Help with Test Question

[omega83]

Hi I recently took the journeymans for Mn and had a question I was wondering if I did it right. Can you guys help?

Question was: You have a 2 story building and its 28*40. How many 15 amp circuit can you put on the building?
answers where: 2, 3, 4, 5, 6

1st)What I did was 28 * 40 * 2( because two stories)=2240 sq ft.
2nd) 2240sq ft * 3va (came off tables in 220) = 6720 va
3rd) 6720va / 240v = 28 amps
4th) 28 amps / 15amps = 1.8 circuits so meaning two 15 amps circuits.

What Im having the issue is, On the third step, Do I divide that by 240v the higher voltage or 120v? Can someone help me if I did this right or not!

 

[iMuse97]

Hi I recently took the journeymans for Mn and had a question I was wondering if I did it right. Can you guys help?

Question was: You have a 2 story building and its 28*40. How many 15 amp circuit can you put on the building?
answers where: 2, 3, 4, 5, 6

1st)What I did was 28 * 40 * 2( because two stories)=2240 sq ft.
2nd) 2240sq ft * 3va (came off tables in 220) = 6720 va
3rd) 6720va / 240v = 28 amps
4th) 28 amps / 15amps = 1.8 circuits so meaning two 15 amps circuits.

What Im having the issue is, On the third step, Do I divide that by 240v the higher voltage or 120v? Can someone help me if I did this right or not!

If you divide by 240 you get the answer for how many 15A 2P breakers you may use to feed the minimum calculated lighting load in the building. I would have divided by 120 to get a minimum of 4 15A 1P breakers.

 

[omega83]

I see so I got it wrong. Yeah I was wondering about that when I got out. So If I divide by 120v, that would give me single and 240v would give me double. now I see. I got a 69% and needed 70%. sucks

 

[kwired]

Question was: You have a 2 story building and its 28*40. How many 15 amp circuit can you put on the building?
answers where: 2, 3, 4, 5, 6

Are you summarizing the question or is this pretty much how it was worded? If that is the actual question or very close to it, it is a horrible question to put on a test. To start with the question does not have a definite answer. As you have it worded every one of the possible answers can be right.

 

[jumper]

Are you summarizing the question or is this pretty much how it was worded? If that is the actual question or very close to it, it is a horrible question to put on a test. To start with the question does not have a definite answer. As you have it worded every one of the possible answers can be right.

Yep.

 

[anbm]

Hi I recently took the journeymans for Mn and had a question I was wondering if I did it right. Can you guys help?

Question was: You have a 2 story building and its 28*40. How many 15 amp circuit can you put on the building?
answers where: 2, 3, 4, 5, 6

1st)What I did was 28 * 40 * 2( because two stories)=2240 sq ft.
2nd) 2240sq ft * 3va (came off tables in 220) = 6720 va
3rd) 6720va / 240v = 28 amps
4th) 28 amps / 15amps = 1.8 circuits so meaning two 15 amps circuits.

What Im having the issue is, On the third step, Do I divide that by 240v the higher voltage or 120v? Can someone help me if I did this right or not!

15A circuit, so use 15A/1P breaker to protect each ckt.
80% of 15A = 12A (80% of breaker size)

If go with NEC, each 15A/1P CB can be loaded max. = 12A x 120V = 1440VA

Total 15/1P circuits needed= (6720 / 1440) = 4.67

So, my answer will be 5

What do you allow for lighting???

Assuming the building doesn't have any equipment use more than 120V, i.e. heater, dryer, oven, AC unit...
Then you need to count those load too :lol:

 

[S'mise]

How many can you put on the building? An infinate amount
What is the minimum? First off they dont say the type of building.
If residental; 15a x 120v=1800va 6720/1800=3.7 round up to 4
(Its best not to convert to amperage it can throw off the answer slightly)
If they mention "continuous" or "more than 4 hours" Then multiply by .8 to get 1440va

 

[kwired]

15A circuit, so use 15A/1P breaker to protect each ckt.
80% of 15A = 12A (80% of breaker size)

If go with NEC, each 15A/1P CB can be loaded max. = 12A x 120V = 1440VA

Total 15/1P circuits needed= (6720 / 1440) = 4.67

So, my answer will be 5

What do you allow for lighting???

Assuming the building doesn't have any equipment use more than 120V, i.e. heater, dryer, oven, AC unit...
Then you need to count those load too :lol:

The 80% factor only applies to continuous loads.

The values in 220.12 are minimum values to use for calculating the load. You could have more load. Individual branch circuits only need to calculate what is actually connected to them. Feeders and services you can apply demand factors where allowed. How many circuits are installed can vary depending on what the actual load is and what size of circuits are run. You can run up to 50 amp lighting circuits for multioutlet circuits (but not in dwelling units)

It is a horrible test question because it is a pretty vague question yet they give specific answers, all of which could be right or wrong depending on details that are not given.

 

[omega83]

How many can you put on the building? An infinate amount
What is the minimum? First off they dont say the type of building.
If residental; 15a x 120v=1800va 6720/1800=3.7 round up to 4
(Its best not to convert to amperage it can throw off the answer slightly)
If they mention "continuous" or "more than 4 hours" Then multiply by .8 to get 1440va

They did not mention continous. I think this question was a general question where they wanted to know the minimum 15amp circuit this building can have. I like how you figure out your answer. Didn't look at it that way.

 

[Little Bill]

They did not mention continous. I think this question was a general question where they wanted to know the minimum 15amp circuit this building can have. I like how you figure out your answer. Didn't look at it that way.

You mentioned it here but in your OP you said "how many can you put ..."
That question has been on all the simulated tests I took plus the "real" test. They are asking for the minimum and as has been stated you divide by 120V.
Many people will completely overthink these type questions because of the way they are worded or a lot of stuff added to the question that actually has nothing to do with the answer they are looking for. These general type questions are not usually looking for cont. loads or other factors that might change the answer in a real world setting. They will however have other questions that will need to have all variables included. You just have to carefully read the question and consider what they are asking.

Sorry to hear you just missed passing. Study hard, it should be somewhat easier the 2nd time because you will know sort of what to expect.

I also think it is sad that they don't let you know what you missed without paying for a review of what you missed. $$$$ game IMO.

 

[anbm]

My home is 2,200 SF, I just went back to the garage and check on the panel... it has (15) 15/1 breakers and not counting 2 pole breakers...
I agree as other said, the exam question is not really clear and the given answers could be a big guess for us sometimes, do not give up,
try it again...Heard even some engineers had 12 attempts to pass his EIT (-:


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