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Load Calculations & Sizing — Electrician Practice Questions

79 questions · 15% of the Electrician exam

Worked questions

1. Using the optional dwelling method on a 2,000 square foot house at 3 volt-amperes per square foot, with two small-appliance circuits and one laundry circuit at 1,500 volt-amperes each, a 12-kilowatt range, a 5-kilowatt dryer, a 4.5-kilowatt water heater, a 1.2-kilowatt dishwasher, and an 0.8-kilowatt disposal, plus a 5-kilowatt air conditioner and 9 kilowatts of fixed electric space heating that cannot run together, what is the calculated service load at 240 volts? Take the first 10 kilovolt-amperes of general load at 100 percent and the remainder at 40 percent, and take the space heating at 65 percent.

  • A127 amperes

    This adds both the air conditioner and the space heating. Loads that cannot operate at the same time are compared, and only the larger of the two is carried into the total.

  • 106 amperes
  • C103 amperes

    This carries the air conditioner and drops the heating. The two must be compared after their own percentages are applied, and heating at 65 percent comes out higher than the air conditioner at 100 percent.

  • D119 amperes

    This takes the space heating at its full 9 kilowatts. The optional method allows fixed electric space heating with fewer than four separately controlled units to be figured at 65 percent.

Why B is correct

The optional method lumps every general load at nameplate value, discounts everything past the first 10 kilovolt-amperes to 40 percent, and then adds back only the larger of the heating or cooling load. Here heating at 65 percent works out to 5,850 VA, which beats the 5,000 VA air conditioner, so 19,600 + 5,850 = 25,450 VA divided by 240 volts gives 106 amperes.

What this question is testing

Tests a complete optional-method dwelling calculation end to end: assembling general loads at nameplate, applying the 10 kilovolt-ampere and 40 percent split, applying the correct percentage to fixed electric space heating, comparing heating against cooling and carrying only the larger, and converting volt-amperes to amperes at 240 volts.

On the job

The optional method is what lets a fully loaded modern house land on a 200-ampere service, and most residential shops use it exclusively. The heating versus cooling comparison is where money is made or lost: apply the 65 percent factor before comparing, and a heat-dominant house in a cold climate often calculates smaller than the contractor expected. Plan reviewers check that comparison line first.

Memory technique

Ten at full fare, the rest at forty percent, then add whichever of heat or cool is bigger.

Exam tip

Apply the heating percentage first, then compare it against the air conditioning value; the comparison is between the adjusted numbers, not the nameplates.

Where to look it up

Article 220, Part IV, optional calculation for a dwelling unit. The heating and air conditioning list is a separate subsection from the general load subsection, so read both before you total anything.

2. A single dwelling unit feeder supplies five fastened-in-place appliances: a 1,200 volt-ampere dishwasher, a 900 volt-ampere disposal, a 4,500 volt-ampere water heater, a 1,000 volt-ampere trash compactor, and an 800 volt-ampere attic fan. No range, dryer, space heating, or air conditioning is included in this group. Applying the demand factor permitted when four or more such appliances are on the same feeder, what is the demand load?

  • A8,400 VA

    This is the connected nameplate total with no demand factor. Once four or more fastened-in-place appliances share a feeder, the code permits the group to be reduced.

  • B3,360 VA

    This applies 40 percent, the factor used for the remainder of the general load in the optional dwelling method. That percentage belongs to a different calculation method entirely.

  • C2,940 VA

    This applies 35 percent, the demand factor for the portion of the dwelling general lighting load above the first block. Lighting demand factors do not apply to fastened-in-place appliances.

  • 6,300 VA

Why D is correct

Five qualifying appliances totaling 8,400 VA meet the four-or-more threshold, so the 75 percent factor applies to the whole group: 8,400 x 0.75 = 6,300 VA. The reduction recognizes that a dishwasher, disposal, water heater, compactor, and attic fan are very unlikely to be drawing full nameplate current at the same instant.

What this question is testing

Tests whether you can identify which appliances belong to the fastened-in-place group, whether you recognize the four-appliance threshold that unlocks the reduction, and whether you apply the correct percentage rather than borrowing a factor from the lighting table or the optional method.

On the job

Counting appliances correctly is worth real money on a service upgrade. Water heaters dominate this group, and adding a fourth qualifying appliance can actually lower the calculated load because the 75 percent factor suddenly applies to everything in the group. The exclusions matter just as much: put the range or dryer in this bucket and you have understated the service, because those have their own demand rules.

Memory technique

Four appliances buy you a quarter off.

Exam tip

Count the qualifying appliances first; at four or more, multiply the nameplate total by 0.75, and leave ranges, dryers, heat, and air conditioning out of the count.

Where to look it up

Article 220, Part III, appliance load for dwelling units. Read the list of excluded equipment in that section before you decide how many appliances you actually have.

3. A 208Y/120-volt three-phase service calculation for a shop totals 84,000 volt-amperes with every motor already counted at 100 percent of its full-load current. The largest motor on the service draws a full-load current of 27 amperes at 208 volts three phase, and the code requires 25 percent of the largest motor load to be added to the total. What is the final calculated service load?

  • A93,727 VA

    This adds the entire largest motor a second time instead of 25 percent of it. The motor is already included at full value in the 84,000 volt-ampere total.

  • 86,432 VA
  • C85,404 VA

    This omits the square root of three when converting the three-phase motor current to volt-amperes, which understates the motor by about 42 percent and therefore understates the addition.

  • D105,000 VA

    This multiplies the entire service total by 125 percent. The 25 percent addition applies only to the largest motor, not to every load on the service.

Why B is correct

The three-phase volt-ampere formula requires the square root of three, so the 27 ampere motor represents 1.732 x 208 x 27 = 9,727 VA. One quarter of that, 2,432 VA, is the additional amount the code requires for the largest motor, giving a final total of 86,432 VA. The other motors are carried at 100 percent with no addition.

What this question is testing

Tests three things at once: converting a three-phase current to volt-amperes with the correct square root of three factor, recognizing that only the single largest motor receives the additional 25 percent, and understanding that the addition supplements a total in which the motors already appear at full value.

On the job

Shops and light industrial buildings live or die on this line. Add the whole motor twice and you sell the customer a service one size too big; forget the square root of three and you undersize the service and the utility transformer request along with it. When several motors are close in size, identify the largest by full-load current from the motor tables, not by horsepower nameplate alone, because voltage changes the current.

Memory technique

Everybody counts once, but the biggest motor counts once and a quarter.

Exam tip

Find the largest motor by full-load current, convert it to volt-amperes with 1.732, then add one quarter of that number to your total.

Where to look it up

Article 220 motor load reference, which sends you to Article 430 for the actual sizing rules. Keep a tab on the three-phase motor full-load current table because you will need it in the same problem.

4. After all demand factors, the maximum unbalanced load on a 120/240-volt single-phase dwelling feeder works out to 320 amperes. The Code permits the portion of the unbalanced load that exceeds 200 amperes to be taken at 70 percent for feeders and services, with the first 200 amperes counted in full. What is the calculated neutral load in amperes?

  • 284 A
  • B320 A

    This ignores the permitted reduction entirely and carries the full unbalanced load. It is conservative and never unsafe, but it is not the calculated value the question asks for and it can force an unnecessarily large grounded conductor.

  • C224 A

    This applies 70 percent to the whole 320 amperes. The first 200 amperes is specifically excluded from the reduction, so discounting it understates the neutral and can leave the grounded conductor undersized.

  • D260 A

    This reverses the split, discounting the first 200 amperes and carrying the excess at full value. The rule is the other way around: full value up to 200 amperes, reduced value above it.

Why A is correct

The permission is written as a two-step split. Two hundred amperes stays at full value, and only the 120 amperes above that line is multiplied by 0.70, giving 84 amperes. Adding the two portions produces 284 amperes as the calculated neutral load, which is what you carry into the conductor table.

What this question is testing

Tests whether you can apply a two-tier demand factor correctly, keeping the base portion at full value and discounting only the excess, and whether you understand that the neutral of a feeder is sized on maximum unbalance rather than on total load. It also probes awareness that the reduction has exclusions.

On the job

Services above 200 amperes are routine on large homes and small commercial buildings, and the grounded conductor is often the most expensive single conductor in the run because paralleled sets multiply the cost. Knowing that only the excess is discounted keeps you from either overspending on copper or, worse, undersizing a conductor that has to carry real unbalanced current all day. Note also that this reduction is off the table for the nonlinear portion of a load fed from a four-wire wye system.

Memory technique

The first 200 pays full fare; the overflow rides at seven-tenths.

Exam tip

Draw a line at 200 amperes. Everything below it is full price; only what sticks out above gets the 70 percent.

Where to look it up

Article 220, feeder and service neutral load. The same section lists the cases where no reduction is permitted, so read the whole subsection before you commit to an answer.

5. Three current-carrying 6 AWG THWN-2 copper conductors are installed in a raceway in an ambient of 55 degrees Celsius. The 90-degree correction factor for that ambient is 0.76. Table 310.16 gives 6 AWG copper as 55 amperes at 60 degrees, 65 amperes at 75 degrees and 75 amperes at 90 degrees. Terminations at both ends are rated 75 degrees Celsius. What is the maximum load these conductors may serve?

  • A65 amperes

    This is the termination limit with no correction applied at all. In a 55-degree ambient the conductor cannot carry its full rated current, so the correction step cannot be skipped.

  • B49 amperes

    This applies the 0.76 factor to the 75-degree column value of 65 amperes. Correction factors belong on the conductor insulation rating, and the termination limit is applied afterward as a ceiling, not as a starting point.

  • 57 amperes
  • D42 amperes

    This applies the correction to the 60-degree column. Nothing in the problem points to 60 degrees; the insulation is rated 90 degrees and the terminations are rated 75 degrees.

Why C is correct

Correction factors are always applied to the ampacity column matching the conductor insulation, which is 90 degrees for THWN-2. Seventy-five amperes times 0.76 gives 57 amperes. That corrected value is then checked against the 65-ampere termination limit, and because 57 is the smaller number it becomes the maximum permitted load.

What this question is testing

Tests whether you know the correct order of operations for conductor sizing: start at the insulation rating, apply ambient correction and any conductor-count adjustment, then compare the result to the termination temperature limit and take the lower value. It also checks that you can recognize when correction rather than termination controls.

On the job

Boiler rooms, rooftop chases, attics above insulation and enclosures next to furnaces all run hot enough to matter, and the ambient is measured at the conductors, not at the thermostat down the hall. Most of the time the 75-degree termination is the binding constraint and electricians get in the habit of stopping there. In a genuinely hot space the corrected 90-degree value drops below the termination value and the whole circuit has to be upsized, which is the situation this problem is built around.

Memory technique

Start high, correct down, then let the lugs cap it.

Exam tip

Correct from the 90-degree column, then cap at the termination column. Whichever number ends up smaller is your ampacity.

Where to look it up

Table 310.16 for the three columns, the ambient correction tables in Article 310, and the termination temperature rules in Article 110.

6. The guest room lighting load for a motel has already been computed at 60,000 volt-amperes. The applicable demand factors for this occupancy are 50 percent for the first 20,000 volt-amperes, 40 percent for the portion from 20,001 through 100,000 volt-amperes, and 30 percent for anything above that. What is the demand lighting load for the service calculation?

  • 26,000 VA
  • B60,000 VA

    This carries the connected load with no demand factor. Hotels and motels get one of the deepest lighting discounts in the Code precisely because guest rooms are rarely occupied and lit all at once.

  • C30,000 VA

    This applies 50 percent to the entire load. The 50-percent factor belongs only to the first 20,000 volt-amperes; everything above that moves into a lower bracket.

  • D24,000 VA

    This applies 40 percent to the entire load. Like the previous error it flattens a bracketed table into a single multiplier, which understates the first slice.

Why A is correct

Bracketed demand factors work like a tax table. The first 20,000 volt-amperes is halved to 10,000, and the remaining 40,000 volt-amperes falls entirely in the 40-percent bracket and becomes 16,000. Adding the two slices gives 26,000 volt-amperes, and the 30-percent bracket never comes into play because the load never exceeds 100,000.

What this question is testing

Tests whether you can apply a multi-tier demand table slice by slice instead of picking one percentage for the whole number, and whether you stop at the correct bracket when the load does not reach the highest tier. It also checks that you recognize the occupancy-specific nature of these factors.

On the job

Hotels, motels, dormitories and any building where the guest rooms do not have permanent cooking provisions get this treatment, and on a large property it can cut the lighting line by more than half. Getting the brackets right matters when the number sits near a service size boundary, because the difference between 800 and 1,000 amperes of switchgear is a six-figure decision. Note that this discount covers the guest room lighting only, not corridors, lobbies, laundry or house loads.

Memory technique

Demand tables stack like tax brackets, slice by slice from the bottom up.

Exam tip

Fill each bracket before you move to the next, and add the slices at the end. Never multiply the total by a single factor.

Where to look it up

Article 220, the lighting load demand factor table. Check the occupancy column carefully, because the dwelling row and the hotel row use different breakpoints and different percentages.

7. A feeder supplies two continuous-duty motors whose table full-load currents are 28 amperes and 14 amperes, plus a continuous lighting load of 24 amperes. Motor feeder conductors are sized at 125 percent of the largest motor full-load current plus the full-load current of the remaining motors, and a continuous nonmotor load is added at 125 percent. What minimum feeder ampacity is required?

  • A66 amperes

    This adds the three loads at face value with no multiplier anywhere. Both the largest-motor rule and the continuous-load rule are ignored, and the feeder ends up undersized by about 20 percent.

  • 79 amperes
  • C73 amperes

    This applies 125 percent to the largest motor but forgets that the lighting is continuous. The lighting multiplier comes from a different rule than the motor multiplier, and both apply on the same feeder.

  • D82.5 amperes

    This multiplies the entire 66 amperes by 125 percent. Only the largest motor and the continuous nonmotor load earn the multiplier; the second motor is counted at its table value.

Why B is correct

The rule adds 25 percent for exactly one motor, the largest, and takes every other motor at its table value. That gives 35 plus 14. The lighting is not a motor load, so it is handled by the ordinary continuous-load rule and contributes 30 amperes. The three pieces sum to 79 amperes of required feeder ampacity.

What this question is testing

Tests whether you know that only one motor on a feeder receives the 25-percent addition, whether you use table full-load currents rather than nameplate values for conductor sizing, and whether you can apply a second, independent continuous-load multiplier to the nonmotor portion of the same feeder.

On the job

Shop and industrial feeders almost always mix motors with lighting and receptacles, and the two load types follow different rules that happen to share the same 125-percent number. Mixing them up in either direction costs you: undersizing brings nuisance tripping on a hot afternoon, oversizing forces bigger lugs and bigger conduit. The rule also assumes you use the table full-load current for the conductor calculation, not the motor nameplate, which is the opposite of what overload protection uses.

Memory technique

Biggest motor plus a quarter, everybody else plain, and the lights bring their own quarter.

Exam tip

One motor gets the bonus, the rest ride at table value, and any continuous nonmotor load carries its own separate 125 percent.

Where to look it up

Article 430 for motor feeder conductors and Article 215 for the continuous nonmotor portion. The motor full-load current tables are in Article 430, not in Article 310.

8. A school building has 60,000 square feet of floor area and a connected load of 12 volt-amperes per square foot. The optional method for schools applies 100 percent to the first 3 volt-amperes per square foot, 75 percent to the portion from 3 up through 20 volt-amperes per square foot, and 25 percent to anything above that. What is the calculated load?

  • A720,000 volt-amperes

    Using the connected total with no demand factors gives 720,000 volt-amperes and defeats the purpose of an optional calculation.

  • B540,000 volt-amperes

    Applying 75 percent to the entire connected load gives 540,000 volt-amperes and ignores the first bracket, which is counted at full value.

  • C405,000 volt-amperes

    Taking only the 75 percent bracket and forgetting to add the first 3 volt-amperes per square foot gives 405,000 volt-amperes. This is the most common slip.

  • 585,000 volt-amperes

Why D is correct

The method brackets the connected load by volt-amperes per square foot rather than by total volt-amperes, which trips people who try to apply the percentages to the raw total. Splitting 12 volt-amperes per square foot into 3 at full value and 9 at three quarters, then multiplying each by the floor area, gives 180,000 plus 405,000, or 585,000 volt-amperes.

What this question is testing

The item tests whether the candidate can apply demand brackets stated in volt-amperes per square foot, keeping the unit conversion straight and adding the brackets rather than choosing between them.

On the job

Schools carry heavy lighting, kitchen, and mechanical loads that never run together, which is why the optional method exists for them. On an addition or a renovation the calculation decides whether the existing service can absorb the new wing. Because the brackets are stated per square foot, the area figure has to be right, and disputes usually come down to whether covered walkways and mechanical penthouses are counted in the floor area.

Memory technique

Three at full, the next seventeen at three quarters, the rest at a quarter.

Exam tip

When brackets are expressed per square foot, do the bracketing first and multiply by area last. It halves the arithmetic.

Where to look it up

Look in the load calculation article, in the optional calculations part, at the section covering schools and its demand factor table.

9. A 277-volt single-phase lighting circuit will carry 20 amperes 300 feet one way in aluminum, and the drop must be held to 3 percent of the supply voltage. Use 21.2 as the aluminum constant and the single-phase relationship in which drop equals 2 times the constant times the current times the one-way length divided by circular mils. Aluminum areas are 16,510 circular mils for 8 AWG, 26,240 for 6 AWG, 41,740 for 4 AWG and 66,360 for 2 AWG. What is the smallest of these conductors that holds the drop within the limit?

  • A6 AWG aluminum

    This runs the calculation with the copper constant of 12.9 on an aluminum conductor. Aluminum has roughly 1.6 times the resistivity of copper, so using copper's constant understates the area needed.

  • 4 AWG aluminum
  • C8 AWG aluminum

    This leaves the factor of 2 out, which halves the required area to about 15,300 circular mils. The multiplier accounts for the return conductor and cannot be dropped.

  • D2 AWG aluminum

    This doubles the length to 600 feet and also keeps the factor of 2, counting the return path twice and roughly doubling the area the calculation appears to demand.

Why B is correct

Three percent of 277 volts is 8.31 volts. Rearranged for area, CM = (2 x K x I x L) / VD = (2 x 21.2 x 20 x 300) / 8.31 = 254,400 / 8.31 = 30,614 circular mils. That is more than the 26,240 of 6 AWG, so the next size up is required, and 4 AWG at 41,740 circular mils satisfies it with margin.

What this question is testing

Tests whether you pick the resistivity constant that matches the conductor material, whether you can rearrange the drop relationship to solve for circular mils, and whether you then round up to a real conductor size rather than reporting the raw calculated area.

On the job

Aluminum is common on long site-lighting and feeder runs because of cost, and it is exactly where drop bites hardest. The material saving evaporates if you have to go two sizes larger, so the calculation is really a cost comparison as much as a Code exercise. Crews who run the numbers before ordering avoid the situation where the conductor is on the truck and the drop calculation says it will not do.

Memory technique

Aluminum needs about two-thirds more copper's worth of metal for the same drop.

Exam tip

Solve for circular mils, then step up to the first listed size that meets or beats it. Never round down.

Where to look it up

Chapter 9, Table 8 lists circular mil areas for both copper and aluminum. The constants come from the exam, not the Code.

70 more in the bank

Answers and explanations for these are in the app.

  • A single-family dwelling measures 2,000 square feet of habitable area figured from the outside dimensions. Using the NEC unit load of 3 volt-amperes per square foot, what is the minimum general lighting and general-use receptacle load that must be entered on the calculation worksheet before any demand factors are applied?
  • A dwelling unit is wired with three 20-ampere small-appliance branch circuits serving the kitchen and dining area and one 20-ampere laundry branch circuit. Using the NEC allowance of 1,500 volt-amperes for each such circuit, what total load must be added to the calculation worksheet for these circuits before demand factors are applied?
  • A dwelling has 2,400 square feet of habitable area, two small-appliance branch circuits, and one laundry branch circuit. The unit load is 3 volt-amperes per square foot and each of those circuits is figured at 1,500 volt-amperes. Applying the dwelling demand factors of 100 percent for the first 3,000 volt-amperes and 35 percent for the remainder, what is the demand load for this group?
  • A dwelling unit is supplied with one 14-kilowatt household electric range. The demand table gives a value of 8 kilowatts for a single range rated 12 kilowatts or less, and a table note requires that value to be increased 5 percent for each kilowatt by which the rating exceeds 12 kilowatts. What demand load must be used for this range?
  • A dwelling unit is supplied with one 240-volt electric clothes dryer having a nameplate rating of 4.5 kilowatts. What value must be entered for that dryer when the feeder and service load for the dwelling is calculated?
  • An office suite has 5,000 square feet of floor area and the applicable unit lighting load is 1.3 volt-amperes per square foot. The lighting is expected to operate for more than three hours at a time, so it is a continuous load. What minimum load must the lighting feeder be sized to carry?
  • A dwelling service calculation shows a lighting and receptacle demand of 6,045 volt-amperes, a range demand of 8,000 volt-amperes, a dryer load of 5,000 volt-amperes, and a 4,500 volt-ampere water heater connected line to line at 240 volts with no neutral. The code permits the range and dryer portions of the neutral load to be taken at 70 percent. What is the calculated neutral load?
  • A one-family dwelling is being supplied by a new 3-wire, 120/240-volt service, and the completed load calculation comes out to 78 amperes. What is the minimum rating permitted for the service disconnecting means serving this dwelling?
  • A dwelling unit has 9 kilowatts of fixed electric space heating and a 5-kilowatt central air conditioner, and the equipment is arranged so the two can never operate at the same time. How does the NEC permit these two loads to be handled when the service load is calculated?
  • Six current-carrying 6 AWG THHN copper conductors are installed in a single raceway in an ambient temperature of 40 degrees Celsius. The 90-degree-Celsius table ampacity of the conductor is 75 amperes and its 75-degree-Celsius ampacity is 65 amperes. The ambient correction factor for the 90-degree-Celsius column at that temperature is 0.91, and the adjustment factor for four to six current-carrying conductors is 0.80. What is the corrected and adjusted ampacity of each conductor?
  • A feeder supplies a noncontinuous load of 95 amperes, and both ends terminate on equipment listed for 75-degree-Celsius terminations. Copper conductor ampacities are: 4 AWG, 70 at 60 degrees, 85 at 75 degrees, and 95 at 90 degrees; 3 AWG, 85 at 60 degrees, 100 at 75 degrees, and 115 at 90 degrees; 2 AWG, 95 at 60 degrees, 115 at 75 degrees, and 130 at 90 degrees; 1 AWG, 110 at 60 degrees, 130 at 75 degrees, and 145 at 90 degrees. What is the smallest conductor permitted?
  • A feeder carries a continuous load of 100 amperes and terminates on equipment listed for 75-degree-Celsius terminations. Copper ampacities in the 75-degree-Celsius column are 100 amperes for 3 AWG, 115 amperes for 2 AWG, 130 amperes for 1 AWG, and 150 amperes for 1/0 AWG. There is no ambient correction or conductor-count adjustment on this run. What is the smallest copper conductor permitted for the feeder?
  • A 12 AWG copper THHN conductor is run in a raceway with two other current-carrying conductors in a 30-degree-Celsius ambient, so no correction or adjustment applies. Its ampacity is 30 amperes in the 90-degree-Celsius column and 25 amperes in the 75-degree-Celsius column, and the terminations are listed for 75 degrees Celsius. Under the small conductor rule, what is the largest overcurrent device permitted to protect this conductor?
  • A 25-horsepower, 460-volt, three-phase squirrel-cage motor runs continuous duty, and the motor full-load current table gives 34 amperes for this motor. Branch-circuit conductors must be sized at 125 percent of that table value. Terminations are listed for 75 degrees Celsius, and copper ampacities in that column are 35 amperes for 10 AWG, 50 amperes for 8 AWG, 65 amperes for 6 AWG, and 85 amperes for 4 AWG. What is the smallest conductor permitted?
  • A continuous-duty motor has a marked service factor of 1.15 and a nameplate full-load current of 32 amps, while the code motor table lists 34 amperes for a motor of that horsepower and voltage. A separate overload device is permitted at 125 percent for a motor with this service factor. What is the maximum rating or setting permitted for the overload device?
  • A 15-horsepower, 230-volt, three-phase motor has a code table full-load current of 42 amperes and is protected by an inverse time circuit breaker. The table of maximum ratings permits an inverse time breaker at 250 percent of full-load current for this motor, and where the result does not correspond to a standard device size the next higher standard size is permitted. What is the largest standard breaker allowed?
  • A feeder supplies two motors. The largest motor is protected by a 110-ampere branch-circuit short-circuit and ground-fault device, and the other motor has a full-load current of 28 amperes. The feeder protective device rating may not be greater than the largest branch-circuit device rating plus the sum of the full-load currents of the other motors. What is the largest standard device permitted to protect this feeder?
  • A 120-volt single-phase branch circuit is wired with 12 AWG copper and carries a steady 16-ampere load. The one-way distance from the panel to the load is 90 feet. Using a copper constant of 12.9 and a circular mil area of 6,530 for 12 AWG, what is the approximate voltage drop on this circuit?
  • A 240-volt single-phase circuit carries 24 amperes over a one-way distance of 150 feet, and voltage drop must be held to 3 percent of the supply voltage. Using a copper constant of 12.9 and circular mil areas of 6,530 for 12 AWG, 10,380 for 10 AWG, 16,510 for 8 AWG, and 26,240 for 6 AWG, what is the smallest copper conductor that keeps the drop within the limit?
  • A 45-kilovolt-ampere transformer has a 208Y/120-volt three-phase secondary. What is the rated secondary line current of this transformer at full load?
  • A balanced three-phase load operating at 480 volts line to line draws 40 amperes per line at a power factor of 0.85. What is the real power consumed by this load?
  • A one-family dwelling is being calculated by the optional method. The house has a 5-kilowatt central air conditioner and a 9-kilowatt central electric space heating unit. The optional method counts air conditioning at 100 percent and central electric space heating at 65 percent, and directs you to include only the largest of those selections. What value goes on the worksheet for this pair of loads?
  • A dwelling unit has a 7.5-kilowatt counter-mounted cooktop and two 3.5-kilowatt wall ovens, all supplied from a single branch circuit. The demand table gives 8 kilowatts for one appliance of 12 kilowatts or less, and requires that value be increased 5 percent for each additional kilowatt or major fraction above 12 kilowatts. What demand load must be entered for this cooking equipment?
  • A dwelling service calculation has already established a range demand of 8,000 volt-amperes and a dryer demand of 5,000 volt-amperes. The Code permits the grounded conductor load for household electric ranges, wall-mounted ovens, counter-mounted cooking units and electric clothes dryers to be taken as 70 percent of the load computed for the ungrounded conductors. What do these two appliances contribute to the neutral load?
  • A branch circuit supplies a lighting load of 34 amperes that will operate continuously for more than three hours. All terminations at both ends are rated for 75 degrees Celsius, the ambient is 30 degrees Celsius, and no adjustment for conductor count is needed. Using the 75-degree column values of 30 amperes for 10 AWG, 50 amperes for 8 AWG and 65 amperes for 6 AWG copper, what combination satisfies the Code?
  • A feeder supplies 80 amperes of continuous load and 60 amperes of noncontinuous load. Terminations at both ends are rated 75 degrees Celsius and no ambient or conductor-count adjustment applies. The 75-degree copper ampacities are 115 amperes for 2 AWG, 150 amperes for 1/0 AWG, 175 amperes for 2/0 AWG and 200 amperes for 3/0 AWG. What conductor and overcurrent device combination is the minimum permitted?
  • A 400-ampere feeder is installed as two parallel sets of copper conductors in two separate raceways, with three current-carrying conductors in each raceway, a 30-degree ambient and 75-degree terminations. The 75-degree copper ampacities are 150 amperes for 1/0, 200 amperes for 3/0, 230 amperes for 4/0 and 420 amperes for 600 kcmil. What is the minimum size for each phase conductor in each raceway?
  • A feeder supplies a continuous load of 88 amperes, and the assembly containing the feeder overcurrent device is listed for operation at 100 percent of its rating, so the conductor ampacity need only equal the load rather than 125 percent of it. Using 75-degree copper ampacities of 85 amperes for 4 AWG, 100 amperes for 3 AWG, 115 amperes for 2 AWG and 130 amperes for 1 AWG, what is the smallest copper feeder conductor permitted?
  • A 277-volt single-phase lighting circuit is wired in 8 AWG copper with an area of 16,510 circular mils and runs 250 feet from the panel to the first fixture. The drop is to be held within 3 percent of the supply voltage. Taking the copper constant as 12.9 and remembering that the current travels out and back over the pair of conductors, what is the greatest steady current this run can carry?
  • A length of trade size 2 electrical metallic tubing 18 inches long is installed between two enclosures as a nipple. Chapter 9, Table 4 gives the total interior cross-sectional area of trade size 2 electrical metallic tubing as 3.356 square inches, and a raceway of this length is permitted a higher fill percentage than a normal run. What is the maximum cross-sectional area the conductors may occupy?
  • Each tenant space in a commercial building that is accessible to pedestrians and has a ground-level entrance must be provided with an outlet for sign or outline lighting. What does the Code require of the branch circuit supplying that outlet, and what load must be entered for it on the calculation?
  • A 45 kVA three-phase transformer has a 480-volt primary and is protected by an overcurrent device on the primary side only. For a transformer of 1000 volts or less with a primary current of 9 amperes or more, the primary device may not exceed 125 percent of the primary full-load current, and where that value does not correspond to a standard rating the next higher standard rating is permitted. What is the maximum device rating?
  • A shop bench in a commercial building has 36 feet of continuous multioutlet assembly along it, and the appliances plugged into it are likely to be used simultaneously. The Code assigns 180 volt-amperes for each 5 feet or fraction where simultaneous use is unlikely, and 180 volt-amperes for each 1 foot or fraction where simultaneous use is likely. What load must be entered for this assembly?
  • A feeder supplies a 100-ampere noncontinuous load and terminates on equipment rated 100 amperes with no temperature marking on the equipment or its terminals. Table 310.16 gives copper values of 95 amperes for 2 AWG, 110 amperes for 1 AWG and 125 amperes for 1/0 AWG in the 60-degree column, and 100 amperes for 3 AWG in the 75-degree column. What is the minimum conductor size?
  • A commercial feeder has a calculated load of 131 amperes, all noncontinuous, and terminates on equipment listed for 75-degree Celsius terminations at both ends. There is no ambient correction or conductor-count adjustment. The 75-degree copper values are 115 amperes for 2 AWG, 130 amperes for 1 AWG, 150 amperes for 1/0 AWG and 175 amperes for 2/0 AWG. What is the minimum conductor size?
  • Ampacity must normally be adjusted downward when more than three current-carrying conductors are bundled or run together in a raceway. Under which of the following conditions is that adjustment not required?
  • A 208Y/120-volt, four-wire feeder supplies a panel whose load is largely electronic ballasts, LED drivers and switch-mode power supplies. How must the grounded conductor of that feeder be handled in the calculation and in the conduit fill and ampacity work?
  • A retail tenant space has 6,000 volt-amperes of sales floor lighting that runs the whole time the store is open, 5,400 volt-amperes of general-purpose receptacle load that is used intermittently, and a 4,000 volt-ampere illuminated sign that runs all day. Continuous loads are figured at 125 percent for feeder and overcurrent sizing. What total volt-amperes must the feeder be sized for?
  • A feeder must carry a 150-ampere noncontinuous load using aluminum conductors, with 75-degree terminations at both ends and no correction or adjustment required. The 75-degree aluminum values are 120 amperes for 1/0, 135 amperes for 2/0, 155 amperes for 3/0, 180 amperes for 4/0 and 205 amperes for 250 kcmil. What is the minimum aluminum conductor size?
  • Two insulated conductors are to be installed in a run of trade size 1 electrical metallic tubing more than 24 inches long. Chapter 9, Table 1 permits 53 percent fill for one conductor, 31 percent for two conductors and 40 percent for more than two conductors. Chapter 9, Table 4 gives the total interior area of trade size 1 electrical metallic tubing as 0.864 square inch. What is the maximum area the two conductors may occupy?
  • An office building has 8,000 square feet of floor area, and at the time of the service calculation the number of general-use receptacle outlets has not been determined. For banks and office buildings the Code requires the receptacle load to be taken as the larger of the outlets counted at 180 volt-amperes each or a unit load of 1 volt-ampere per square foot. What receptacle load must be entered?
  • A commercial building has a total calculated load of 150,000 volt-amperes supplied at 208Y/120 volts, three phase. Standard overcurrent device ampere ratings above 300 amperes include 350, 400, 450, 500 and 600 amperes. What is the smallest standard service rating that will carry this load?
  • A 240-volt single-phase feeder is wired with 2 AWG aluminum having 66,360 circular mils and carries 80 amperes over a one-way distance of 150 feet. Using the single-phase relationship in which drop equals 2 times K times current times one-way length divided by circular mils, and using an aluminum K value of 21.2, what is the approximate voltage drop?
  • An existing dwelling is being evaluated using the optional method for adding a load, and no additional air conditioning or space heating is involved. The existing and new loads together total 26 kilovolt-amperes. The method applies 100 percent to the first 8 kilovolt-amperes and 40 percent to the remainder. What is the calculated demand?
  • A multifamily building has 20 dwelling units, each with electric cooking and electric space heating, and qualifies for the optional multifamily calculation. The total connected load for the building is 400 kilovolt-amperes, and the demand factor table assigns 38 percent for 20 units. What is the calculated service load?
  • A feeder supplies two household electric ranges, each rated 14 kilowatts. The demand table gives 11 kilowatts for two ranges, and its note increases that value 5 percent for each kilowatt by which the range rating exceeds 12 kilowatts. What demand load must be used for the two ranges?
  • A feeder supplies four household electric ranges, each rated 6 kilowatts. Because these appliances fall in the 3-1/2 through 8-3/4 kilowatt range, the demand table column for that range applies, and it assigns 66 percent for four appliances. What demand load must be used?
  • A dwelling unit has 42 general-use receptacle outlets on its 15- and 20-ampere general-purpose branch circuits, in addition to the required small-appliance and laundry circuits. Using the standard calculation method, what volt-ampere load must be added for those 42 outlets?
  • Two feeder calculations produce results of 63.4 amperes and 76.6 amperes. How does the Code permit those fractional results to be handled?
  • A commercial kitchen contains four units of electric cooking and warming equipment rated 22, 16, 3, and 3 kilovolt-amperes. The demand table permits 80 percent for four such units, but the demand carried into the feeder calculation may never be less than the sum of the two largest units. What demand load must be used for this equipment?
  • A dwelling feeder calculation includes a dishwasher, a disposer, a water heater, a trash compactor, an electric range, and an electric clothes dryer, all fastened in place. Which of these may be included in the group to which the fastened-in-place appliance demand factor is applied?
  • A motor drives a short-time duty application and has a nameplate current rating of 28 amperes. The duty-cycle table assigns 85 percent of the nameplate current rating for a 5-minute rated motor in short-time duty. What minimum ampacity must the branch-circuit conductors have?
  • A continuous-duty motor rated more than 1 horsepower has a nameplate full-load current of 24 amperes. Its nameplate shows no service factor and no temperature rise marking. What is the maximum rating or setting permitted for its separate overload protection?
  • A three-phase squirrel-cage motor has a table full-load current of 34 amperes and will be protected by dual-element time-delay fuses. The table permits 175 percent of full-load current for that device type, and the next higher standard size is permitted where the calculated value does not correspond to a standard rating. What is the largest standard fuse permitted?
  • A single insulated 1/0 AWG copper conductor is installed in free air, isolated from other conductors, and terminates on equipment listed for 75 degree Celsius terminations. The 75 degree column gives 150 amperes for not more than three conductors in a raceway or cable, and 205 amperes for a single insulated conductor in free air. Which value applies?
  • A raceway contains three 120/240-volt, three-wire single-phase circuits: six ungrounded conductors, three neutrals, and three equipment grounding conductors. Each neutral carries only the unbalanced current of its own circuit, and all loads are linear. How many conductors are counted as current-carrying for ampacity adjustment?
  • A 208-volt single-phase branch circuit is taken from two ungrounded conductors of a 208Y/120-volt system. It carries 24 amperes over a one-way distance of 160 feet using 10 AWG copper, which has 10,380 circular mils. Using 12.9 as the resistivity constant for copper, what is the approximate voltage drop?
  • A commercial building has a calculated load of 240 amperes of continuous load and 90 amperes of noncontinuous load, supplied at 480Y/277 volts three phase. Before any temperature correction or conductor adjustment, what minimum ampacity must the service conductors have?
  • A fixed storage-type electric water heater with a capacity of 120 gallons or less is rated 4,500 watts at 240 volts and is supplied by its own branch circuit. What is the smallest standard overcurrent device rating that may protect that circuit?
  • A three-phase load operating at 208 volts draws 90 amperes per line. Standard three-phase dry-type transformer sizes available are 15, 30, 45, 75, and 112.5 kilovolt-amperes. What is the smallest standard transformer that will carry this load?
  • Three current-carrying 3/0 AWG THHN copper conductors run in a raceway where the ambient temperature is 14 degrees Celsius. The 90 degree column gives 225 amperes for this conductor, the 75 degree column gives 200 amperes, the correction factor for this ambient in the 90 degree column is 1.12, and the terminations at both ends are listed for 75 degrees Celsius. What ampacity may be used?
  • A 277-volt single-phase lighting circuit is wired in 10 AWG copper with an area of 10,380 circular mils and carries a steady 15 amperes to fixtures 200 feet from the panel. Using a copper constant of 12.9 and the single-phase relationship in which the drop equals 2 times the constant times the current times the one-way length divided by the circular mil area, what percentage of the supply voltage is lost in the conductors?
  • A 208-volt feeder supplies a subpanel, and a branch circuit runs from that subpanel out to a load. The informational notes in the Code suggest holding a branch circuit to 3 percent and the feeder and branch circuit together to 5 percent of the supply voltage. The feeder portion has already been measured at 3.2 volts of drop. Working to the combined figure, how much drop is left for the branch circuit?
  • A 480-volt three-phase feeder carries 50 amperes on 2 AWG copper conductors having an area of 66,360 circular mils, and the drop is to be held to 3 percent of the supply voltage. Using a copper constant of 12.9 and the three-phase relationship in which the drop equals 1.732 times the constant times the current times the one-way length divided by the circular mil area, what is the greatest one-way distance this feeder may run?
  • A 120-volt single-phase circuit in 12 AWG copper with an area of 6,530 circular mils leaves a panel, reaches a first 10-ampere load 60 feet out, then continues another 60 feet to a second 10-ampere load. Using a copper constant of 12.9 and the single-phase relationship in which drop equals 2 times the constant times the current times the length divided by circular mils, what is the total voltage drop measured at the far load?
  • A 240-volt single-phase motor draws 30 amperes while running, and its locked-rotor current is 6 times that value. The motor is fed with 8 AWG copper of 16,510 circular mils over a one-way distance of 150 feet. Using a copper constant of 12.9 and the single-phase relationship in which drop equals 2 times the constant times the current times the one-way length divided by circular mils, what is the approximate drop at the instant the motor starts?
  • A 240-volt single-phase feeder carries 200 amperes 250 feet one way. Each leg is made up of two 1/0 AWG copper conductors in parallel, and 1/0 AWG has an area of 105,600 circular mils. Using a copper constant of 12.9 and the single-phase relationship in which drop equals 2 times the constant times the current times the one-way length divided by circular mils, what is the approximate voltage drop on this feeder?
  • A 208-volt three-phase feeder must carry a 55-ampere noncontinuous load 200 feet one way, and the drop is to be held to 3 percent. Copper ampacities in the 75 degree Celsius column are 50 amperes for 8 AWG, 65 for 6 AWG, 85 for 4 AWG, 100 for 3 AWG and 150 for 1/0 AWG. Circular mil areas for those same sizes are 16,510, 26,240, 41,740, 52,620 and 105,600. Using a copper constant of 12.9 and the three-phase relationship with the 1.732 multiplier, what is the smallest conductor that satisfies both?
  • An existing feeder is protected by a 100-ampere device. It supplies a motor whose branch-circuit short-circuit and ground-fault device is rated 60 amperes, and a second motor with a full-load current of 15 amperes. The rule caps a feeder device at the largest motor branch-circuit device rating plus the full-load currents of the remaining motors plus any other load carried. How much nonmotor load may still be added without exceeding the device already installed?
  • A continuous-duty three-phase motor has a table full-load current of 40 amperes, and its branch-circuit conductors must have an ampacity of at least 125 percent of that value. Nine current-carrying conductors share the raceway, which calls for an adjustment factor of 70 percent. The conductors are rated 75 degrees Celsius and the ambient is 30 degrees Celsius, so no temperature correction applies. Copper ampacities at 75 degrees are 50 amperes for 8 AWG, 65 for 6 AWG, 85 for 4 AWG and 100 for 3 AWG. What is the smallest conductor permitted?

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