Written against the 2026 NEC and checked section by section, September 2026
Motor Feeder Sizing Practice Questions
A feeder to a group of motors is a sum with four parts, and only one motor in the group gets the 125 percent. These 25 items work that sum, the separate and differently shaped sum that sizes the feeder protective device, the rounding that runs the opposite way from the branch circuit, how the highest rated motor is picked, the four exceptions, and the packaged equipment section that stops the calculation before it starts. Full-load currents come from Table 430.248 and Table 430.250.
Pick an answer and the explanation opens with the section it comes from. These are original questions written for this site, not questions from the exam. Have your own code book open: the Texas exam is open book, and looking the section up is the skill being tested.
Read the motor feeder sizing study page first if the topic is new to you.
1. Conductors supplying several motors are sized from a sum of how many parts, and what are they?
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A. Four: 125 percent of the highest rated motor, 100 percent of the other motors, 100 percent of the noncontinuous non-motor load, and 125 percent of the continuous non-motor load
Two of the four parts carry a 125 percent and two carry 100 percent, and they alternate, which is what makes the sum easy to misremember. Sorting the group and labeling each term before you multiply anything is what keeps it straight. Every motor current in the sum comes from the Article 430 tables rather than from a nameplate.
NEC 2026 430.24
2. How many motors in a group of five get the 125 percent in the feeder conductor sum?
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B. One, the highest rated motor
The 125 percent is the same margin that sizes the branch-circuit conductors for a single motor, and the code applies it once to the group rather than to every member. It is not a starting current allowance, since a motor draws several times its full-load current on start and that is a different device's problem. Giving every motor the bump produces an answer that looks careful and is wrong.
NEC 2026 430.24, 430.22
3. A feeder supplies three continuous duty motors on a 460 volt three-phase system: 20 horsepower at 27 amperes, 15 horsepower at 21 amperes and 10 horsepower at 14 amperes. What is the minimum feeder conductor ampacity?
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C. 68.75 amperes
The 20 horsepower motor has the largest table current, so it takes 27 times 1.25, or 33.75 amperes, and the other two come in at their table values. That gives 33.75 plus 21 plus 14, or 68.75 amperes. The 80.75 answer comes from bumping the wrong motor or bumping more than one, and 62 is the plain sum with no margin at all.
NEC 2026 430.24, Table 430.250
4. That same feeder runs copper conductors to 75 degree C terminations with no correction or adjustment. What is the smallest conductor?
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D. 4 AWG
The required ampacity is 68.75 amperes. In the 75 degree column, 6 AWG copper carries 65 amperes and 4 AWG carries 85, so 6 AWG falls just short and 4 AWG is the answer. This is the kind of margin where a 3.75 ampere error in the sum changes the wire size, which is why the arithmetic is worth writing down.
NEC 2026 430.24, Table 310.16
5. A feeder supplies a 25 horsepower motor at 34 amperes, a 10 horsepower motor at 14 amperes, and a 20 ampere continuous lighting load, all on a 460 volt system. What is the minimum feeder conductor ampacity?
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A. 81.5 amperes
The largest motor takes 125 percent, the other motor comes in whole, and the lighting is continuous so it takes 125 percent as well. That is 42.5 plus 14 plus 25, or 81.5 amperes. The 76.5 answer treats the lighting as noncontinuous, which is the version of this item with one word changed in the stem.
NEC 2026 430.24
6. Take the same two motors, but the 20 ampere non-motor load is noncontinuous. What is the minimum feeder conductor ampacity now?
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B. 76.5 amperes
The two motor terms do not move, so the sum is 42.5 plus 14 plus 20, or 76.5 amperes. The continuous and noncontinuous split lives on the non-motor side of the sum only, because the motors are already handled by the first two terms. Candidates who apply the continuous rule to the motors as well add the same margin twice.
NEC 2026 430.24
7. A feeder supplies two 30 horsepower motors at 40 amperes each and one 5 horsepower motor at 7.6 amperes, on a 460 volt three-phase system. What is the minimum feeder conductor ampacity?
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C. 97.6 amperes
Nothing in the sum breaks a tie and nothing needs to. Take one of the 40 ampere motors as the highest rated and the other joins the rest, giving 50 plus 40 plus 7.6, or 97.6 amperes. Giving 125 percent to both tied motors produces 107.6 and is what a careful person does when they think a tie has to be resolved.
NEC 2026 430.24, 430.17
8. A feeder supplies a 10 horsepower single-phase motor at 230 volts, whose table current is 50 amperes, and a 15 horsepower three-phase motor at 230 volts, whose table current is 42 amperes. What is the minimum feeder conductor ampacity?
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D. 104.5 amperes
The highest rated motor is the one with the larger table full-load current, which here is the smaller horsepower on the single-phase supply. So the sum is 50 times 1.25 plus 42, or 104.5 amperes. Sorting by horsepower instead gives 102.5, which is the wrong answer this item exists to catch.
NEC 2026 430.24, 430.17, Table 430.248, Table 430.250
9. How does the code settle which motor in a group is the highest rated?
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A. By rated full-load current selected from the Article 430 tables
The answer sits in its own short section rather than inside the feeder sum, which is part of why people miss it. That section also governs the smallest rated motor where other parts of the article ask for one. On a group at one voltage and phase the table current and the horsepower agree, and on a mixed group they do not.
NEC 2026 430.17
10. The 20 horsepower motor at 27 amperes on that first feeder is protected by an inverse time circuit breaker sized to the maximum the table permits, and it is the largest such device in the group. The other two motors draw 21 and 14 amperes. What is the largest feeder protective device permitted?
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B. 100 amperes
The branch-circuit device comes first: 27 times 2.5 is 67.5 amperes, and the next higher standard rating is 70. The feeder sum is then 70 plus 21 plus 14, or 105 amperes, which is not a standard rating, so the next size down applies and the answer is 100. Carrying the branch-circuit rounding permission forward gives 110 and puts the device above the maximum.
NEC 2026 430.62(A), Table 430.52(C)(1), 240.6(A)
Ten down. If you want these timed and scored by content area, the practice exam runs thirty questions free, and the full bank of 178 plus the calculation ladder is $79 once.
11. The feeder device calculation lands between two standard ampere ratings. Which way does it go?
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C. Down to the next standard rating, because the section states a maximum
Two rounding rules in the same problem run in opposite directions, and this is the one that goes down. The next higher permission belongs to the branch-circuit section and does not reach the feeder. Know which article and which part you are standing in before you round anything.
NEC 2026 430.62(A), 240.6(A)
12. Two motors on a feeder are protected by branch-circuit devices of the same rating, and that rating is the largest in the group. How does the feeder device calculation handle it?
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D. One of the two devices is treated as the largest, and the other motor contributes only its full-load current
The section says this outright, which is a small mercy on an item that otherwise looks ambiguous. Adding both device ratings would put a feeder device well above what the rule allows. The rest of the sum is unaffected, since the second motor still contributes its own full-load current like every other motor in the group.
NEC 2026 430.62(A)
13. In the feeder conductor sum, which term does the continuous versus noncontinuous distinction apply to?
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A. The non-motor load only
The motors are already accounted for by the first two terms of the sum, so the continuous rule never touches them. Only the non-motor part of the load splits into a 100 percent term and a 125 percent term. Applying the continuous rule to a motor is double counting a margin the sum has already provided.
NEC 2026 430.24
14. A chiller arrives from the factory with a minimum circuit ampacity marked on it. How are its supply conductors sized?
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B. From the marked minimum circuit ampacity, without calculating the internal motors
Multimotor and combination-load equipment has its own section, and the marking already accounts for which internal loads run at the same time. That marking is a floor for the conductor rather than a device rating, so it is selected at or above. This is a separate section rather than one of the exceptions to the several-motors rule, which is worth knowing before you go looking for it under time pressure.
NEC 2026 430.25
15. When does multimotor equipment send you back to the several-motors sum?
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C. Where the equipment is not factory wired and the individual motor nameplates are visible
Without factory wiring there is no assembly rating to rely on, so the marking that would have answered the question does not exist and you calculate. The visible nameplates are what make that calculation possible. Both conditions have to be present, which is why an outdoor location or a suspicion about the marking does not open this door.
NEC 2026 430.25, 430.7(D)
16. What does the marking rule put on the nameplate of factory-wired multimotor and combination-load equipment?
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D. The minimum supply circuit conductor ampacity and the maximum rating of the circuit short-circuit and ground-fault protective device
The nameplate also carries the manufacturer name, the voltage, the frequency and the number of phases. It gives you a current rather than a wire size, because the size still depends on your material, your ambient and your terminations. One of the two numbers is a floor and the other is a ceiling, and swapping them is the usual miss.
NEC 2026 430.7(D)
17. One motor in a group runs on short-time duty. What does the first exception to the several-motors rule do with it?
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A. Its ampere rating for the sum comes from the duty-cycle table instead of its table full-load current
The exception swaps the current that motor contributes, and it also says what to do where that motor is the highest rated one, taking the greater of its duty-cycle rating or 1.25 times the largest continuous duty motor current. Only the duty-cycle motor changes, and the rest of the group is untouched. Read the exception in your own book, because a paraphrase does not carry its conditions.
NEC 2026 430.24, 430.22(E)
18. A group includes motor-operated fixed electric space-heating equipment. What does the second exception do?
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B. It sends the conductor ampacity for that equipment to the fixed electric space-heating article, where the equipment and its motors are treated as a continuous load
The pointer runs out of Article 430 and into the space-heating article, and in the 2026 edition the section it lands on states that the equipment and any associated motors are a continuous load. That reclassification is the whole content of the exception. Working the equipment as an ordinary motor in the sum misses the continuous treatment.
NEC 2026 430.24, 424.5
19. A feeder serves six motors, but the control wiring is interlocked so that only four of them can run at once. What does the third exception permit?
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C. Sizing on the combination of motors and other loads that can operate at the same time and produces the highest total current
The permission is real but it is bounded, since you still have to find the worst combination the interlock allows rather than a convenient one. That means checking every group the controls can energize together. The exception turns on how the circuitry is arranged, not on how the equipment is normally operated.
NEC 2026 430.24
20. Which exception to the several-motors conductor rule arrived with the 2026 edition?
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D. A conductor temperature provision covering Design BE and Design CE motors
The new exception keeps the four-part sum and then adds a temperature condition on 90 degree rated copper or aluminum conductors supplying a group that contains those motor designs. It is written around locked-rotor current drawn for a stated time with the conductors already at their maximum operating temperature. Those designations also appear in the branch-circuit device table, which is where most people meet them first.
NEC 2026 430.24
21. A designer wants to size a motor feeder below what the several-motors sum gives, because the motors never all run at once and there is no interlock. What does the code say?
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B. The authority having jurisdiction may grant permission where reduced heating results, provided the conductors carry the maximum load determined from the sizes, number and duties of the motors
There is a demand factor section for exactly this, and it is written as a permission the authority having jurisdiction grants rather than a factor you apply on your own. That makes it a design conversation and not a calculation step. A stem that mentions duty cycle or motors not running together without describing an interlock is pointing here.
NEC 2026 430.26
22. Where do the motor currents used in the several-motors sum come from?
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B. The Article 430 full-load current tables, by horsepower, voltage and phase
The feeder rule points at the general ampacity determination section for every motor in the group, and that section sends general motor applications to the tables. Nameplate values belong to the overload step and nowhere in this sum. A question that prints both currents for each motor is checking whether you noticed.
NEC 2026 430.24, 430.6(A)(1)
23. Feeder tap conductors run to a motor controller. What does the motor article require of them?
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C. An ampacity not less than Article 430 Part II requires, termination in an overcurrent device, and compliance with one of the listed length and enclosure conditions
The motor article carries its own tap section, so the general tap rules are not the last word on a motor feeder. Three requirements stack: an ampacity floor from the conductor part of the article, a termination in a device, and one of the enclosure and length options. Answering from the general tap rules alone leaves out the ampacity floor the motor article sets.
NEC 2026 430.28
24. Where does the 125 percent in the feeder conductor sum come from?
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D. The single-motor conductor rule, applied once to the group
It is the same multiplier that sizes conductors for one motor, and the code applies it a single time no matter how many motors are on the feeder. A motor draws several times its full-load current on start rather than a quarter more, so inrush is not the reason and reasoning from inrush leads people to apply it everywhere. Trace the number back to where it came from and the sum stops looking arbitrary.
NEC 2026 430.24, 430.22
25. Two motors are connected to the same branch circuit rather than to a feeder. Which section governs that arrangement?
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A. The several motors or loads on one branch circuit section, which sets its own conditions and requires fuses or inverse time circuit breakers
The feeder sections reach conductors on the supply side of the branch-circuit devices, and two motors sharing one branch circuit is a different arrangement with its own conditions. That section also names the device types permitted for the shared circuit. Reading a group of motors as a feeder problem when they share one circuit sends you to the wrong sum entirely.
NEC 2026 430.53
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