Dwelling, Feeder, Box Fill, And Motor Calculations - MT

Dwelling, Feeder, Box Fill, and Motor Calculations - MT Author: David Burtt

Dwelling, Feeder, Box Fill, and Motor Calculations - MT

Single Family Dwelling Calculations

Question 1: 220.12 Lighting Load for Specified Occupancies.Question ID#: 3196.0

The general lighting load for various types of buildings or occupancies is given inTable 220.12, and is shown in volt/amps per square foot. When determining thesquare footage of a building the outside dimensions of each floor of the building areused. For dwelling units, the calculated floor area shall not include open porches,garages, or unused or unfinished spaces not adaptable for future use.

For example, an attic space with a pull-down ladder is not adaptable for future use. An attic space with stairs would be adaptable for future use and should be includedin the general lighting load calculation.

According to 220.14(J), for dwelling units, receptacles and lighting outlets areincluded in the general lighting load. The unit load for dwellings is 3 VA per sq. ft.

Question 1: What is the general lighting load, without applying demand factors, for a single family dwelling thatmeasures 60 ft. by 30 ft.? A: 1800 VA.B: 3600 VA.C: 5400 VA.D: 6000 VA.

Question 2: 220.12 Lighting Load for Specified Occupancies.Question ID#: 3197.0

The general lighting load for various types of buildings or occupancies is given inTable 220.12, and is shown in volt/amps per square foot. When determining thesquare footage of a building the outside dimensions of each floor of the building areused. For dwelling units, the calculated floor area shall not include open porches,garages, or unused or unfinished spaces not adaptable for future use.

For example an attic space with a pull-down ladder is not adaptable for future use. An attic space with stairs would be adaptable for future use and should be includedin the general lighting load calculation.

According to 220.14(J), for dwelling units, receptacles and lighting outlets areincluded in the general lighting load. The unit load for dwellings is 3 VA per sq. ft.

Question 2: What is the general lighting load, without applying demand factors, for a 2500 sq. ft. single familydwelling that has an additional 400 sq. ft. in an attached garage? A: 2500 VA.B: 2900 VA.C: 7500 VA.D: 8700 VA.

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Question 3: 220.14 (J) Other Loads – All Occupancies. Dwelling Occupancies.Question ID#: 3198.0

In dwellings, including one-family, two-family and multifamily dwellings, all general-use receptacle outlets are included in thegeneral lighting load calculations. This includes the receptacle outlets in the living area, bathroom, outdoors, in thebasement, garage, or an accessory building. The required lighting outlets in each habitable room, hallways, stairways,storage spaces and at outdoor entrances, are also included in the general lighting load for dwelling units. No additional loadfor these receptacle and lighting outlets is included in the general lighting load. The general lighting load for dwelling units iscalculated by the square footage of the dwelling, and includes all general purpose receptacle and lighting outlets.

Question 3: Without applying any demand factors, what is the general lighting load for a 3000 sq. ft. dwelling witha total of 60 receptacle outlets? A: 9000 VA.B: 6000 VA.C: 3500 VA.D: 3000 VA.

Question 4: 220.14 (J) Other Loads – All Occupancies. Dwelling Occupancies.Question ID#: 3199.0

In dwellings, including one-family, two-family and multifamily dwellings, all general-use receptacle outlets are included in thegeneral lighting load calculations. This includes the receptacle outlets in the living area, bathroom, outdoors, in thebasement, garage, or an accessory building. The required lighting outlets in each habitable room, hallways, stairways,storage spaces and at outdoor entrances, are also included in the general lighting load for dwelling units. No additional loadfor these receptacle and lighting outlets is included in the general lighting load. The general lighting load for dwelling units iscalculated by the square footage of the dwelling, and includes all general purpose receptacle and lighting outlets.

Question 4: Without applying any demand factors, what is the general lighting load for a two-story house that has1400 sq. ft. on the first floor, 900 sq. ft. on the 2nd floor, and 1000 VA of general purpose lighting? A: 3400 VA.B: 5400 VA.C: 7900 VA.D: 6900 VA.

Question 5: Part IV. Optional Feeder and Service Load Calculation. 220.82 Dwelling Unit.Question ID#: 3200.0

The optional method of calculating the service and feeder load on the ungrounded conductors for a single family dwelling,multifamily dwelling, schools, and restaurants is described in Part IV. It is permitted to use this optional method, rather thanthe standard method as described in Part III.

When calculating the load on the grounded neutral conductor, refer to section 220.61.

In Informative Annex D, Examples D2(a), D2(b), and D2(c) show how to make a single family dwelling calculation. Examples D4(b), D5(a) show how to make a multifamily dwelling calculation.

There are two major components in a single family dwelling calculation using the optional method: (1) Calculating theGeneral Loads. (2) Calculating the Heating and Air-Conditioning Load. The two parts must be done separately and addedtogether to get the total load for the dwelling.

220.82(B) General Loads.

The general load consists of:

(1) 3 volt-amperes per sq. ft. for general lighting and general-use receptacles. The floor area does not include openporches, garages, or unused or unfinished spaces not adaptable for future use.

(2) 1500 volt-amperes for each 20 ampere small appliance branch circuit and each laundry branch circuit.

(3) The nameplate rating of



(a) all appliances fastened in place

(b) Ranges, wall mounted ovens, counter-mounted cooking units

(c) Clothes dryers that are not connected to the laundry branch circuit

(d) Water heaters

After these loads are added together, the first 10 KVA is taken at 100%. The remainder of the total load (after subtractingthe first 10 KVA) is taken at 40%.

Here is a sample calculation:

Using the optional calculation, what is the general load for a single family dwelling that is 1800 sq. ft. and has a 10 kWrange, 5 kW clothes dryer and 5.5 kW water heater?Here is how to calculate it:1800 sq. ft. x 3 VA per sq. ft. 5,400 VA

2 small appliance branch circuits 3,000 VA

1 laundry circuit 1,500 VA

Range 10,000 VA

Clothes dryer 5,000 VA

Water Heater 5,500 VA

Total 30,400 VA

1st 10 KVA at 100% 10,000 VA

Remainder (30,400 VA – 10,000 VA = 20,400 VA) at 40% 8,160 VA

Total general load: 18,160 VA.

Question 5: Using the optional calculation, what is the general load for a single family dwelling that is 2500 sq. ft.and has a 12 kW range, 5 kW clothes dryer, 5 kW water heater, 1.5 kW dishwasher and a 9 ampere, 240 volt wholehouse fan? A: 37,660 VA.B: 15,064 VA.C: 25,064 VA.D: 21,064 VA.

Question 6: 220.82(B) General Loads.Question ID#: 3201.0

The general load consists of:









(d) Water heaters


Question 6: Using the optional calculation, what is the general load for a single family dwelling that is 2000 sq. ft.and has an 8 kW range, 5kW dryer, 4 kW water heater, 1.2 kW trash compactor, .8 KW garbage disposer, and a 5amp, 240 volt well pump? A: 18,280 VA.B: 12,280 VA.C: 22,820 VA.D: 30,700 VA.

Question 7: 220.82(C) Heating and Air-Conditioning Load.Question ID#: 3202.0

The largest of the following six types of heating and air-conditioning loads is selected and added to the general load. Thepercentages of the nameplate rating for each type of load are used in determining which one is the largest load.

» 100 percent of the nameplate rating(s) of the air conditioning and cooling. » 100 percent of the nameplate rating(s) of the heat pump when the heat pump is used without any supplemental electricheating. » 100 percent of the nameplate rating(s) of the heat pump compressor and 65 percent of the supplemental electric heatingfor central electric space-heating systems. If the heat pump compressor is prevented from operating at the same time asthe supplementary heat, it does not need to be added to the supplementary heat for the total central space heating load. » 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units. » 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units. » 100 percent of the nameplate ratings of electric thermal storage and other heating systems where the usual load isexpected to be continuous at the full nameplate value. Systems qualifying under this selection shall not be calculated underany other selection in 220.82(C).

Question 7: What is the calculated load for a heat pump when the nameplate rating of the compressor is 18 ampsat 240 volts, and the heat strips are rated at 45 amps? The compressor and supplemental heat strips can operate atthe same time. A: 11,340 VA.B: 12,460 VA.C: 14,800 VA.D: 15,120 VA.



» 100 percent of the nameplate rating(s) of the air conditioning and cooling. » 100 percent of the nameplate rating(s) of the heat pump when the heat pump is used without any supplemental electricheating. » 100 percent of the nameplate rating(s) of the heat pump compressor and 65 percent of the supplemental electric heatingfor central electric space-heating systems. If the heat pump compressor is prevented from operating at the same time asthe supplementary heat, it does not need to be added to the supplementary heat for the total central space heating load. » 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units.



» 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units. » 100 percent of the nameplate ratings of electric thermal storage and other heating systems where the usual load isexpected to be continuous at the full nameplate value. Systems qualifying under this selection shall not be calculated underany other selection in 220.82(C).

Question 8: What is the calculated load for a heat pump when the nameplate rating of the compressor is 25 ampsat 240 volts, and the heat strips are rated at 60 amps? The compressor and supplemental heat strips can operate atthe same time. A: 11,340 VA.B: 12,460 VA.C: 14,800 VA.D: 15,360 VA.



» 100 percent of the nameplate rating(s) of the air conditioning and cooling. » 100 percent of the nameplate rating(s) of the heat pump when the heat pump is used without any supplemental electricheating. » 100 percent of the nameplate rating(s) of the heat pump compressor and 65 percent of the supplemental electric heatingfor central electric space-heating systems. If the heat pump compressor is prevented from operating at the same time asthe supplementary heat, it does not need to be added to the supplementary heat for the total central space heating load. » 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units. » 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units. » 100 percent of the nameplate ratings of electric thermal storage and other heating systems where the usual load isexpected to be continuous at the full nameplate value. Systems qualifying under this selection shall not be calculated underany other selection in 220.82(C).

Question 9: What is the calculated load for six separately controlled electric space heaters when the nameplaterating of each heater is 3 kW, 240 volts? A: 18,000 VA.B: 11,700 VA.C: 7,200 VA.D: 6,300 VA.



» 100 percent of the nameplate rating(s) of the air conditioning and cooling. » 100 percent of the nameplate rating(s) of the heat pump when the heat pump is used without any supplemental electricheating. » 100 percent of the nameplate rating(s) of the heat pump compressor and 65 percent of the supplemental electric heatingfor central electric space-heating systems. If the heat pump compressor is prevented from operating at the same time asthe supplementary heat, it does not need to be added to the supplementary heat for the total central space heating load. » 65 percent of the nameplate rating(s) of electric space heating if less than four separately controlled units. » 40 percent of the nameplate rating(s) of electric space heating if four or more separately controlled units.



» 100 percent of the nameplate ratings of electric thermal storage and other heating systems where the usual load isexpected to be continuous at the full nameplate value. Systems qualifying under this selection shall not be calculated underany other selection in 220.82(C).

Question 10: Which of the following loads is the largest load: 8 kW of air-conditioning or 10 kW of electric spaceheat with two thermostats? A: 8 kW of air conditioning.B: 10 kW of electric space heat.C: Apply demand factors to both heating and air conditioning.D: No demand factors apply to either load.

Question 11: 220.82 Optional Calculation for a Dwelling Unit.Question ID#: 3206.0

Use 220.82(B) and (C) as a guideline when doing an optional dwelling service calculation. Use Example D2(a) inInformative Annex D as an example of how to perform the calculation.

Calculate the general load first:







(d) Water heaters


Next, calculate the heating or cooling load separately, using the larger load and dropping the smaller load. When makingthe comparison of which is the largest load, use the percents in 220.82(C).

Add the general load and the largest of the heating or cooling load to find the total calculated load for the dwelling.

Question 11: Use the optional calculation to find the load on the following single family dwelling:2800 sq. ft., 14 kW range, 3 kW water heater, 5 kW clothes dryer, 1.5 kW dishwasher, 15 kW central heat, 29 amp,240 volt air conditioning. A: 20,560 VA.B: 28,830 VA.C: 36,400 VA.D: 30,310 VA.

Question 12: Informative Annex D, Example D2(a) and D2(b).Question ID#: 3207.0

To find the load in amperes for a single family dwelling, divide the load in volt-amps by the voltage.

Question 12: What is the rating in amps for a load on a single family dwelling that is calculated at 45,500 VA with



a 120/240 volt service? A: 150 amps.B: 165 amps.C: 190 amps.D: 200 amps.

Question 13: Informative Annex D, Example D2(a) and D2(b).Question ID#: 3208.0

To find the load in amperes for a single family dwelling, divide the load in volt-amps by the voltage.

Question 13: What is the rating in amps for a load on a single family dwelling that is calculated at 28,800 VA witha 120/240 volt service? A: 100 amps.B: 120 amps.C: 125 amps.D: 135 amps.

Question 14: 220.61 Feeder or Service Neutral Load (A) Basic Calculation.Question ID#: 3209.0

220.61. The feeder or service neutral load shall be the maximum unbalance of the load determined by this article. TheMaximum unbalanced load shall be the maximum net calculated load between the neutral conductor and any oneungrounded conductor.

The feeder neutral load is the maximum unbalance between the neutral and the largest load connected between the neutraland any one phase conductor. In other words, the neutral must be sized to carry the largest load that it will ever see. Thecurrent that returns on the neutral cancels out when loads on both phases are energized. For example, in a feeder to asingle family dwelling where there is 85 amps connected between Phase A and the neutral, and 100 amps connectedbetween Phase B and the neutral, if both loads were on at the same time the unbalanced neutral current would only be 15amps. But in a worst case scenario, the 85 amp load would be turned off and the 100 amp load would continue to run. Thenall 100 amps on phase B would have to return on the neutral. The neutral must be sized for this worst case scenario so thatit can carry the entire load that is connected between the neutral and any one phase conductor if the loads on the otherphase conductor are turned off.

Question 14: What is the feeder or service load on the neutral of a 120/240 volt, single-phase feeder where allloads operate at 120 volts, and the load on phase A is 60 amps and the load on Phase B is 75 amps? A: 15 amps.B: 60 amps.C: 75 amps.D: 135 amps.



The feeder neutral load is the maximum unbalance between the neutral and the largest load connected between the neutraland any one phase conductor. In other words, the neutral must be sized to carry the largest load that it will ever see. Thecurrent that returns on the neutral cancels out when loads on both phases are energized. For example, in a feeder to asingle family dwelling where there is 85 amps connected between Phase A and the neutral, and 100 amps connectedbetween Phase B and the neutral, if both loads were on at the same time the unbalanced neutral current would only be 15amps. But in a worst case scenario, the 85 amp load would be turned off and the 100 amp load would continue to run. Thenall 100 amps on phase B would have to return on the neutral. The neutral must be sized for this worst case scenario so thatit can carry the entire load that is connected between the neutral and any one phase conductor if the loads on the otherphase conductor are turned off.



Question 15: What is the feeder neutral load on a 120/240 volt, single-phase feeder where all loads operate at 120volts, and the load on phase A is 150 amps and the load on Phase B is 125 amps? A: 275 amps.B: 150 amps.C: 125 amps.D: 25 amps.



Loads that are not connected to the neutral are not added to the total neutral load.

Question 16: What is the maximum unbalanced load on the neutral conductor for a 120/240 volt, single-phasefeeder with the following loads: 45 amps connected between phase A and the neutral; 50 amps connectedbetween phase B and the neutral; 30 amps connected between phase A and phase B? A: 50 amps.B: 45 amps.C: 95 amps.D: 125 amps.

Multi-Family Dwelling Calculations

Question 17: 220.84 Multifamily Dwelling.Question ID#: 3213.0

Use section 220.84 and Table 220.84 to figure the load on the service for a group of apartments. Add all loads together at100% (only the larger of the heating or A/C) for one apartment, multiply by the number of apartments and multiply thatnumber by the demand factors in Table 220.84.

The optional calculation for multi-family dwellings can be used when (1) no dwelling unit is supplied by more than one feeder(2) each dwelling unit is equipped with electric cooking equipment and (3) each dwelling unit is equipped with either electricspace heating or air conditioning, or both.

When calculating the load on each unit add 1500 VA for each of two small appliance circuits and 1500 for the laundry circuit.

For example:

What is the service load in VA and amps on the phase conductors for this apartment building?

Six apartments, 950 sq. ft. each, supplied by a 120/240 single phase service with the following loads:

9.5 kW Central Heat

4 kW A/C

6 kW Cooktop

4 kW Oven



4500 VA Water heater

1500 VA Dishwasher

1000 VA Disposer

Solution:

950 sq. ft. x 3 2,850 VA

2 small appliance branch circuits 3,000 VA

1 laundry circuit 1,500 VA

Heat 9,500 VA

Cook top 6,000 VA

Oven 4,000 VA

Water heater 4,500 VA

Dishwasher 1,500 VA

Disposal 1,000 VA

Total 33,850 VA

33,850 VA x 6 apartments = 203,100 VA203,100 VA x .44 = 89,364 VA

(From Table 220.84)

TOTAL:89,364 VA / 240 volts = 372 amps.

Question 17: What is the service load in VA and amps on the phase conductors for this apartment building? Three apartments, 875 sq. ft. each, supplied by a 120/240 single-phase service with the following loads: 8 kWSpace heat, 10 kW Range, 5,000 VA Water heater, 1,200 VA Dishwasher, 900 VA Disposal, 850 VA Compactor. A: 44,651 VA, 186 amps.B: 44,651 VA, 215 amps.C: 99,225 VA, 276 amps.D: 99,225 VA, 477 amps.





Question 18: What is the service load in VA and amps on the phase conductors for this apartment building? Four apartments, 825 sq. ft. each, supplied by a 120/240 single phase service with the following loads: 6.5 KWHeating unit, 4 KW A/C, 3 KW Cooktop, 3 KW Oven, 4000 VA Water heater, 1500 VA Dishwasher, 1000 VA Disposer. A: 37,796 VA, 157 amps.



B: 38,425 VA, 160 amps.C: 40,125 VA, 167 amps.D: 46,755 VA, 195 amps.





Question 19: What is the service load in VA and amps on the phase conductors for this duplex?Six apartments, 1375 sq. ft. each, supplied by a 120/240 single phase service with the following loads: 3.5 KWHeating unit, 6.5 KW A/C, 3 KW Cooktop, 3.5 KW Oven, 5,000 VA Water heater, 1,700 VA Dishwasher, 1,200 VADisposer. A: 29,525 VA, 123 amps.B: 64,955 VA, 271 amps.C: 77,946 VA, 325 amps.D: 177,150 VA, 738 amps.

Feeder and Neutral Calculations

Question 20: 220.61 Feeder or Service Neutral Load.Question ID#: 3217.0

The maximum unbalance of the load on the neutral conductor is the largest loadconnected between any one phase and the neutral. Plus 70% of that part of theneutral load over 200 amps. To find the neutral load ignore any loads that areconnected phase-to-phase.

220.61(B) Permitted Reductions

A service or feeder supplying the following loads shall be permitted to have anadditional demand factor of 70 percent applied to the amount in 220.61(B)(1) or theportion of the amount in 220.61(B)(2) determined by the basic calculation.

220.61(B)(2)

That portion of the unbalanced load in excess of 200 amperes where the feeder orservice is supplied from a 3-wire dc or single-phase ac system; or a w-wire,3-phase, 3-wire, 2-phase system; or a 5-wire, 2-phase system.

Follow these steps to answer the following question.

» Find the largest load connected between any one ungrounded conductor andthe neutral. » Divide the load in kW by the phase-to-neutral voltage (120V).



» Take the first 200 amps at 100% and the remainder over 200 amps at 70%. » Add the first 200 amps and the result of applying the 70% demand factor to the portion over 200 amps to find the totalload on the neutral.

Question 20: What is the maximum load on Conductor N? Voltage is 120/240 volts single-phase. A: 458 amps.B: 292 amps.C: 264 amps.D: 146 amps.

Question 21: 220.61 Feeder or Service Neutral Load.Question ID#: 3218.0

The maximum unbalance of the load on the neutral conductor is the largest loadconnected between any one phase and the neutral. Plus 70% of that part of theneutral load over 200 amps. To find the neutral load ignore any loads that are notconnected phase-to-phase.

220.61(B) Permitted Reductions

A service or feeder supplying the following loads shall be permitted to have anadditional demand factor of 70 percent applied to the amount in 220.61(B)(1) or theportion of the amount in 220.61(B)(2) determined by the basic calculation.

220.61(B)(2)

That portion of the unbalanced load in excess of 200 amperes where the feeder orservice is supplied from a 3-wire dc or single-phase ac system; or a w-wire,3-phase, 3-wire, 2-phase system; or a 5-wire, 2-phase system.

Follow these steps to answer the following question.

» Find the largest load connected between any one ungrounded conductor andthe neutral. Check to see if the sum of two separate loads connected between aphase conductor and the neutral is larger than a single load. » Divide the load in kW by the phase-to-neutral voltage (120V).

Question 21: What is the maximum load on Conductor N? Voltage is 120/240 volts. A: 133 amps.B: 125 amps.C: 167 amps.D: 292 amps.



Question 22: Annex D, Example D3(a) Ungrounded Feeder Conductors.Question ID#: 3219.0

When more than one load is connected to an ungrounded conductor, add the loadson the conductor together.

To find the load on a conductor connected to a 3-phase system, divide the load inkW by the phase-to-phase voltage x 1.73.

For a 120/208 Y, 3-phase, 4-wire system, the voltage for 3-phase loads is 208 volts:208 X 1.73 = 359.84 rounded to 360 .

For a 277/480Y, 3-phase, 4-wire system, the voltage for 3-phase loads is 480 volts:480 X 1.73 = 830.4 rounded to 830.

For example, the load in amps for a 3-phase, 100 kW load connected to a 480Y3-phase system is: 100,000 VA ÷ (480 X 1.73) = 120.48 amps.

The load in amps for a 85 kW, 3-phase load connected to a 208Y, 3-phase systemis: 85,000VA ÷ (208 X 1.73) = 236.111 amps.

Question 22: What is the load on Conductor A? Voltage is 277/480 volts, 3-phase. A: 71 amps.B: 105 amps.C: 183 amps.D: 41 amps.


When more than one load is connected to an ungrounded conductor, add the loadson the conductor together.

To find the load on a conductor connected to a 3-phase system, divide the load inkW by the phase-to-phase voltage x 1.73. For a 120/208 Y, 3-phase, 4-wire system,the voltage for 3-phase loads is 360 volts. For a 277/480Y, 3-phase, 4-wire system,the voltage for 3-phase loads is 830 volts.

For example, the load in amps for 3-phase, 100 kW load connected to a 480Y3-phase system is: 100,000 VA ÷ 830 volts = 120.5 amps. The load in amps for a85 kW, 3-phase load connected to a 208Y, 3-phase system is: 85,000VA ÷ 360volts = 236 amps.

Question 23: What is the load on conductor C? Voltage is 120/208 3-phase. A: 97 amps.B: 144 amps.C: 83 amps.D: 168 amps.




To find the load on any one conductor, when the conductor is connected tosingle-phase loads, phase-to-phase loads, and 3-phase loads, divide the load in kWby the voltage that the load operates at.

For example, if a 1500 VA load operates at 120 volts, divide 1500 VA by 120 volts.1500 ÷ 120 volts = 12.5 amps. If a 30 kW load operates at 208 volts, divide 30kWby 208 volts. 30,000VA ÷ 208 volts = 144 amps. If a 3-phase 50kW load operatesat 208 volts 3-phase, divide 50 kW by (208 x 1.73). 50,000 ÷ 360 volts = 139 amps. Add all the loads together to get the total load on a single ungrounded conductor. 12.5 amps + 144 amps + 139 amps = 296 amps.

Question 24: What is the load on conductor A? Voltage is 120/208 volts, 3-phase? A: 291 amps.B: 337 amps.C: 194 amps.D: 255 amps.


To find the load on any one conductor, when the conductor is connected tosingle-phase loads, phase-to-phase loads, and 3-phase loads, divide the load in kWby the voltage that the load operates at.

For example, if a 1500 VA load operates at 120 volts, divide 1500 VA by 120 volts.

1500 ÷ 120 volts = 12.5 amps. If a 30 kW load operates at 208 volts, divide 30kWby 208 volts. 30,000VA ÷ 208 volts = 144 amps. If a 3-phase 50kW load operatesat 208 volts 3-phase, divide 50 kW by (208 x 1.73). 50,000 ÷ 360 volts = 139 amps. Add all the loads together to get the total load on a single ungrounded conductor. 12.5 amps + 144 amps + 139 amps = 296 amps.

If a load is continuous, multiply the load on the nameplate x 1.25 to find thecontinuous load.

Question 25: What is the load on conductor B? Voltage is 120/240 volts, single phase. All loads are continuous. A: 161 amps.B: 129 amps.C: 121 amps.D: 97 amps.

Box Fill Calculations



Question 26: Table 314.16(A) Metal Boxes.Question ID#: 3224.0

Use Table 314.16(A) to find how many conductors can fit in a standard metal box.

Question 26: How many No. 12 conductors are permitted in a 4 11/16 in. x 2 1/8 in.square metal box? A: 21.B: 18.C: 16.D: 14.


Use Table 314.16(A) to find how many conductors can fit in a standard metal box

Question 27: How many No. 14 conductors are permitted in a 4 x 1 1/2 round metal box? A: 10.B: 8.C: 7.D: 6.




Use Table 314.16(A) to select a box based on the number of conductors in the box.

Question 28: What size of octagonal box is required for eight, No. 12 conductors? A: 4 x 1 1/4 inches.B: 4 x 1 1/2 inches.C: 4 x 2 1/8 inches.D: 4 x 2 1/2 inches.


Use Table 314.16(A) the select the correct size box based on the volume in cubicinches (in.3 ) required.

Question 29: What size device box is required if a minimum volume of 16 in3 is needed? A: 3" x 2" x 3 1/2".B: 4" x 2 1/8" x 1 1/2".C: 4" x 2 1/8" x 1 7/8".D: 4" x 2 1/8" x 2 1/8".




Use Table 314.16(A) to find the minimum volume of box covers.

Question 30: What is the minimum volume for a single FS box cover (1 3/4)? A: 24 in.3.B: 21 in.3.C: 18 in.3.D: 13.5 in.3.

Question 31: Table 314.16(B) Volume Allowance Required per Conductor.Question ID#: 3229.0

Use table 314.16(B) to find the minimum volume allowance required per conductor. To find the total volume required in the box, multiple the number of conductors x thevolume required per conductor.

Question 31: What is the minimum in3 volume required within a box for six, No. 12 conductors? A: 2.25 in.3.B: 2.50 in.3.C: 6.75 in.3.D: 13.50 in.3.





Question 32: What is the minimum in3 volume required within a box for twelve, No. 14 conductors? A: 12 in.3.B: 18 in.3.C: 24 in.3.D: 30 in.3.



Question 33: What is the minimum in3 volume required within a box for three, No. 10 and four No. 12conductors? A: 16.5 in.3.B: 15 in.3.C: 14.75 in.3.D: 13.5 in.3.





Question 34: What is the minimum in3 volume required within a box for six, No. 14 and two No. 12conductors? A: 14.5 in.3.B: 16.5 in.3.C: 12 in.3.D: 18 in.3.

Question 35: 314.28(A)(1) Minimum Size.Question ID#: 3233.0

In pull and junction boxes that contain 4 AWG or larger conductors that are required to be insulated, the pull or junction boxmust be large enough so that when the conductor is pulled into the box it is not damaged. The required dimensions of thebox differ depending on whether the conductors are pulled straight into the box, or if an angle or U pull is made.

In straight pulls, the length of the box or conduit body shall not be less than eight times the trade size of the largest raceway. Other raceways that enter the box are not counted, only the largest raceway.

For example, if a 2 inch and 4 inch conduit containing No. 4 or larger conductors enters a box the minimum length of the boxis 32 inches. Eight times the diameter of the largest raceway. 4 in. x 8 = 32 inches.

Question 35: What is the minimum length of a box used for a straight pull with a 3 inch conduit containing 4 AWGconductors entering the box? A: 24 inches.B: 18 inches.C: 12 inches.D: 10 inches.

Question 36: 314.28(A)(1) Minimum Size.Question ID#: 3234.0


In straight pulls, the length of the box or conduit body shall not be less than eight times the trade size of the largest raceway. Other raceways that enter the box are not counted, only the largest raceway.

For example, if a 2 inch and 4 inch conduit containing No. 4 or larger conductors enters a box the minimum length of the boxis 32 inches. Eight times the diameter of the largest raceway. 4 in. x 8 = 32 inches.

Question 36: What is the minimum length of a box used for a straight pull when two, 2 1/2 inch conduits



containing 4 AWG conductors on the same row enter the box? A: 18 inches.B: 20 inches.C: 24 inches.D: 25 inches.

Question 37: 314.28(A)(2) Angle or U Pulls, or Splices.Question ID#: 3235.0


An angle pull is when the conductor is not pulled straight through the box, but exits the box on a wall that is not opposite thewall where it entered the box, in a U shape. In an angle pull, the distance between each raceway entry inside the box orconduit body and the opposite wall of the box or conduit body cannot be less than six times the trade size of the largestraceway in a row. This dimension is increased by the sum of the diameters of the other raceways in the same row on thesame wall of the box.

For example, if the conductors in a 1 inch and 3 inch conduit make an angle pull in the box, the minimum dimension fromthe side of the box where the conduits enter to the opposite wall of the box is 19 inches. Six times the 3 in. conduit plus the1 inch conduit. (6 x 3 in.) + 1 in. = 19 inches.

Question 37: What are the minimum dimensions of a box used for an angle pull where a 1 1/4 inch conduit and a 2inch conduit containing 4 AWG conductors on the same row enter and leave the box? A: 12 inches x 12 inches.B: 13 1/4 inches x 13 1/4 inches.C: 14 1/2 inches x 13 1/2 inches.D: 15 1/2 inches x 15 1/2 inches.

Question 38: 314.28(A)(2) Angle or U Pulls, or Splices.Question ID#: 3236.0

An angle pull is when the conductor is not pulled straight through the box, but exits the box on a wall that is not opposite thewall where it entered the box, in a U shape. In an angle pull, the distance between each raceway entry inside the box orconduit body and the opposite wall of the box or conduit body cannot be less than six times the trade size of the largestraceway in a row. This dimension is increased by the sum of the diameters of the other raceways in the same row on thesame wall of the box.

For example, if the conductors in a 1 inch and 3 inch conduit make an angle pull in the box, the minimum dimension fromthe side of the box where the conduits enter to the opposite wall of the box is 19 inches. Six times the 3 in. conduit plus the1 inch conduit. (6 x 3 in.) + 1 in. = 19 inches.

Question 38: What are the dimensions of a box used for an angle pull where a 2 inch conduit and a 3 inch conduitcontaining 4 AWG conductors on the same row enter and leave the box? A: 30 inches x 30 inches.B: 26 inches x 26 inches.C: 24 inches x 24 inches.D: 20 inches x 20 inches.

Motors



Question 39: Table 430.248 Full Load Currents in Amperes, Single-Phase Alternating-CurrentMotors.

Question ID#: 3238.0

Table 430.248 lists the full load current for single-phase AC motors, based on theoperating voltage of the motor.

Question 39: What is the full-load current for a single-phase, 115 volts, 3/4 HP motor? A: 5.6 amps.B: 9.8 amps.C: 13.8 amps.D: 16 amps.

Question 40: Table 430.250 Full-Load Current, Three-Phase Alternating-Current Motors.Question ID#: 3239.0

Table 430-250 lists the full-load current for three-phase AC motors, based on theoperating voltage of the motor.

Question 40: What is the full-load current for a 3-phase, 208 volts, 5 HP motor? A: 16.7 amps.B: 22 amps.C: 24.2 amps.D: 28 amps.



Question 41: Table 430.250 Full-Load Current, Three-Phase Alternating-Current Motors.Question ID#: 3240.0

Table 430.250 lists the full-load current for three-phase AC motors, based on theoperating voltage of the motor.

Question 41: What is the full-load current for a 3-phase, 460 volts, 40 HP motor? A: 27 amps.B: 32 amps.C: 40 amps.D: 52 amps.

Question 42: Table 430.247 Full-Load Current in Amperes, Direct-Current Motors.Question ID#: 3241.0

Table 430.247 lists the full-load current for DC motors, based on the operatingvoltage of the motor.

Question 42: What is the full-load current for a direct-current, 240 volts, 10 HP motor? A: 20 amps.B: 27 amps.C: 38 amps.D: 55 amps.



Question 43: 430.22 Single Motor.Question ID#: 3242.0

Conductors that supply a single motor are selected based on 125 percent of themotor full-load rating.

Question 43: What is the minimum ampacity for conductors that supply a single motor, rated 208 volts, 3-phase,10 HP? A: 24.2 amps.B: 30.8 amps.C: 38.5 amps.D: 46.2 amps.

Question 44: 430.24 Several Motors or a Motor(s) and Other Load(s).Question ID#: 3243.0

Conductors that supply a group of motors are selected based on 125 percent of thehighest rated motor and the sum of the full-load current rating for other motors in thegroup.

Question 44: What is the minimum ampacity for conductors that supply a group of two motors, rated 3-phase, 208volts, 3 HP and 5 HP? A: 27.3 amps.B: 31.48 amps.C: 34.13 amps.D: 37.25 amps.




Conductors that supply a group of motors are selected based on 125 percent of thehighest rated motor and the sum of the full-load current rating for other motors in thegroup plus 100 percent of a noncontinuous load and 125% of continuous loads.

Question 45: What is the minimum ampacity for conductors that supply a group of two 3-phase motors, rated 230volts, 1/2 HP and 1 HP, and a continuous load rated at 12 amps? A: 21.45 amps.B: 26.9 amps.C: 24.9 amps.D: 22.45 amps.

Question 46: 430.22. Single Motor. Table 310.15(B)(16) Allowable Ampacities of InsulatedConductors.


Calculate the minimum ampacity of a conductor to supply a single motor bymultiplying the full-load current x 125%. Select a conductor that is equal to orgreater than the ampacity of conductors listed in Table 310.15(B)(16).

Question 46: What is the minimum wire size, THWN cu., that can be used to supply a 115 volts, single phase, 3 HPmotor? A: No. 6 THWN.B: No. 8 THWN.C: No. 10 THWN.D: No. 12 THWN.






Question 47: What is the minimum wire size, THWN cu., that can be used to supply a 208 volts, 3-phase, 15 HPmotor? Assume all terminals are rated for 75 degree C. A: No. 8 THWN.B: No. 6 THWN.C: No. 4 THWN.D: No. 3 THWN.




Question 48: What is the minimum wire size, THWN cu., that can be used to supply a 240 volt, DC, 71/2 HP motor? A: No. 12 THWN.B: No. 10 THWN.C: No. 8 THWN.D: No. 6 THWN.




Calculate the minimum ampacity of a conductor to supply a group of motors bymultiplying the full-load current of the highest rated motor x 125% and the sum ofthe full-load currents of other motors in the group. Select a conductor that is equalto or greater than the ampacity of conductors listed in Table 310.15(B)(16).

Question 49: What is the minimum wire size, THWN cu., that can be used to supply a group of three motors, rated460 volts, 10 HP, 15 HP, and 20 HP? A: No. 8 THWN.B: No. 6 THWN.C: No. 4 THWN.D: No. 3 THWN.


Calculate the minimum ampacity of a conductor to supply a group of motors andother loads by multiplying the full-load current of the highest rated motor x 125%plus the sum of the full-load currents of other motors in the group, plus 100% of thenoncontinuous loads and 125% of the continuous loads. Select a conductor that isequal to or greater than the ampacity of conductors listed in Table 310.15(B)(16).

Question 50: What is the minimum wire size, THWN cu., that can be used to supply a group of two motors rated230 volts, 3-phase, 25 HP and 30 HP, and a continuous load of 75 amps? A: 250 kcmil.B: 300 kcmil.C: 350 kcmil.D: 400 kcmil.

Documents

Dwelling, Feeder, Box Fill, And Motor Calculations - MT