PE 2017 NEC CODE QUESTION SAMPLES

Answer this question in accordance with the 2017 National Electrical Code@ The minimum size THWN copper conductors rated at 750C installed in conduit required to serve a and noncontinuous I-kW resistance heater on a circuit operating at 240-V single phase are: (A) 4/0 AWG (B) 2/0 AWG (C) 4 AWG (D) 6 AWG

(C) 4 AWG 1. 10-hp load -> NEC Table 430.248 Full-Load Currents in Amperes, Single-Phase Alternating-Current Motors pg. 321 = 50A 2. 430.24(4) Several Motors or a Motor(s) and Other Load(s) pg. 303 a. 50A*1.25 = 62.5A 3. Resistance Heating Load (kVA=kW) -> 1000W/240V=4.2A 4. Total Load = 62.5A + 4.2A = 66.7A Table 310.15(B)(16) Allowable Ampacities of Insulated Conductors Rated Up to and Including 2000 Volts, 60°C Through 90°C (140°F Through 194°F), Not More Than Three Current-Carrying Conductors in Raceway, Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30°C (86°F)* pg. 150 -> 4 WG

106. You have sized branch-circuit conductors supplying a 3-phase squirrel-cage induction motor in accordance with the 2017 National Electrical Code@ (NEC@). Now you must size the fuses to protect the conductors against short circuit or ground faults. What is the correct approach? (A) The maximum fuse rating is the ampacity of the conductors, except the next larger size fuses can be used if the conductor ampacity does not match a standard fuse size. (B) The2.maximum fuse rating is 100% of the locked-rotor current of the motor. (D) None of the above. SOL: (D) None of the above. NEC 430.52(C) Rating or Setting pg. 307, this code does not mention any of these conditions, it should be 300%

(D) None of the above. NEC 430.52(C) Rating or Setting pg. 307, this code does not mention any of these conditions, it should be 300%

75. Determine the minimum size service entrance conductor permitted according to the 2017 National Electrical Code® for a small three-phase industrial facility with a maximum load of 375 amps if three sets of 75º C rated conductors are connected in parallel for each phase. Assume each set of phase conductors are run in separate raceways. (A) 2 AWG (B) 1 AWG (C) 1/0 AWG (D) This application does not meet the requirements to parallel conductors and therefore is not permitted to code.

The ampacity of service entrance conductors is governed by NEC® 230.42(A) as not less than the maximum load being served. NEC® 310.10(H) permits conductors to be connected in parallel in sizes 1/0 AWG or greater as long as the conditions of NEC® 310.10(H)(a), (b), and (c) are met. If three sets of conductors are being connected in parallel for the service entrance conductors, that means that 1/3rd of the total load will flow through each phase conductor which equals 125A per phase conductor: 375A / 3 = 125A Since each set of phase conductors are being run in a separate raceway, we can ignore the adjustment factors required by NEC® 310.10(H)(4) that point to NEC® 310.15(B)(3)(a). Since there are no additional adjustment factors being applied, and not more than three current carrying conductors in each raceway, the ampacity table we will use is NEC® Table 310.15(B)(16). According to NEC® Table 310.15(B)(16), 125A does not match a specific size in the 75º C column, so the next size up is 1 AWG. However, NEC® 310.10(H) only permits size 1/0 AWG or greater for connecting conductors in parallel since service entrance conductors do not meet NEC® 310.10(H) Exception No. 1. Because of this, we will have to round up

105. A squirrel-cage induction motor is rated 40 hp, 460 Vs 3 phase, 60 Hz, 52 A, 0.87 lagging power factor at full load, Design-P of the 2017 National Electrical Code@ (NEC@). The maximum locked-rotor current (amperes) for selection of the motor disconnect is most nearly: (A) 412 (B) 356 (C) 290 (D) 240

(C) 290 1able NEC 430.251(B) Conversion Table of Polyphase Design B, C, and D Maximum Locked Rotor Currents for Selection of Disconnecting Means Controllers as Determined from Horsepower and Voltage Rating and Design Letter, pg 323

53. A temporary receptacle is installed at a residential construction site to be used by on site qualified construction personnel only. Select the answer below that is in accordance with the 2017 National Electrical Code®. (A) The temporary receptacle must have ground fault circuit interrupter protection provided by a ground fault circuit interrupter receptacle, a ground fault circuit interrupter circuit breaker, or a ground fault circuit interrupter cord plugged directly into the temporary receptacle. (B) The temporary receptacle must have ground fault circuit interrupter protection that may only be provided by a ground fault circuit interrupter receptacle or a ground fault circuit interrupter circuit breaker. (C) The temporary circuit is only required to have ground fault circuit interrupter protection if it is installed in a wet location. (D) The temporary circuit is not required to have ground fault circuit interruption protection as long as the temporary circuit is only used by qualified personnel.

The answer is: (A) The temporary receptacle must have ground fault circuit interrupter protection provided by a ground fault circuit interrupter receptacle, a ground fault circuit interrupter circuit breaker, or a ground fault circuit interrupter cord plugged directly into the temporary receptacle. According to NEC®590.6(A), a temporary circuit requires ground fault circuit interrupter protection for personnel. There are three types of devices for ground fault circuit interrupter protection: ● A ground fault circuit interrupter (GFCI) protection receptacle. This type of receptacle has GFCI protection built into the receptacle and is commonly found in wet areas and bathrooms with a "push to test" button. A GFCI receptacle provides ground fault circuit interrupter protection for the receptacle itself, and all other receptacles wired to it. A picture of a GFCI receptacle is located in the Handbook Edition of the 2017 National Electrical Code® in article 210.8 Exhibit 210.8. ● A ground fault circuit interrupter (GFCI) protection circuit breaker. This is a 120V panel circuit breaker with GFCI protection build into the circuit breaker. It is common to see non GFCI branch circuit receptacles that are fed from a GFCI panel circuit breaker to have a sticker that says "protected by GFCI" on the outlet to indicate that it is fed from a GFCI circuit breaker. A GFCI circuit breaker provides ground fault circuit interrupter protection for the entire circuit it feeds. A picture of a GFCI circuit breaker is located in the Handbook Edition of the 2017 National Electrical Code® in article 210.8 Exhibit 210.7. ● A portable GFCI cord. GFCI cords can be plugged directly into non-GFCI receptacles to provide ground fault circuit interrupter protection. A GFCI cord provides ground fault circuit interrupter protection for any equipment that is plugged into the GFCI cord. NEC®590.6(A)(1) specifically mentions the permitted use of GFCI cords. A picture of a portable GFCI cord is located in the Handbook Edition of the 2017 National Electrical Code® in article 590.5 Exhibit 590.1.

Select the minimum working space clearance in feet for an energized conductor that is 277volts to ground with exposed energized parts on one side and grounded parts on the other side according to the 2017 National Electrical Safety Code®. (A) 3 ½ (B) 3 (C) 2 1/2 (D) 277 volts to ground does not require a minimum working space clearance.

The answer is: (A) 3 ½ Working space clearances for 600 volts or less is given in NESC® Table 125-1. Exposed energized parts on one side and grounded parts on the other side is classified as Condition 2. 277 Volts to ground falls in the 151 - 600 volts to ground category. The table value for condition 2 clearance and 151 - 600 volts to ground is 3 ½ feet.

20. Three 250 kcmil uncoated copper conductors per phase are ran in PVC conduit to provide power to a three-phase load 75 feet away from a 480 volt power source. Calculate the percent voltage drop of the system in accordance with the 2017 National Electrical Code® if the load draws 650 amps at a lagging 0.80 power factor. (A) 0.2% (B) 0.4% (C) 0.6% (D) None of the above.

The answer is: (B) 0.4% Use NEC® Chapter 9 Table 9 to look up the resistance and reactance for one 250 kcmil uncoated copper conductor in PVC conduit: R = 0.052 Ω/1,000ft XL = 0.041 Ω/1,000ft According to note 2 at the bottom of NEC® Chapter 9 Table 9, we cannot use the "Effective Z at 0.85 PF" values directly from the table since the power factor is not 0.85. Instead, we will have to calculate the new approximate effective impedance value for this conductor at a lagging power factor of 0.80 using the equation given in note 2: The effective impedance of one 250 kcmil uncoated copper conductor in PVC conduit (shown as "250 kcmil" below) at 0.80 lagging power factor is 0.0662 Ω/1,000ft. Next we will need to calculate the impedance of each phase taking into consideration that there are three parallel conductors per phase: Three conductors run in parallel will have an equivalent impedance equal to 1/3rd of the original value. This means that each phase only sees 0.0662 ÷ 3 = 0.0221 Ω/1,000ft of impedance per phase. 75 ft 75 ft Now let's calculate the voltage drop across each phase conductor. Don't forget to use the circuit length to convert from ohms per 1,000 ft to ohms: It's important to note that even though this is a three-phase system, the voltage drop across each phase is a single-phase, or "per phase" value. You'll notice we use the subscript of 1ø. This becomes important in the next step. Voltage drop in percent (VD%) is the ratio of the magnitude of the voltage drop across the conductor, to the magnitude of the voltage supplied at the source. We can calculate it two different ways. The first way (a) is to compare the per phase voltage drop to the per phase voltage at the source. The second way (b) is to compare the line voltage drop to the line voltage supplied by the source. Just like a regular per phase voltage quantity, we can convert the magnitude to a line quantity by multiplying by √3. Notice that both methods results in the same percent voltage drop. The closest answer is a voltage drop of 0.4%.

56. Calculate the minimum ampacity rating in amperes for the disconnecting means of the runway conductors for an indoor bridge crane according to the 2017 National Electrical Code® if the crane has one hoist motor, one trolley motor, and two bridge motors. The nameplate ratings of the motors are given below. Nameplate data Hoist motor: Three-phase, 60 Hz, 460 V, 75 HP, 92 A Trolley motor: Three-phase, 60 Hz, 460 V, 25 HP, 31 A Bridge motor: Three-phase, 60 Hz, 460 V, 20 HP, 24 A (A) 108 (B) 116 (C) 125 (D) 142

The answer is: (B) 116 Typically, when sizing a motor disconnect, we use the tables in NEC® Article 430 based on motor voltage and horsepower for the full load current used in calculations and not the nameplate full load amps. However, when sizing a disconnect specifically for cranes, according to NEC® 610.14(E) the motor nameplate full load amps are used instead. Generally there are two disconnects installed for cranes. The first is between the power supply and the runway conductors NEC® 610.31, and the second is between the runway conductors and the crane equipment NEC® 610.32. For this question, we are tasked with sizing the disconnecting means of the runway conductors from the power supply. According to NEC® 610.31, the minimum rating of the disconnect will be not less NEC® 610.14(E)(2) for multiple motors. For multiple motors, the motor load is calculated as the sum of the largest motor (or group of motors) followed by 50% of the sum of the next largest motor (or group of motors). For a crane, a group of motors typically refers to motors that are switched on at the same time to move the crane in one direction, such as a pair of bridge motors that move the crane up and down the crane bridge. Because the crane in this question has two bridge motors, the pair of bridge motors is counted as a single motor load of one group. 2 Bridge motors = 2 X 24A = 48A. The largest motor load to be used for the calculation from NEC® 610.14(E)(2) is therefore the 92A hoist motor, followed by the 48A group of bridge motors as the next largest motor load. The 31A trolley motor will not be used in the calculation since it is smaller than the second largest motor (or group of motors). The minimum ampacity rating of the runway conductor disconnect can be calculated as: Answer: 92A + 50%(48A) = 116A. Note that the problem asked for the minimum ampacity rating of the disconnect, and not the actual standard size of the disconnect. This means we are just calculating what the minimum ampacity rating of the of the disconnect needs to be and not rounding up to meet a standard disconnect rating.

62. Select the minimum standard trade size liquidtight flexible metal conduit in inches required according to the 2017 National Electrical Code® for the following RHH conductors with the outer covering removed: three 4/0 AWG, and one 1/0 AWG. (A) 2 (B) 2 1/2 (C) 3 (D) 3 1/2

The answer is: (B) 2 1/2 For simpler raceway sizing problems when the type and size of the conductors are the same, we can jump straight to the tables in NEC® Annex C to determine the minimum size raceway based on raceway type. However, when the conductors in the raceway have different sizes and/or insulation types, the total individual cross section area of each conductor in the raceway must first be calculated using the Tables in NEC® Chapter 9. Step 1: Determine the cross section area of each conductor in the raceway. For RHH without the outer covering appears in NEC® Ch. 9 Table 5 as RHH* (RHH without the asterisk is with the outer covering on): RHH* 1/0 AWG = 0.2223 in2 RHH* 4/0 AWG = 0.3718 in2 According to the problem, there are three 4/0 AWG RHH* conductors, and one 1/0 AWG RHH* conductor in the raceway. Calculate the total conductor area: Total conductor cross section area in raceway = 3(0.3718 in2) + 0.2223 in2 Total conductor cross section area in raceway = 1.3377 in2 Step 2: Check with NEC® Ch. 9 Table 1 to determine the how much bigger the raceway needs to be compared to the total conductor cross section area according to how many conductors are in the raceway: For more than 2 conductors in the raceway, the raceway must be 40% larger than the total conductor cross section area. Step 3: Find the correct table in NEC® Ch. 9 Table 4 for the raceway type. The raceway type in the problem is liquidtight flexible metal conduit, so we want to use the table titled: Article 350 — Liquidtight Flexible Metal Conduit (LFMC). Using the Over 2 Wires 40% column in inches squared (in.2), round up to the next area size compared to the total conductor cross section area of 1.3377 in2. We are rounding up instead of down since this represents the minimum size diameter raceway, as in the raceway cannot be any smaller than this number, and the calculated number does not match a standard size. The next size up is 1.953 in2 which corresponds to the maximum 40% total conductor area for size 2 1/2 in. LFMC raceway. Step 3 Alternative: Another way to solve this is to calculate the actual minimum size area of the raceway based off the total conductor cross section area and the multiplier from NEC® Ch. 9 Table 1: Total conductor cross section area in raceway from step 1 = 1.3377 in2 NEC® Ch. 9 Table 1 = raceway must be 40% larger than the total conductor cross section area from step 2. Minimum cross section area of the raceway = 1.3377 in2 / 40% Minimum cross section area of the raceway = 3.3443 in2 Now go back to NEC® Ch. 9 Table 4 Article 350 — Liquidtight Flexible Metal Conduit (LFMC). This time use the Total Area 100% column in inches squared (in.2) and round up to the next area size compared to the minimum cross section area of the raceway 3.3443 in2. The next size up is 4.881 in2 which corresponds

Select the control method from the choices below that has the highest hierarchy for preventive and protective risk according to the NFPA® Standard for Electrical Safety in the Workplace. (A) Personal protective equipment (PPE) (B) Awareness (C) Approach boundary (D) Administrative controls

The answer is: (B) Awareness The risk control method hierarchy is the order of control methods used to reduce risk ranked in the order of the most effective to the least effective: 1. Elimination - Remove the hazard altogether if possible. 2. Substitution - Complete the task with a different method or tool that is less hazardous than the current method. 3. Engineering controls - Redesign the equipment so that it is less hazardous. 4. Awareness - Bring more attention to the hazard by alerting personnel mostly through the use of visual aids and barricades. 5. Administrative controls - Improve the training and job planning to help decrease the hazard if possible. 6. Personal protective equipment (PPE) - Equipment such as hard hats, flame retardant clothing, and insulated tools to increase the protection of personnel. Elimination should always be the first control implemented for a hazard if possible. If it is not possible, then the next highest control method should be implemented followed by the remaining control methods. Personal protective equipment (PPE), is generally considered the "last resort" control method to protect personal as much as possible should they be exposed to the hazard. The Hierarchy of Risk Control Methods can be found in the following location in the Standard for Electrical Safety in the Workplace®: 2018 NFPA® 70E 110.1(H)(3).

A square metal junction box is needed for the following: a single yoke three-way light switch. with three 12 AWG conductors, and a 12 AWG uninsulated ground wire terminated on the switch held in place by a cable clamp inside the box, one 12 AWG and 14 AWG conductor spliced together inside the box, and one 12 AWG conductor that passes through the box. Select the minimum box size that may be used according to the 2017 National Electrical Code® if all conductors originate outside the box. (A) 4 × 2 1⁄2 inch (B) 4 × 2 1⁄8 inch (C) 4 11⁄16 × 1 1⁄4 inch (D) 4 11⁄16 × 1 1⁄2 inch

The answer is: (C) 4 11⁄16 × 1 1⁄4 inch This will be a lot easier with a quick sketch of the box: 1. A yoke is the structural frame of a device, receptacle, or switch that is mounted to the box with screws. For conductor fill volume, each yoke counts as two conductors based on the largest conductor that is connected to it [NEC® 314.16(B)(4)]. Since size 12 AWG wires connect to the switch, the switch counts as two 12 AWG conductors. 2. Every conductor that originates outside of the box and terminates inside the box counts as one conductor based on its size [NEC® 314.16(B)(1)]. Three 12 AWG conductors from outside the box are run to the switch. This counts as three 12 AWG conductors. 3. Except for equipment grounding conductors that are used for isolated ground receptacles, only the largest equipment grounding conductor that enters the box is counted [NEC® 314.16(B)(5)]. There is one 12 AWG EGC in this box. This counts as one 12 AWG conductor (and would still only count as one 12 AWG conductor even if there were more than one EGCs entering the box). 4. Similar to EGCs, only one conductor volume allowance is given to clamps if there is one or more present in the box but this time it is based on the largest conductor inside the box [NEC® 314.16(B)(2)]. There is one clamp present and the largest conductor in the box is 12 AWG. The clamp counts as one 12 AWG conductor (and would still only count as one 12 AWG conductor even if there was more than one clamp present inside the box). COPYRIGHT © 2019 · ELECTRICAL PE REVIEW, INC - Not Authorized for Distribution 72 5. Every conductor that originates outside the box and passes through the box without being spliced (uncut) or termination, counts as one conductor. The 12 AWG conductor that passes through the box counts as one 12 AWG conductor. 6. Every conductor that originates outside the box and terminates or is spliced within the box counts as one conductor [NEC® 314.16(B)(1)]. The 12 AWG conductor that is spliced (wire nutted) to the 14 AWG conductor counts as one 12 AWG conductor. 7. Just like the previous step except now we are counting the smaller 14 AWG wire. The 14 AWG conductor spliced (wire nutted) to the 12 AWG conductor counts as one 14 AWG conductor. Use NEC® Table 314.16(B) to count the total minimum fill volume of the box based on the conductor volume allowances from the 7 steps above: 1: 2 X 12 AWG Conductor = 2(2.25 in3) 2: 3 X 12 AWG Conductor = 3(2.25 in3) 3: 1 X 12 AWG Conductor = 1(2.25 in3) 4: 1 X 12 AWG Conductor = 1(2.25 in3) 5: 1 X 12 AWG Conductor = 1(2.25 in3) 6: 1 X 12 AWG Conductor = 1(2.25 in3) 7: 1 X 14 AWG Conductor = 1(2.00 in3) The total minimum volume requirement of the box: 2(2.25 in3) + 3(2.25 in3) + 1(2.25 in3) + 1(2.25 in3) + 1(2.25 in3) + 1(2.25 in3) + 1(2.00 in3) = 22.25 in3 Use NEC® Table 314.16(A) to look at the minimum volume size compared to the box dimensions given in the answers. The smallest box out of the possible answers that is at least 22.25 in3 is the square metal 4 11⁄16 × 1 1⁄4 inch box that has a minimum volume of 25.5 in3 according to NEC® Table 314.16(A).

4. Select the minimum size copper 90 degree Celsius rated conductor run in conduit that is permitted to provide power to a three-phase, 60 Hz, 208 Volt, 60 HP, continuous duty, NEMA D induction motor according to the 2017 National Electrical Code®. (A) 2/0 AWG (B) 3/0 AWG (C) 4/0 AWG (D) 250 kcmil

The answer is: (C) 4/0 AWG The application of different temperature ratings of conductors according to code is commonly misunderstood. According to temperature limitations NEC® 110.14(C), conductor temperature ratings must be selected in order to "not exceed the lowest temperature rating of any connected termination, conductor, or device." If a conductor with a greater temperature rating (for example, 90 C) is being used on equipment with a lower temperature rating (for example, 75 C), then ampacity of the conductor must be selected based on the conductor ampacity values at the lower temperature rating of the equipment (75 C), instead of the conductor ampacity values at the actual higher temperature rating of the conductor (90 C). This may seem confusing at first. Let's work through this problem to have a better idea of how to put this into practice. Step 1: The FLC of a three-phase, 208V, 60 HP motor according to NEC® Table 430.250 is 169A. Step 2: According NEC® 430.22, conductors for a continuous duty motor must have a minimum ampacity of 125% of the motor FLC: 169A(125%) = 211.3A Step 3: The four possible answer choices are 90 C rated conductors according to the problem. We need to check the standard temperature rating of the motor according to the equipment provisions in the NEC®: NEC® 110.14(C)(1)(b)(1) circuits rated for greater than 100A will be rated for 75 C (our circuit is rated for 211.3A). Step 4: Even though all of the possible answer choices are conductors with 90 C rated insulation, we will have to select the conductor size based on the 75 C rated ampacities in the NEC® conductor ampacity tables since the circuit is rated for 75 C Conductors. This is what the code means by "conductors with higher temperature ratings, provided the ampacity of such conductors does not exceed the 75°C (167°F) ampacity of the conductor size used" in NEC® 110.14(C)(1)(b)(2). 4. The answer is: (C) 4/0 AWG ( ← continued from previous page) Now we may use NEC® Table 310.15(B)(16) to identify the correct size for a copper, 75 C (instead of 90 C due to the temperature provisions of the motor) rated conductor with a minimum ampacity of 211.3A . 211.3A is the minimum required ampacity of the conductor, which means as long as the conductor is rated for 211.3A, or greater, then it is applicable to code. Since 211.3A does not correspond to an actual conductor size, and falls between copper 75 C 3/0 AWG (Rated for 200A) and copper 75 C 4/0 AWG rated for (230A), we must round up to 75 C 4/0 AWG. 4/0 AWG 75 C Copper has an ampacity of 230A, and is greater than the minimum required ampacity of 211.3A for this circuit. The answer is 4/0 AWG. Note 1: Answer (D) 250 kcmil would still be applicable to code since it has a greater ampacity rating than the minimum requirement of 211.3A for this circuit, however, it is still the incorrect answer for this problem since the question asks for the "minimum" size conductor, meaning the smallest possible size out of the answer choices that still meets code. Note 2: If you ignored temperature provisions and sized the copper cable at the 90 C ampacity rating, you would have incorrectly chosen 3/0 AWG which at 90 C has an ampacity rating of 225A, but is undersized for this application when sized at 75 C (200A). Since the NEMA motor has conductor insulation ratings of 75 C, we may not use the 90 C ampacity of the cable, even though the conductor itself is rated for 90 C. NEMA D Motor 75 C Rating NEC® 110.14(C)(1)(b) Must be 75 C Min. If greater than 75 C, size based off of 75 C ampacity ratings NEC® 110.14(C)(1)(b)(2)

34. Select the maximum standard size amperage rating of an instantaneous circuit breaker that may be used for the short circuit protection of the following induction motor according to the 2017 National Electrical Code®. The motor nameplate is given below. ELECTRIC MOTOR MANUFACTURER ORDER NO. PQS 4817-135 PHASE: 3 FRAME: 286T HP: 30 SF: 1.15 FLC: 35 VOLTS: 460 RPM: 1780 HZ: 60 DESIGN: B AMB TEMP: 40 C CODE: G ENCL: TEFC EFF %: 82 PF: 0.84 (A) 385 (B) 400 (C) 440 (D) 450

The answer is: (D) 450 One of the biggest misunderstandings of applying the NEC® to motors is when to use the amperage of the motor that is marked on the nameplate, and when to use the full load current based on voltage and horsepower given in NEC® tables 430.247-250. According to NEC® 430.6(1), the full load current given in NEC® tables 430.247-250 is to be used for sizing conductors, switches, short circuit protection, and ground fault protection for motors, instead of the full load amps marked directly on the motor name plate unless it is a motor that is less than 1200 RPM, high torque, or a multi-speed motor in which case the motor nameplate full load current is used for sizing. In general, the only time you use the motor nameplate full load current is for sizing a separate motor overload device (see NEC® 430.32(A)(1) - "percent of motor nameplate full load current"). In this question, we are sizing the short circuit protection for a three-phase, 460 volt, 30 HP motor using an instantaneous trip circuit breaker. Step 1: Use NEC® Table 430.250 for three-phase alternating motors to look up the full load current to use for sizing according to the NEC® instead of using the full load current given in the name plate. According to the table, the FLC is 40 Amps. Step 2: Use NEC® Table 430.52 for the max rating of short circuit protection as a percentage of the full load current. According to the problem, we are sizing an instantaneous trip circuit breaker for a NEMA Design B (shown in the nameplate) motor. According to the table, we are permitted to use a maximum of 1,100% of the motor full load current: 1,100%(40A) = 440A. Step 3: Exception No. 1 of NEC® 430.52(C)(1) in accordance with NEC® Table 430.54, instructs that if the value from Step 2 does not match a standard size, you may round up to a value that does not exceed the next standard size. If you missed this exception in the code book, there was a clue left in the problem to select the maximum "standard size" amperage rating. Standard sizes are listed in NEC® 240.6(A). Step 4: 440 Amps does not match a standard size listed in NEC® 240.6(A), It falls between 400A and 450A. According to the exception, we are permitted to round up to a standard size of 450A as the maximum rating for an instantaneous trip circuit breaker used for the short circuit protection of this motor. The answer is 450 amps. If you incorrectly used the motor FLC, or incorrectly failed to round up to a standard size, you will choose the wrong answer.

According to the 2017 National Electrical Code®, select the minimum size copper AWG equipment grounding conductor required for a branch circuit that provides power to 208 volt, three-phase, combination non-motor load consisting of 15 kVA continuous and 5 kVA non-continuous. Assume that the minimum standard size overcurrent protection device is protecting the circuit. (A) 14 (B) 12 (C) 10 (D) 8

The answer is: (D) 8 The minimum size for an equipment grounding conductor (EGC) is selected based off of the rating of the overcurrent device in the circuit according to NEC® Table 250.122, so first we must properly size the overcurrent device. The minimum rating of the overcurrent device for continuous and non-continuous non-motor loads is the sum of 125% of the continuous loads and 100% of the non-continuous loads NEC® 210.20(A). Let's calculate the minimum rating of the overcurrent device: 125% • 15kVA/(√3•208V) + 5kVA/(√3•208V) = 66A Since according to code this is the minimum OCPD rating for the combination load, we will round up to the next standard size listed in NEC® Table 240.6(A) since the calculated version falls between two standard sizes. The next standard size is 70 amps. We can now look up the value of the OCPD in NEC® Table 250.122 to select the minimum EGC size. Notice not every standard size OCPD is listed in the table. The table describes the OCPD sizes (ratting/setting) as not exceeding. The rating of the OCPD for this circuit (70A) falls between two sizes in the table, 60 and 100 amperes. 70A exceeds 60A, so the next table size up of 100A is used. The copper EGC size in the table for 100A is 8 AWG.