G
Guest
Guest
What is the voltage drop of a 90' run of:
14 AWG NM
12 AWG NM
10 AWG NM
Thank you for your help!
14 AWG NM
12 AWG NM
10 AWG NM
Thank you for your help!
- Status
charlie
Senior Member
- Location
- Indianapolis
Re: Voltage drop
What answers did you get and how did you do your calculations?
What answers did you get and how did you do your calculations?
G
Guest
Guest
Re: Voltage drop
My experience is dated. I last worked full-time in the electrical trades in 1994. From 1994 to 1999 I was the RMO licensee for the local chapter of Habitat for Humanity. After a 1999 workplace injury I am now easing back into contracting. Anything you can do to help me is greatly apprecited. My 2002 NEC books & references are on order. Thank you for your previous answers. They have all been helpful and informative during this time of new beginnings.
I am doing some electrical work at home this weekend. I want to run a couple of dedicated circuits with minimal voltage drop but I don't want to go overboard. It's 90' from my meter/main combo panel to my dedicated receptacles. I have 14 AWG and 12 AWG, NM on hand, but wondering what the voltage drop is for those. Another option is 10 AWG NM and wondering if it's worth buying a box this weekened. I usually work with pipe, but I don't want to snake a 90' pipe thorough my garage and attic.
On a 90' run what is the voltage drop for:
14 AWG NM
12 AWG NM
10 AWG NM
Thank you for helping me while I get my legs again. I apologize for my brevity in my initial post.
My experience is dated. I last worked full-time in the electrical trades in 1994. From 1994 to 1999 I was the RMO licensee for the local chapter of Habitat for Humanity. After a 1999 workplace injury I am now easing back into contracting. Anything you can do to help me is greatly apprecited. My 2002 NEC books & references are on order. Thank you for your previous answers. They have all been helpful and informative during this time of new beginnings.
I am doing some electrical work at home this weekend. I want to run a couple of dedicated circuits with minimal voltage drop but I don't want to go overboard. It's 90' from my meter/main combo panel to my dedicated receptacles. I have 14 AWG and 12 AWG, NM on hand, but wondering what the voltage drop is for those. Another option is 10 AWG NM and wondering if it's worth buying a box this weekened. I usually work with pipe, but I don't want to snake a 90' pipe thorough my garage and attic.
On a 90' run what is the voltage drop for:
14 AWG NM
12 AWG NM
10 AWG NM
Thank you for helping me while I get my legs again. I apologize for my brevity in my initial post.
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
Re: Voltage drop
It is related to the voltage and current load.
It is related to the voltage and current load.
- Location
- Massachusetts
Re: Voltage drop
The reason for the odd responses where due to your lack of information.
To figure voltage drop you need to know the voltage and amperage of the circuit.
For 120 volts at 90 feet.
12 amps on 14 awg will result in a loss of 6.7 volts
16 amps on 12 awg will result in a loss of 5.8 volts.
24 amps on 10 awg will result in a loss of 5.2 volts.
The reason for the odd responses where due to your lack of information.
To figure voltage drop you need to know the voltage and amperage of the circuit.
For 120 volts at 90 feet.
12 amps on 14 awg will result in a loss of 6.7 volts
16 amps on 12 awg will result in a loss of 5.8 volts.
24 amps on 10 awg will result in a loss of 5.2 volts.
Re: Voltage drop
To calculate voltage drop you must know two of the three values that make up the circuit.
The size and length of the conductors will determine the resistance. One more value must be known.
Voltage drop is "eye times are". You only furnished the "are".
[ August 31, 2003, 04:32 PM: Message edited by: bennie ]
To calculate voltage drop you must know two of the three values that make up the circuit.
The size and length of the conductors will determine the resistance. One more value must be known.
Voltage drop is "eye times are". You only furnished the "are".
[ August 31, 2003, 04:32 PM: Message edited by: bennie ]
G
Guest
Guest
Re: Voltage drop
A couple of things. First I apologize for my brevity. I assumed that certain things would go without saying. I was not aware that amperage was a factor in voltage drop. That's a new concept for me. That you for bringing that to my attention so I can explore the amperage vs. voltage drop component in more detail. Again, it's news to me but that's why I'm here in this forum. It would be puzzling to me at this juncture why fluctuating load would change the voltage drop. I thought voltage drop percentage was a function of wire gauge and distance.
Secondly, I assumed 120 volts per hot leg. Regardless, in my experience voltage drop is a constant percentage regardless of actual voltage. Does the percentage change as the voltage changes? I thought the percentage was a constant and that the would not vary as a percentage value. If the voltage drop is 2% for a certain gauge of wire running a certain distance the voltage drop percentage would not change if the amperage or voltage were changed.
Percentage values would have been adequate. I am capable of doing simple math can do my own voltage times percentage calculations.
If you need further information to help me it would be helpful to me if you would just ask the question straight out. I don't always get it when questions are veiled in cryptic answers.
All that said, thank you for your input and help. I'm all ears. Bring it on.
Here are my details restated:
14AWG Type NM 120 volts running 90'
12AWG Type NM 120 volts running 90'
10AWG Type NM 120 volts running 90'
What is the voltage drop in either percentage or volts? The amperage will vary depending on the load. The maximum load on the 14 AWG will be 12.5 amps; 12 AWG max will be 16.67 amps; 10 AWG max will be 25 amps-- if amperage is indeed a voltage loss factor.
In my training the smaller the wire the more the voltage drop. After a certain number of feet the drop becomes excessive and a higher gauge of wire must be selected to maintain an adequate voltage. That's the mission here. I would like to buy my wire today if I need to buy more wire.
A couple of things. First I apologize for my brevity. I assumed that certain things would go without saying. I was not aware that amperage was a factor in voltage drop. That's a new concept for me. That you for bringing that to my attention so I can explore the amperage vs. voltage drop component in more detail. Again, it's news to me but that's why I'm here in this forum. It would be puzzling to me at this juncture why fluctuating load would change the voltage drop. I thought voltage drop percentage was a function of wire gauge and distance.
Secondly, I assumed 120 volts per hot leg. Regardless, in my experience voltage drop is a constant percentage regardless of actual voltage. Does the percentage change as the voltage changes? I thought the percentage was a constant and that the would not vary as a percentage value. If the voltage drop is 2% for a certain gauge of wire running a certain distance the voltage drop percentage would not change if the amperage or voltage were changed.
Percentage values would have been adequate. I am capable of doing simple math can do my own voltage times percentage calculations.
If you need further information to help me it would be helpful to me if you would just ask the question straight out. I don't always get it when questions are veiled in cryptic answers.
All that said, thank you for your input and help. I'm all ears. Bring it on.
Here are my details restated:
14AWG Type NM 120 volts running 90'
12AWG Type NM 120 volts running 90'
10AWG Type NM 120 volts running 90'
What is the voltage drop in either percentage or volts? The amperage will vary depending on the load. The maximum load on the 14 AWG will be 12.5 amps; 12 AWG max will be 16.67 amps; 10 AWG max will be 25 amps-- if amperage is indeed a voltage loss factor.
In my training the smaller the wire the more the voltage drop. After a certain number of feet the drop becomes excessive and a higher gauge of wire must be selected to maintain an adequate voltage. That's the mission here. I would like to buy my wire today if I need to buy more wire.
Re: Voltage drop
The voltage is proportionate to the current. In simple terms the voltage drop is current times resistance or I*R. Under the conditions you stated the maximum voltage drop would be.
#14 AWG @ 90? @ 12.5-amps = 7 volts
#12 AWG @ 90? @ 16.7-amps = 6 volts
# 10 AWG @ 90? @ 25 amps = 5.4 volts
The voltage is proportionate to the current. In simple terms the voltage drop is current times resistance or I*R. Under the conditions you stated the maximum voltage drop would be.
#14 AWG @ 90? @ 12.5-amps = 7 volts
#12 AWG @ 90? @ 16.7-amps = 6 volts
# 10 AWG @ 90? @ 25 amps = 5.4 volts
websparky
Senior Member
- Location
- Cleveland, Ohio
Re: Voltage drop
awwt,
We are all trying to help. However, you need to realize that voltage drop is proportional and inverse to the amp draw at the distance determined.
Like Bennie said, you have only supplied two of the requirements, volts and distance. You are missing the amp load at the distance and volts stated.
If you are asking the amount of volts based on the resistance of the conductor at the distance stated, it would be negligible or a moot point.
awwt,
We are all trying to help. However, you need to realize that voltage drop is proportional and inverse to the amp draw at the distance determined.
Like Bennie said, you have only supplied two of the requirements, volts and distance. You are missing the amp load at the distance and volts stated.
If you are asking the amount of volts based on the resistance of the conductor at the distance stated, it would be negligible or a moot point.
hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
Re: Voltage drop
Well, I don't know what kind of training you had but this is simple ohm's law that you should have had on day one. There is NO way that voltage drop on a piece of wire can be expressed as a percentage. None. Percentage implies a reference to 0% or 100%. What would 0% or 100% in this case be?
To determine voltage drop on a piece of wire you use the formula E=IxR, where E= voltage drop, I= the current drawn and R= the resistance of the 90ft run (180 ft)length of wire in ohms. You multiply the current drawn by the resistance of the wire. From this you should see that the voltage drop is going to increase or decrease depending on the current. Zero current equals zero voltage drop. Increase the current and the voltage drop goes up until the wire burns up. NOTHING to do with the voltage guys!
Perhaps you are confusing allowable voltage drop which is often given as a percentage. You would use the voltage drop from the above calculation and if it exceeds the allowable you need to use a larger wire size.
Well, I don't know what kind of training you had but this is simple ohm's law that you should have had on day one. There is NO way that voltage drop on a piece of wire can be expressed as a percentage. None. Percentage implies a reference to 0% or 100%. What would 0% or 100% in this case be?
To determine voltage drop on a piece of wire you use the formula E=IxR, where E= voltage drop, I= the current drawn and R= the resistance of the 90ft run (180 ft)length of wire in ohms. You multiply the current drawn by the resistance of the wire. From this you should see that the voltage drop is going to increase or decrease depending on the current. Zero current equals zero voltage drop. Increase the current and the voltage drop goes up until the wire burns up. NOTHING to do with the voltage guys!
Perhaps you are confusing allowable voltage drop which is often given as a percentage. You would use the voltage drop from the above calculation and if it exceeds the allowable you need to use a larger wire size.
websparky
Senior Member
- Location
- Cleveland, Ohio
Re: Voltage drop
hbiss,
I'm confused by what you mean when you say "There is NO way that voltage drop on a piece of wire can be expressed as a percentage. None. Percentage implies a reference to 0% or 100%. What would 0% or 100% in this case be?"
What is meant by this below?
hbiss,
I'm confused by what you mean when you say "There is NO way that voltage drop on a piece of wire can be expressed as a percentage. None. Percentage implies a reference to 0% or 100%. What would 0% or 100% in this case be?"
What is meant by this below?
charlie
Senior Member
- Location
- Indianapolis
Re: Voltage drop
Awwt, I hope this will not dissuade you from posting in the future. We are attempting to help you and others and the posts from Hbliss, Iwire, Bennie, and all the others, including me, are not meant to be malicious.
Actually Hbliss, voltage drop is expressed as a percentage all the time. The meaning and understanding is as a percentage of the nominal voltage. Additionally, if you take a look at Awwt's past posts, he makes a lot of sense on them, this is just an area where he needs some work. I am considered an expert on the Code but, I assure you, I don't know it all. I know of no one who does know it all. Many of us have a good handle on parts of the Code and theory.
Awwt, I hope this will not dissuade you from posting in the future. We are attempting to help you and others and the posts from Hbliss, Iwire, Bennie, and all the others, including me, are not meant to be malicious.
david
Senior Member
- Location
- Pennsylvania
Re: Voltage drop
Approximate K = 12.9 for copper.
VD= 2 x k x I x L / CM area 3% of 120volts = 3.6 volts
CM area : 14awg= 4110
12awg=6530
10awg=10380
VD= 2 X 12.9 X 15 amps X 90ft / 4110
34830/ 4110= 8.47
VD= 8.47 8.47 / 120volts = .07= 7% VD
[ September 01, 2003, 02:09 AM: Message edited by: david ]
Approximate K = 12.9 for copper.
VD= 2 x k x I x L / CM area 3% of 120volts = 3.6 volts
CM area : 14awg= 4110
12awg=6530
10awg=10380
VD= 2 X 12.9 X 15 amps X 90ft / 4110
34830/ 4110= 8.47
VD= 8.47 8.47 / 120volts = .07= 7% VD
[ September 01, 2003, 02:09 AM: Message edited by: david ]
Re: Voltage drop
awwt: I have a wierd sense of humor, please forgive my flippant remarks. They are not intended
to offend.
Besides that, my answer is wrong. It should have been "none of the voltage will be dropped until a load and source are connected.
The value numbers of the load, source, and wire length, will produce the correct answer as illustrated by the others.
[ September 01, 2003, 06:56 PM: Message edited by: bennie ]
awwt: I have a wierd sense of humor, please forgive my flippant remarks. They are not intended
to offend.
Besides that, my answer is wrong. It should have been "none of the voltage will be dropped until a load and source are connected.
The value numbers of the load, source, and wire length, will produce the correct answer as illustrated by the others.
[ September 01, 2003, 06:56 PM: Message edited by: bennie ]
hbiss
EC, Westchester, New York NEC: 2014
- Location
- Hawthorne, New York NEC: 2014
- Occupation
- EC
Re: Voltage drop
Actually Hbliss, voltage drop is expressed as a percentage all the time. The meaning and understanding is as a percentage of the nominal voltage.
That's exactly what I said. You need to find the voltage drop across the wire first then you subtract it from the nomimal voltage and convert it to a percentage if you want. There is no such thing as a percent voltage drop given only a wire gauge and length as AWWT believed.
I apologize if I came across as being mean. This is basic stuff and I think everybody is just getting confused.
Actually Hbliss, voltage drop is expressed as a percentage all the time. The meaning and understanding is as a percentage of the nominal voltage.
That's exactly what I said. You need to find the voltage drop across the wire first then you subtract it from the nomimal voltage and convert it to a percentage if you want. There is no such thing as a percent voltage drop given only a wire gauge and length as AWWT believed.
I apologize if I came across as being mean.
This is basic stuff and I think everybody is just getting confused.
G
Guest
Guest
Re: Voltage drop
I am recalling now that in long range music applications (sending a signal from an amplifier to a speaker) there was significant signal loss at the speaker end. The workaround is to BOOST the voltage, go the distance, then REDUCE the voltage at the speaker end. This prevents the loss (and avoids having to run 3/0 speaker wire).
Thank you for bearing with me. I was frustrated by the initial lack of help here. Without being rude maybe it would be better to hold off with the humor and suspicious comments until at least one good answer is posted.
I'm new here and I already get alarmed when a DIY comes in to this forum and talks about doing something really hairy and wants major advice. I understand your reluctance to answer basic questions. I always try to find an answer myself first but sometimes I have to ask for help. Somebody said asking for help is a sign of strength not a sign of weakness. I'm not so sure about that, but I'll keep asking anyway!
In conclusion: I got the message. I got my answer. Thank you very much for your help in getting my legs back.
[Note: This post was spell-checked prior to submission by ieSpell]
ieSpell get it here FREE
OK, Bennie: Your answer caused me the most confusion, but it also somehow alerted me that I needed to furnish amps too. My knowledge is deep and long, but it's old, dusty, and rusty. Thank you for bearing with me while I shake off the cobwebs. I know all this stuff, I just need to re-visit it and refresh my memory.
I am recalling now that in long range music applications (sending a signal from an amplifier to a speaker) there was significant signal loss at the speaker end. The workaround is to BOOST the voltage, go the distance, then REDUCE the voltage at the speaker end. This prevents the loss (and avoids having to run 3/0 speaker wire).
Thank you for bearing with me. I was frustrated by the initial lack of help here. Without being rude maybe it would be better to hold off with the humor and suspicious comments until at least one good answer is posted.
I'm new here and I already get alarmed when a DIY comes in to this forum and talks about doing something really hairy and wants major advice. I understand your reluctance to answer basic questions. I always try to find an answer myself first but sometimes I have to ask for help. Somebody said asking for help is a sign of strength not a sign of weakness. I'm not so sure about that, but I'll keep asking anyway!
In conclusion: I got the message. I got my answer. Thank you very much for your help in getting my legs back.
[Note: This post was spell-checked prior to submission by ieSpell]
ieSpell get it here FREE
karl riley
Senior Member
Re: Voltage drop
It would be good for everyone who has posted on this thread to read the preceding message carefully. Perhaps it is the summer tensions leading up to Labor Day, but it has seemed that some members have been quick to slam new or occasional posters recently. This is ego, folks. At others expense.
Someone say something about a kinder, gentler nation?
Karl
It would be good for everyone who has posted on this thread to read the preceding message carefully. Perhaps it is the summer tensions leading up to Labor Day, but it has seemed that some members have been quick to slam new or occasional posters recently. This is ego, folks. At others expense.
Someone say something about a kinder, gentler nation?
Karl
Re: Voltage drop
Karl: Thanks for the statements. You should save your response and re-post it every month.
Sometimes my attempts at humor are not funny. I need to be reminded of that often.
My firm apology to anyone taking offense with some of my remarks.
Best regards to everyone: Bennie
Karl: Thanks for the statements. You should save your response and re-post it every month.
Sometimes my attempts at humor are not funny. I need to be reminded of that often.
My firm apology to anyone taking offense with some of my remarks.
Best regards to everyone: Bennie
B
bthielen
Guest
Re: Voltage drop
AWWT,
2002 NEC, Art. 210 (A) FPN No.4
Conductors for branch circuits as defined in article 100, sized to prevent a voltage drop exceeding 3% at farthest outlet of power, heating and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and brach circuits to the farthest outlets dones not exceed 5%, provide reasonable efficientcy of operation. See 215.2 for voltage drop on feeder conductors.
It sounds to me that you are wondering what is an acceptable voltage drop and how to calculate your branch circuit accordingly. It sounds to me that you could assume full load at the farthest outlet to determine your current and the distance of such for the length of wire. Now refer to chapter 9, table 8 for nominal resistance per kFT. Accordingly, 14ga. single strand uncoated copper has 3.07 ohms/kFT, 12ga. is 1.93 ohms/kFT, and 10ga. is 1.21 ohms/kFT. Hope I have steered you accurately.
Bob
AWWT,
2002 NEC, Art. 210 (A) FPN No.4
Conductors for branch circuits as defined in article 100, sized to prevent a voltage drop exceeding 3% at farthest outlet of power, heating and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and brach circuits to the farthest outlets dones not exceed 5%, provide reasonable efficientcy of operation. See 215.2 for voltage drop on feeder conductors.
It sounds to me that you are wondering what is an acceptable voltage drop and how to calculate your branch circuit accordingly. It sounds to me that you could assume full load at the farthest outlet to determine your current and the distance of such for the length of wire. Now refer to chapter 9, table 8 for nominal resistance per kFT. Accordingly, 14ga. single strand uncoated copper has 3.07 ohms/kFT, 12ga. is 1.93 ohms/kFT, and 10ga. is 1.21 ohms/kFT. Hope I have steered you accurately.
Bob
- Status