Motor Calculation Practice Questions
Work through 32 original, unofficial motor calculation practice questions for U.S. electrician exam preparation. Use each question’s supplied values and rules, then open the worked explanation beneath it; these guided exercises are not installation instructions.
Practice questions
Question 1 of 32 · Which current to start from
A 15 hp, 460 V, three-phase squirrel-cage motor has a nameplate current of 19.5 A. What current do you start from to size its branch-circuit conductors?
Supplied reference: For this continuous-duty exercise, start conductor sizing from table full-load current (FLC), not nameplate current. Supplied lookup: 15 hp, three-phase, 460 V → 21 A.
- A. 21 A
- B. 19.5 A
- C. 24.4 A
- D. 26.25 A
Show answer and explanation
Answer: A. 21 A
For this ordinary motor exercise, conductors start from the table, not the nameplate. The supplied three-phase lookup gives 21 A for 15 hp at 460 V. The 19.5 A nameplate value is for overloads. 24.4 A is 19.5 × 1.25 rounded from 24.375 A; 26.25 A is 21 × 1.25. Both already apply the conductor multiplier, but the question only asks where you start.
Current-basis guidance: Minnesota DLI, pp. 18–19. The lookup value is supplied above.
Related NEC study lookup: 430.6(A)(1); Table 430.250.
Question 2 of 32 · Which current to start from
Same motor: 15 hp, 460 V, three-phase, nameplate current 19.5 A. What current is the starting point for sizing a separate overload device?
Supplied reference: Supplied current-basis rule: separate overload calculations start from nameplate current, not table current. The supplied three-phase 15 hp, 460 V table value is 21 A.
- A. 19.5 A
- B. 21 A
- C. 22.4 A
- D. 24.15 A
Show answer and explanation
Answer: A. 19.5 A
Overloads protect this particular motor, so they start from its nameplate current: 19.5 A. In these ordinary supplied-rule cases, the 21 A table value is the starting current for conductors and branch short-circuit/ground-fault protection. 22.4 A is 19.5 × 1.15 rounded from 22.425 A; 24.15 A is 21 × 1.15. Both apply a percentage, a step the question hasn't reached yet.
Current-basis guidance: Minnesota DLI, pp. 18–19. The lookup value is supplied above.
Related NEC study lookup: 430.6(A)(2); 430.32(A)(1).
Question 3 of 32 · Which current to start from
Using the supplied lookup, what is the table full-load current of a 3 hp, 230 V, single-phase motor?
Supplied reference: Supplied 3 hp lookup: single-phase 115 V → 34 A; single-phase 208 V → 18.7 A; single-phase 230 V → 17 A; three-phase 230 V → 9.6 A.
- A. 9.6 A
- B. 18.7 A
- C. 34 A
- D. 17 A
Show answer and explanation
Answer: D. 17 A
In the supplied single-phase lookup, 3 hp at 230 V is 17 A. 9.6 A is the supplied three-phase value, which is the wrong table. 18.7 A is the supplied 208 V entry, and 34 A is the 115 V entry.
Lookup inputs are supplied above. The 115/230 V single-phase and 230 V three-phase entries can also be compared with the historical manufacturer tables, pp. 149 and 154–155; the 208 V single-phase entry is a supplied exercise value.
Related NEC study lookup: 430.6(A)(1); Table 430.248.
Question 4 of 32 · Which current to start from
A 10 hp, three-phase squirrel-cage motor is rated 208 V. What table full-load current do you use from the supplied lookup?
Supplied reference: Supplied three-phase 10 hp lookup: 200 V → 32.2 A; 208 V → 30.8 A; 230 V → 28 A; 460 V → 14 A.
- A. 30.8 A
- B. 14 A
- C. 28 A
- D. 32.2 A
Show answer and explanation
Answer: A. 30.8 A
The supplied three-phase lookup has its own 208 V column: 10 hp is 30.8 A. 28 A is the 230 V value. 32.2 A is the 200 V column and 14 A is 460 V. In these supplied entries, lower voltage corresponds to a higher table current for the same horsepower.
Lookup cross-check: historical manufacturer tables, pp. 146, 148–149 and 151. These are supplied lookup entries, not measured motor currents.
Related NEC study lookup: Table 430.250.
Question 5 of 32 · Branch-circuit conductors
What is the minimum branch-circuit conductor ampacity for a 10 hp, 230 V, three-phase, continuous-duty motor?
Supplied reference: Table FLC = 28 A. Minimum branch-conductor ampacity = 125% of table FLC; no other adjustment applies.
- A. 28 A
- B. 32.2 A
- C. 70 A
- D. 35 A
Show answer and explanation
Answer: D. 35 A
Table FLC is 28 A, and 28 × 125% = 35 A. 28 A skips the 125%. 32.2 A uses 115%, not the supplied conductor percentage. 70 A is 28 × 250%, a different exercise factor for inverse-time branch protection, not the wire.
Underlying conductor/current-basis method: Minnesota DLI, pp. 18–19. Apply it to the supplied values above.
Related NEC study lookup: Table 430.250; 430.22.
Question 6 of 32 · Branch-circuit conductors
A 5 hp, 230 V, single-phase, continuous-duty motor is supplied by copper THWN conductors. The motor and controller terminals are listed and identified for 75 °C. What is the minimum conductor size?
Supplied reference: Table FLC = 28 A. All terminations are identified for 75 °C. Supplied copper ampacities: 12 AWG → 25 A; 10 AWG → 35 A; 8 AWG → 50 A; 6 AWG → 65 A. AWG means American Wire Gauge. Use copper at 30 °C, no more than three current-carrying conductors, and no other correction or adjustment. Choose the smallest listed size whose supplied ampacity meets 125% of table FLC.
- A. 12 AWG
- B. 8 AWG
- C. 6 AWG
- D. 10 AWG
Show answer and explanation
Answer: D. 10 AWG
The supplied single-phase lookup gives 28 A, and 28 × 125% = 35 A. With 75 °C terminals, 10 AWG copper is rated 35 A in Table 310.16, which is exactly enough. 12 AWG (25 A) is too small. 8 AWG and 6 AWG work but aren't the minimum. This answer checks the stated conductor-ampacity requirement, not a complete circuit or a required breaker rating.
Ampacity cells and adjustment factor: 2023-labeled ampacity extract, pp. 1–2. Motor-conductor method: Minnesota DLI, pp. 18–19.
Related NEC study lookup: Table 430.248; 430.22; Table 310.16; 110.14(C).
Question 7 of 32 · Branch-circuit conductors
Same 5 hp, 230 V, single-phase motor, but the terminals are rated only 60 °C. What is the minimum copper conductor size?
Supplied reference: Continuous-duty motor; table FLC = 28 A. All terminations require the 60 °C column. Supplied copper ampacities: 12 AWG → 20 A; 10 AWG → 30 A; 8 AWG → 40 A; 6 AWG → 55 A. Use copper at 30 °C, no more than three current-carrying conductors, and no other correction or adjustment. Choose the smallest listed size whose supplied ampacity meets 125% of table FLC.
- A. 12 AWG
- B. 8 AWG
- C. 10 AWG
- D. 6 AWG
Show answer and explanation
Answer: B. 8 AWG
You still need 35 A (28 × 1.25), but now you read the 60 °C column. 10 AWG is 30 A there, which is too small. 8 AWG is 40 A, which is enough. The higher insulation rating cannot override this problem's 60 °C termination limit. 12 AWG supplies only 20 A; 6 AWG supplies 55 A but is larger than needed.
Ampacity cells and adjustment factor: 2023-labeled ampacity extract, pp. 1–2. Motor-conductor method: Minnesota DLI, pp. 18–19.
Related NEC study lookup: 430.22; 110.14(C)(1)(a); Table 310.16.
Question 8 of 32 · Branch-circuit conductors
What is the minimum copper THWN conductor size for a 50 hp, 460 V, three-phase, continuous-duty motor with 75 °C terminals?
Supplied reference: Table FLC = 65 A. All terminations use 75 °C. Supplied copper ampacities: 6 AWG → 65 A; 4 AWG → 85 A; 3 AWG → 100 A; 2 AWG → 115 A. Use copper at 30 °C, no more than three current-carrying conductors, and no other correction or adjustment. Choose the smallest listed size whose supplied ampacity meets 125% of table FLC.
- A. 6 AWG
- B. 3 AWG
- C. 4 AWG
- D. 2 AWG
Show answer and explanation
Answer: C. 4 AWG
65 A (supplied table current) × 125% = 81.25 A. In the 75 °C column, 6 AWG is 65 A, which is too small, and 4 AWG is 85 A, the smallest that works. 3 AWG and 2 AWG work but aren't the minimum. A mistake is stopping at 6 AWG because its 65 A matches the FLC.
Ampacity cells and adjustment factor: 2023-labeled ampacity extract, pp. 1–2. Motor-conductor method: Minnesota DLI, pp. 18–19.
Related NEC study lookup: Table 430.250; 430.22; Table 310.16.
Question 9 of 32 · Branch-circuit conductors
A 25 hp, 460 V, three-phase, continuous-duty motor has a nameplate current of 32 A. What minimum ampacity must its branch-circuit conductors have?
Supplied reference: Table FLC = 34 A. Use 125% of table FLC, not nameplate current; no other adjustment applies.
- A. 32 A
- B. 34 A
- C. 40 A
- D. 42.5 A
Show answer and explanation
Answer: D. 42.5 A
Conductors use the table FLC of 34 A, not the nameplate: 34 × 1.25 = 42.5 A. 40 A is the trap, the 32 A nameplate × 1.25. 32 A and 34 A both skip the 125%.
Underlying conductor/current-basis method: Minnesota DLI, pp. 18–19. Apply it to the supplied values above.
Related NEC study lookup: 430.6(A)(1); Table 430.250; 430.22.
Question 10 of 32 · Branch-circuit conductors
A 20 hp, 460 V, three-phase, continuous-duty motor is supplied by copper THHN (90 °C) conductors. They are in a dry location and share a raceway longer than 24 inches with other circuits, for 9 current-carrying conductors in total. Ambient temperature is 30 °C (86 °F), and all terminals are rated 75 °C. What is the minimum conductor size?
Supplied reference: Table FLC = 27 A; required ampacity = 125% of table FLC. Apply a 0.70 adjustment factor to 90 °C ampacity, then cap the result at the 75 °C termination ampacity. No ambient correction or exception applies. Copper values, 90 °C / 75 °C: 12 AWG → 30 / 25 A; 10 AWG → 40 / 35 A; 8 AWG → 55 / 50 A; 6 AWG → 75 / 65 A.
- A. 8 AWG
- B. 12 AWG
- C. 10 AWG
- D. 6 AWG
Show answer and explanation
Answer: A. 8 AWG
You need 27 A × 125% = 33.75 A. Apply the supplied adjustment to the 90 °C ampacity. The factor for this nine-conductor exercise is 70%. 10 AWG gives 40 × 0.70 = 28 A, which is too small. 8 AWG gives 55 × 0.70 = 38.5 A, which is enough. Then check the terminals: 8 AWG in the 75 °C column is 50 A, still above 33.75 A. At 30 °C there's no temperature correction. Picking 10 AWG means you skipped the adjustment, because its 35 A at 75 °C looks like enough on its own. The usable ampacity is the lesser of adjusted 90 °C ampacity and the 75 °C termination ampacity. For 12 AWG that is 21 A, too small; for 6 AWG it is 52.5 A, enough but larger than the required 8 AWG.
Ampacity cells and adjustment factor: 2023-labeled ampacity extract, pp. 1–2. Motor-conductor method: Minnesota DLI, pp. 18–19.
Related NEC study lookup: 430.22; Table 310.16; Table 310.15(C)(1); 110.14(C).
Question 11 of 32 · Branch-circuit conductors
After any correction and adjustment, a copper conductor has an allowable ampacity of 65 A. It will supply one continuous-duty motor, and the terminal rating is not a limit. Under the supplied 125% conductor rule, what is the largest table full-load current this conductor can serve?
Supplied reference: Usable conductor ampacity must be at least 1.25 × table FLC. The stated 65 A already includes every conductor and termination limitation.
- A. 56.5 A
- B. 65 A
- C. 81.25 A
- D. 52 A
Show answer and explanation
Answer: D. 52 A
The supplied rule requires ampacity of at least 125% of the FLC, so the FLC can be no more than 65 ÷ 1.25 = 52 A. 81.25 A multiplies when the rule needs division. 56.5 A is 65 ÷ 1.15 rounded to one decimal place, using the wrong factor. 65 A ignores the 125% entirely. As a check, 52 A is the supplied table current for a 40 hp, 460 V, three-phase motor.
Underlying conductor/current-basis method: Minnesota DLI, pp. 18–19. Apply it to the supplied values above.
Related NEC study lookup: 430.22; Table 430.250.
Question 12 of 32 · Overload protection
A continuous-duty motor rated more than 1 hp has a nameplate current of 26 A and a marked service factor of 1.15. What is the calculated overload trip-current ceiling under the supplied basic rule?
Supplied reference: For this continuous-duty, over-1-hp motor with a separate overload device, use 125% of nameplate current when marked service factor is at least 1.15 OR marked temperature rise is 40 °C or less; otherwise use 115%. No increased-setting exception or rounding above the ceiling applies. This asks for a trip-current limit, not a relay-dial value.
- A. 29.9 A
- B. 35 A
- C. 36.4 A
- D. 32.5 A
Show answer and explanation
Answer: D. 32.5 A
Under the supplied basic rule, a service factor of 1.15 or higher selects 125% of nameplate: 26 × 1.25 = 32.5 A. 29.9 A (115%) is for motors without a qualifying marking. 36.4 A is 140%, which is not a factor permitted by this question. 35 A is a rounded-up size, but an overload value is a ceiling, so you don't round it up.
The factor and qualification rule are supplied in this exercise. Nameplate-current basis: Minnesota DLI, p. 18. Why a trip-current limit is not automatically a dial setting: manufacturer instructions, pp. 9–10.
Related NEC study lookup: 430.6(A)(2); 430.32(A)(1).
Question 13 of 32 · Overload protection
A continuous-duty motor rated more than 1 hp has a nameplate current of 26 A, a service factor of 1.0, and a marked temperature rise of 50 °C. What is the calculated overload trip-current ceiling under the supplied basic rule?
Supplied reference: For this continuous-duty, over-1-hp motor with a separate overload device, use 125% of nameplate current when marked service factor is at least 1.15 OR marked temperature rise is 40 °C or less; otherwise use 115%. No increased-setting exception or rounding above the ceiling applies. This asks for a trip-current limit, not a relay-dial value.
- A. 26 A
- B. 29.9 A
- C. 30 A
- D. 32.5 A
Show answer and explanation
Answer: B. 29.9 A
The service factor is below 1.15 and the temperature rise is above 40 °C, so the supplied rule selects 115%. 26 × 1.15 = 29.9 A. 32.5 A applies 125% without a qualifying marking. 30 A rounds up past the maximum. 26 A leaves out the percentage.
The factor and qualification rule are supplied in this exercise. Nameplate-current basis: Minnesota DLI, p. 18. Why a trip-current limit is not automatically a dial setting: manufacturer instructions, pp. 9–10.
Related NEC study lookup: 430.32(A)(1).
Question 14 of 32 · Overload protection
A continuous-duty motor rated more than 1 hp has a nameplate current of 40 A, a service factor of 1.0, and a marked temperature rise of 40 °C. What is the calculated overload trip-current ceiling under the supplied basic rule?
Supplied reference: For this continuous-duty, over-1-hp motor with a separate overload device, use 125% of nameplate current when marked service factor is at least 1.15 OR marked temperature rise is 40 °C or less; otherwise use 115%. No increased-setting exception or rounding above the ceiling applies. This asks for a trip-current limit, not a relay-dial value.
- A. 40 A
- B. 46 A
- C. 56 A
- D. 50 A
Show answer and explanation
Answer: D. 50 A
A marked rise of 40 °C or less qualifies for 125% on its own, even with a 1.0 service factor. Either condition is enough. 40 × 1.25 = 50 A. 46 A (115%) misses the temperature-rise condition. 56 A is 140%. "Temperature rise" is the motor's marked rating, not the room temperature. 40 A leaves out the percentage.
The factor and qualification rule are supplied in this exercise. Nameplate-current basis: Minnesota DLI, p. 18. Why a trip-current limit is not automatically a dial setting: manufacturer instructions, pp. 9–10.
Related NEC study lookup: 430.32(A)(1).
Question 15 of 32 · Overload protection
A 10 hp, 230 V, three-phase motor has a nameplate current of 27 A and a marked service factor of 1.15. What is the calculated overload trip-current ceiling under the supplied basic rule?
Supplied reference: For this continuous-duty, over-1-hp motor with a separate overload device, use 125% of nameplate current when marked service factor is at least 1.15 OR marked temperature rise is 40 °C or less; otherwise use 115%. No increased-setting exception or rounding above the ceiling applies. This asks for a trip-current limit, not a relay-dial value. Supplied table FLC for the comparison in the explanation: 28 A.
- A. 31.05 A
- B. 33.75 A
- C. 35 A
- D. 70 A
Show answer and explanation
Answer: B. 33.75 A
Overloads use the nameplate: 27 × 1.25 = 33.75 A. 35 A is the trap. It's the 28 A table FLC × 1.25, which is the conductor calculation. 31.05 A applies 115% despite the 1.15 service factor. 70 A applies a 250% factor to the 28 A table value, not the supplied overload rule.
The factor and qualification rule are supplied in this exercise. Nameplate-current basis: Minnesota DLI, p. 18. Why a trip-current limit is not automatically a dial setting: manufacturer instructions, pp. 9–10.
Related NEC study lookup: 430.6(A)(2); 430.32(A)(1).
Question 16 of 32 · Overload protection
A 7½ hp, 460 V, three-phase motor has a nameplate current of 10.4 A and a marked service factor of 1.25. What is the calculated overload trip-current ceiling under the supplied basic rule?
Supplied reference: For this continuous-duty, over-1-hp motor with a separate overload device, use 125% of nameplate current when marked service factor is at least 1.15 OR marked temperature rise is 40 °C or less; otherwise use 115%. No increased-setting exception or rounding above the ceiling applies. This asks for a trip-current limit, not a relay-dial value. Supplied table FLC for the comparison in the explanation: 11 A.
- A. 13 A
- B. 11.96 A
- C. 13.75 A
- D. 16.25 A
Show answer and explanation
Answer: A. 13 A
A 1.25 service factor is "1.15 or greater," so the percentage is still 125%: 10.4 × 1.25 = 13 A. The service-factor number isn't the multiplier. 16.25 A stacks it on top (10.4 × 1.25 × 1.25). 13.75 A uses the 11 A table FLC, the wrong base. 11.96 A is 115%.
The factor and qualification rule are supplied in this exercise. Nameplate-current basis: Minnesota DLI, p. 18. Why a trip-current limit is not automatically a dial setting: manufacturer instructions, pp. 9–10.
Related NEC study lookup: 430.32(A)(1).
Question 17 of 32 · Branch-circuit breaker and fuse sizing
What is the largest listed inverse-time circuit-breaker rating allowed by the supplied rules for the branch circuit of a 10 hp, 230 V, three-phase squirrel-cage motor?
Supplied reference: Table FLC = 28 A; inverse-time breaker factor = 250%. Listed ratings: 35, 40, 45, 50, 60, 70, 80, 90 A. Use the stated table FLC and device factor. Select the raw result if it is listed; otherwise the next higher listed rating is permitted. Separate overload protection is assumed, with no increased-starting allowance or lower equipment-specific limit in this exercise.
- A. 35 A
- B. 50 A
- C. 80 A
- D. 70 A
Show answer and explanation
Answer: D. 70 A
28 A × 250% = 70 A, which is already a listed rating, so you don't round. 80 A goes past it, because next-size-up only applies when the result isn't listed. 35 A is the conductor ampacity. 50 A comes from 28 × 175% = 49 A rounded up, which is the time-delay fuse percentage, not the breaker's.
The numerical result follows from the current values, factors and rating-selection rule supplied above. Use the related study lookup to locate this topic in your codebook.
Related NEC study lookup: Table 430.250; Table 430.52(C)(1); 240.6(A).
Question 18 of 32 · Branch-circuit breaker and fuse sizing
What is the largest listed inverse-time circuit-breaker rating allowed by the supplied rules for a 5 hp, 230 V, three-phase squirrel-cage motor?
Supplied reference: Table FLC = 15.2 A; inverse-time breaker factor = 250%. Listed ratings: 30, 35, 40, 45, 50 A. Use the stated table FLC and device factor. Select the raw result if it is listed; otherwise the next higher listed rating is permitted. Separate overload protection is assumed, with no increased-starting allowance or lower equipment-specific limit in this exercise.
- A. 35 A
- B. 40 A
- C. 38 A
- D. 45 A
Show answer and explanation
Answer: B. 40 A
15.2 A × 250% = 38 A. That isn't a listed rating, so the next higher listed rating, 40 A, is permitted. 38 A is the raw calculation, not one of this question's listed ratings. 35 A is below the permitted maximum; this exercise does not decide whether a smaller device meets every installation requirement. 45 A skips past the next listed rating.
The numerical result follows from the current values, factors and rating-selection rule supplied above. Use the related study lookup to locate this topic in your codebook.
Related NEC study lookup: Table 430.52(C)(1); 430.52(C)(1)(a); 240.6(A).
Question 19 of 32 · Branch-circuit breaker and fuse sizing
What is the largest listed dual-element (time-delay) fuse rating allowed by the supplied rules for a 25 hp, 460 V, three-phase squirrel-cage motor?
Supplied reference: Table FLC = 34 A; dual-element time-delay fuse factor = 175%. Listed ratings: 40, 45, 50, 60, 70, 80, 90 A. Use the stated table FLC and device factor. Select the raw result if it is listed; otherwise the next higher listed rating is permitted. Separate overload protection is assumed, with no increased-starting allowance or lower equipment-specific limit in this exercise.
- A. 60 A
- B. 50 A
- C. 59.5 A
- D. 90 A
Show answer and explanation
Answer: A. 60 A
34 A × 175% = 59.5 A, so the next higher listed rating is 60 A. 90 A uses the 250% breaker percentage (85 A) and rounds up. 50 A rounds down when it doesn't need to. 59.5 A is the raw result, not one of this question's listed ratings.
Time-delay calculation example: Minnesota DLI, p. 19. The rating list and permitted selection step are supplied above.
Related NEC study lookup: Table 430.52(C)(1); 430.52(C)(1)(a); 240.6(A).
Question 20 of 32 · Branch-circuit breaker and fuse sizing
What is the largest listed nontime-delay fuse rating allowed by the supplied rules for a 15 hp, 208 V, three-phase squirrel-cage motor?
Supplied reference: Table FLC at 208 V = 46.2 A; the comparison value at 230 V is 42 A. Nontime-delay fuse factor = 300%. Listed ratings: 100, 110, 125, 150, 175, 200 A. Use the stated table FLC and device factor. Select the raw result if it is listed; otherwise the next higher listed rating is permitted. Separate overload protection is assumed, with no increased-starting allowance or lower equipment-specific limit in this exercise.
- A. 125 A
- B. 138.6 A
- C. 150 A
- D. 175 A
Show answer and explanation
Answer: C. 150 A
Use the 208 V column: 46.2 A × 300% = 138.6 A, so the next listed rating is 150 A. The supplied 230 V value (42 A) would give 126 A and happen to land on 150 A too, but only the 208 V value is the right starting point. 175 A skips a size, and 125 A is below the calculated value. 138.6 A is the raw result, not a listed rating. Write the raw result as well as the final rating so a wrong voltage lookup does not go unnoticed.
The complete rule is supplied above. Historical context only: manufacturer column explanations, p. 145; this older document is not a current-code certification.
Related NEC study lookup: Table 430.250; Table 430.52(C)(1); 240.6(A).
Question 21 of 32 · Branch-circuit breaker and fuse sizing
What is the largest listed dual-element (time-delay) fuse rating allowed by the supplied rules for a 3 hp, 230 V, single-phase motor?
Supplied reference: Table FLC = 17 A; dual-element time-delay fuse factor = 175%. Listed ratings: 20, 25, 30, 35, 40, 45, 50 A. Use the stated table FLC and device factor. Select the raw result if it is listed; otherwise the next higher listed rating is permitted. Separate overload protection is assumed, with no increased-starting allowance or lower equipment-specific limit in this exercise.
- A. 25 A
- B. 35 A
- C. 30 A
- D. 45 A
Show answer and explanation
Answer: C. 30 A
The supplied single-phase lookup gives 17 A, and 17 × 175% = 29.75 A, so the next listed rating is 30 A. 45 A uses the 250% breaker percentage (42.5 A) and rounds up. 35 A overshoots the next listed rating. 25 A is below the calculated value.
Time-delay calculation example: Minnesota DLI, p. 19. The rating list and permitted selection step are supplied above.
Related NEC study lookup: Table 430.248; Table 430.52(C)(1); 240.6(A).
Question 22 of 32 · Branch-circuit breaker and fuse sizing
A 25 hp, 460 V, three-phase squirrel-cage motor is protected by dual-element time-delay fuses. The ordinary 60 A selection won't let the motor start. Under the supplied increased-starting limit, what is the largest listed time-delay fuse rating?
Supplied reference: Table FLC = 34 A. For this increased-starting exercise, the fuse rating must not exceed 225% of table FLC; do not take the next size above that ceiling. Listed ratings: 60, 70, 80, 90 A. Separate overload protection is assumed.
- A. 76.5 A
- B. 80 A
- C. 70 A
- D. 90 A
Show answer and explanation
Answer: C. 70 A
For this exercise, the supplied increased-starting limit is 225% of table FLC: 34 × 2.25 = 76.5 A. That's a hard ceiling with no next-size-up, so the largest listed fuse that doesn't exceed 76.5 A is 70 A. 80 A goes over it. 90 A also exceeds the ceiling. 76.5 A is the calculated ceiling, not a rating in the supplied list.
The complete rule is supplied above. Historical context only: manufacturer column explanations, p. 145; this older document is not a current-code certification.
Related NEC study lookup: 430.52(C)(1)(b); 240.6(A).
Question 23 of 32 · Feeders with several motors
A feeder supplies three continuous-duty, 460 V, three-phase motors: 20 hp, 15 hp, and 10 hp. What is the minimum feeder conductor ampacity?
Supplied reference: Supplied three-phase 460 V table currents: 20 hp → 27 A; 15 hp → 21 A; 10 hp → 14 A. All motors operate simultaneously. Minimum feeder-conductor ampacity = all table currents + 25% of one largest table current. No other loads, correction, adjustment or exception applies.
- A. 62 A
- B. 77.5 A
- C. 68.75 A
- D. 85.94 A
Show answer and explanation
Answer: C. 68.75 A
The FLCs are 27, 21, and 14 A. Take 125% of the largest (27 × 1.25 = 33.75 A), then add 21 + 14: 68.75 A. 62 A adds them all at 100%. 77.5 A applies 125% to every motor. 85.94 A applies 125% a second time to the correct total, then rounds 85.9375 A to two decimal places.
Underlying conductor/current-basis method: Minnesota DLI, pp. 18–19. Apply it to the supplied values above.
Related NEC study lookup: Table 430.250; 430.24.
Question 24 of 32 · Feeders with several motors
A feeder supplies two 20 hp motors and one 10 hp motor, all continuous-duty, 460 V, three-phase. What is the minimum feeder conductor ampacity?
Supplied reference: Supplied three-phase 460 V table currents: each 20 hp motor → 27 A; 10 hp → 14 A. All motors operate simultaneously. Minimum feeder-conductor ampacity = all table currents + 25% of one largest table current. No other loads, correction, adjustment or exception applies.
- A. 68 A
- B. 74.75 A
- C. 81.5 A
- D. 85 A
Show answer and explanation
Answer: B. 74.75 A
When two motors tie for largest, apply 125% to only one of them: 27 × 1.25 + 27 + 14 = 74.75 A. 81.5 A applies 125% to both 20 hp motors. 68 A skips the 125%. 85 A is 68 × 1.25.
Underlying conductor/current-basis method: Minnesota DLI, pp. 18–19. Apply it to the supplied values above.
Related NEC study lookup: 430.24.
Question 25 of 32 · Feeders with several motors
A feeder supplies three continuous-duty, 230 V, three-phase motors: 15 hp, 10 hp, and 5 hp. Using copper THWN with 75 °C terminals, what is the minimum feeder conductor size?
Supplied reference: Supplied three-phase 230 V table currents: 15 hp → 42 A; 10 hp → 28 A; 5 hp → 15.2 A. Copper ampacities at the specified 75 °C terminations: 4 AWG → 85 A; 3 AWG → 100 A; 2 AWG → 115 A; 1 AWG → 130 A. All motors operate simultaneously. Minimum feeder-conductor ampacity = all table currents + 25% of one largest table current. No other loads, correction, adjustment or exception applies. Choose the smallest listed conductor meeting the result.
- A. 4 AWG
- B. 2 AWG
- C. 3 AWG
- D. 1 AWG
Show answer and explanation
Answer: C. 3 AWG
42 × 1.25 + 28 + 15.2 = 95.7 A. In the 75 °C column, 4 AWG is 85 A, which is too small, and 3 AWG is 100 A, which is enough. 2 AWG and 1 AWG work but aren't the minimum.
Ampacity cells and adjustment factor: 2023-labeled ampacity extract, pp. 1–2. Motor-conductor method: Minnesota DLI, pp. 18–19.
Related NEC study lookup: Table 430.250; 430.24; Table 310.16.
Question 26 of 32 · Feeders with several motors
A feeder supplies only three 230 V, three-phase squirrel-cage motors: 10 hp, 7½ hp, and 5 hp. Each branch circuit is protected by an inverse-time breaker. What is the largest listed inverse-time breaker rating allowed by the supplied rules for the feeder?
Supplied reference: Table currents: 10 hp → 28 A; 7½ hp → 22 A; 5 hp → 15.2 A. Inverse-time branch factor = 250%. Listed branch and feeder ratings: 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150 A. For these fixed motor-only loads, assume feeder conductors meet the motor sum-plus-25% rule. Calculate each branch maximum using its stated factor and the next listed rating only when needed. The feeder ceiling is the largest branch maximum plus the table currents of the other motors. Select the largest listed feeder rating not exceeding that ceiling; no feeder next-size-up permission applies.
- A. 107.2 A
- B. 110 A
- C. 100 A
- D. 125 A
Show answer and explanation
Answer: C. 100 A
Start with the largest permitted branch-device rating in this case: 28 A × 250% = 70 A, which is already a listed rating. Add the other motors' FLCs: 70 + 22 + 15.2 = 107.2 A. The supplied feeder rule does not allow a rating greater than that, so select the largest listed rating below the ceiling: 100 A. 110 A is the branch-circuit habit of rounding up. 107.2 A is the raw ceiling, not a listed rating; 125 A exceeds it.
The numerical result follows from the current values, factors and rating-selection rule supplied above. Use the related study lookup to locate this topic in your codebook.
Related NEC study lookup: 430.62(A); Table 430.52(C)(1); 240.6(A).
Question 27 of 32 · Feeders with several motors
A feeder supplies only three 460 V, three-phase squirrel-cage motors: 30 hp, 15 hp, and 10 hp. Each branch circuit is protected by an inverse-time breaker. What is the largest listed inverse-time breaker rating allowed by the supplied rules for the feeder?
Supplied reference: Table currents: 30 hp → 40 A; 15 hp → 21 A; 10 hp → 14 A. Inverse-time branch factor = 250%. Listed branch and feeder ratings: 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150 A. For these fixed motor-only loads, assume feeder conductors meet the motor sum-plus-25% rule. Calculate each branch maximum using its stated factor and the next listed rating only when needed. The feeder ceiling is the largest branch maximum plus the table currents of the other motors. Select the largest listed feeder rating not exceeding that ceiling; no feeder next-size-up permission applies.
- A. 110 A
- B. 135 A
- C. 125 A
- D. 150 A
Show answer and explanation
Answer: C. 125 A
The largest permitted branch-device rating in this case is 40 A × 250% = 100 A, a listed rating. Adding 21 + 14 gives 135 A. That isn't a listed rating, and the feeder rule is a ceiling, so choose 125 A. 150 A rounds the wrong way. 110 A is below the permitted maximum; this exercise does not establish every other requirement for a smaller device. 135 A is the raw ceiling, not a listed rating.
The numerical result follows from the current values, factors and rating-selection rule supplied above. Use the related study lookup to locate this topic in your codebook.
Related NEC study lookup: 430.62(A); 240.6(A).
Question 28 of 32 · Feeders with several motors
A feeder supplies only three 460 V, three-phase squirrel-cage motors: 30 hp, 25 hp, and 10 hp. Each branch circuit is protected by dual-element time-delay fuses. What is the largest listed time-delay fuse rating allowed by the supplied rules for the feeder?
Supplied reference: Table currents: 30 hp → 40 A; 25 hp → 34 A; 10 hp → 14 A. Dual-element time-delay branch factor = 175%. Listed branch and feeder ratings: 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150 A. For these fixed motor-only loads, assume feeder conductors meet the motor sum-plus-25% rule. Calculate each branch maximum using its stated factor and the next listed rating only when needed. The feeder ceiling is the largest branch maximum plus the table currents of the other motors. Select the largest listed feeder rating not exceeding that ceiling; no feeder next-size-up permission applies.
- A. 100 A
- B. 110 A
- C. 118 A
- D. 125 A
Show answer and explanation
Answer: B. 110 A
The largest permitted branch-device rating in this case is 40 × 175% = 70 A, a listed rating. 70 + 34 + 14 = 118 A, so round down to 110 A. 125 A rounds up, which the feeder rule doesn't allow. 118 A isn't a listed rating. 100 A is below the largest listed rating that stays within the ceiling.
The numerical result follows from the current values, factors and rating-selection rule supplied above. Use the related study lookup to locate this topic in your codebook.
Related NEC study lookup: 430.62(A); Table 430.52(C)(1); 240.6(A).
Question 29 of 32 · Disconnecting means
Under the supplied 115% rule, what is the calculated minimum ampere-rating threshold for the disconnecting means for a 40 hp, 460 V, three-phase motor?
Supplied reference: Table FLC = 52 A. For this exercise, calculate only the minimum disconnect ampere-rating threshold at 115% of table FLC. No special equipment or alternate disconnect rule is included.
- A. 59.8 A
- B. 52 A
- C. 65 A
- D. 130 A
Show answer and explanation
Answer: A. 59.8 A
The supplied disconnect rule requires at least 115% of table FLC: 52 × 1.15 = 59.8 A. 65 A is 52 × 1.25, the conductor habit. 130 A is 52 × 250%. 52 A skips the percentage. This is an ampere-rating threshold, not a complete disconnect selection or a horsepower-rating check.
The complete rule is supplied above. Historical context only: manufacturer column explanations, p. 145; this older document is not a current-code certification.
Related NEC study lookup: Table 430.250; 430.110(A).
Question 30 of 32 · Three-part motor circuit case
Use this motor for Questions 30–32: a 20 hp, 208 V, three-phase squirrel-cage motor with a nameplate current of 56 A and a marked service factor of 1.15. It is supplied by copper THWN conductors with terminations rated 75 °C and protected by an inverse-time circuit breaker. A separate overload device is used. What is the minimum branch-circuit conductor size?
Supplied reference: Continuous-duty motor; table FLC = 59.4 A. Supplied copper ampacities at 75 °C: 6 AWG → 65 A; 4 AWG → 85 A; 3 AWG → 100 A; 2 AWG → 115 A. Use copper at 30 °C, no more than three current-carrying conductors, and no other correction or adjustment. Choose the smallest listed size whose supplied ampacity meets 125% of table FLC.
- A. 4 AWG
- B. 6 AWG
- C. 3 AWG
- D. 2 AWG
Show answer and explanation
Answer: A. 4 AWG
The table FLC at 208 V is 59.4 A, and 59.4 × 1.25 = 74.25 A. 6 AWG (65 A) is too small, and 4 AWG (85 A) works. Using the nameplate (56 × 1.25 = 70 A) also lands on 4 AWG here, but only because the numbers happen to line up. 3 AWG and 2 AWG also meet the supplied ampacity requirement but are larger than necessary. Record 74.25 A before choosing the conductor to catch the wrong-current method.
Ampacity cells and adjustment factor: 2023-labeled ampacity extract, pp. 1–2. Motor-conductor method: Minnesota DLI, pp. 18–19.
Related NEC study lookup: Table 430.250; 430.22; Table 310.16.
Question 31 of 32 · Three-part motor circuit case
Same motor as Question 30: 20 hp, 208 V, three-phase, with 56 A nameplate current and a 1.15 service factor. What is the calculated overload trip-current ceiling under the supplied basic rule?
Supplied reference: For this continuous-duty, over-1-hp motor with a separate overload device, use 125% of nameplate current when marked service factor is at least 1.15 OR marked temperature rise is 40 °C or less; otherwise use 115%. No increased-setting exception or rounding above the ceiling applies. This asks for a trip-current limit, not a relay-dial value. Supplied table FLC for comparison: 59.4 A.
- A. 64.4 A
- B. 70 A
- C. 74.25 A
- D. 150 A
Show answer and explanation
Answer: B. 70 A
The nameplate is 56 A and the supplied rule selects 125% for its 1.15 service factor: 56 × 1.25 = 70 A. 74.25 A uses the table FLC. 64.4 A uses 115%. 150 A is the separate branch-breaker answer in Question 32.
The factor and qualification rule are supplied in this exercise. Nameplate-current basis: Minnesota DLI, p. 18. Why a trip-current limit is not automatically a dial setting: manufacturer instructions, pp. 9–10.
Related NEC study lookup: 430.6(A)(2); 430.32(A)(1).
Question 32 of 32 · Three-part motor circuit case
Same motor as Question 30: 20 hp, 208 V, three-phase, with 56 A nameplate current and a 1.15 service factor. What is the largest listed inverse-time circuit-breaker rating allowed by the supplied rules?
Supplied reference: Table FLC = 59.4 A; inverse-time branch factor = 250%. Listed ratings: 100, 110, 125, 150, 175, 200 A. Use the stated table FLC and device factor. Select the raw result if it is listed; otherwise the next higher listed rating is permitted. Separate overload protection is assumed, with no increased-starting allowance or lower equipment-specific limit in this exercise.
- A. 125 A
- B. 150 A
- C. 140 A
- D. 175 A
Show answer and explanation
Answer: B. 150 A
59.4 × 250% = 148.5 A, so the next listed rating is 150 A. 140 A is the 56 A nameplate × 250%, the wrong base. 175 A skips a size. 125 A is below the calculated value.
The numerical result follows from the current values, factors and rating-selection rule supplied above. Use the related study lookup to locate this topic in your codebook.
Related NEC study lookup: Table 430.52(C)(1); 430.52(C)(1)(a); 240.6(A).
Review by topic
Your number correct is feedback on these 32 questions. It isn't an official score or a prediction of how you'll do on exam day. Where it helps is showing which step keeps tripping you up. Count only answers you chose before revealing the explanation; leave unattempted questions separate.
| If you missed… | Questions | Check this step | Related study lookup |
|---|---|---|---|
| Which current to start from | 1–4 | Nameplate versus table current, then the right phase and voltage | 430.6(A); Tables 430.248 and 430.250 |
| Branch-circuit conductors | 5–11 | The 125% factor, the termination column, and any adjustment | 430.22; 110.14(C); Table 310.16; Table 310.15(C)(1) |
| Overload protection | 12–16 | Nameplate current, the qualifying marking, and a ceiling rather than a dial setting | 430.32(A)(1) |
| Breaker and fuse sizing | 17–22 | Device factor and the stated rating-selection direction | 430.52; 240.6(A) |
| Feeders | 23–28 | Apply 125% to only one largest motor for conductors; identify the controlling branch device for feeder protection | 430.24; 430.62(A) |
| Disconnecting means | 29 | The supplied 115% ampere threshold, not a complete switch selection | 430.110(A) |
| Three-part motor circuit case | 30–32 | Keep the conductor, overload and breaker calculations separate | All of the above |
For a missed item, write the starting current, factor, raw result and selection rule before trying it again. A repeat attempt is useful practice, not a fresh measurement of exam readiness.
Reference values used in these questions
These are the lookup values used by the questions, their distractors, and the worked example below. They're here so you can work the set without a codebook. An em dash means that this set does not use that cell; it does not mean zero current.
Supplied three-phase motor full-load current
Related study table: 430.250. Values are amperes.
| Horsepower | 200 V | 208 V | 230 V | 460 V |
|---|---|---|---|---|
| 3 | — | — | 9.6 | — |
| 5 | — | — | 15.2 | — |
| 7½ | — | — | 22 | 11 |
| 10 | 32.2 | 30.8 | 28 | 14 |
| 15 | — | 46.2 | 42 | 21 |
| 20 | — | 59.4 | — | 27 |
| 25 | — | — | — | 34 |
| 30 | — | — | 80 | 40 |
| 40 | — | — | — | 52 |
| 50 | — | — | — | 65 |
Supplied single-phase motor full-load current
Related study table: 430.248. Values are amperes.
| Horsepower | 115 V | 208 V | 230 V |
|---|---|---|---|
| 3 | 34 | 18.7 | 17 |
| 5 | — | — | 28 |
The three-phase entries and the 115/230 V single-phase entries were cross-checked against the historical manufacturer motor tables, printed pp. 146, 148–149, 151 and 154–155. That document is dated 2005, not 2023 or 2026. The 18.7 A single-phase 208 V entry is supplied exercise data. These limited lookups are not complete or edition-certified motor tables.
Copper conductor ampacity
Table 310.16 copper entries, in amperes, at 30 °C ambient and before any stated adjustment for additional current-carrying conductors. Use the termination column specified by the question.
| Size | 60 °C | 75 °C | 90 °C |
|---|---|---|---|
| 12 AWG | 20 | 25 | 30 |
| 10 AWG | 30 | 35 | 40 |
| 8 AWG | 40 | 50 | 55 |
| 6 AWG | 55 | 65 | 75 |
| 4 AWG | 70 | 85 | 95 |
| 3 AWG | 85 | 100 | 115 |
| 2 AWG | 95 | 115 | 130 |
| 1 AWG | 110 | 130 | 145 |
Source: HELUKABEL’s extract labeled 2023 NEC, PDF p. 1, Table 310.16. The adjacent free-air table is a different table and is not used here.
Other supplied values
| Item | Value used in this set |
|---|---|
| Adjustment factor for the nine-current-carrying-conductor problem | 70%, from the 7–9 row |
| Listed device ratings used across the exercises | 15, 20, 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150, 175, 200 A; each question prints its relevant list |
| Ordinary branch-device factors supplied for these cases | Nontime-delay fuse 300% · dual-element time-delay fuse 175% · inverse-time breaker 250% |
| Increased-starting time-delay fuse ceiling supplied in Question 22 | 225%; no step above that ceiling |
The 70% adjustment row is in the 2023-labeled ampacity extract, PDF p. 2. The device lists and factors are the stated premises of the exercises, not a list of every available product or every Code permission.
Which current, which percentage, which way to round
Most motor questions come down to three decisions: which current you start from, which percentage applies, and whether the answer is a floor or a ceiling. This table puts all three side by side for the supplied-rule cases on this page.
| You're sizing | Start from | Multiply by or combine | How to select the answer | Related study lookup |
|---|---|---|---|---|
| Branch-circuit conductors, one motor | Table FLC | 125% | Pick a conductor whose usable ampacity is at least the result | 430.6(A)(1); 430.22 |
| Feeder conductors, several motors | Table FLC | 125% of one largest motor + 100% of the others | Conductor at least the result | 430.24 |
| Separate overload trip-current ceiling, basic rule | Nameplate current | 125% if service factor is 1.15+ or marked rise is 40 °C or less; otherwise 115% | The result is a ceiling, so don't round up; it is not automatically a physical dial setting | 430.6(A)(2); 430.32(A)(1) |
| Ordinary branch breaker or fuse | Table FLC | 250% inverse-time breaker · 175% dual-element time-delay fuse · 300% nontime-delay fuse | Keep a listed result; otherwise select the next higher listed rating only where the question permits it | 430.52; 240.6(A) |
| Increased-starting time-delay fuse, Question 22 | Table FLC | Up to 225% | Select the largest listed rating at or below the ceiling | 430.52 |
| Feeder breaker or fuse, Questions 26–28 | The largest permitted branch-device rating in that case, plus table FLC of the other motors | No additional blanket multiplier | Select the largest listed rating not greater than the ceiling | 430.62(A) |
| Disconnect ampere-rating threshold | Table FLC | 115% | At least the result; other disconnect-selection checks are outside this calculation | 430.110(A) |
The conductor and current-basis methods are illustrated in the Minnesota DLI guide, pp. 18–19. Other factors and rating-selection directions above are the explicit exercise rules; apply them only to the cases stated here.
Three things this table can't show on its own:
- The table current does almost all the work in these exercises. The nameplate supplies the overload-current basis. That is not a claim that nameplate current is irrelevant to every other type of motor or equipment calculation.
- “Maximum” doesn't mean “required.” The 250% inverse-time breaker factor is the initial calculation in the ordinary examples, followed by the stated rating-selection step. A smaller rating being below that maximum does not, by itself, establish that the device is suitable for an installation.
- Next size up is a permission, not a habit. In this set it doesn't apply when the result is already listed, to overload ceilings, to Question 22's increased-starting ceiling, or to the feeder-device ceilings.
How to work these motor questions in four steps
- Find the starting current. Match the motor type, phase, voltage column, and horsepower in the supplied lookup. For the overload questions, use the nameplate current instead.
- Name what you're sizing. Conductor, overload, branch device, feeder, or disconnect. That tells you which supplied rule to use.
- Calculate. Apply the supplied relationship; divide when solving the reverse-ampacity question. Keep the decimals until the last step.
- Select in the right direction. Conductor ampacity and disconnect thresholds meet a floor. Overload and feeder-device ratings stay under their stated ceilings. Only use a higher listed branch-device rating when the question permits that step.
Worked example. A 30 hp, 230 V, three-phase squirrel-cage motor has a nameplate current of 76 A and a 1.15 service factor. It uses copper THWN with 75 °C terminals and an inverse-time breaker.
Use 80 A as its supplied table FLC and the ordinary rules in the map above. Assume continuous duty, 30 °C ambient, no more than three current-carrying conductors, separate overload protection, and no other correction, adjustment or special limit. The supplied device list includes 200 A.
| What you're sizing | Work | Answer |
|---|---|---|
| Table current | Supplied three-phase lookup, 30 hp at 230 V | 80 A |
| Branch conductors | 80 × 1.25 = 100 A → 75 °C column | 3 AWG (100 A) |
| Overload trip-current ceiling (nameplate) | 76 × 1.25 = 95 A | 95 A maximum under the supplied basic rule |
| Inverse-time breaker | 80 × 2.50 = 200 A, already listed | 200 A under the supplied ordinary branch rule |
| Disconnect ampere threshold | 80 × 1.15 = 92 A | At least 92 A |
Notice that the same motor produces five different numbers. That's normal, because each one answers a different question.
Common questions
Why does the conductor sometimes look too small for its breaker? In these ordinary motor-circuit exercises, the overload calculation and the branch short-circuit/ground-fault calculation are separate. The conductor-ampacity answer is not automatically the branch breaker's rating. The Minnesota DLI guide, pp. 18–19, distinguishes these purposes. A complete installation still needs its applicable conductor-protection and equipment checks; the question's arithmetic is not an approval of a complete circuit.
Should I put the calculated overload limit directly on a relay dial? Not automatically. A relay can already incorporate a trip multiplier. For example, the E100 manufacturer's manual, pp. 9–10, distinguishes the dial setting from a trip rating of 120% of that setting. That illustrates why Questions 12–16 and 31 ask for a calculated trip-current ceiling, not instructions to adjust equipment. Actual settings depend on the device's instructions.
My exam bulletin says to ignore exceptions unless the question says otherwise. Does next-size-up still count? This set explicitly states when a higher listed rating is permitted and when a ceiling cannot be exceeded. For your examination, read your own bulletin's wording and the provision in the edition it specifies; do not assume every “next-size-up” instruction has the same scope or placement across editions.
Do these exercises cover air-conditioning and refrigeration equipment? No. Hermetic refrigerant motor-compressors and their Article 440 calculations are outside this set. So are torque, multispeed, wound-rotor and fire-pump motors, drive-system selection, special motor-design provisions, and complete installation design. Do not carry these supplied factors into an equipment case the question does not cover.
Which NEC edition should you use?
Use the edition and permitted references named in your current candidate bulletin. These are guided, supplied-data exercises, not a full-length licensing examination or a fully validated 2023 or 2026 NEC question bank. The related NEC lookups are study-navigation aids; they do not establish that every exception or edition change has been covered.
The copper ampacity extract used here is explicitly labeled 2023 NEC. The source notes distinguish that extract from the historical motor-current reference and the government teaching examples. Do not transfer this answer key automatically to another edition or a differently worded motor problem. NFPA's 2023 code-reader route identifies the reference used for the navigation labels.
Master candidates looking for broader study can use the master electrician exam prep guide. To finish this set, return to the review table, choose a missed question, and explain the starting current and selection direction before looking at its answer again.
Sources
- Minnesota Department of Labor and Industry, Electrical license examination guide. Printed p. 16 illustrates conductor adjustment; pp. 18–19 explain motor-current basis, conductor ampacity, motor-only feeder ampacity and a time-delay fuse example. These teaching examples do not establish a national exam blueprint.
- HELUKABEL, Allowable Ampacity Tables. Extract labeled 2023 NEC; PDF pp. 1–2 provide the conductor cells and adjustment factor used here.
- Cooper Bussmann, Motor Circuit Protection Tables, hosted by Eaton. 2005 historical technical reference, printed pp. 145–155. Used for limited cross-checks, not as evidence that these exercises match every current Code requirement.
- Rockwell Automation, E100 Electronic Overload Relay User Manual. Publication 193-UM013C-EN-P, November 2022, printed pp. 9–10. Supports the distinction between a device's dial value and trip rating, not a universal relay setting.
- NFPA 70, 2023 official reference route. Use the applicable edition for complete provisions and exceptions; the related section labels on this page are navigation aids.
Castleport Test Prep Editorial Team · Last checked: September 29, 2026. The checks cover the cited source passages and table images, the supplied reference data as scoped above, and the arithmetic for all 32 answers. This is not a claim of full-edition or licensed-professional review. Developed with AI assistance under our methodology and editorial standards. Spot an error? Tell us.
Castleport Test Prep is an independent exam prep publisher. We're not affiliated with, endorsed by, or approved by the National Fire Protection Association, any state licensing board, or any testing company. National Electrical Code, NEC, and NFPA 70 identify the code published by NFPA; trademarks belong to their respective owners. These are original, unofficial practice questions, not real exam items. They don't guarantee an exam result or a license. Independence policy.