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A 900 sq ft. single-family dwelling has the following loads: four kitchen appliances (2,500 VA each), range (15 kW), two small appliances, laundry, electric heating system (10,000 VA), HVAC (12,000 VA), and an electric vehicle charging station (6,000 VA). Calculate the minimum service size required for this dwelling.
The correct answer is
Step 1: Identify and sum up all loads
General lighting: 900 x 3 VA = 2,700 VA
Kitchen appliances: 4 x 2,500 VA = 10,000 VA
Range: 1 x 15,000 kW = 15,000 VA
Small appliances: 2 x 1500 VA = 3000 VA
Laundry = 1 x 1500 VA = 1500 VA
Electric heating system: 10,000 VA
HVAC: 12,000 VA
EV charging station: 6,000 VA
Step 2: Apply demand factors
General lighting, laundry, and small appliances:
First 3,000 VA at 100%, 3001 to 120,000 VA at 35% (NEC 220.45)
Total General lighting, laundry & small appliances = 2700 + 3000 + 1500
= 7,200 VA
Load with demand factors: 3000 + (7,200 - 3000) x 0.35 = 4,470 VA
Kitchen appliances:
More than four appliances means apply demand factor of 75% on total load (220.53)
Load with demand factors = 10,000 VA x 0.75 = 7,500 VA
Range
More than 8.75 kW apply demand factors based on Table 220.55 Column C and more than 12 kW apply demand factors based on Table 220.55 Note 1 where the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.
Base demand is 8 kW.
Additional demand: 5% x (15 / 12) = 15%
Load with demand factors is 8,000 x 1.15 = 9,200 VA
Electric heating system & HVAC
Non coincidental loads where they will not run at the same time so select the largest of the two (220.60)
Load with demand factors for both heat and AC: 12,000 VA
EV charging:
Either 7200 watts (volt-amperes) or the nameplate rating of the equipment, whichever is larger (220.57)
Since 6000 VA less than 7200 VA, use 7,200 VA
Step 3: Add up all of the calculated loads
General lighting, laundry, and small appliances: 4,470 VA
Kitchen appliances: 7,500 VA
Range: 9,200 VA
Electric heating system & HVAC: 12,000 VA
EV charging: 7,200 VA
Adjusted total: 40,370 VA
Step 3: Convert VA to amperes
Assuming 240V single-phase service: 40,370 VA / 240V = 168 A
Step 4: Determine minimum service size
According to NEC 240.6, the next standard size up is 175 A. Minimum service size is 175 A.
That's correct! Way to go
Step 1: Identify and sum up all loads
General lighting: 900 x 3 VA = 2,700 VA
Kitchen appliances: 4 x 2,500 VA = 10,000 VA
Range: 1 x 15,000 kW = 15,000 VA
Small appliances: 2 x 1500 VA = 3000 VA
Laundry = 1 x 1500 VA = 1500 VA
Electric heating system: 10,000 VA
HVAC: 12,000 VA
EV charging station: 6,000 VA
Step 2: Apply demand factors
General lighting, laundry, and small appliances:
First 3,000 VA at 100%, 3001 to 120,000 VA at 35% (NEC 220.45)
Total General lighting, laundry & small appliances = 2700 + 3000 + 1500
= 7,200 VA
Load with demand factors: 3000 + (7,200 - 3000) x 0.35 = 4,470 VA
Kitchen appliances:
More than four appliances means apply demand factor of 75% on total load (220.53)
Load with demand factors = 10,000 VA x 0.75 = 7,500 VA
Range
More than 8.75 kW apply demand factors based on Table 220.55 Column C and more than 12 kW apply demand factors based on Table 220.55 Note 1 where the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.
Base demand is 8 kW.
Additional demand: 5% x (15 / 12) = 15%
Load with demand factors is 8,000 x 1.15 = 9,200 VA
Electric heating system & HVAC
Non coincidental loads where they will not run at the same time so select the largest of the two (220.60)
Load with demand factors for both heat and AC: 12,000 VA
EV charging:
Either 7200 watts (volt-amperes) or the nameplate rating of the equipment, whichever is larger (220.57)
Since 6000 VA less than 7200 VA, use 7,200 VA
Step 3: Add up all of the calculated loads
General lighting, laundry, and small appliances: 4,470 VA
Kitchen appliances: 7,500 VA
Range: 9,200 VA
Electric heating system & HVAC: 12,000 VA
EV charging: 7,200 VA
Adjusted total: 40,370 VA
Step 3: Convert VA to amperes
Assuming 240V single-phase service: 40,370 VA / 240V = 168 A
Step 4: Determine minimum service size
According to NEC 240.6, the next standard size up is 175 A. Minimum service size is 175 A.
The circuit breaker protecting a motor-compressor with a rated-load current of 55A can have a maximum rating of ________ amps.
The correct answer is
NEC 2017: 440.22(A)
NEC 2020: 440.22(A)
NEC 2023: 440.22(A)
Step 1: Calculate 175% of the rated-load current: 55A x 1.75 = 96.25A.
Step 2: Round up to the next standard circuit breaker size: 100A.
Step 3: This 100A is the maximum permissible rating for the circuit breaker protecting this motor-compressor.
That's correct! Way to go
NEC 2017: 440.22(A)
NEC 2020: 440.22(A)
NEC 2023: 440.22(A)
Step 1: Calculate 175% of the rated-load current: 55A x 1.75 = 96.25A.
Step 2: Round up to the next standard circuit breaker size: 100A.
Step 3: This 100A is the maximum permissible rating for the circuit breaker protecting this motor-compressor.
A residence has a 3-ton central air conditioner with a nameplate rating of 3,500 VA, and an electric furnace rated at 3.6 kW. The air handler for the AC system draws 800 VA. Calculate the total VA that should be included in the service calculations.
The correct answer is
Step 1: Identify and convert all loads to VA
Air conditioner compressor: 3,500 VA
Air handler: 800 VA
Electric furnace: 3.6 kW = 3,600 VA (1 kW = 1000 VA)
Step 2: Calculate total cooling load
Total cooling load = Compressor VA + Air handler VA
Total cooling load = 3,500 VA + 800 VA = 4,300 VA
Step 3: Compare the loads
We should use the larger of the two loads since heating and cooling won't operate simultaneously.
Cooling load: 4,300 VA
Heating load: 3,600 VA
Step 4: Determine the final load
The electric furnace load (4,300 VA) is larger, so this is the value that should be included in the service calculations.
That's correct! Way to go
Step 1: Identify and convert all loads to VA
Air conditioner compressor: 3,500 VA
Air handler: 800 VA
Electric furnace: 3.6 kW = 3,600 VA (1 kW = 1000 VA)
Step 2: Calculate total cooling load
Total cooling load = Compressor VA + Air handler VA
Total cooling load = 3,500 VA + 800 VA = 4,300 VA
Step 3: Compare the loads
We should use the larger of the two loads since heating and cooling won't operate simultaneously.
Cooling load: 4,300 VA
Heating load: 3,600 VA
Step 4: Determine the final load
The electric furnace load (4,300 VA) is larger, so this is the value that should be included in the service calculations.
An installation consists of one 5-hp, one 3-hp, and two 1/2-hp motors, all rated 230 V, 3-phase. All motors are Design B motors. Determine the size of the safety switch required for this combination load.
The correct answer is
Step 1: Identify Full-Load and Locked-Rotor Currents
Using Table 430.250 for FLC and Table 430.251(B) for LRC:
5-hp motor:
FLC = 15.2A
LRC = 92A
3-hp motor:
FLC = 9.6A
LRC = 64A
Two 1/2-hp motors:
FLC = 2.2A each
LRC = 20A each
Step 2: Calculate Combined Load Currents
Calculate total full-load current (FLC):
Total FLC = 15.2A + 9.6A + (2 x 2.2A) = 29.2A
Calculate total locked-rotor current (LRC):
Total LRC = 92A + 64A + (2 x 20A) = 196A
Step 3: Determine Minimum Ampere Rating
Per 430.110:
Minimum ampere rating = FLC x 115%
Minimum ampere rating = 29.2A x 1.15 = 33.58A
Step 4: Determine Required Horsepower Rating
Using Table 430.251(B):
Combined LRC of 196A corresponds to a 15 hp motor
Table 430.250: A 15 hp, 230V motor has FLC of 42A
42A > 33.58A (calculated minimum ampere rating)
Therefore, 15 hp rating is sufficient
Step 5: Final Safety Switch Requirements
The safety switch must be:
Rated minimum 15 hp
Have ampere rating not less than 33.58A
Listed for use as motor disconnect
Rated for 230V, 3-phase service
That's correct! Way to go
Step 1: Identify Full-Load and Locked-Rotor Currents
Using Table 430.250 for FLC and Table 430.251(B) for LRC:
5-hp motor:
FLC = 15.2A
LRC = 92A
3-hp motor:
FLC = 9.6A
LRC = 64A
Two 1/2-hp motors:
FLC = 2.2A each
LRC = 20A each
Step 2: Calculate Combined Load Currents
Calculate total full-load current (FLC):
Total FLC = 15.2A + 9.6A + (2 x 2.2A) = 29.2A
Calculate total locked-rotor current (LRC):
Total LRC = 92A + 64A + (2 x 20A) = 196A
Step 3: Determine Minimum Ampere Rating
Per 430.110:
Minimum ampere rating = FLC x 115%
Minimum ampere rating = 29.2A x 1.15 = 33.58A
Step 4: Determine Required Horsepower Rating
Using Table 430.251(B):
Combined LRC of 196A corresponds to a 15 hp motor
Table 430.250: A 15 hp, 230V motor has FLC of 42A
42A > 33.58A (calculated minimum ampere rating)
Therefore, 15 hp rating is sufficient
Step 5: Final Safety Switch Requirements
The safety switch must be:
Rated minimum 15 hp
Have ampere rating not less than 33.58A
Listed for use as motor disconnect
Rated for 230V, 3-phase service
The minimum ampacity is __________ for a neutral conductor feeding a 600 A load containing no fluorescent lighting and 100 A of non-linear loads.
The correct answer is
NEC 2017: 220.61(B), (C)(2)
NEC 2020: 220.61(B), (C)(2)
NEC 2023: 220.61(B), (C)(2)
According to 220.61(B), A service or feeder supplying the following loads shall be permitted to have an additional demand factor of 70 percent applied to the amount in 220.61(B)(1) and a portion of the amount in 220.61(B)(2). It states that unbalanced load in excess of 200 Amperes. As a result, you must minus the non-linear loads and 200 As from the load and mutliple that by 0.7.
Minimum neutral conductor ampacity: 200 + 100 + (0.7 X 300) = 510 A
That's correct! Way to go
NEC 2017: 220.61(B), (C)(2)
NEC 2020: 220.61(B), (C)(2)
NEC 2023: 220.61(B), (C)(2)
According to 220.61(B), A service or feeder supplying the following loads shall be permitted to have an additional demand factor of 70 percent applied to the amount in 220.61(B)(1) and a portion of the amount in 220.61(B)(2). It states that unbalanced load in excess of 200 Amperes. As a result, you must minus the non-linear loads and 200 As from the load and mutliple that by 0.7.
Minimum neutral conductor ampacity: 200 + 100 + (0.7 X 300) = 510 A
An aluminum bonding conductor with a minimum size of __________ is required to interconnect a grounding electrode used for signal circuits to a grounding electrode used for lightning protection.
The correct answer is
NEC 2017: Table 250.66
NEC 2020: Table 250.66
NEC 2023: Table 250.66
That's correct! Way to go
NEC 2017: Table 250.66
NEC 2020: Table 250.66
NEC 2023: Table 250.66
Nine No. 10 AWG UF conductors have a minimum current carrying capacity of __________ when installed in a run of rigid conduit in an area with an ambient temperature of 45 degrees C.
The correct answer is
NEC 2017: Table 310.15(B)(3)(a)
NEC 2020: Table 310.15(C)(1)
NEC 2023: Table 310.15(C)(1)
Maximum conductor ampacity of No. 10 AWG UF - 30 A; 310.16.
Correction Factor for 9 conductor in the conduit - 0.7 (310.15(C)(1))
Correction Factor for 45 degrees C ambient temperature - 0.71 (310.15(B)(1))
Maximum conductor ampacity: 30 x 0.7 x 0.71 = 14.91 A
That's correct! Way to go
NEC 2017: Table 310.15(B)(3)(a)
NEC 2020: Table 310.15(C)(1)
NEC 2023: Table 310.15(C)(1)
Maximum conductor ampacity of No. 10 AWG UF - 30 A; 310.16.
Correction Factor for 9 conductor in the conduit - 0.7 (310.15(C)(1))
Correction Factor for 45 degrees C ambient temperature - 0.71 (310.15(B)(1))
Maximum conductor ampacity: 30 x 0.7 x 0.71 = 14.91 A
For fire alarm wiring in a combustible building, __________ cannot be used.
The correct answer is
NEC 2017: 760.49
NEC 2020: 760.49
NEC 2023: 760.49
That's correct! Way to go
NEC 2017: 760.49
NEC 2020: 760.49
NEC 2023: 760.49
The least likelihood of flammable gas being present at __________.
The correct answer is
NEC 2017: 505.5(C)
NEC 2020: 505.5(C)
NEC 2023: 505.5(C)
That's correct! Way to go
NEC 2017: 505.5(C)
NEC 2020: 505.5(C)
NEC 2023: 505.5(C)
Minimum of __________ overload devices are required for a 2 hp, 600 V, 1-phase, dual capacitor motor.
The correct answer is
NEC 2017: 430.37
NEC 2020: 430.37
NEC 2023: 430.37
That's correct! Way to go
NEC 2017: 430.37
NEC 2020: 430.37
NEC 2023: 430.37
In a dwelling unit, a receptacle located 24 in away from a hydro massage bathtub can be protected by __________
An isolating transformer
Class 'A' ground fault circuit interrupter
An independent bare ground extended back to the system ground
None of the above; cannot be installed in this location
The correct answer is
NEC 2017: 680.71
NEC 2020: 680.71
NEC 2023: 680.71
That's correct! Way to go
NEC 2017: 680.71
NEC 2020: 680.71
NEC 2023: 680.71
Determine the minimum cubic inch capacity for a square junction box accommodating three 16 AWG conductors, two 14 AWG conductors, three 14 AWG equipment grounding conductor, two internal cable clamps, a single-gang device yoke connected to 14 AWG conductors and a 16 AWG pigtail. The required capacity is ________ cubic inches.
The correct answer is
Step 1: Identify all items inside of the questions, so for this question:
3 x 16 AWG conductor
2 x 14 AWG conductor
3 x 14 AWG equipment grounding conductor
2 x internal cable clamps
1 x device yoke connected to 14 AWG
1 x 12 AWG pigtail
Step 2: Determine the count towards box fill for each item
3 x 16 AWG conductor -> 3 x 16 AWG
2 x 14 AWG conductor -> 2 x 14 AWG
3 x 14 AWG equipment grounding conductor -> 1 x 14 AWG (counts as one & not above 4 conductors)
2 x internal cable clamps -> 1 x 14 AWG (counts as one)
1 x device yoke connected to 14 AWG -> 2 x 14 AWG (counts as two)
1 x 12 AWG pigtail -> Not applicable as it originates and does not leave the box
Step 3: Sum up the different conductors and multiply by the volume in Table 314.16(B)(1)
3 x 1.75 in3 (16 AWG volume allowance) = 5.25 in3
(2 + 1 + 1 + 2) x 2.00 in3 (14 AWG volume allowance) = 12.00 in3
Step 4: Max box size calculation
5.25 + 12.00 = 17.25 in3
That's correct! Way to go
Step 1: Identify all items inside of the questions, so for this question:
3 x 16 AWG conductor
2 x 14 AWG conductor
3 x 14 AWG equipment grounding conductor
2 x internal cable clamps
1 x device yoke connected to 14 AWG
1 x 12 AWG pigtail
Step 2: Determine the count towards box fill for each item
3 x 16 AWG conductor -> 3 x 16 AWG
2 x 14 AWG conductor -> 2 x 14 AWG
3 x 14 AWG equipment grounding conductor -> 1 x 14 AWG (counts as one & not above 4 conductors)
2 x internal cable clamps -> 1 x 14 AWG (counts as one)
1 x device yoke connected to 14 AWG -> 2 x 14 AWG (counts as two)
1 x 12 AWG pigtail -> Not applicable as it originates and does not leave the box
Step 3: Sum up the different conductors and multiply by the volume in Table 314.16(B)(1)
3 x 1.75 in3 (16 AWG volume allowance) = 5.25 in3
(2 + 1 + 1 + 2) x 2.00 in3 (14 AWG volume allowance) = 12.00 in3
Step 4: Max box size calculation
5.25 + 12.00 = 17.25 in3
A residential kitchen is equipped with two ranges: one rated at 14 kW and another at 18 kW. Determine the demand load for these appliances on the service entrance conductors.
The correct answer is
Step 1: Identify the ranges and their ratings to determine which column of Table 220.55 to use
Range 1: 14 kW (Column C)
Range 2: 18 kW (Column C)
Column C for both ranges as they're above 8 & 3/4 kW
Step 2: Calculate the total connected load
Total connected load = 14 kW + 18 kW = 32 kW
Step 3: Apply Note 2 under Table 220.55
Note 2 states: "Over 12 kW through 27 kW ranges of unequal ratings. For ranges individually rated more than 12 kW but not more than 27 kW, an average value of rating shall be calculated by adding together the ratings of all ranges to obtain the total connected load (32 kW) and dividing by the number of ranges (2). Then the maximum demand in Column C shall be increased 5% for each kW or major fraction thereof by which this average value exceeds 12 kW."
Average rating = 32 kW / 2 = 16 kW
Step 4: Calculate the increase based on the average rating
Increase = (16 kW - 12 kW) x 5% = 4 x 5% = 20%
Step 5: Apply the increase to the base demand from Column C
Base demand for 2 appliances from Column C = 11 kW
Final demand load = 11 kW + (11 kW x 20%) = 11 kW + 2.2 kW = 13.2 kW
Therefore, the calculated demand load for the two ranges (14 kW and 18 kW) on the service entrance conductors is 13.2 kW.
That's correct! Way to go
Step 1: Identify the ranges and their ratings to determine which column of Table 220.55 to use
Range 1: 14 kW (Column C)
Range 2: 18 kW (Column C)
Column C for both ranges as they're above 8 & 3/4 kW
Step 2: Calculate the total connected load
Total connected load = 14 kW + 18 kW = 32 kW
Step 3: Apply Note 2 under Table 220.55
Note 2 states: "Over 12 kW through 27 kW ranges of unequal ratings. For ranges individually rated more than 12 kW but not more than 27 kW, an average value of rating shall be calculated by adding together the ratings of all ranges to obtain the total connected load (32 kW) and dividing by the number of ranges (2). Then the maximum demand in Column C shall be increased 5% for each kW or major fraction thereof by which this average value exceeds 12 kW."
Average rating = 32 kW / 2 = 16 kW
Step 4: Calculate the increase based on the average rating
Increase = (16 kW - 12 kW) x 5% = 4 x 5% = 20%
Step 5: Apply the increase to the base demand from Column C
Base demand for 2 appliances from Column C = 11 kW
Final demand load = 11 kW + (11 kW x 20%) = 11 kW + 2.2 kW = 13.2 kW
Therefore, the calculated demand load for the two ranges (14 kW and 18 kW) on the service entrance conductors is 13.2 kW.
The largest solid aluminum conductor is __________ that can be drawn into a raceway.
The correct answer is
NEC 2017: 310.106(A)
NEC 2020: 310.3(C)
NEC 2023: 310.3(C)
That's correct! Way to go
NEC 2017: 310.106(A)
NEC 2020: 310.3(C)
NEC 2023: 310.3(C)
The flexible metal conduit contains the THWN Nylon motor copper supply conductors for a 30 hp, 600 V, 3-phase, S.F. 1.1 synchronous motor-generator and the motor- generator's is protected by time-delay fuses. A copper equipment grounding conductor with a minimum size of __________ is required in the conduit.
The correct answer is
NEC 2017: Table 250.122, Table 430.250, Table 430.52(C)(1)
NEC 2020: Table 250.122, Table 430.250, Table 430.52(C)(1)
NEC 2023: Table 250.122, Table 430.250, Table 430.52(C)(1)
Step 1: Find the motor FLA
30 hp, 600 V synchoronous is rated at 26 A
Step 2: Find the overcurrent value of the motor for time delay fuse
26 A x 175% = 45.5 A
Step 3: Find the equipment grounding conductor size
45.5 A is below 60 A and furthermore it should be 10 AWG
That's correct! Way to go
NEC 2017: Table 250.122, Table 430.250, Table 430.52(C)(1)
NEC 2020: Table 250.122, Table 430.250, Table 430.52(C)(1)
NEC 2023: Table 250.122, Table 430.250, Table 430.52(C)(1)
Step 1: Find the motor FLA
30 hp, 600 V synchoronous is rated at 26 A
Step 2: Find the overcurrent value of the motor for time delay fuse
26 A x 175% = 45.5 A
Step 3: Find the equipment grounding conductor size
45.5 A is below 60 A and furthermore it should be 10 AWG
In a 3-phase transformer bank of three 75 kVA, 4160 V / 600 V transformers protected by a time-limit circuit breaker (terminating at 75 C), the minimum size of XHHW copper conductor that can be used in flexible metal conduit on the primary side, assuming continuous use, is __________.
The correct answer is
NEC 2017: Table 310.15(B)(16), Section 210.19(A)(1)
NEC 2020: Table 310.16, Section 210.19(A)(1)
NEC 2023: Table 310.16, Section 210.19(A)(1)
Step 1: Calculate the full load current on the primary side
Total kVA = 3 x 75 kVA = 225 kVA
Primary voltage = 4160 V
I = kVA / (1.73 x V) = 225 / (1.73 x 4160) = 31.22 A
Step 2: Determine the minimum conductor ampacity
Assuming continuous load:
Minimum ampacity = 31.22 A x 1.25 = 39.03 A
Step 3: Select the conductor size
Based on NEC Table 310.16 for XHHW copper conductors, the next standard size up that can handle at least 39.03 A would be 8 AWG (50 A at 75C).
That's correct! Way to go
NEC 2017: Table 310.15(B)(16), Section 210.19(A)(1)
NEC 2020: Table 310.16, Section 210.19(A)(1)
NEC 2023: Table 310.16, Section 210.19(A)(1)
Step 1: Calculate the full load current on the primary side
Total kVA = 3 x 75 kVA = 225 kVA
Primary voltage = 4160 V
I = kVA / (1.73 x V) = 225 / (1.73 x 4160) = 31.22 A
Step 2: Determine the minimum conductor ampacity
Assuming continuous load:
Minimum ampacity = 31.22 A x 1.25 = 39.03 A
Step 3: Select the conductor size
Based on NEC Table 310.16 for XHHW copper conductors, the next standard size up that can handle at least 39.03 A would be 8 AWG (50 A at 75C).
Class A GFCI interrupts the load at a predetermined value of _________ mA.
The correct answer is
NEC 2017: Article 100 - Definitions
NEC 2020: Article 100 - Definitions
NEC 2023: Article 100 - Definitions
That's correct! Way to go
NEC 2017: Article 100 - Definitions
NEC 2020: Article 100 - Definitions
NEC 2023: Article 100 - Definitions
What is the maximum primary overcurrent rating allowed without requiring separate secondary overcurrent protection for a 150 kVA, dry-type, 600 V-208 V, 3-phase, delta delta 3-wire transformer?
The correct answer is
NEC 2017: 240.21(C)(1)
NEC 2020: 240.21(C)(1)
NEC 2023: 240.21(C)(1)
Based on 240.21(C)(1), we do not need secondary protection if the primary protection does not exceed the value determined by multiplying the secondary conductor ampacity by the secondary-to-primary transformer voltage ratio.
Step 1: Calculate the secondary full-load current
3-Phase Current Formula:
I = 150 x 1,000 / (208 x 1.732) = 150,000 VA / (208 x 1.732) = 150,000 / 360.25 = 416.67 amps
Step 2: Calculate the secondary-to-primary voltage ratio
208V / 600V = 0.347
Step 3: Calculate the equivalent primary current
416.67A x 0.347 = 144.58 A
Step 4: Round down to the nearest standard overcurrent device size
The nearest standard size below 144.58A is 125A
That's correct! Way to go
NEC 2017: 240.21(C)(1)
NEC 2020: 240.21(C)(1)
NEC 2023: 240.21(C)(1)
Based on 240.21(C)(1), we do not need secondary protection if the primary protection does not exceed the value determined by multiplying the secondary conductor ampacity by the secondary-to-primary transformer voltage ratio.
Step 1: Calculate the secondary full-load current
3-Phase Current Formula:
I = 150 x 1,000 / (208 x 1.732) = 150,000 VA / (208 x 1.732) = 150,000 / 360.25 = 416.67 amps
Step 2: Calculate the secondary-to-primary voltage ratio
208V / 600V = 0.347
Step 3: Calculate the equivalent primary current
416.67A x 0.347 = 144.58 A
Step 4: Round down to the nearest standard overcurrent device size
The nearest standard size below 144.58A is 125A
A 90 degree bend in a run of 53 EMT used for a feeder conductor consisting of four 3/0 XHHW with 1000 V insulation has a minimum radius of __________.
The correct answer is
NEC 2017: Section 358.24, Table 2
NEC 2020: Section 358.24, Table 2
NEC 2023: Section 358.24, Table 2
That's correct! Way to go
NEC 2017: Section 358.24, Table 2
NEC 2020: Section 358.24, Table 2
NEC 2023: Section 358.24, Table 2
In spaces where a failure of equipment or systems could lead to serious injury or death for patients, a copper bonding conductor with a minimum size of __________ is needed to connect receptacles to the patient equipment grounding point.
The correct answer is
NEC 2017: 517.13(B)
NEC 2020: 517.13(B)
NEC 2023: 517.13(B)
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NEC 2017: 517.13(B)
NEC 2020: 517.13(B)
NEC 2023: 517.13(B)
A single-phase, 240-volt circuit supplies a noncontinuous load of 30 amperes. The load is located 300 feet from the source. Using copper THWN conductors and assuming K = 12.9, the smallest wire size that will keep the voltage drop at or below 3.5% is ________.
The correct answer is
Step 1: Calculate the maximum allowable voltage drop. Voltage drop = 3.5% of 240 V = 0.035 x 240 V = 8.4 V.
Step 2: Plug in the values and calculate. CM = (2 x K x I x L) / Vd. CM = (2 x 12.9 x 30 x 300) / 8.4. CM = 27,643
Step 3: Round up to the next available wire size using Table 8. The next standard size above 27,643 circular mils for copper conductors is 4 AWG, which has 41,740 circular mils.
That's correct! Way to go
Step 1: Calculate the maximum allowable voltage drop. Voltage drop = 3.5% of 240 V = 0.035 x 240 V = 8.4 V.
Step 2: Plug in the values and calculate. CM = (2 x K x I x L) / Vd. CM = (2 x 12.9 x 30 x 300) / 8.4. CM = 27,643
Step 3: Round up to the next available wire size using Table 8. The next standard size above 27,643 circular mils for copper conductors is 4 AWG, which has 41,740 circular mils.
The minimum ampacity for insulated conductor from small wind turbines shall not be less than __________.
The sum of all the name plate ratings of the turbines connected to the system
Not less than the rating of the overcurrent device
125% of nameplate current
All of the above
The correct answer is
NEC 2017: 694.12(B)
NEC 2020: 694.12(B)
NEC 2023: 694.12(B)
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NEC 2017: 694.12(B)
NEC 2020: 694.12(B)
NEC 2023: 694.12(B)
An electrician needs to install a single-phase circuit with a neutral and ground in a trade size 1/2 RMC conduit. If using XHHW conductors what is the largest wire size that can be used?
The correct answer is
Chapter 9 Articles, Table 5
Chapter 9 Articles, Table 5
Chapter 9 Articles, Table 5
Step 1: Find the maximum area for conductors within the conduit; Maximum percent of area of Trade size 1/2 (16) RMC conduit that can be used for conductor fill = 40 percent (Chapter 9, Table 1); Cross-sectional Area of 16 RMC conduit at 40 percent = 0.125 in2 (81 mm2) (Chapter 9, Article 344);
Step 2: Calculate the area available per conductor; Total conductors = 2 (phase and neutral) + 1 (ground) = 3; Area per conductor = 0.125 in2 / 3 = 0.0417 in2;
Step 3: Find the largest wire size that fits; From Chapter 9 Table 5, the largest XHHW wire size with an area less than or equal to 0.0417 in2 is 10 AWG (0.0243 in2)
That's correct! Way to go
Chapter 9 Articles, Table 5
Chapter 9 Articles, Table 5
Chapter 9 Articles, Table 5
Step 1: Find the maximum area for conductors within the conduit; Maximum percent of area of Trade size 1/2 (16) RMC conduit that can be used for conductor fill = 40 percent (Chapter 9, Table 1); Cross-sectional Area of 16 RMC conduit at 40 percent = 0.125 in2 (81 mm2) (Chapter 9, Article 344);
Step 2: Calculate the area available per conductor; Total conductors = 2 (phase and neutral) + 1 (ground) = 3; Area per conductor = 0.125 in2 / 3 = 0.0417 in2;
Step 3: Find the largest wire size that fits; From Chapter 9 Table 5, the largest XHHW wire size with an area less than or equal to 0.0417 in2 is 10 AWG (0.0243 in2)
For a 460-volt, three-phase system with two 20 hp and two 15 hp continuous-duty motors, the feeder ampacity must be no less than _____ amps.
The correct answer is
NEC 2017: 430.24, Table 430.250
NEC 2020: 430.24, Table 430.250
NEC 2023: 430.24, Table 430.250
Using Table 430.250, the FLC for each 20 hp, 460V motor is 27A, and for each 15 hp, 460V motor is 21A. We apply 125% to one of the largest motors (27A * 1.25 = 33.75A) and add the FLC of the other three motors (27A + 21A + 21A = 69A). The total is 33.75A + 69A = 102.8A. This calculation demonstrates how to handle multiple motors of the same size in a circuit.
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NEC 2017: 430.24, Table 430.250
NEC 2020: 430.24, Table 430.250
NEC 2023: 430.24, Table 430.250
Using Table 430.250, the FLC for each 20 hp, 460V motor is 27A, and for each 15 hp, 460V motor is 21A. We apply 125% to one of the largest motors (27A * 1.25 = 33.75A) and add the FLC of the other three motors (27A + 21A + 21A = 69A). The total is 33.75A + 69A = 102.8A. This calculation demonstrates how to handle multiple motors of the same size in a circuit.
A 25 hp, 230-volt, three-phase, Design B motor has a high starting current. If the initial nontime-delay fuse rating is insufficient, the maximum permitted fuse rating becomes _____ amperes.
The correct answer is
NEC 2017: Table 430.250, Table 430.52(C)(1), 430.52(C)(1)(a)
NEC 2020: Table 430.250, Table 430.52(C)(1), 430.52(C)(1)(a)
NEC 2023: Table 430.250, Table 430.52(C)(1), 430.52(C)(1)(a)
Step 1: Find the full-load current (FLC) for a 25 hp, 230-volt, three-phase motor in Table 430.250. The FLC is 68 amperes.
Step 2: Calculate 400% of FLC: 68 x 400% = 272 amperes.
Step 3: Since 272A doesn't correspond to a standard fuse size, round up to the next standard size, which is 300A.
Step 4: However, 300A exceeds 400% of FLC, so we must use the next lower standard size, which is 250A.
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NEC 2017: Table 430.250, Table 430.52(C)(1), 430.52(C)(1)(a)
NEC 2020: Table 430.250, Table 430.52(C)(1), 430.52(C)(1)(a)
NEC 2023: Table 430.250, Table 430.52(C)(1), 430.52(C)(1)(a)
Step 1: Find the full-load current (FLC) for a 25 hp, 230-volt, three-phase motor in Table 430.250. The FLC is 68 amperes.
Step 2: Calculate 400% of FLC: 68 x 400% = 272 amperes.
Step 3: Since 272A doesn't correspond to a standard fuse size, round up to the next standard size, which is 300A.
Step 4: However, 300A exceeds 400% of FLC, so we must use the next lower standard size, which is 250A.