motors 2-6 motor branch circuits

Using NEC Part XIV of Article 430, match the following motor data to the correct full-load current.

1/2 hp, 208 V, single-phase → 5.4 A 5 hp, 230 V, single-phase → 28 A 5 hp, 460 V, 3-phase → 7.6 A 15 hp, 200 V, 3-phase → 48.3 A

Using NEC Part XIV of Article 430, match the following induction motor data to the correct full-load current.

2 hp, 115 V, 3-phase → 13.6 A 7-1/2 hp, 480 V, 3-phase → 11 A 10 hp, 200 V, single-phase → 57.5 A 60 hp, 575 V, 3-phase → 62 A

Match the motor conductor ampacity to the correct THWN (75°C) copper wire size using Table 310.16.HINT: Section 240.4(D) does not apply to motor circuit conductors.

21 A → 12 AWG 50 A → 8 AWG 155 A → 2/0 AWG 264 A → 300 kcmil

Match the motor conductor ampacity to the correct THWN (75°C) copper wire size using Table 310.16.HINT: Section 240.4(D) does not apply to motor circuit conductors.

32 A → 10 AWG 64 A → 6 AWG 69 A → 4 AWG 216 A → 4/0 AWG

A single 10-horsepower, 240-volt, single-phase compressor motor with a rated load current of 50 amperes and a branch-circuit selection current (BCSC) of 54.5 amperes is supplied by a refrigeration equipment branch circuit. Calculate the minimum ampacity for the branch-circuit conductors.

440.32, Use branch-circuit selection currentBranch circuit ampacity = BCSC × 125% = 54.5 × 1.25 = 68.13 A

A 25-horsepower, 3-phase compressor motor with a full-load rating of 42 amperes and a branch-circuit selection current of 45 amperes is installed on an individual 3-phase, 460-volt branch circuit. Calculate the minimum ampacity for the single refrigeration branch-circuit conductors.

440.32, Use branch-circuit selection currentBranch circuit ampacity = BCSC × 125%= 45 × 1.25= 56.25 A The correct answer is: 56.25 A

One branch circuit supplies two 15-horsepower, 240-volt, 3-phase compressor motors. Both motors can operate at the same time. Each motor has a full-load current of 43 amperes; and a branch-circuit selection current of 46 amperes. Calculate the minimum ampacity for the single refrigeration branch-circuit conductors.

440.33, Use branch-circuit selection currentBranch circuit ampacity = (largest BCSC × 125%) + other load(s)= (46 × 1.25) + 46= 57.5 + 46= 103.5 A The correct answer is: 103.5 A

Match the conductor ampacity to the correct THWN (75°C) copper wire size using Table 310.16.

54 A → 6 AWG 68 A → 4 AWG 76 A → 4 AWG 104 A → 2 AWG

Adjust the conductor ampacity, using Tables 310.15(B)(1) and 310.15(C)(1) as appropriate, and then match to the correct THWN (75°C) copper wire size using Table 310.16.

56 A at 30°C → 6 AWG,93.5 A at elevated ambient of 40°C → 2 AWG 93.5 A at elevated ambient of 40°C with all six current-carrying conductors in one raceway → 1/0 AWG 96.25 A at elevated ambient of 108°F → 1 AWG

Match the motor conductor ampacity to the correct THWN (75°C) copper wire size using Table 310.16.HINT: Section 240.4(D) does not apply to motor circuit conductors.

90 A → 3 AWG 103 A → 2 AWG 120 A → 1 AWG 211 A → 4/0 AWG

A 7 1/2-horsepower, 220-volt, single-phase compressor motor with a full-load current rating of 38 amperes and a branch-circuit selection current of 42.5 amperes is installed with 5,000 watts of noncontinuous supplementary heat on a single-branch circuit. Calculate the minimum ampacity for the single refrigeration/heater branch-circuit conductors.

I = W ÷ E for the heating load= 5,000 ÷ 220= 22.73 A440.33 and 440.34, Use branch-circuit selection currentBranch circuit ampacity = (BCSC × 125%) + other load(s)= (42.5 × 1.25) + 22.73= 53.13 + 22.73= 75.86 A The correct answer is: 75.86 A

A 10-horsepower, 208-volt, single-phase, continuous-duty, induction type Design B motor is supplied by a motor branch circuit. Calculate the minimum ampacity for the motor branch-circuit conductors.

Table 430.248, single-phase, 10 hp 208 V = 55 A FLCBranch circuit ampacity = FLC × 125%= 55 × 1.25= 68.75 A The correct answer is: 68.75 A

A single 3-horsepower, 240-volt, single-phase, continuous-duty, induction type Design B motor is supplied by a motor branch circuit. Calculate the minimum ampacity for the motor branch-circuit conductors.

Table 430.248, single-phase, 3 hp, 240 V = 17 A FLC430.22(A) branch circuit ampacity = FLC × 125% = 17 × 1.25 = 21.25 A

Three single-phase motors, rated 1/2 horsepower, 3/4 horsepower, and 2 horsepower, are to be installed on the same 120-volt branch circuit. Calculate the minimum ampacity of the branch-circuit conductors.

Table 430.248, single-phase1/2 hp 120 V = 9.8 A FLC3/4 hp 120 V = 13.8 A FLC2 hp, 120 V = 24 A FLC430.24, branch circuit ampacity = (largest FLC × 125%) + other motors= (24 × 1.25) + 13.8 + 9.8= 30 + 13.8 + 9.8= 53.6 A The correct answer is: 53.6 A

One 2-horsepower, 120-volt, single-phase continuous-duty, induction type Design B motor and an additional noncontinuous load of 7,200 watts are supplied by a single branch circuit. Calculate the minimum branch-circuit ampacity to supply the motor load and the additional noncontinuous load.

Table 430.248, single-phase2 hp, 120 V, single-phase = 24 A FLCI = W ÷ E= 7,200 ÷ 120= 60 A430.24, branch circuit ampacity = (largest motor FLC × 125%) + other load(s)= (24 × 1.25) + 60= 30 + 60= 90 A The correct answer is: 90 A

One 3-horsepower, 240-volt, single-phase, continuous-duty, induction-type Design B motor; one 71/2-horsepower, 240-volt, single-phase, continuous-duty, induction-type Design B motor; and one 10-horsepower, 240-volt, single-phase, continuous-duty, induction-type Design B motor are supplied by a single motor branch circuit. Calculate the minimum ampacity for the single motor branch-circuit conductors supplying all three motors.

Table 430.248, single-phase3 hp, 240 V = 17 A FLC7 1/2 hp, 240 V = 40 A FLC10 hp, 240 V = 50 A FLC430.24, branch circuit ampacity = (largest FLC × 125%) + other motor(s)= (50 × 1.25) + 40 + 17= 62.5 + 40 + 17= 119.5 A The correct answer is: 119.5 A

One branch circuit supplies a piece of fixed electric space heating equipment (motor-operated) with one 5-horsepower, 230-volt, single-phase motor and one unit of electric heat rated at 12,500 watts at 230 volts. Calculate the minimum branch-circuit ampacity to supply the combined load.

Table 430.248, single-phase5 hp 230 V = 28 A FLCI = W ÷ E = 12,500 ÷ 230 = 54.35 A430.24, Exception 2 and 424.4(B)Branch circuit ampacity = (motor FLC + heat FLC) × 125% = (28 + 54.35) × 1.25 = 82.35 × 1.25 = 102.94 A

Two individual motor branch circuits are installed in individual conduits for two 25-horsepower, 3-phase, 208-volt, continuous duty motors using THWN copper conductors. Calculate the minimum ampacity for each motor branch-circuit conductor.

Table 430.250 3-phase25 hp 208 V = 74.8 A FLCBranch circuit ampacity = FLC × 125%= 74.8 × 1.25= 93.5 A The correct answer is: 93.5 A

A 230-volt, 3-phase, 25-horsepower squirrel-cage motor and a 230-volt, 30-horsepower synchronous motor are installed on the same motor branch circuit. What is the minimum ampacity of the conductors for this single branch circuit?

Table 430.250, 3-phase 25 hp squirrel-cage = 68 A FLC30 hp synchronous = 63 A FLC430.24, branch circuit ampacity = (largest FLC × 125%) + other(s)= (68 × 1.25) + 63= 85 + 63= 148 A The correct answer is: 148 A

Two 10-horsepower, 440-volt, 3-phase, continuous-duty, induction type Design B motors are supplied by a single motor branch circuit. Calculate the minimum ampacity for the single motor branch-circuit conductors supplying both motors.

Table 430.250, 3-phase, 10 hp, 440 V = 14 A FLC430.24, branch circuit ampacity = (largest FLC × 125%) + other motor(s)= (14 × 1.25) + 14= 17.5 + 14= 31.5 A The correct answer is: 31.5 A

A single 30-horsepower, 460-volt, 3-phase, continuous-duty, induction type Design B motor is supplied by a motor branch circuit. Calculate the minimum ampacity for the motor branch-circuit conductors.

Table 430.250, 3-phase, 30 hp 460 V = 40 A FLC430.22(A), branch circuit ampacity = FLC × 125% = 40 × 1.25 = 50 A

One 10-horsepower, 208-volt, 3-phase, continuous-duty, induction type Design B motor; one 25-horsepower, 208-volt, 3-phase, continuous-duty, induction type Design B motor; and one 30-horsepower, 208-volt, 3-phase, continuous-duty, induction type Design B motor are supplied by a single motor branch circuit. Calculate the minimum ampacity for the single motor branch-circuit conductors supplying all three motors.

Table 430.250, 3-phase10 hp, 208 V = 30.8 A FLC25 hp, 208 V = 74.8 A FLC30 hp, 208 V = 88 A FLC430.24, branch circuit ampacity = (largest FLC × 125%) + other motor(s)= (88 × 1.25) + 74.8 + 30.8= 110 + 74.8 + 30.8= 215.6 A The correct answer is: 215.6 A

One 15-horsepower, 440-volt, 3-phase, continuous-duty, induction type Design B motor and one 25-horsepower, 440-volt, 3-phase, continuous-duty, induction type Design B motor are supplied by a single motor branch circuit. Calculate the minimum ampacity for the single motor branch-circuit conductors supplying both motors.

Table 430.250, 3-phase15 hp, 440 V = 21 A FLC25 hp, 440 V = 34 A FLC430.24, branch circuit ampacity = (largest FLC × 125%) + other motor(s)= (34 × 1.25) + 21= 42.5 + 21= 63.5 A The correct answer is: 63.5 A

A 60-horsepower, 208-volt, 3-phase, continuous duty, squirrel-cage motor is supplied by a motor branch circuit. Calculate the minimum ampacity for the motor branch-circuit conductors.

Table 430.250, 3-phase60 hp, 208 V = 169 A FLC430.22, branch circuit ampacity = FLC × 125%= 169 × 1.25= 211.25 A The correct answer is: 211.25 A

A 60-horsepower, 3-phase, 440-volt, continuous duty motor is installed on an individual branch circuit. Calculate the minimum ampacity for the motor branch-circuit conductors.

Table 430.250, 3-phase60 hp, 440 V = 77 A FLC430.22, branch circuit ampacity = FLC × 125%= 77 × 1.25= 96.25 A The correct answer is: 96.25 A

A single 100-horsepower, 440-volt, 3-phase, continuous-duty, induction type Design B motor is supplied by a motor branch circuit. Calculate the minimum ampacity for the motor branch-circuit conductors.

table 430.250, 3-phase, 100 hp, 440 V = 124 A FLC 430.22(A), branch circuit ampacity = FLC × 125%= 124 × 1.25= 155 A The correct answer is: 155 A

A single 75-horsepower, 208-volt, 3-phase, continuous-duty, induction type Design B motor is supplied by a motor branch circuit. Calculate the minimum ampacity for the motor branch-circuit conductors.

table 430.250, 3phase 75 hp 208v 430.22(A), branch circuit ampacity = FLC × 125% = 211 × 1.25 = 263.75 A