 0 W

load. RR= R= R= R=

If current remains the same and the resistance increases, then the energy consumed will increase. Example: P = I2 x R P = 1OA2 x 5 ohms P = 500W P -1 0 A 2 x10ohm s

E2/P 230V2/1,500W (230V x 230V)/1,500W 52,900/1,500 35.27 ohms

Step 2: Determine the power consumed by a 35.27 ohm load connected to a 208V source. P = E2/R P = 208V2/35.27 ohms P = (208V X 208V)/35.27 ohms

P = 1.000W

P = 43,264/35.27 P = 1,227W

10. (c) 46W P - l2/R P - 16A2 x0.18ohm s P -(l6 A x 1 6 A )x 0 .1 8 o h m s

13. (b) 29W P = I2 x R P = 12A2 x 0.20 ohms P = 29W

P = 256W x 0.18 ohms P = 46W 11. (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/P R — 115V2/100W R = (115V x 115V)/100W R = 13,225/100 R = 132.25 ohms

Unit 2— Electrical Circuits Unit 2— Review Questions 1.

(a) True

2.

(c) a and b

3.

(b) False

4.

(a) True

5.

(a) True

6.

(a) True

7.

(a) True

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (632.2633)

5

Answer Key t Unit 2— Electrical Circuits

8.

(d) the same

45. (b) False

9.

(b) False

46. (a) True

10. (a) True

47. (b) 6.40V

11. (a) True

= 1x R I = 16A R = 2 ohms/1,000 ft = 0.002 ohms per ft Resistance of Conductors = 0.002 ohms x 100 ft x 2 conductors Resistance of Two Conductors = 0.40 ohms Voltage Drop of Conductors: EVD = I x R Evo = 16Ax0.40 ohms EVo

12. (a) True 13. (a) True 14. (b) False 15. (b) False 16. (a) True 17. (b) False

Eun - 6.40V ^vu

18. (a) True 19. (a) True

48. (a) 3.20V

Eyp = I X R

20. (b) False

I = 16A R = 2 ohms/1,000 ft = 0.002 ohms per ft Resistance of Conductors = 0.002 ohms x 100 ft Resistance of One Conductor = 0.20 ohms Voltage Drop of Conductors: EVD - I x R Evd = 16Ax 0.20 ohms Eun vu = 3.20V

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

49. (d) b and c

28. (a) True

50. (a) 2-wire

29. (a) True

51. (b) grounded

30. (a) True 31. (b) neutral point 32. (a) True 33. (b) neutral point 34. w

Unit 2-—Challenge Questions 1.

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

True

35. (d) 100 percent

I = E/R E - 12V R = 12 ohms I - 12V/12 ohms I = 1A

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

6

(a) 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.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Unit 2-— Electrical Circuits I Answer Key

10. (b) 1 ohm

3.(d) all of these

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

4.(d) 10V Step 1: Determine the current of the circuit. I - e/R t E - 30V Rt = 22.50 ohms I = 30V/22.50 ohms I - 1 ,33A

Rt = 1/(1/R1 + 1/R2 + 1/R3) Rt = 1/(1/z + 1/3 + 1/5) Rt = 1/(0.50 + 0.33 + 0.20) Rt = 1/1 Rt - 1 ohm

Step 2: Determine the voltage of Resistor 2.

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

E2 = I x R2 I = 1.33A R2 = 7.50 ohms E2 = 1.33A x 7.50 ohms E2 - 9.98V 5.

A1 a3

A2

ip j

(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

Figure 2 -5 3

11. (c) A3

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

E2/R 30V2/10 ohms 90W for each resistor in series 90W x 4 360W

P = 1,440W x 4 P = 5.760W

R = E2/P E - 24V P= RRR=

Parallel Example: If the four resistors are connected in parallel, each resistor will operate at 120V. P = E2/R P = 120V2/10 ohms p = 1,440W for each resistor in parallel

12. (d) 48 ohms

12W 24V2/12W 576/12 48 ohms

13. (a) 22 ohms ^ (Bell 1) - Et/IT E, - 30V lT = 0.75A R, = 30V/0.75A R1 - 40 ohms R2 (Bell 2) = 48 ohms Rt = Product/Sum Rt = (R, x R2)/(R-| + R2) Rt - (40 ohms x 48 ohms)/(40 ohms + 48 ohms) RT = 1,920/88 Rt = 21.82 ohms

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (632.2633)

Answer Key I Unit 2— Electrical Circuits

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

16. (b) 3V lT = 0.75A (last answer) ^ 3 ,4 .5 =

^ 3 ,4 +

^5

R3 4 5 - 2 ohms + 2 ohms R3 4 5 - 4 ohms E4 - lT x R3i4i5 ^3,4,5 = 0.75A x 4 ohms E 3 ;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 Rv CrvT/'ijnt VO'i. *v/.vNKeHoit oyn Figure 2 -5 4

Resistance of one resistor/number of resistors - 15 ohms/3 resistors = 5 ohms. Note: The three parallel resistors can be thought of as a single 5-ohm resistor in series with resistor R1.

14. (d) all of these 15. (a) 1.50V Step 1: Determine the Resistance of the two parallel resistors R3 and R4: RT = Product/Sum Rt = Rt = RT = RT =

(R3 x R4)/{R3 + R4) (4 ohms x 4 ohms)/(4 ohms + 4 ohms) 16/8 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 R34).

RT = Rj + ^ 2,3,4) RT = 5 ohms + 5 ohms RT - 1 0 ohms 18. (c) 60V Voltage drop across R1 is determined by: Et = lT x R, >t =

e s /r t

lT = 120V/10 ohms lT = 12A R1 = 5 ohms E1 = lT x Rt E] = 12Ax 5 ohms Et = 60V

>T = E s /R t

Es - 6V RT = 2 ohms + 2 ohms + 2 ohms + 2 ohms Rt = 8 ohms lT = 6V/8 ohms lT = 0.75A Step 3: Calculate the voltage across R3 4 E3 4 ~ I x R3 4 E34 = 0.75Ax 2 ohms E3|4 = 1.50V

8

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Unit 3— Understanding Alternating Current I Answer Key

17. (a) lead

19. (b) 0.58A If the neutral is opened, the multiwire branch circuit becomes one 240V series circuit,

18. (b) lags

I r = £s/Rt

20. r

Calculated Load = 5 kW x 10 Units x 0.50 [Table 220.54] Calculated Load = 25 kW

40

l ot a!

60 70 +80 210

15 units = 30 kW

Waste Disposal Dishwasher Water Heater Total Unit Connected Load ! 0 Uii

Step 1: Determine the total connected load (using 12 kW

Column C.

[220.53]

4.(b)

33 kW

General Lighting and Receptacles (840 sq ft x 3 VA) 2,520 VA Small-Appliance Circuits (1,500 VAx 2 circuits) 3,000 VA Laundry Circuit + OVA Unit Load 5,520 VA Total Connected Load 5,520 VA x 20 units = 110,400 VA

First 3.000 VA at 100%: 3,000 VA x 1.00 = Next 117,000 VA at 35%: 107,400 VA x 0.35 = Total Calculated Load 3.(b)

i.osiu ■■■ C n iu m i i C l / a l i i o >; M a K ij ili o i

laundry circuit.

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Unit 10— M ultifam ily Dwelling Calculations I Answer Key

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

9.

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

Step 1: Determine the total connected load (using 12 kW as minimum*).

The word “ maximum” in a range question is asking for the larger of Column B or C.

Ten ranges at 12 kW (10 units x 12 kW) Eight ranges at 14 kW {8 units x 14 kW) *Two ranges at 9 kW (2 units x 12 kW) Total Connected Load

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

120 kW 112 kW + 24 kW 256 kW

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

Step 2: Determine the average range kW. A vera u e H aiu jo

Step 2: Apply the Column B demand factor to the total connected load.

fatal k W / N u m b e r of H anges

256 kW/20 units - 12.80 kW 120 kW x 0.32 = 38.40 kW Step 3: Determine the calculated load using Table 220.55 Column C.

Column C for 15 units = 30 kW minimum calculated load. 10. (c) 70 [220.61(B)(1)]

20 units - 35 kW Step 4: Increase Table 220.55 Column C load. 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 (12.80 kW) exceeds 12 kW by 1 kW; there­ fore, Column C must be increased by 5%, which results in 105%, or a 1.05 multiplier.

Column C, 15 units: 30 kW Neutral Calculated Load ■■■Hanue Calculator! Load x 0.70 [220.61 (B)(1)]

Neutral Calculated Load = 21 kW

C a lcula te d l.oad ■ C o lum n DMaluo x M u ltip lie r Calculated Load = 35 kW x 1.05 Calculated Load = 36.75 kW

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

Note 3 to Table 220.55 Calculations (d)

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

Oryer Calculated l oad -••• Nameplate x N um ber of Units x

19.60 kW [Table 220.55, Note 3]

Demand Factor [fable 220.5-1]

Dryer Calculated Load = 5 kW x 10 Units x 0.50 Dryer Calculated Load = 25 kW

Step 1: Determine the total connected load. 5 units rated 5 kW 2 units rated 4 kW 4 units rated 7 kW Total Connected Load

25 kW 8 kW + 28 kW 61 kW

Neutral Calculated Load - Dry or Calculated I oaf I x Demand Factor [220.61 (R)(1}j

Neutral Calculated Load = 25 kW x 0.70 Neutral Calculated Load - 17.50 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.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (632.2633)

Answer Key I Unit 10— Multifamily Dwelling Calculations

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 0.47 Dryer Calculated Load = 25.85 kW Neutral Calculated Load = Dryer Calculated Load x Demand Factor [220.61(B)(1)] Neutral Calculated Load = 25.85 kWx 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, 10 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 15. (a)

8 kW [220.84(A)(2) Ex]

16. (b) 205 kVA [220.84] If there is no electric range in a multifamily dwelling, we can use the optional method, including the calculation for an 8 kW electric range, and compare the results to a standard calculation with no range. We are allowed to use the lesser of these two results [220.84(A)(2) Ex]

Step 2: Calculated load [Table 220.84]. 538,400 VA x 0.38 = 204,592 VA Note: Service size: I = VA/E I = 204,592 VA/240V I = 852A Compare this to the Standard Calculation: Step 1: Determine the general lighting and receptacle load. General Lighting 900 sq ft x 3 VA 2,700 VA Two Small-Appliance Circuits (required) 3,000 VA Laundry Circuit (required) 1,500 VA 7,200 VA 7,200VAX 20 units = 144,000 VA Demand Factors [Table 220.42] 1st 3,000 VA at 100% 3,000 VA Next 117,000 VA at 35% 117,000 VAx 35% Above 120,000 VA at 25% 24,000 VA x 25% General Lighting and Receptacle Calculated Load

6,000 VA 49,950 VA

Step 2: Determine appliance load (75% for 4 or more) [220.53]. Water Heater

5,000 VA X 20 units = 100,000 VA 100,000 VAx 7 5% = 75,000 VA

Step 3: No dryer.

Step 1: Determine the total connected load.

Step 4: No range.

General Lighting 900 sq ft x 3 VA 2,700 VA Two Small-Appliance Circuits (required) 3,000 VA Laundry Circuit (required) 1,500 VA Water FIeater 5,000 VA Range (minimum required 220.84(A)(2) Ex) 8,000 VA Air-Conditioning (240V x 28A)* 6,720 VA Heat 5 kW, (omit) [220.60] + OVA Total Connected Load 26,920 VA x 20 units = 538,400 VA

Step 5: Determine the larger of electric heat or air-conditioning. Air-Conditioning (240V x28A)* (a/c is larger than heat) 6,720 VA x 20 units = 134,400 VA Heat 5 kW, (omit) [220.60] + OVA Total Calculated Load 259,350 VA *5 hp, 230V motor FLC - 28A [Table 430.248]

*5 hp, 230V motor FLC = 28A [Table 430.248]

42

40,950 VA

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Unit 11— Commercial Calculations I Answer Key

Step 6: Determine the calculated load.

The receptacle load (180 VA for each receptacle) is permitted

538.400 VAX 0.38 = 204.592 VA

to be included with the general lighting for storage ware­ houses, see 220.44 of the NEC for details. If they were

Mote: Service size

combined and calculated using only Table 220.42, the result would be:

I - 259.350 VA/240V

[Tables 220.12 and 220.42]

I -- 1081A

General lighting (250,000 sq ft x 0.25 VA) 62,500 VA Receptacles (200 x 180 VA)* + 36,000 VA Total Connected Load 98,500 VA First 12,500 VA at 100% -12,50 0 VA x 1.00 Remainder at 50% 86,000 VA x 0.50 Total Calculated Load

The lesser result of the two methods was the optional method, so it is allowed [220.84(A)(2) Ex],

1.

(b)

15 kVA

3.

12.500 VA + 43,000 VA 55.500 VA

(c) 8,000 VA

[Tables 220.12 and 220.42]

[215.2(A), 215.3, 230.42(A), and Table 220.12]

24 motel rooms 16,440 sq ft x 685 sq ft --16,440 sq ft x 2 VA 32.880 VA First 20,000 VA at 50% -2 0 .000 VA x 0.50 = 12.880 VA x 0.40 ~ Next 80,000 VA at 40% Total Calculated Load

General Lighting Load (3,200 sq ft x 2 VA x 1.25) = 8,000 VA 4.

[215.2(A), 215.3, 230.42(A), and Table 220.12]

10,000 VA + 5,152 VA 15,152 VA

General Lighting (90,000 sq ft x 3 VA x 1.25) - 337,500 VA 5.

61 kVA

The feeder calculated load for one sign circuit: 1,200 VA x

General lighting (250,000 sq ft x 0.25 VA) 62,500 VA Total Connected Lighting Load 62;500VA First 12,500 VA at 100% -1 2 ,5 0 0 VA x 1.00 Remainder at 50% 50,000 VA x 0.50 Total Calculated Lighting Load

Total Calculated Lighting Load Total Calculated Receptacle Load Total General Lighting and Receptacle Calculated Load

(b) 1,500 VA [220.14(F), 60 0 .5 ,2 1 5.2(A), and 230.42(A)]

[Tables 220.12. 220.42, and 220.44]

Receptacle Load (200 x 180 VA) 36.000 VA First 10 kVA at 100% —10,000 VA Remainder at 50% 26,000 VA Total Calculated Receptacle Load

(d) 338 kVA

1.25 = 1,500 VA 6.

(b) 33 kVA [220.43(A), 210.19(A)(1), and 210.20(A)] Total VA - 130 ft x 200 VA per foot

12.500 VA + 25,000 VA 37.500 VA

Total VA - 26,000 VA This is a continuous load, so when calculating the service conductor size, take the Article 220 load times 125%: Total VA for Continuous Load on Feeder/Service

10.000 VA +13,000 VA 23.000 VA 37,500 VA +23,000 VA

Conductor • VA x 1.25 Total VA for Continuous Load on Feeder/Service Conductor = 26,000 VA x 1.25 Total VA for Continuous Load on Feeder/Service Conductor = 32,500 VA. 32,500

60,500 VA

VA/1,000 = 32.50 kVA

See Annex D Example D3.

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (632.2633)

43

^ J jjj

Answer Key j Unit 11— Commercial Calculations

(a) 3,600 VA

11. (d)

Multioutlet Receptacle Assembly [220.14(H)]

[220.12, 220.14(K), 215.2(A), 230.42(A), and 220,44]

VA per 5 ft = 180 VA when unlikely to be used simultaneously:

General Lighting (10.000 sq ft x 3.50 VAx 1.25) 43.750 VA

Number of 180 VA Sections = ten 5-foot sections

Receptacle Demand: Larger of 220.14(K)(1) or (2) 220.14(K)(1): 75 receptacles x 180 VA 13,500 VA 220.14(K){2): Load at 1 VA per sq ft: 10,000 VA (omit) Apply 220.44 13,500 VA 1st 10,000 VA at 100% -10,00 0 VA x 1.00 - 10,000 VA Remainder at 50% 3,500 VA x 0.50 = + 1.750 VA Receptacle Calculated Load 11.750 VA

VA Load = Number of sections x 180 VA VALoad - 10 x 180 VA VALoad - 1,800 VA VA per 1 ft = 180 VA when likely to be used simultaneously: Number of 180 VA Sections - Number of ft/1 ft per section Number of 180 VA Sections = 10 ft/1 ft per section Number of 180 VA Sections = 10 sections VA Load = Number of sections x 180 VA VA Load = 10 x 180 VA VA Load= 1,800 VA

Total General Lighting and Receptacle Calculated Load = 43.750 VA + 11.750 VA Total General Lighting and Receptacle Calculated Load = 55,500 VA

Total multioutlet receptacle assembly load = 3,600 VA (b)

56 kVA

13 [220.14(1)] i'ili. liii. V/\

C il'iilii '/ -iis /, C il'U ii, ■'i i.;:::' r

12. (c)

Circuit VA - 120V x2 0A

644A [Table 550.31]

Circuit VA - 2,400 VA

The feeder and service for manufactured home parks are

Receptacle Yokes per Circuit = Circuit VA/180VA

sized using the demand factors of Table 550.31.

Receptacle Yokes per Circuit = 2,400 VA/180 VA

Be careful. The demand factors of Table 550.31 apply to the

Receptacle Yokes per Circuit = 13 receptacles. Note: Receptacles aren’t considered a continuous load.

larger of: (1) 16,000 VA for each manufacture home lot [550.31(1)] or

9.

(c)

15 kVA

(2) The calculated load for the largest typical manufactured

[220.44 and Table 220.44] 110 receptacles x 180 VA 19,800 VA First 10,000 VA at 100%: -10 ,00 0 VA x 1.00 Remainder at 50%: 9,800 VA x 0.50 Total Receptacle Calculated Load 10. (c)

home each lot will accept [550.31 (2)]. - 10,000 VA = + 4,900 VA 14,900 VA

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

162 kVA [220.12, 220.14(K)(2), 220.44, and 215.3] General Lighting (30,000 sq ft x 3.50 VA x 1.25) Receptacle load [Table 220.12, Note and 220.14(K)] 1 VA per sq ft Total General Lighting and Receptacle Calculated Load

44

I -- 154,560 VA/240V 131.250 VA

I = 644A

30.000 VA 161,250 VA

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Unit 11— Commercial Calculations I Answer Key

13. (b) 400A

16. (a)

[551.73(A)]

210 kVA Calculated Load = 50% x (amount over 325 kVA)

Recreational vehicle parks are calculated using the demand

+ 172.50 kVA [Table 220.88]

factors of Table 551.73. Step 1: Determine the VA load per site.

Calculated Load =

0.50 x

(400 kVA - 325 kVA) + 172.50 kVA

Calculated Load =

0.50 x

75 kVA + 172,50 kVA

Calculated Load = 37.50 kVA + 172.50 kVA

Each 20A receptacle site dedicated to tents is rated a

Calculated Load = 210 kVA

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)],

17. (b) 297 kVA Calculated Load = 45% x (amount over 325 kVA)

Step 2: The total connected load is:

+ 262.50 kVA [Table 220.88]

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 VA

Calculated Load =

0.45 x

(400 kVA - 325 kVA) + 262.50 kVA

Calculated Load =

0.45 x

75 kVA + 262.50 kVA

Calculated Load = 33.75 kVA + 262.50 kVA

Total Load = 232,200 VA

Calculated Load = 296.25 kVA

Step 3: Determine the demand factor. The demand factor for 62 sites listed in Table 551.73 is 0.41. Step 4: Determine the total calculated load. 232,200 VAX 0.41 = 95,202 VA

1.

(a) 23 kVA [Table 220.12 and Table 220.42]

Step 5: Calculate the total ampere rating.

I - 95,202 VA/240V I = 396.70A 14. (b) 600A Section 555.12 and Table 555.12 permit a demand factor according to the number of receptacles. 54 receptacles = 40% demand factor. 24 receptacles x 20A 480A 30 receptacles x 30A + 900A 1,380A x 0.40 (demand factor) = 552A 15. (c)

Urn* i 1- ■■■■■titiislkitige (luustUms

300 kcmil 570A/2 raceways = 260A

Hotel: 12 x 15 ft = 180 sq ft x 2 VA x 100 Rooms 36,000 VA First 20,000 VA at 50% -2 0 ,0 0 0 VA x 0.50 Remainder at 40% 16,000 VA x 0.40 Office: 60 x 20 ft = 1,200 sq ft x 3.50 VAX 1.25 (continuous ioad)

5,250 VA

Office Receptacles: 1,200 sq ft x 1 VA*

1,200 VA

Hallway Lighting: 120 sq ft x 0.50 VA x 1.25 (continuous load) + 75 VA 6.525 VA

[Table 310.15(B)(16)] 300 kcmil conductors have a rating of

10,000 VA 6,400 VA

6,525 VA 22,925 VA

285A each x 2 conductors = 570A combined. Because the

22,925 V A /1 ,000 = 22.93 kVA

load is less than 800A. the conductor must [240.4(B)]:

*[220.12(K)(2) and Table 220.12, Note b] add 1 VA per sq ft

(1) have an ampacity of at least 570A,

for unknown receptacle count.

(2) be protected by a 600A protection device, and (3) must be sized according to Table 310.15(B)(16), 300 kcmil, using the 75°C terminal rating [110.14(C)(1 )(b)]. Note: 250 kcmil THHN conductors have an ampacity of 290A each, but we can’t use the 90°C ampacity rating for sizing conductors [110.14(C)(1)(b)],

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (632.2633)

Answer Key I Unit 11 — Commercial Calculations

(c)

14,000 VA

6.

(a)

83 kVA

[220.44]

[551.71 and 551.73(A)]

Table 220.44; nondweilirtg unit receptacles at 180 VA each

Note: A minimum of 70% of the sites must have a 30A or

100 Receptacles x 180 VA 18,000 VA First 10,000 VA at 100%: 10,000 VA x 1.00 Remainder at 50%: 8,000 VA x 0.50 Total Calculated Load

20A facility (3.600 VA per site) and a minimum of 20% of the 10,000 VA + 4,000 VA 14.000 VA

sites must have a 50A facility (9,600 VA per site). Check: 42 sites x 0.70 = 29 minimum 3,600 VA sites 42 sites x 0.20 = 8.40 or 9 minimum 9,600 VA sites

(b)

20,000 VA

Step 1: Determine the total connected load.

[220.14(K){2) and Table 220.12]

*Allow 1 VA per sq ft for receptacles when the number of recep­

9 sites at 50A {9 sites x 9,600 VA) 30 sites at 30A (30 sites x 3.600 VA) 3 sites at 20A (3 sites x 2.400 VA) Total Connected Load

tacles is unknown in banks and office buildings. The demand

Step 2: Determine the demand factor for 42 sites [Table

factors of 220.42 apply when the receptacle calculation is

551.73],

General-Use Receptacles: 20.000 sq ft x 1 VA per sq ft* 20,000 VA

based on 220.14(l), but not when based on 220.14(K) [220.44], (c)

41%

162 kVA

Step 3: Determine the calculated load.

[220.14(K)(2), Table 220.12, and Table 220.44] General Lighting (30,000 sq ft x 3.50 VAx 1.25)

Calculated Load = 201,600 VA x 0.41 Calculated Load ~ 82,656 VA 131,250

General-Use Receptacles: 30.000 sq ft x 1 VA per sq ft* 30,000 Total Lighting and Receptacles 161,250 Calculated Load

86.400 VA 108.000 VA ■+ 7.200 VA 201.600 VA

Calculated Load in kVA - 82.656 VA/1,000 = 82.66 kVA

VA 7.

(b)

630A [555.12]

VA

Step 1: Determine the total connected receptacle ratings. 20 receptacles rated 20A (20 x 20A) 17 receptacles rated 30A {17 x 30A) 7 receptacles rated 50A (7 x 50A) Total Connected Receptacle Rating

VA

161,250 VA /1 ,000 - 161.25 kVA

*Allow 1 VA per sq ft for receptacles when the number of

400A 510A + 350A 1,260A

receptacles is unknown in banks and office buildings. The demand factors of 220.42 apply when the receptacle calcula­

Step 2: Determine the demand factor from Table 555.12.

tion is based on 220.14(1), but not when based on 220.14(K)(2)

Table 555.12 is 50% for 44 receptacles.

[220.44],

Step 3: Determine the calculated load.

(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 [Table 550.31],

22% Step 3: Determine the total calculated load. 1.200.000 VA x 0.22 = 264,000 VA

1 ,260A x 0.50 = 630A 8.

(d) 64 kVA [2 2 0 .88]

Step 1: Determine the connected load, General Lighting Dishwasher Coffee Makers (2 kW x 2 units) Kitchen Appliances (2 kW x 5 units) Small-Appliances Circuits (1.50 kVA x 10 units) Total Connected Load

264,000VA/1,000 = 264 kVA

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

30 kVA 5 kW 4 kW 10 kVA t J 5_kVA 64 kVA

Unit 12— Transformer Calculations I Answer Key

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

13. (d) 61A I = VA/(E x 1.732) I = 22,000 VA/(208V x 1.732) I = 22,000 VA/360V I = 61A

Unit 12— Transformer Calculations Unit 12— Review Questions 1.

(a) series

2.

(a) A Delta system.

3.

(b) line

4.

(b) Line

5.

(d) 208V

14. (a) 125 (Table 450.3(B)] 15. (b) 125 gable 450.3(B)] 16. (b) 4 AWG I = VA/(E x 1.732) I = 45,000 VA/(480V x 1.732) I = 54.13A 54.13Ax 1.25 = 67.67A

High Leg Voltage = Voltage-to-Ground x 1.732 High Leg Voltage = 120V x 1.732 High Leg Voltage = 207.84V = 208V

4 AWG has an ampacity of 85A at 75°C [Table 310.15(B)(16)]. 17. (c) 2 AWG

6.

(d) orange [110.15]

7.

(b) Wye

8.

(b) False

9.

(b) 54A l= V A /(E x 1.732) I = 45,000 VA/(480V x 1.732) I = 45,000 VA/831V I = 54A

10. (c) 108A I = VA/(E x 1.732) I = 45,000 VA/(240V X 1.732) I = 45,000 VA/416V I = 108A 11. (a) True 12. (b) 26A I = VA/(E x 1.732) I = 22,000 VA/(480V x 1.732) I = 22,000 VA/831V I = 26A

I = VA/(E x 1.732) I = 75,000 VA/(480Vx 1.732) 1= 90.21A 90.21 x 1.25 = 112.77A 2 AWG has an ampacity of 115A at 75°C [Table 310.15(B){16)]. 18. (a) 2/0 AWG 2/0 has an ampacity of 175A at 75°C [240.21 (C) and Table 310.15(B)(16)]. 19. (c) 350 kcmil 350 kcmil has an ampacity of 310A at 75°C [240.21(C) and Table 310.15(B)(16)]. 20. (c) at any single point [250.30(A)(1)] 21. (a) True [250.30(A)(1)(a)] 22. (d) a orb [250.30(A)(4)] 23. (a) 250.66 [250.30(A)(5)] 24. (d) 4 AWG [250.66(B)]

Mike Holt Enterprises, Inc. • www.MikeHolt.com • 888.NEC.CODE (632.2633)

Answer Key I Unit 12 — Transformer Calculations

Unit 12— Challenge Questions

4.

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

1.(a) 45A

secondary VA.

I - VA/(E x 1.732)

Step 1: Determine the transform er VA.

I - (37.50 kVA x 1000)/(480V x 1.732)

2.(c)

I = (37.50 kVAx 1000)/831 V

Secondary VA ■- 1: x I >' 1.73?

I = 45.13A

Secondary VA = 208V x 200A x 1.732 Secondary VA = 72,051 VA

108A There are two ways to calculate this problem.

Step 2: Determine the prim ary line current.

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

h - VA/(h X 1,732) i = 72,051 VA/(480x 1.732)

determine the secondary line current:

1= 72,051 VA/831V I = 86.70A

Primary Line Current ■■■-■Primary Phase Power/ (Primary Phase Volts x 1.732)

Primary Line Current = 45,000 VA/(480V x 1.732)

5.(c)

t ine Current, for Wye System

Primary Line Current = 54.21A

(Line Volts x 1.732)

Next, use the ratio to determine the secondary voltage, and

If the phase voltage is 277V, the line voltage is 277V x 1.732

then calculate the secondary line current.

- 480V.

The ratio is 2:1, therefore the secondary voltage is one-half of

Line Current - 10,000 VA/(480V x 1.732)

the primary voltage and the current will be twice the primary

Line Current - 10,000 VA/831V

line current.

Line Current — 12A

I ■■VA/'(!:. x 1.732)

(2) The voltage ratio is 2:1. If the primary voltage is 480V, the

I = (37.50 kVA x 1000)/(480V x 1.732)

secondary is 240V

I = (37.50 kVAx 1000)/831 V

Secondary Line Current = 45,000 VA/(240V x 1.732)

I = 45.13A

Secondary Line Current = 108.25A

[Table 450.3(B)] not more than 125% of primary current,

17A

45.13Ax 1.25 - 56.41A. [Table 450.3(b), Note 1] next size up, [240.6(A)], 60A device

Step 1: Determine the prim ary line power. A. The primary line power is the same as the secondary line

7.

54A x 1.25 = 67.50A, next size up is 70A

B. The secondary line power is calculated as:

[240.6(A), Table 450.3(B), Note 1]

VA - Secondary Line Volts >; Secondary Line Amperes

8.

x 1./3X

(c) 70A 45,000VA/(480Vx 1.732) = 54A

power, assuming 100% efficiency.

(c) 125A

Secondary Line Power = 240V x 300A x 1.732

75,000VA/(480Vx 1.732) = 90A

Secondary Line Power = 124,704 VA

90A x 1.25 — 112.50A, next size up is 125A [240.6(A), Table 450.3(B), Note 1]

Step 2: Determine the prim ary line current. Prinlaiy Line Current

Primal y Line Power/

(P rimaiy Line Voltage x 1.732)

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

48

Line Power/

6.(d) 60A

Secondary Line Current =108.43A

(a)

12A

Primary Line Current = 45,000 VA/830V

Secondary Line Current ~ 54.21 A x 2

3.

(d) 87A

9.

(d) 175A 112,500VA/(480V x 1.732) = 135A 135A x 1 .2 5 = 169A, next size up is 175A [240.6(A), Table 450.3(B), Note 1]

Mike Holt’s Illustrated Guide to Electrical Exam Preparation, based on the 2014 NEC

Practice Quiz 1— Straight Order [Articles 90-110] I Answer Key Question #

Answer

NEC Section #

Question #

CHAPTER 4— NEC PRACTICE QUIZZES AND EXAMS Practice Quiz 1 — Straight Order [Articles 9 0 - 1 1 0 ] 1. 2. 3. 4.

(c) (C) (d) (d)

5. 6.

(a)

7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

(d) (d)

23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41.

(b)

(c) (d)

(c) (b)

(c) (c) (b)

(a) (b) (d) (d) (d) (b) (b) (d) (b) (d) (d)