Electrician (ELE)

Third Period Package (25 Modules) Comments

Date: 10/11/2019 11:36:57 AM
Module: 030301d
Version: 21
Page: 47
Comment: The last paragraph, below Figure 53, references Rule 4-038 (4) of the Canadian Electrical Code. Please change the Rule to 4-032 4) to reflect the change in the 2018 edition. Thank you.
Status: Approved for Review

Date: 5/31/2019 11:16:23 PM
Module: 030302a
Version: 7
Page: 26
Comment: My apologies, I recently sent this feedback with the wrong phone number. My current phone number is 780-982-7394. This module conflates the direction of induced current flow with the direction of electron flow. Electrons have negative charge, and so the direction they flow in is the direction of negative current, which is opposite to the direction of positive current. Consider for example questions 2 and 3 of Objective Three exercise. They are essentially the same question, but the answer given for question 3 is the opposite of the answer given for question 2. Please correct this conflation; direction of induced current is *not* the same as the direction of electron flow
Status: Approved for Review

Date: 10/9/2019 4:05:11 PM
Module: 030302a
Version: 7
Page: 26
Comment: This module conflates the direction of induced current flow with the direction of electron flow. Electrons have negative charge, and so the direction they flow in is the direction of negative current, which is opposite to the direction of positive current. Consider for example questions 2 and 3 of Objective Three exercise. They are essentially the same question, but the answer given for question 3 is the opposite of the answer given for question 2. Please correct this conflation; direction of induced current is *not* the same as the direction of electron flow.
Status: Approved for Review

Date: 11/7/2019 12:10:03 PM
Module: 030302a
Version: 21
Page: 33
Comment: The formula to calculate inrush VA is incorrect. Based on the the formula, the larger the motor the smaller the inrush VA would be. Example is if you have a 10 HP motor the inrush will be 10 times less than a 1 HP motor. The KVA required to start a motor is simply KVA multiplied by it's HP.
Status: Approved for Review

Date: 5/31/2019 11:17:01 PM
Module: 030302b
Version: 8
Page: 7
Comment: The reactance of the rotor is mostly inductive reactance because the rotor's capacitive reactance is miniscule" An induction motor does not have any capacitive reactance because it doesn't have a capacitor. Cap-start motors haven't been introduced yet other than a brief sentence in 2nd year for series RLC. Please remove or change to the rotor's reactance is all inductive reactance.
Status: Approved for Review

Date: 2/3/2020 9:26:07 AM
Module: 030302b
Version: 21
Page: 25
Comment: 030302b – Page 25 – Table 5 – Breakdown torque values are backwards. In order to be consistent with Module A and the info in module B, it should be: Design A - Normal Breakdown Torque Design B - Normal Breakdown Torque Design C - High Breakdown Torque Design D - Very High Breakdown Torque
Status: Approved for Review

Date: 2/3/2020 9:26:20 AM
Module: 030302b
Version: 21
Page: 37
Comment: Page 27 – Question 3 – Answer Key Page 37 – The answer for question 3 should be D
Status: Approved for Review

Date: 10/11/2019 11:36:47 AM
Module: 030303d
Version: 21
Page: 17
Comment: The second last bullet point references Rule 4-038(4) of the CEC. Please change it to Rule 4-032 4) to reflect the change in the 2018 edition. Thank you.
Status: Approved for Review

Date: 10/11/2019 11:36:35 AM
Module: 030303e
Version: 21
Page: 8
Comment: The first paragraph provides a reference to CEC Rule 4-038, subrules (4) and (5). Please change it to Rule 4-032 to reflect the change in the 2018 edition. The subrules remain the same. Thank you.
Status: Approved for Review

Date: 5/31/2019 11:16:29 PM
Module: 030304b
Version: 11
Page: 43
Comment: Question 6 is okay. The 5 to 1 ratio is not needed to answer the question but having it there just makes the student think more. The ratio indicates that the transformer is a delta-wye connection. If you work through the question using the current values you will still get the correct answer. However, the code states that the secondary conductor ampacity is calculated by multiplying the primary conductor ampacity by the voltage ratio which is 600/208 = 2.885.
Status: Approved for Review

Date: 5/31/2019 11:16:29 PM
Module: 030304b
Version: 11
Page: 43
Comment: I would like to ammend my comment that starts "Question 6 is okay". I think the 5 to 1 ratio should be removed because this is a code question not a theory question. Knowing that the transformer is a delta-wye connection is not needed to determine the conductor sizes.
Status: Approved for Review


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