4/22/2026 11:56:12 AM Module: 030405a Version: 1st Edition Page: 40 Comment: The answer for question #2, activity #1, has an incorrect explanation. Because there is no temperature rating for the equipment and it's under 100A, we must assume 60 degrees (4-006 (2)), not 4-006 (1) as stated in the answer, and the conductor would be sized off of the 60 degree column, not the 75 degree column as the answer states. Status: Implemented
2/27/2026 10:50:04 AM Module: 030404h Version: 1st Edition Page: 67 Comment: 404h: The answer for Objective 6 question 3 seems to be incorrect, I a 400A rated panel is connected to 500A of total supply capacity from the utility and the renewable system, then the connection of the 2 sources must be at opposite ends of the bus bars. This is only true in a residential application in a commercial building then the supply must be no more than 120% of the bus rating. rule 64-112 - 4 articles b, e, and f Status: Implemented
2/27/2026 9:23:22 AM Module: 030401eA Version: 1st Edition Page: 13 Comment: The ILM uses 9.80 as the gravitational constant, but the AIT formula sheet uses 9.81 which is the standard value used in most physics courses, can we use the same value in the ILMs to avoid confusion for the students? Status: Implemented
2/18/2026 2:43:49 PM Module: 030404e Version: 1st Edition Page: 28 Comment: Module states 'As a good practice, you should ensure that the inverter or UPS system output rating is at least 125% greater than the maximum critical load demand.' The wording of the calculation is incorrect. '125% greater than' means that the critical load demand value is increased by an amount equal to 125% of itself. Eg. 100 + (100X1.25) = 225 This can be understood more easily by using the following examples: 100% greater than: 20+ (20x1) = 40 30% greater than: 20+ (20x0.3) = 26 The intent of the original statement is to increase the value by 25%. This can be correctly conveyed by either of the following statements: '...the inverter or UPS system output rating is at least 25% greater than the maximum critical load demand.' '...the inverter or UPS system output rating is at least 125% of the maximum critical load demand.' Status: Implemented
2/12/2026 3:28:15 PM Module: 030401eB Version: 1st Edition Page: 64 Comment: The paragraph says the error is determined by subtracting the setpoint from the current process value, and then the formula underneath has it backwards. It should read Error= Process Value- Setpoint. This is reiterated by the formulas on Page 67. Status: Implemented
2/12/2026 2:38:05 PM Module: 030401eB Version: 1st Edition Page: 73 Comment: Question 16: The question answer key says d) 6.67 and the question has b) 6.67. Status: Implemented
2/9/2026 11:52:51 AM Module: 030404f Version: 1st Edition Page: 33 Comment: The answer to question #1 is listed as C) but written out as text offered in option B), which is correct. Status: Implemented
1/27/2026 11:40:04 AM Module: 030405e Version: 1st Edition Page: 13,14 Comment: In the feeder overcurrent protection, the module states that based on rule 42-008 3) the second step is to add the remaining calculated values from 42-006 2) to the rating of the overcurrent device from the first step. In 2024 CEC (October 2024) 42-008 3) states that the second step is to add the sum of ampacities as calculated by rule 42-006 1) for all other welders in the group. The module calculations continue on using the calculated sum from 42-006 2) and based on this discrepancy, do not match the actual text of the codebook. Whether this is an error in the codebook or an interpretation error by ILM is difficult to determine as no related errata have been noted as far as I have been able to find. Currently, this is being treated by instructors as an error in the CEC in order to best prepare students for a test based on content in the ILM materials. Status: Implemented
1/8/2026 1:38:49 PM Module: 030405a Version: 1st Edition Page: 15 Comment: The answer for objective one activity, question 1 states that the answer should be obtained from table 2, the question is referencing an aluminum conductor so the ampacity is taken from table 3. The 75A stated is the correct answer when table 3 is used. the answer needs to have table 2 changed to table 3 Status: Implemented
1/8/2026 1:34:27 PM Module: 030405a Version: 1st Edition Page: 31-32 Comment: In example 4.8 the load is listed as 63A but in the calculation the load is 60A. The math works with 63A, not 60. Status: Implemented
1/8/2026 1:32:31 PM Module: 030405a Version: 1st Edition Page: 27, 28 Comment: In the voltage drop calculation in example 4.5 starting on page 27 and ending on page 28 the example starts out with a 12A load, and part way through the question it changes to a 10A load. in order for the example to be correct the load needs to be 10A Status: Implemented
7/15/2025 10:09:34 AM Module: 030401eA Version: 1st Edition Page: 13 Comment: 1kg of mass with earth gravity is 9.81 N, not 9.80 N. 9.81 is the number on the AIT formula sheet, as well as the "Work, Power and Efficiency" module in first year. This is also the number used in the 310103a24 module used in instrumentation. Status: Implemented
5/21/2025 1:49:07 PM Module: 030404g Version: Version 24 Page: 42 Comment: Question 1, Obj 4. The question is from the 2018 CEC. Status: Implemented
12/14/2021 1:26:07 PM Module: 030405o Version: 21 Page: 44 Comment: Objective Five Exercise Answers: Question 1 needs to be changed to FT6. The FT4 rating is based on the 2015, 23rd Edition of the CEC. See Rule 2-130, Appendix B of the 2018 or 2021 CEC. Status: Implemented
12/10/2021 10:12:10 AM Module: 030404g Version: 21 Page: 35 Comment: The reference to (Rule 10-702) in the first diagram is incorrect and needs to be changed to (Rule10-104). Rule 10-702 deals with equipotential bonding and the diagram is referring to the interconnecting conductor between grounding electrodes which is covered within Rule 10-104. Status: Implemented
12/10/2021 10:04:32 AM Module: 030404f Version: 21 Page: 9 Comment: There are three references to the 2015 CEC. My suggestion is to reference the CEC and omit the year or it will have to be changed every 3 years. Status: Implemented
12/10/2021 9:21:38 AM Module: 030405n Version: 22 Page: 19 Comment: There is just a typo: word is "corona" NOT "coronal". Status: Implemented
12/8/2021 12:42:10 PM Module: 030405l Version: 22 Page: 36 Comment: Objective Four Exercise: Question 5. Wet wells shall be classified as what location? For Zone 2 to be the correct answer, the wet well would have to be provided with adequate continuous positive pressure ventilation. The Rule 22-704 is correct. Sewer gas(hydrogen sulphide- H2S) is not only poisonous but also corrosive and flammable; thus without adequate continuous positive pressure ventilation a wet well could be classified as a Zone 1 or Zone 0 location depending on gas concentrations and how often they are present. Status: Implemented
2/8/2021 1:04:20 PM Module: 030405i Version: 21 Page: Comment: Objective 5 for this module states “Determine the minimum ampacity of the feeder overcurrent device required for a group of motors”; however, the phrase “minimum ampacity” is used incorrectly here. We are always looking for the maximum overcurrent device, not the minimum. This is stated right in the first sentence of Objective 5. Also, as per Section 0, the term “ampacity” is used for conductors not for equipment such as overcurrent devices. This objective should be reworded as “Determine the rating of the feeder overcurrent device required for a group of motors”. This wording matches well with the wording of Objective 3 in the same module. Status: Declined
2/8/2021 12:59:01 PM Module: 030405a Version: 21 Page: 1 Comment: The outcome for 030405a is to "Determine the size of conductors and conduit"; however, conduit sizing was moved to module 030405c during the last outline revision. Reword this outcome to state "Determine the size of conductors. Status: Declined
4/17/2020 10:26:27 AM Module: 030401c Version: 21 Page: Comment: The differential compound motor is noted as a motor used when a "nearly constant speed" is required. This is only true when operating at near full load, from no load to full load the overall speed regulation of a shunt motor is better than a differential compound but the speed curve of the differential is almost perfectly flat at near full load. I think this is confusing to students and could be explained better. Adding the differential compound motors speed curve to figure 37 and the torque curve to figure 38 would also be helpful. Status: Declined
2/24/2020 8:08:51 AM Module: 030401e Version: 21 Page: 18 Comment: The answer to question four is incorrect. Although this is how it is explained throughout this module it is fundamentally way off. If context is need please see Electrical Machines, Drives, and Power Systems text by Theodore Wildi, page 382. Status: Declined
2/12/2020 9:46:44 AM Module: 030405b Version: 21 Page: Comment: 030405b Grounding and Bonding and Distribution Layout – Section 10 • Objective 1, p. 2 – Figure 1 delete Circuit from picture Circuit grounding. Leave in just Grounding. • Objective 1, p.12 – The second bullet point should read “connect the grounded conductor…” • Objective 1, Figure 13 – Cross-out “Bonding Screw Installed” text in top left of diagram. Status: Implemented
2/12/2020 9:44:39 AM Module: 030405a Version: 21 Page: Comment: 030405a Conductors – Section 4 • Objective 1, Figure 2 – Should say “Ampacity = Table 1 Value…” (145 A is correct). • Objective 1, Example #3 (p.6, 7) –Add to picture for Panel board to wall greater than 1.2m. Also add from JB to wall greater than 1.2 m. • Objective 1, Example #4 (p.7) – Add to question 1.2 meters from panelboard to area and 1.2 meters at least from area to equipment. Or Change the question to below (from RW 90 conductor) TW 75 wire and Solution should indicate that the ampacity is reduced from the 75o degree column of Table 5A according to insulation: 80 A ÷ 0.75 = 106.6 A. Select a conductor from the 75o degree column of Table 2. A No. 2 AWG would be selected: 115 A x 0.75 = 86.2 A. A No 2 AWG with an allowable ampacity of 86.2 is acceptable for the 80 A load. • Objective 1 Exercise, Q.3 – Add 1.2 meters out of ambient area on either end of the conductor or Change questions to Conductor to (from RW 90) TW 75 Then change answer to #3 AWG. • Objective 2, Example (p. 14) Add based on service OC before Table 16 at bottom of page 14 • Objective 3, (p. 18) 4th line down add underline portion 4-008 1) c) 2) and 4-008 Appendix B • Objective 3 Title for Figure 9 should be changed to underline Figure 9 – Flow chart based on Rules 4-004, 4-006, 4-008 1) 2), and 4-008 Appendix B and Appendix D. • Objective 4, (p.21) Third paragraph down after Voltage Drop. Change last sentence to underlined. Table D4 is handy when applying Rule 46-306 for the remote lamps of unit equipment. • Objective 4, Example #2 (p.22) – Assume all equipment has 90o ratings. • Objective 4, Example #3 (p.24) – A #1 AWG has an allowable ampacity of 130 A at 75o. Status: Implemented
2/5/2020 11:48:55 AM Module: 030405h Version: 21 Page: 7 Comment: In the heading for row housing.....it should refer to 8-200 2) not 8-202 2) Status: Implemented
2/5/2020 11:46:00 AM Module: 030405h Version: 21 Page: 5 Comment: On page , for step 8 add information. Add rule 4-004 22) appendix B and include information that in order to use Table 39 ...markings and labels are required on the equipment to allow you to use Table 39. If no labels must use 14-104 and table 13 and table 2. Status: Declined
12/2/2019 2:08:29 PM Module: 030405e Version: 21 Page: 40 Comment: The answer to Question 1 should be FT6, not FT4. According to 2-130 Appendix B, FT4 rated conductor is suitable for combustible and non-combustible buildings. FT6 is also rated for the above-mentioned construction and may be used in a plenum as the question mentions. Consider supplying code rule and reference to the appendix as feedback for the question. Status: Declined
11/12/2019 9:25:55 AM Module: 030405n Version: 21 Page: Comment: Objective 5 has many formatting errors. Status: Implemented
11/12/2019 9:00:15 AM Module: 030405n Version: 21 Page: 12 Comment: When listing the two types of insulation shielding, bullet 1 "Non-metallic insulating shielding" is italicized while bullet 2 is not. Make consistent. Status: Implemented
10/22/2019 10:24:28 AM Module: 030405g Version: 21 Page: 19 Comment: The code rule in the second paragraph needs to be changed from 8-104 5) a) to 8-104 6) a) Status: Implemented
10/11/2019 3:11:31 PM Module: 030405h Version: 21 Page: 21 Comment: For the service calculation, everything is correct with the calculations. The issue is we should be saying T-4 and T-39 both use a 4/0awg with a 200A service. By just mentioning T-4 you may assume T-39 is not considered. If I’m incorrect in my interpretation please get back to me. Status: Declined
1/1/2019 12:00:00 AM Module: 030405a Version: 12 Page: 24 Comment: Would you please check on the DCF (Table D3) formula and add clarification to the ILM as to whether you use the rated conductor temperature or the Conductor allowable ampacity, based on termination temperature. The Masters course offered by NAIT uses Termination Temperature to determine the conductor ampacity and this makes sense based on rule 4-006. Our ILM uses the 90 deg ampacity based on Table 2, yet gives termination temperature of 75 deg 0 Status: Implemented
1/1/2019 12:00:00 AM Module: 030405g Version: 12 Page: 5 Comment: First paragraph line five states subrule (2) subrule(3) should be subrule (3) subrule(4) Status: Implemented
1/1/2019 12:00:00 AM Module: 030405d Version: 12 Page: 4 Comment: first sentance of page 4 reads If a conductor is 3m or less and fulfills the requirements of Rule 14-100(c), it may be smaller than required by Rule 14-100(b) and still be acceptable. if we are going to make a statement like this an example would have to be provided to back this up as this is a very unlikely event and may cause confusion for student Status: Declined
1/1/2019 12:00:00 AM Module: 030405g Version: 12 Page: 16 Comment: second paragraph line 2 states a no. 10AWG 90C conductors has an allowable ampacity of 35 A A No 10 AWG from the 75 degree colomn has an allowable ampacity of 35A from the 90 degree colomn the allowable ampacity is 40A Suggest changing too- from the 75 degree colomn the allowable ampacity for the No. !0 AWG 90 degree would be 35A Status: Implemented
1/1/2019 12:00:00 AM Module: 030405i Version: 14 Page: 24 Comment: 3rd paragraph If locked rotor current is not listed on the equipment nameplate, it is assumed to be six times the full load current rating. (we do not have to assume cec gives us a value to use) should read If locked rotor current is not listed on the equipment Refer to Rule 28-010 Locked rotor current rating where not marked, deemed to be equal to six times the FLA rating from the nameplate Status: Implemented
1/1/2019 12:00:00 AM Module: 030405l Version: 12 Page: 20 Comment: Question 5, obj. 2 The term 'may be' denotes permission to do something. The given answer is False for this question. It should be True. You may install equipment approved for Class1, Div 1 locations in a Zone 0. The question does not state any requirements. So, as long as specific requirements are met, YOU MAY INSTALL class1 div 1 equipment in zone 0 Status: Implemented