Electrician Math Practice Test

2025 Edition

Take a free Electrician Math Practice Test to see what kind of math questions are on actual electrician license exams. Most states require an electrician to pass an exam to receive a journeyman or master electrician license. State exams cover such areas as power and current formulas, branch circuit load calculations, voltage drop, raceway fill and sizing and more.

For help with math calculations on electrician license exams, the following two books written by Ray Holder (Master Electrician and Certified Electrical Trade Instructor) focus on math and show you, in a step-by-step way, how to solve electrician math problems:

Electricians Practice Calculation Exams
Practical Calculations for Electricians



1.	A single-phase, 240-volt, 15 kVA standby generator may be loaded to a MAXIMUM of _____ per line.
		a. 62.50 amperes
To solve this problem apply the single-phase current formula: I = P ÷ E 15 kVA x 1,000 = 15,000 = 62.50 amperes 240 volts 240 Incorrect answer. Please choose another answer.
		b. 31.25 amperes
To solve this problem apply the single-phase current formula: I = P ÷ E 15 kVA x 1,000 = 15,000 = 62.50 amperes 240 volts 240 Incorrect answer. Please choose another answer.
		c. 41.66 amperes
To solve this problem apply the single-phase current formula: I = P ÷ E 15 kVA x 1,000 = 15,000 = 62.50 amperes 240 volts 240 Incorrect answer. Please choose another answer.
		d. 52.50 amperes
To solve this problem apply the single-phase current formula: I = P ÷ E 15 kVA x 1,000 = 15,000 = 62.50 amperes 240 volts 240 Incorrect answer. Please choose another answer.



2.	When installing NM type cable in a residential construction project, you find that you are compelled to use a shallow square metal box, 4" x 1.5" deep. If you are to use this box as a junction box for wiring in a finished area wall, how many cubic inches will the plaster ring have to be, minimum, if you have five (5) number twelve two conductor cables with ground wires (#12/2 +gnd) coming into this box?
		a. 12.75 cu. in.
You have 10 current carrying conductors and 1 equipment grounding conductor to count in your conductor fill volume allowance, 314.16 (B)(1),(5). Table 314.16(B) shows an allowance of 2.25 cu. in./ #12. So, 11 conductors multiplied by 2.25 equals 24.75. Table 314.16(A) shows the box having a capacity of 21 cu. in. With 24.75 cu. in. required, subtracting the available 21 cu. in. leaves 3.75 cu. in. minimum to be accommodated with the plaster ring. Incorrect answer. Please choose another answer.
		b. 3.75 cu. in.
You have 10 current carrying conductors and 1 equipment grounding conductor to count in your conductor fill volume allowance, 314.16 (B)(1),(5). Table 314.16(B) shows an allowance of 2.25 cu. in./ #12. So, 11 conductors multiplied by 2.25 equals 24.75. Table 314.16(A) shows the box having a capacity of 21 cu. in. With 24.75 cu. in. required, subtracting the available 21 cu. in. leaves 3.75 cu. in. minimum to be accommodated with the plaster ring. Incorrect answer. Please choose another answer.
		c. 1 cu. in.
You have 10 current carrying conductors and 1 equipment grounding conductor to count in your conductor fill volume allowance, 314.16 (B)(1),(5). Table 314.16(B) shows an allowance of 2.25 cu. in./ #12. So, 11 conductors multiplied by 2.25 equals 24.75. Table 314.16(A) shows the box having a capacity of 21 cu. in. With 24.75 cu. in. required, subtracting the available 21 cu. in. leaves 3.75 cu. in. minimum to be accommodated with the plaster ring. Incorrect answer. Please choose another answer.
		d. 1.5 cu. in.
You have 10 current carrying conductors and 1 equipment grounding conductor to count in your conductor fill volume allowance, 314.16 (B)(1),(5). Table 314.16(B) shows an allowance of 2.25 cu. in./ #12. So, 11 conductors multiplied by 2.25 equals 24.75. Table 314.16(A) shows the box having a capacity of 21 cu. in. With 24.75 cu. in. required, subtracting the available 21 cu. in. leaves 3.75 cu. in. minimum to be accommodated with the plaster ring. Incorrect answer. Please choose another answer.



3.	Based on the circumstances described in the previous question, what is the minimum cubic inch size of the plaster ring required if you were to install a standard duplex outlet at this location?
		a. 12.75 cu. in.
314.16(B)(4) requires a double volume allowance based on the size of wire connected to the device - #12 wire being 2.25 cu. in. as per Table 314.16(B). Adding 2 x 2.25 = 4.5 cu.in., added to the 3.75 cu. in. from the previous question, equals 8.25 cu. in. Incorrect answer. Please choose another answer.
		b. 3.75 cu. in.
314.16(B)(4) requires a double volume allowance based on the size of wire connected to the device - #12 wire being 2.25 cu. in. as per Table 314.16(B). Adding 2 x 2.25 = 4.5 cu.in., added to the 3.75 cu. in. from the previous question, equals 8.25 cu. in. Incorrect answer. Please choose another answer.
		c. 8.25 cu. in.
314.16(B)(4) requires a double volume allowance based on the size of wire connected to the device - #12 wire being 2.25 cu. in. as per Table 314.16(B). Adding 2 x 2.25 = 4.5 cu.in., added to the 3.75 cu. in. from the previous question, equals 8.25 cu. in. Incorrect answer. Please choose another answer.
		d. 6 cu. in.
314.16(B)(4) requires a double volume allowance based on the size of wire connected to the device - #12 wire being 2.25 cu. in. as per Table 314.16(B). Adding 2 x 2.25 = 4.5 cu.in., added to the 3.75 cu. in. from the previous question, equals 8.25 cu. in. Incorrect answer. Please choose another answer.



4.	Where an 80-ampere, 240-volt, single-phase load is located 200 feet from a panelboard and is supplied with size 3 AWG copper conductors with THWN insulation, what is the approximate voltage-drop on the circuit? (K = 12.9)
		A. 6.0 volts
To determine the voltage-drop, first find the circular mil area of the conductors as shown in Chapter 9, Table 8. Then apply the single-phase voltage-drop formula as follows: VD = 2KID VD = 2 x 12.9 x 80 amps x 200 ft. = 7.84 volts CM 52,620 CM Incorrect answer. Please choose another answer.
		B. 4.0 volts
To determine the voltage-drop, first find the circular mil area of the conductors as shown in Chapter 9, Table 8. Then apply the single-phase voltage-drop formula as follows: VD = 2KID VD = 2 x 12.9 x 80 amps x 200 ft. = 7.84 volts CM 52,620 CM Incorrect answer. Please choose another answer.
		C. 9.2 volts
To determine the voltage-drop, first find the circular mil area of the conductors as shown in Chapter 9, Table 8. Then apply the single-phase voltage-drop formula as follows: VD = 2KID VD = 2 x 12.9 x 80 amps x 200 ft. = 7.84 volts CM 52,620 CM Incorrect answer. Please choose another answer.
		D. 7.84 volts
To determine the voltage-drop, first find the circular mil area of the conductors as shown in Chapter 9, Table 8. Then apply the single-phase voltage-drop formula as follows: VD = 2KID VD = 2 x 12.9 x 80 amps x 200 ft. = 7.84 volts CM 52,620 CM Incorrect answer. Please choose another answer.



5.	You are installing a conduit on a rooftop to supply power to an air conditioning unit. The unit requires a 60 ampere 240 volt supply, with a nameplate load amperage of 45.5A. The rooftop will regularly reach 145°F in the summer, as you are installing above dark rooftop material, though your conduit will be 40" above the rooftop. What is the minimum size THWN-2 copper conductor that you can use?
		a. #8
Table 310.15 (B)(2)(a) shows that a correction factor multiplier of 0.65 applies to this environment. Table 310.15(B)(16) shows that a #6 THWN-2 copper conductor is rated for 75 amps, therefore: 75 x 0.65 = 48.75A. Incorrect answer. Please choose another answer.
		b. #6
Table 310.15 (B)(2)(a) shows that a correction factor multiplier of 0.65 applies to this environment. Table 310.15(B)(16) shows that a #6 THWN-2 copper conductor is rated for 75 amps, therefore: 75 x 0.65 = 48.75A. Incorrect answer. Please choose another answer.
		c. #4
Table 310.15 (B)(2)(a) shows that a correction factor multiplier of 0.65 applies to this environment. Table 310.15(B)(16) shows that a #6 THWN-2 copper conductor is rated for 75 amps, therefore: 75 x 0.65 = 48.75A. Incorrect answer. Please choose another answer.
		d. #3
Table 310.15 (B)(2)(a) shows that a correction factor multiplier of 0.65 applies to this environment. Table 310.15(B)(16) shows that a #6 THWN-2 copper conductor is rated for 75 amps, therefore: 75 x 0.65 = 48.75A. Incorrect answer. Please choose another answer.



6.	You are installing 1.5" metal conduit on the ceiling of an unfinished basement to connect a sub-panel to the service equipment. In the conduit you will install 4- #1 THWN copper conductors plus an uninsulated #6 equipment grounding conductor. For ease of installation, you will add a junction box in a straight section of the raceway, where you will also splice the wires. This is the only raceway in this box. What is the minimum required length of the box?
		a. 9"
314.28(A)(2) states that the distance from the raceway entry to the opposite side of the box shall not be less than 6 times the size of the largest conduit. 6 x 1.5" = 9" Incorrect answer. Please choose another answer.
		b. 12"
314.28(A)(2) states that the distance from the raceway entry to the opposite side of the box shall not be less than 6 times the size of the largest conduit. 6 x 1.5" = 9" Incorrect answer. Please choose another answer.
		c. 14"
314.28(A)(2) states that the distance from the raceway entry to the opposite side of the box shall not be less than 6 times the size of the largest conduit. 6 x 1.5" = 9" Incorrect answer. Please choose another answer.
		d. 6"
314.28(A)(2) states that the distance from the raceway entry to the opposite side of the box shall not be less than 6 times the size of the largest conduit. 6 x 1.5" = 9" Incorrect answer. Please choose another answer.



7.	When a retail furniture store has 80 continuous linear feet of display show window, the NEC mandates at LEAST _____ receptacle outlets be provided within 18 inches of the top of the show window for the show window lighting.
		A. six
To prevent the use of floor receptacles and extension cords, receptacles are required to be installed directly above a show window. One receptacle is required for every 12 linear ft. or "major fraction thereof" (6 ft. or more). Refer to Section 210.62. The calculation for 80 continuous feet of show window is as follows: 80 ft. (show window) = 6.7 = 7 receptacles 12 ft. (per receptacle) Incorrect answer. Please choose another answer.
		B. seven
To prevent the use of floor receptacles and extension cords, receptacles are required to be installed directly above a show window. One receptacle is required for every 12 linear ft. or "major fraction thereof" (6 ft. or more). Refer to Section 210.62. The calculation for 80 continuous feet of show window is as follows: 80 ft. (show window) = 6.7 = 7 receptacles 12 ft. (per receptacle) Incorrect answer. Please choose another answer.
		C. eight
To prevent the use of floor receptacles and extension cords, receptacles are required to be installed directly above a show window. One receptacle is required for every 12 linear ft. or "major fraction thereof" (6 ft. or more). Refer to Section 210.62. The calculation for 80 continuous feet of show window is as follows: 80 ft. (show window) = 6.7 = 7 receptacles 12 ft. (per receptacle) Incorrect answer. Please choose another answer.
		D. nine
To prevent the use of floor receptacles and extension cords, receptacles are required to be installed directly above a show window. One receptacle is required for every 12 linear ft. or "major fraction thereof" (6 ft. or more). Refer to Section 210.62. The calculation for 80 continuous feet of show window is as follows: 80 ft. (show window) = 6.7 = 7 receptacles 12 ft. (per receptacle) Incorrect answer. Please choose another answer.



8.	You are preparing to pull in the wires from service equipment to a sub-panel through parallel conduits that you have already installed. It is a 400 ampere three phase four wire feeder, plus the equipment grounding conductor. Due to the distance, to accommodate voltage drop you have to increase your ungrounded conductors from 3/0 cu to #250 kcmil. What minimum size will your copper (cu) equipment grounding conductor need to be?
		a. #1 al
250.122(B) requires the EGC to be increased proportionately to the ungrounded conductors. Table 250.122 gives the EGC size for a 400a feed to be a #3 cu wire. Ch. 9 Table 8 gives the conductor circular mil area, for a #3/0 - 167,800, a #3 - 52,620. To find the ratio of increase in conductor size: 250,000 kcmil / 167,800 = 1.49. Apply the ratio to the #3 area: 52,620 x 1.49 = 78,404. Ch.9 Table 8 - 78,404 is larger than a #2, but smaller than a #1. Incorrect answer. Please choose another answer.
		b. #3 cu
250.122(B) requires the EGC to be increased proportionately to the ungrounded conductors. Table 250.122 gives the EGC size for a 400a feed to be a #3 cu wire. Ch. 9 Table 8 gives the conductor circular mil area, for a #3/0 - 167,800, a #3 - 52,620. To find the ratio of increase in conductor size: 250,000 kcmil / 167,800 = 1.49. Apply the ratio to the #3 area: 52,620 x 1.49 = 78,404. Ch.9 Table 8 - 78,404 is larger than a #2, but smaller than a #1. Incorrect answer. Please choose another answer.
		c. #2/0 cu
250.122(B) requires the EGC to be increased proportionately to the ungrounded conductors. Table 250.122 gives the EGC size for a 400a feed to be a #3 cu wire. Ch. 9 Table 8 gives the conductor circular mil area, for a #3/0 - 167,800, a #3 - 52,620. To find the ratio of increase in conductor size: 250,000 kcmil / 167,800 = 1.49. Apply the ratio to the #3 area: 52,620 x 1.49 = 78,404. Ch.9 Table 8 - 78,404 is larger than a #2, but smaller than a #1. Incorrect answer. Please choose another answer.
		d. #1 cu
250.122(B) requires the EGC to be increased proportionately to the ungrounded conductors. Table 250.122 gives the EGC size for a 400a feed to be a #3 cu wire. Ch. 9 Table 8 gives the conductor circular mil area, for a #3/0 - 167,800, a #3 - 52,620. To find the ratio of increase in conductor size: 250,000 kcmil / 167,800 = 1.49. Apply the ratio to the #3 area: 52,620 x 1.49 = 78,404. Ch.9 Table 8 - 78,404 is larger than a #2, but smaller than a #1. Incorrect answer. Please choose another answer.



9.	Determine the MAXIMUM number of 125-volt, general-purpose receptacles the NEC permits to be protected by a 20-ampere, 120-volt, single-pole inverse time circuit breaker in a commercial occupancy.
		A. 18
Section 220.14(I) requires receptacles on a single strap to be considered a load of 180 VA each. To determine how many receptacles are permitted to be connected to a circuit, first find the VA of the circuit (volts x amperes) and divide the result by 180 VA. Thus, 120 volts x 20 amperes = 2,400 VA (circuit) 2,400 VA (circuit) ÷ 180 VA (one receptacle) = 13 outlets Incorrect answer. Please choose another answer.
		B. 15
Section 220.14(I) requires receptacles on a single strap to be considered a load of 180 VA each. To determine how many receptacles are permitted to be connected to a circuit, first find the VA of the circuit (volts x amperes) and divide the result by 180 VA. Thus, 120 volts x 20 amperes = 2,400 VA (circuit) 2,400 VA (circuit) ÷ 180 VA (one receptacle) = 13 outlets Incorrect answer. Please choose another answer.
		C. 13
Section 220.14(I) requires receptacles on a single strap to be considered a load of 180 VA each. To determine how many receptacles are permitted to be connected to a circuit, first find the VA of the circuit (volts x amperes) and divide the result by 180 VA. Thus, 120 volts x 20 amperes = 2,400 VA (circuit) 2,400 VA (circuit) ÷ 180 VA (one receptacle) = 13 outlets Incorrect answer. Please choose another answer.
		D. 10
Section 220.14(I) requires receptacles on a single strap to be considered a load of 180 VA each. To determine how many receptacles are permitted to be connected to a circuit, first find the VA of the circuit (volts x amperes) and divide the result by 180 VA. Thus, 120 volts x 20 amperes = 2,400 VA (circuit) 2,400 VA (circuit) ÷ 180 VA (one receptacle) = 13 outlets Incorrect answer. Please choose another answer.



10.	Where a rooftop mounted air-conditioning unit is supplied with three (3) size 8 AWG THWN copper conductors, enclosed in an electrical metallic tubing (EMT) within three (3) inches of the rooftop, and exposed to direct sunlight and an ambient temperature of 100 degrees F, the allowable ampacity of the conductors is _____ .
		A. 50 amperes
To solve this problem, first find the temperature adder, 40ºF, from Table 310.15(B)(3)(c) due to the proximity of the conduit to the rooftop exposure to sunlight then, add this value to the outdoor ambient temperature: outdoor ambient temperature = 100 deg. F adder (3" above roof) + 40 deg. F TOTAL = 140 deg. F (for derating) Table 310.15(B)(16) shows the allowable ampacity of the 8 AWG THWN copper conductors to be 50 amperes before derating, which is multiplied by 0.58 [taken from the ambient temperature correction factors in Table 310.15(B)(2)(a)]. Thus the allowable ampacity is reduced to 29 amperes: Size 8 AWG THWN ampacity (before derating) = 50 amperes 50 amperes x .58 (temperature correction) = 29 amperes Incorrect answer. Please choose another answer.
		B. 44 amperes
To solve this problem, first find the temperature adder, 40ºF, from Table 310.15(B)(3)(c) due to the proximity of the conduit to the rooftop exposure to sunlight then, add this value to the outdoor ambient temperature: outdoor ambient temperature = 100 deg. F adder (3" above roof) + 40 deg. F TOTAL = 140 deg. F (for derating) Table 310.15(B)(16) shows the allowable ampacity of the 8 AWG THWN copper conductors to be 50 amperes before derating, which is multiplied by 0.58 [taken from the ambient temperature correction factors in Table 310.15(B)(2)(a)]. Thus the allowable ampacity is reduced to 29 amperes: Size 8 AWG THWN ampacity (before derating) = 50 amperes 50 amperes x .58 (temperature correction) = 29 amperes Incorrect answer. Please choose another answer.
		C. 29 amperes
To solve this problem, first find the temperature adder, 40ºF, from Table 310.15(B)(3)(c) due to the proximity of the conduit to the rooftop exposure to sunlight then, add this value to the outdoor ambient temperature: outdoor ambient temperature = 100 deg. F adder (3" above roof) + 40 deg. F TOTAL = 140 deg. F (for derating) Table 310.15(B)(16) shows the allowable ampacity of the 8 AWG THWN copper conductors to be 50 amperes before derating, which is multiplied by 0.58 [taken from the ambient temperature correction factors in Table 310.15(B)(2)(a)]. Thus the allowable ampacity is reduced to 29 amperes: Size 8 AWG THWN ampacity (before derating) = 50 amperes 50 amperes x .58 (temperature correction) = 29 amperes Incorrect answer. Please choose another answer.
		D. 25 amperes
To solve this problem, first find the temperature adder, 40ºF, from Table 310.15(B)(3)(c) due to the proximity of the conduit to the rooftop exposure to sunlight then, add this value to the outdoor ambient temperature: outdoor ambient temperature = 100 deg. F adder (3" above roof) + 40 deg. F TOTAL = 140 deg. F (for derating) Table 310.15(B)(16) shows the allowable ampacity of the 8 AWG THWN copper conductors to be 50 amperes before derating, which is multiplied by 0.58 [taken from the ambient temperature correction factors in Table 310.15(B)(2)(a)]. Thus the allowable ampacity is reduced to 29 amperes: Size 8 AWG THWN ampacity (before derating) = 50 amperes 50 amperes x .58 (temperature correction) = 29 amperes Incorrect answer. Please choose another answer.



11.	You need to install Sch. 80 PVC conduit on an exposed exterior wall in straight runs with a junction box at each end. This area is subject to a daily temperature fluctuation that is most extreme in the winter, when it can drop to -30° F overnight, and then rise to 45° F in the early afternoon. How long can the conduit runs be, individually, before you need an expansion coupling? (Give answer in whole feet)
		a. 8'
352.44 states that an expansion coupling is needed if the conduit will change more than 1/4" in length in a straight run. For this environment, Table 352.44 shows a length change of 3.04"/100'. 3.04 ÷ 100 = .0304"/ foot 0.0304 × 8 = 0.2432", just less than 1/4" (.25"). Incorrect answer. Please choose another answer.
		b. 26'
352.44 states that an expansion coupling is needed if the conduit will change more than 1/4" in length in a straight run. For this environment, Table 352.44 shows a length change of 3.04"/100'. 3.04 ÷ 100 = .0304"/ foot 0.0304 × 8 = 0.2432", just less than 1/4" (.25"). Incorrect answer. Please choose another answer.
		c. 64'
352.44 states that an expansion coupling is needed if the conduit will change more than 1/4" in length in a straight run. For this environment, Table 352.44 shows a length change of 3.04"/100'. 3.04 ÷ 100 = .0304"/ foot 0.0304 × 8 = 0.2432", just less than 1/4" (.25"). Incorrect answer. Please choose another answer.
		d. None needed
352.44 states that an expansion coupling is needed if the conduit will change more than 1/4" in length in a straight run. For this environment, Table 352.44 shows a length change of 3.04"/100'. 3.04 ÷ 100 = .0304"/ foot 0.0304 × 8 = 0.2432", just less than 1/4" (.25"). Incorrect answer. Please choose another answer.



12.	You need to install a feeder to a subpanel in a newly constructed remote workshop at a single family dwelling. The source power is 120/240V single phase. The load in the workshop is engineered to be 85 amperes continuous at 240V, and a non-continuous 240V load of 35 amperes. The wire length will be 245' from the main service termination point to the workshop termination point. Using the Informational notes in Article 215 and the following formula, size the THWN copper feeder wires to provide reasonable efficiency of operation by preventing excessive voltage drop (keeping the wires as small as possible). The formula is: VD = 2 x L x R x I ÷ 1000. VD = voltage drop, L = length, R = resistance in /1000'.
		a. #1/0 THWN copper
215.2(A)(1)(a,b) directs the load to be calculated as follows: 35A @ 100%, 85A @ 125%. 35 x 106.25 = 141.25A. Inf. note 2 shows the need to maintain the voltage drop to less than 3%. The wire resistance is found in Ch. 9 Table 8, and the wire ampacity is from Table 310.15(B)(16). Using the resistance from #350 kcmil: 2 x 245' x 0.0382 x 141.25A = 2643.9175 ÷ 1000 = 2.643V, just under the 3V maximum. Incorrect answer. Please choose another answer.
		b. #4/0 THWN copper
215.2(A)(1)(a,b) directs the load to be calculated as follows: 35A @ 100%, 85A @ 125%. 35 x 106.25 = 141.25A. Inf. note 2 shows the need to maintain the voltage drop to less than 3%. The wire resistance is found in Ch. 9 Table 8, and the wire ampacity is from Table 310.15(B)(16). Using the resistance from #350 kcmil: 2 x 245' x 0.0382 x 141.25A = 2643.9175 ÷ 1000 = 2.643V, just under the 3V maximum. Incorrect answer. Please choose another answer.
		c. #350 kcmil copper
215.2(A)(1)(a,b) directs the load to be calculated as follows: 35A @ 100%, 85A @ 125%. 35 x 106.25 = 141.25A. Inf. note 2 shows the need to maintain the voltage drop to less than 3%. The wire resistance is found in Ch. 9 Table 8, and the wire ampacity is from Table 310.15(B)(16). Using the resistance from #350 kcmil: 2 x 245' x 0.0382 x 141.25A = 2643.9175 ÷ 1000 = 2.643V, just under the 3V maximum. Incorrect answer. Please choose another answer.
		d. #500 kcmil copper
215.2(A)(1)(a,b) directs the load to be calculated as follows: 35A @ 100%, 85A @ 125%. 35 x 106.25 = 141.25A. Inf. note 2 shows the need to maintain the voltage drop to less than 3%. The wire resistance is found in Ch. 9 Table 8, and the wire ampacity is from Table 310.15(B)(16). Using the resistance from #350 kcmil: 2 x 245' x 0.0382 x 141.25A = 2643.9175 ÷ 1000 = 2.643V, just under the 3V maximum. Incorrect answer. Please choose another answer.



13.	In the circumstances described in the previous question, the breaker feeding the workshop is a 150A 240V 2 pole device. What is the minimum required size for the copper Equipment Grounding Conductor in THWN insulation?
		a. #8 THWN
Table 250.122 shows a #6 required; 250.122(B) states that if the ungrounded conductors are increased in size, the ECG must be increased proportionately. Per Table 310.15(B)(16), #1/0 would be sufficient for 150A, that was increased to #350 kcmil. Ch. 9 Table 8 gives the circular mil area of #1/0 to be 105,600. To determine ratio: 350,000 ÷ 105,600 = 3.314. Circular mil area of #6 is 26,240, so applying the ratio = 26,240 x 3.314 = 86,959 circ. mils = #1/0. Incorrect answer. Please choose another answer.
		b. #6 THWN
Table 250.122 shows a #6 required; 250.122(B) states that if the ungrounded conductors are increased in size, the ECG must be increased proportionately. Per Table 310.15(B)(16), #1/0 would be sufficient for 150A, that was increased to #350 kcmil. Ch. 9 Table 8 gives the circular mil area of #1/0 to be 105,600. To determine ratio: 350,000 ÷ 105,600 = 3.314. Circular mil area of #6 is 26,240, so applying the ratio = 26,240 x 3.314 = 86,959 circ. mils = #1/0. Incorrect answer. Please choose another answer.
		c. #3 THWN
Table 250.122 shows a #6 required; 250.122(B) states that if the ungrounded conductors are increased in size, the ECG must be increased proportionately. Per Table 310.15(B)(16), #1/0 would be sufficient for 150A, that was increased to #350 kcmil. Ch. 9 Table 8 gives the circular mil area of #1/0 to be 105,600. To determine ratio: 350,000 ÷ 105,600 = 3.314. Circular mil area of #6 is 26,240, so applying the ratio = 26,240 x 3.314 = 86,959 circ. mils = #1/0. Incorrect answer. Please choose another answer.
		d. #1/0 THWN
Table 250.122 shows a #6 required; 250.122(B) states that if the ungrounded conductors are increased in size, the ECG must be increased proportionately. Per Table 310.15(B)(16), #1/0 would be sufficient for 150A, that was increased to #350 kcmil. Ch. 9 Table 8 gives the circular mil area of #1/0 to be 105,600. To determine ratio: 350,000 ÷ 105,600 = 3.314. Circular mil area of #6 is 26,240, so applying the ratio = 26,240 x 3.314 = 86,959 circ. mils = #1/0. Incorrect answer. Please choose another answer.



14.	A local appliance store, as part of a publicity campaign, has donated 12 electric cooktop stoves to your local school's Home Economics classroom. That's great, but the school's current cooktops are all gas appliances. The school would therefore like to know if their existing electrical system has the capacity to add these cooktops. Given the following details, what load will be added to the school's electrical system? Each unit is 8,600 watts @ 240V, which fortunately matches the school's single phase system.
		a. 116 amperes
220.55 applies Table 220.55 column B, which shows a demand factor of 32% to be applied to the total appliance combined wattage. 8.6kw x 12 units = 103.2kw 103.2kw x 32% = 33.024kw 33.024kw ÷ 240V = 137.6A. Incorrect answer. Please choose another answer.
		b. 138 amperes
220.55 applies Table 220.55 column B, which shows a demand factor of 32% to be applied to the total appliance combined wattage. 8.6kw x 12 units = 103.2kw 103.2kw x 32% = 33.024kw 33.024kw ÷ 240V = 137.6A. Incorrect answer. Please choose another answer.
		c. 280 amperes
220.55 applies Table 220.55 column B, which shows a demand factor of 32% to be applied to the total appliance combined wattage. 8.6kw x 12 units = 103.2kw 103.2kw x 32% = 33.024kw 33.024kw ÷ 240V = 137.6A. Incorrect answer. Please choose another answer.
		d. 430 amperes
220.55 applies Table 220.55 column B, which shows a demand factor of 32% to be applied to the total appliance combined wattage. 8.6kw x 12 units = 103.2kw 103.2kw x 32% = 33.024kw 33.024kw ÷ 240V = 137.6A. Incorrect answer. Please choose another answer.



15.	For each individual cooktop from the previous question, there will be a dedicated branch circuit. What is each circuit's individual ampacity?
		a. 72 amperes
210.19(A)(4) requires an ampacity sufficient for the load served, and as a non-commercial cooking facility, in a classroom setting, the continuous load rating would not apply. 8.6kw ÷ 240V = 35.8A. Even though the breaker and wiring will be rated at 40 amperes, the minimum ampacity needed is 36A. Incorrect answer. Please choose another answer.
		b. 36 amperes
210.19(A)(4) requires an ampacity sufficient for the load served, and as a non-commercial cooking facility, in a classroom setting, the continuous load rating would not apply. 8.6kw ÷ 240V = 35.8A. Even though the breaker and wiring will be rated at 40 amperes, the minimum ampacity needed is 36A. Incorrect answer. Please choose another answer.
		c. 40 amperes
210.19(A)(4) requires an ampacity sufficient for the load served, and as a non-commercial cooking facility, in a classroom setting, the continuous load rating would not apply. 8.6kw ÷ 240V = 35.8A. Even though the breaker and wiring will be rated at 40 amperes, the minimum ampacity needed is 36A. Incorrect answer. Please choose another answer.
		d. 11.5 amperes
210.19(A)(4) requires an ampacity sufficient for the load served, and as a non-commercial cooking facility, in a classroom setting, the continuous load rating would not apply. 8.6kw ÷ 240V = 35.8A. Even though the breaker and wiring will be rated at 40 amperes, the minimum ampacity needed is 36A. Incorrect answer. Please choose another answer.



16.	You have 132 #14 THWN copper wires used for signaling in a Class 1 system, and 5 #12 THWN spare copper conductors. To get the wiring from the PLC to the machinery, you will use EMT conduit. What is the minimum size that the EMT will be required to be?
		a. 2.5"
725.51(A) refers you to 300.17, which directs you to refer to the appropriate wiring method section of the NEC. In this case, you would use 358.22. Here it refers you to the appropriate Tables in Ch. 9, you will need to use Tables 1, 4, and 5 for the following numbers: #14 = 0.0097 circ. mil, #12 = 0.0133 circ. mil: 0.0097 x 132 = 1.2804, 0.0133 x 5 = 0.0665, 1.2804 + 0.0665 = 1.3469"². Table 4 shows 2.5" EMT to be the smallest available with sufficient area. Incorrect answer. Please choose another answer.
		b. 2"
725.51(A) refers you to 300.17, which directs you to refer to the appropriate wiring method section of the NEC. In this case, you would use 358.22. Here it refers you to the appropriate Tables in Ch. 9, you will need to use Tables 1, 4, and 5 for the following numbers: #14 = 0.0097 circ. mil, #12 = 0.0133 circ. mil: 0.0097 x 132 = 1.2804, 0.0133 x 5 = 0.0665, 1.2804 + 0.0665 = 1.3469"². Table 4 shows 2.5" EMT to be the smallest available with sufficient area. Incorrect answer. Please choose another answer.
		c. 4"
725.51(A) refers you to 300.17, which directs you to refer to the appropriate wiring method section of the NEC. In this case, you would use 358.22. Here it refers you to the appropriate Tables in Ch. 9, you will need to use Tables 1, 4, and 5 for the following numbers: #14 = 0.0097 circ. mil, #12 = 0.0133 circ. mil: 0.0097 x 132 = 1.2804, 0.0133 x 5 = 0.0665, 1.2804 + 0.0665 = 1.3469"². Table 4 shows 2.5" EMT to be the smallest available with sufficient area. Incorrect answer. Please choose another answer.
		d. 3"
725.51(A) refers you to 300.17, which directs you to refer to the appropriate wiring method section of the NEC. In this case, you would use 358.22. Here it refers you to the appropriate Tables in Ch. 9, you will need to use Tables 1, 4, and 5 for the following numbers: #14 = 0.0097 circ. mil, #12 = 0.0133 circ. mil: 0.0097 x 132 = 1.2804, 0.0133 x 5 = 0.0665, 1.2804 + 0.0665 = 1.3469"². Table 4 shows 2.5" EMT to be the smallest available with sufficient area. Incorrect answer. Please choose another answer.



17.	At the job from the previous question, the site engineer informs you that 4 of the spare #12 wires need to be used as power supply conductors for components on the same machinery. The maximum current carried by any of the signaling conductors is 0.5 amperes. What is the maximum ampacity of the power supply conductors, assuming no temperature or voltage drop limitations apply?
		a. 10 amperes
725.51(B)(1,2) clarifies that the adjustment factors of 310.15(B)(3)(a) apply only to the power supply conductors, which would be an 80% adjustment factor. From 310.15(B)(16) we find these wires have an ampacity of 25, therefore: 25A x 80% = 20A. Other adjustment factors could apply, but are here unstated. Incorrect answer. Please choose another answer.
		b. 16 amperes
725.51(B)(1,2) clarifies that the adjustment factors of 310.15(B)(3)(a) apply only to the power supply conductors, which would be an 80% adjustment factor. From 310.15(B)(16) we find these wires have an ampacity of 25, therefore: 25A x 80% = 20A. Other adjustment factors could apply, but are here unstated. Incorrect answer. Please choose another answer.
		c. 20 amperes
725.51(B)(1,2) clarifies that the adjustment factors of 310.15(B)(3)(a) apply only to the power supply conductors, which would be an 80% adjustment factor. From 310.15(B)(16) we find these wires have an ampacity of 25, therefore: 25A x 80% = 20A. Other adjustment factors could apply, but are here unstated. Incorrect answer. Please choose another answer.
		d. 25 amperes
725.51(B)(1,2) clarifies that the adjustment factors of 310.15(B)(3)(a) apply only to the power supply conductors, which would be an 80% adjustment factor. From 310.15(B)(16) we find these wires have an ampacity of 25, therefore: 25A x 80% = 20A. Other adjustment factors could apply, but are here unstated. Incorrect answer. Please choose another answer.



18.	In the interior of a finished commercial facility, you are asked to add a new 100 ampere subpanel, fed with a 100A 3 phase 4 wires plus ground feeder from the main service equipment, which is sitting 50' away. Both the new subpanel and the existing main are surface mount units, and you can install surface mounted EMT conduit. You are given no details of what the load will be. Without applying any exceptions or special conditions requirements, what is the minimum THWN copper wire size that you can use for the current carrying conductors?
		a. #4 THWN
310.15(B)(3)(a) requires an ampacity adjustment for more than 3 current carrying conductors, and since you do not know what the loads are, you must count the neutral. Table 310.15(B)(3)(a) shows an 80% adjustment factor needing to be applied, Table 310.15(B)(16) shows #2 THWN having an ampacity of 115A: 115A x 80% = 92A. 240.4(B) allows you to use the next size up if ampacity is not equal to a standard size, therefore this wire can be connected to a 100 breaker. Incorrect answer. Please choose another answer.
		b. #2 THWN
310.15(B)(3)(a) requires an ampacity adjustment for more than 3 current carrying conductors, and since you do not know what the loads are, you must count the neutral. Table 310.15(B)(3)(a) shows an 80% adjustment factor needing to be applied, Table 310.15(B)(16) shows #2 THWN having an ampacity of 115A: 115A x 80% = 92A. 240.4(B) allows you to use the next size up if ampacity is not equal to a standard size, therefore this wire can be connected to a 100 breaker. Incorrect answer. Please choose another answer.
		c. #3 THWN
310.15(B)(3)(a) requires an ampacity adjustment for more than 3 current carrying conductors, and since you do not know what the loads are, you must count the neutral. Table 310.15(B)(3)(a) shows an 80% adjustment factor needing to be applied, Table 310.15(B)(16) shows #2 THWN having an ampacity of 115A: 115A x 80% = 92A. 240.4(B) allows you to use the next size up if ampacity is not equal to a standard size, therefore this wire can be connected to a 100 breaker. Incorrect answer. Please choose another answer.
		d. #1 THWN
310.15(B)(3)(a) requires an ampacity adjustment for more than 3 current carrying conductors, and since you do not know what the loads are, you must count the neutral. Table 310.15(B)(3)(a) shows an 80% adjustment factor needing to be applied, Table 310.15(B)(16) shows #2 THWN having an ampacity of 115A: 115A x 80% = 92A. 240.4(B) allows you to use the next size up if ampacity is not equal to a standard size, therefore this wire can be connected to a 100 breaker. Incorrect answer. Please choose another answer.



19.	You need to install some EMT conduit for feeder conductors. The conductors are all copper with THWN insulation. There are 3 #500 kcmil wires and 1 #3 wire. What is the smallest size of EMT that the NEC will allow?
		a. 2.5"
Using Ch.9 Table 5, you find the following wire area numbers: #500 kcmil THWN = 0.7073"², #3 THWN = 0.0973"². 3 x 0.7073 = 2.1219 + 0.0973 = 2.2192"². Ch. 9 Table 4 shows 21/2" EMT with a 40% fill capacity to have an available area of 2.343"². - Caution - though 3 would clearly fit, you would not do that because of the reasoning found in Inf. Note 2. Incorrect answer. Please choose another answer.
		b. 3"
Using Ch.9 Table 5, you find the following wire area numbers: #500 kcmil THWN = 0.7073"², #3 THWN = 0.0973"². 3 x 0.7073 = 2.1219 + 0.0973 = 2.2192"². Ch. 9 Table 4 shows 21/2" EMT with a 40% fill capacity to have an available area of 2.343"². - Caution - though 3 would clearly fit, you would not do that because of the reasoning found in Inf. Note 2. Incorrect answer. Please choose another answer.
		c. 3.5"
Using Ch.9 Table 5, you find the following wire area numbers: #500 kcmil THWN = 0.7073"², #3 THWN = 0.0973"². 3 x 0.7073 = 2.1219 + 0.0973 = 2.2192"². Ch. 9 Table 4 shows 21/2" EMT with a 40% fill capacity to have an available area of 2.343"². - Caution - though 3 would clearly fit, you would not do that because of the reasoning found in Inf. Note 2. Incorrect answer. Please choose another answer.
		d. 4"
Using Ch.9 Table 5, you find the following wire area numbers: #500 kcmil THWN = 0.7073"², #3 THWN = 0.0973"². 3 x 0.7073 = 2.1219 + 0.0973 = 2.2192"². Ch. 9 Table 4 shows 21/2" EMT with a 40% fill capacity to have an available area of 2.343"². - Caution - though 3 would clearly fit, you would not do that because of the reasoning found in Inf. Note 2. Incorrect answer. Please choose another answer.



20.	From the previous question, you find that you have to change the conduit routing, and now a portion will have to use PVC conduit, which has to be 3" to have sufficient size. To transition between the two conduits, you will use a pull box, entering from opposite sides. What minimum size will it have to be - without splicing the wires or applying any exceptions?
		a. 18" between conduits
314.28(A)(1) requires 8x the largest raceway minimum: 8 x 3" = 24" Incorrect answer. Please choose another answer.
		b. 20" between conduits
314.28(A)(1) requires 8x the largest raceway minimum: 8 x 3" = 24" Incorrect answer. Please choose another answer.
		c. 24" between conduits
314.28(A)(1) requires 8x the largest raceway minimum: 8 x 3" = 24" Incorrect answer. Please choose another answer.
		d. 30" between conduits
314.28(A)(1) requires 8x the largest raceway minimum: 8 x 3" = 24" Incorrect answer. Please choose another answer.