Code Ques.
- Thread starter Flight987
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- Connecticut
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- Engineer
T220.12 requires 3VA per Square Foot for lighting and general-use receptacles in a Dwelling Unit.
A 15A, 120V circuit would give you: 15A x 120V = 1800VA.
1800VA (per 15A ckt) divided by 3VA/Sf would give you: 1800VA/(3VA/SF)= 600SF per circuit.
A 15A, 120V circuit would give you: 15A x 120V = 1800VA.
1800VA (per 15A ckt) divided by 3VA/Sf would give you: 1800VA/(3VA/SF)= 600SF per circuit.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
I don't know where you are getting your study materials, but I suggest you look somewhere else. The code never requires a 15 amp circuit. It sometimes allows us to use one, but that is not the same thing. I believe the author of this question took the 15 amps, multiplied it by 120 volts, got a result of 1800 VA, and divided it by the factor of (3 VA per square foot, per Table 220.12). But that is nonsense. The question is badly written, in that the code book is not a design manual, and does not tell us how many circuits we need, nor does it say we must use 15 amp circuits for lighting, nor for that matter does the question clearly state that the load under consideration is lighting. Add to that the fact that the answer does not take into account that the same dwelling unit will also need circuits for receptacles, some of which can be on 15 amp circuits, and none of which can be determined on the basis of square footage.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
This is not a code question, and it (like the earlier question) does not make sense. What do you mean by "from B to ground"? What is "B"? To that I will add that a 120/240 volt, single phase system will not have 208 volts involved anywhere.
Code Ques.
These Ques. come from a Tech. type school here in MN.. They are Ques. that the school's training
program have to prepare you for the state. I did not go to the Tech. course, but a good friend of
got me a of this study book. It has all kind's different Ques. in it and they tried to set the Ques.
up as close to MN state test as it be to prepare us for what we need to know.
It has all the law's of MN. Department of Labor & Industry in it. This is a 2005 code deal, but
these are the thing's that we need to know. I feel blessed that I have this to use.
Take care
These Ques. come from a Tech. type school here in MN.. They are Ques. that the school's training
program have to prepare you for the state. I did not go to the Tech. course, but a good friend of
got me a of this study book. It has all kind's different Ques. in it and they tried to set the Ques.
up as close to MN state test as it be to prepare us for what we need to know.
It has all the law's of MN. Department of Labor & Industry in it. This is a 2005 code deal, but
these are the thing's that we need to know. I feel blessed that I have this to use.
Take care
Code Ques.
If you have three resistors connected in series (R1,R2,andR3) across a 180 volt supply,and the total voltage drop across R1 and R2 is 140 volt's,and the total across R2 and R3 is 100 volt's, and a Voltmeter
across R3 reads 40 volt's,how many volt's would you have across R1:_______; and alsoR2:______?
I have not started to work on the Answ. yet.
See how many guy's get this one right. For me, I have a hard time with these. 220.55 Is
easy to me.
If you have three resistors connected in series (R1,R2,andR3) across a 180 volt supply,and the total voltage drop across R1 and R2 is 140 volt's,and the total across R2 and R3 is 100 volt's, and a Voltmeter
across R3 reads 40 volt's,how many volt's would you have across R1:_______; and alsoR2:______?
I have not started to work on the Answ. yet.
See how many guy's get this one right. For me, I have a hard time with these. 220.55 Is
easy to me.
- Location
- Connecticut
- Occupation
- Engineer
Don't overthink it. This is nothing more than simple math:
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- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
We are not going to be able to help you, if you are not clearer in the description of the question. Your post #6 addresses a three phase, 4-wire, grounded-WYE system. In that type of system, the phase-to-phase voltage will be higher than the phase-to-ground voltage by a factor of the square root of 3 (approximately 1.732).
In your post #3 (and again in post #8), I am wondering if you are addressing a high-leg delta system, with phase B being the high leg. You didn't say. And if the original question did not say, then I will say again that the question was badly written. For that type of system, and with phase B being the high leg,
(1) The voltage from phase A is equal to the voltage from phase C to ground,
(2) The voltage from phase A to Phase C is twice the voltage from either phase to ground, and
(3) The voltage from phase B to ground is equal to the square root of 3 (approximately 1.732) times the voltage from either phase A or phase C to ground.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Then please don't ask us to help you with it. We are not here to do your work for you.
Also, these are not code questions.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
The math is shown in post #11.
Did I mention that this is NOT A CODE QUESTION?
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Yes you did.
Roger
Yep. Questions are complete BS, not the OPs fault. Anyone remember INEPT.....
kwired
Electron manager
- Location
- NE Nebraska
More information is needed. You will have differernt answer for a wye system than for a delta system. Also a poorly written question and has nothing to do with codes, but does involve theory.
Coed Ques.
Coed Ques.
Thank's for the help on the square foot Ques. If we reverse the equation, I see how that work's
On the B to ground Ques., I went back on the Mike's DVD's and found it. First I guessed and
took 120/240 Ex., which I knew was 208. Did my math 208/240=.86666. Or 87 percent. I used
that as a multplyer on the Sec. voltage. Did the rest of the Ex. and it worked out just fine. Then
I found it on of Mike's DVD's where Dennis and Mike talked for a short time about around 87 percent
of the voltage of B phase to ground would show up on the meter.
On this Ques. with the resistors, I know that I am thinking to hard. I know that the total Res. of each
Res. equil's the total of applyed voltage. I have looked at Mike's Ex. and there isn't to much that help's
on this type of Ques.. I know that the Inf. on the VD. between the Res. is confusing me. A little help
(not the answer) to get me thinking not so hard. Thank's
Coed Ques.
Thank's for the help on the square foot Ques. If we reverse the equation, I see how that work's
On the B to ground Ques., I went back on the Mike's DVD's and found it. First I guessed and
took 120/240 Ex., which I knew was 208. Did my math 208/240=.86666. Or 87 percent. I used
that as a multplyer on the Sec. voltage. Did the rest of the Ex. and it worked out just fine. Then
I found it on of Mike's DVD's where Dennis and Mike talked for a short time about around 87 percent
of the voltage of B phase to ground would show up on the meter.
On this Ques. with the resistors, I know that I am thinking to hard. I know that the total Res. of each
Res. equil's the total of applyed voltage. I have looked at Mike's Ex. and there isn't to much that help's
on this type of Ques.. I know that the Inf. on the VD. between the Res. is confusing me. A little help
(not the answer) to get me thinking not so hard. Thank's
kwired
Electron manager
- Location
- NE Nebraska
You have the bold part wrong. Sum of voltage drops in a circuit equals voltage applied. Hope tha is enough of a hint to get you going in right direction.
Code Ques.
I understand that the sum of the voltage drop's, equil's the sum of the Res. Voltages. If we have
40volt's on #3, then it is easy to say that there is 140 left over. How do we figure out with the given
volt metter reading's how this work's? Answ. Key say's R3 at 40v,R2 at 60v,R1 at 80v. I can see the
140 leftover, but the basic math is not coming to me. Thank's I ment to say in the past thread that
the total VD of the Res. equil's the 180 volt's. Typo problem. Sorry!
I understand that the sum of the voltage drop's, equil's the sum of the Res. Voltages. If we have
40volt's on #3, then it is easy to say that there is 140 left over. How do we figure out with the given
volt metter reading's how this work's? Answ. Key say's R3 at 40v,R2 at 60v,R1 at 80v. I can see the
140 leftover, but the basic math is not coming to me. Thank's I ment to say in the past thread that
the total VD of the Res. equil's the 180 volt's. Typo problem. Sorry!
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