Bad Exam Prep Book By Craftsman Book Co

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T-minator

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Found some serious errors in this exam prep book.
Save your money on this one. Thankfully I just copied the CD from the library.
I am taking the WI Master Electrician Exam. Think I will buy Mike Holt's material. Any recommendations for good materials?

Electrician's Exam Preparation Guide to the 2011 NEC
By: Dale Brickner

Published by: Craftsman Book Company
ISBN: 978-1-57218-255-43

Chapter 3 questions:
60. A department store is illuminated with 215 fluorescent lighting fixtures and connected to a 120-volt supply. Each lighting unit draws 2.2 amperes. How many 20-amp branch circuits are necessary to feed these fixtures if each branch circuit must not exceed 80% of the branch circuit rating?
A: 20 x .8 = 16 amps, 16/2.2=7.27 or 7 per circuit (7*2.2=15.4amp). ( 215fixtures*2.2=473amp ) but?? 215 fixtures / only 7 per circuit = 30.714 or 31 circuits (15.4*30=only 462amp ) 30 x 7 is only 210 fixtures. 5 fixtures still remain. You need 31 circuits. This book says 30 circuits!!!! 473amp / 16amp = 29.56 exact 16amp circuits they are looking for which is not possible

64. A warehouse requires 60 kw of general illumination using 150 watt incandescent lamps connected to a 120-volt source. How many 20-ampere branch circuits are required to feed the lamps if the circuits are to be kept at 80% of their maximum rating?
A: 20 x .8 = 16amps. 16/1.25=12.8 (12 x 1.25= 15amp per circuit ) 60,000 / 150 = 400 lamps and only 12 per circuit. So, 400 / 12 = 33.3333333 or 34 circuits. The wrong answer they want is 60,000/120 = 500amp 500/16=31.25 or 32 circuits ( they are calc. as if exact 16amp circuits. Won't they be surprised when they buy a 32 space panel!!!


Chapter 4 questions:
11. What is the smallest copper wire size allowed for service-entrance conductors?
A: (the wrong answer they are looking for is 8 awg cu and 6 awg al 230.31(B) is the reference which is for service-lateral conductors not service entrance conductors! Wow)


12. The smallest grounded or neutral conductor for an electric service using 1100 kcmil copper conductor is:
A: (the wrong answer they are looking for is 2/0 from Table 250.66 which is the grounding electrode conductor not the neutral WOW!!!)

I stopped after this question.
 
230.23(B) size of the smallest copper service entrance conductor is #8

230.23(C) refers you to 250.24(C) then it refers you to 250.24(C)(1) which refers you to 250.66 so the neutral can not be smaller than 2/0 for a service that has 1100 KCMIL service conductors.

Unless someone sees something i don't, I think the book is good so far. I have not done the calculations.
 
T-minator said:
Found some serious errors in this exam prep book.
Save your money on this one. Thankfully I just copied the CD from the library.
I am taking the WI Master Electrician Exam. Think I will buy Mike Holt's material. Any recommendations for good materials?

Electrician's Exam Preparation Guide to the 2011 NEC
By: Dale Brickner
Published by: Craftsman Book Company
ISBN: 978-1-57218-255-43

Chapter 3 questions:
60. A department store is illuminated with 215 fluorescent lighting fixtures and connected to a 120-volt supply. Each lighting unit draws 2.2 amperes. How many 20-amp branch circuits are necessary to feed these fixtures if each branch circuit must not exceed 80% of the branch circuit rating?
A: 20 x .8 = 16 amps, 16/2.2=7.27 or 7 per circuit (7*2.2=15.4amp). ( 215fixtures*2.2=473amp ) but?? 215 fixtures / only 7 per circuit = 30.714 or 31 circuits (15.4*30=only 462amp ) 30 x 7 is only 210 fixtures. 5 fixtures still remain. You need 31 circuits. This book says 30 circuits!!!! 473amp / 16amp = 29.56 exact 16amp circuits they are looking for which is not possible

64. A warehouse requires 60 kw of general illumination using 150 watt incandescent lamps connected to a 120-volt source. How many 20-ampere branch circuits are required to feed the lamps if the circuits are to be kept at 80% of their maximum rating?
A: 20 x .8 = 16amps. 16/1.25=12.8 (12 x 1.25= 15amp per circuit ) 60,000 / 150 = 400 lamps and only 12 per circuit. So, 400 / 12 = 33.3333333 or 34 circuits. The wrong answer they want is 60,000/120 = 500amp 500/16=31.25 or 32 circuits ( they are calc. as if exact 16amp circuits. Won't they be surprised when they buy a 32 space panel!!!


Chapter 4 questions:
11. What is the smallest copper wire size allowed for service-entrance conductors?
A: (the wrong answer they are looking for is 8 awg cu and 6 awg al 230.31(B) is the reference which is for service-lateral conductors not service entrance conductors! Wow)


12. The smallest grounded or neutral conductor for an electric service using 1100 kcmil copper conductor is:
A: (the wrong answer they are looking for is 2/0 from Table 250.66 which is the grounding electrode conductor not the neutral WOW!!!)

I stopped after this question.
Click to expand...

Also check out these section of his site....:thumbsup:
Free Quizzes and Exams


Free Videos


Exam Preparation



.
.
 
edward said:
230.23(B) size of the smallest copper service entrance conductor is #8

230.23(C) refers you to 250.24(C) then it refers you to 250.24(C)(1) which refers you to 250.66 so the neutral can not be smaller than 2/0 for a service that has 1100 KCMIL service conductors.

Unless someone sees something i don't, I think the book is good so far. I have not done the calculations.
Click to expand...

service drop or service lateral supply the service-entrance conductors. it sounds like you are saying these are the same wires.
 
edward said:
230.23(B) size of the smallest copper service entrance conductor is #8

230.23(C) refers you to 250.24(C) then it refers you to 250.24(C)(1) which refers you to 250.66 so the neutral can not be smaller than 2/0 for a service that has 1100 KCMIL service conductors.

Unless someone sees something i don't, I think the book is good so far. I have not done the calculations.
Click to expand...

code book committees must get paid by the word. thank you for showing me that.. There must be a reason they are never actually sized that small
 
T-minator said:
Thankfully I just copied the CD from the library.
Click to expand...
Speaking on behalf of libraries everywhere and on behalf of the librarians who work there (my wife being one of them), I am appalled. Speaking on behalf of the Mike Holt Code Forum, we do not condone copyright violations. :rant:

 
kwired said:
Maybe original is good, and they added security features that change it if it is copied:cool:
Click to expand...
This may have been an urban myth, I don't really know but, I remember hearing about a pay per view fight years ago (seems like it was Holmes vs Clooney) and those who were getting it on illegal cable connections saw an add for a free Tshirt, cap, or something. All they had to do was send their name and address and they would receive the gift, I don't know how many fell for it but I am sure some did. :D

Roger
 
Shame on me if I want electricians to have real world knowledge and not this textbook fantasy math. This book was from fiction "free" section of the library.
 
T-minator said:
Shame on me if I want electricians to have real world knowledge and not this textbook fantasy math.
Click to expand...
I don't see anything wrong with the books math

For the first question you should have done this,

  • 215 X 2.2 = 473
  • 473 / 16 = 29.56
  • 29.56 round up to 30


Roger
 
roger said:
I don't see anything wrong with the books math

For the first question you should have done this,

  • 215 X 2.2 = 473
  • 473 / 16 = 29.56
  • 29.56 round up to 30


Roger
Click to expand...

But to be clear, I do know why you answered the way you did and I agree with it.

Roger
 
roger said:
I don't see anything wrong with the books math

For the first question you should have done this,

  • 215 X 2.2 = 473
  • 473 / 16 = 29.56
  • 29.56 round up to 30


Roger
Click to expand...
when i am an inspector and roger puts more than 7- 2.2 amp fixtures on a 20 amp circuit which is rated at 80%. i will fail his work
 
T-minator said:
when i am an inspector and roger puts more than 7- 2.2 amp fixtures on a 20 amp circuit which is rated at 80%. i will fail his work
Click to expand...

But if he figures out a way to actually put 7.27 fixtures on a circuit in real life, I would heap him in praise and send him off to get a patent. :)
Would you say that this problem is a concrete example of quantum theory at work?

Tapatalk!
 
T-minator said:
This book was from fiction "free" section of the library.
Click to expand...
Regardless of how the library made the book available to its patrons, the book itself was copyrighted material. That means that someone owns the rights to make copies, and that someone is not a library patron. If you make a copy of a book that is published on a CD, then you are saving yourself money by not giving money to someone who has a legal right to get that money.

As I said earlier, the owner of this forum does not condone the violation of someone's copyright. :thumbsdown:

 
By the way, I agree with your math on the first two questions. :happyyes: These are not theoretical math problems; they are talking about installing actual circuits.
 
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