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Byron Trantham
09-06-2003, 3:08 PM
I have a circuit that is dropping voltage (lamp dims) when any appreciable load is applied - an iron for example. I removed all outlets and checked for loose or corroded connections. This is a GFI circuit controlling bathroom, outside and garage outlets. I have not replaced the breaker because it seems to be working. I did replace the two outlets on the outside of the house. Any ideas on how to track this problem down? :confused:

Don Abele
09-06-2003, 3:46 PM
Let me start by saying I am not an electrician, but have a lot of electrical experience (I've worked with electricians for the past 10 years and have done a lot of inspected home wiring).

Voltage drop in a line (dimming lights) is caused by too little supply for an particular demand. This is usually caused by undersized wiring.

A 120v 15A circuit uses 14g wire for up to 50 feet of run or 12g wire for lengths of 50-100 feet. For a 20A circuit it's 12g wire up to 50 feet and 10g for 50-100 feet.

If the wiring is undersized for a particular load, not only will you see diming lights, but under full load the wiring will get hot. This is a major fire waiting to happen. The only way to fix it is remove loads from the line or rewire it.

You don't state whether this is a new problem or something that has been around a while. If it's new, I would tend to think that there may be a short in the line somewhere which is decreasing line voltage. If you are familiar (and comfortable) with electricity, you can use a mult-meter to check line voltage and potential across the legs and this will help some. If you're not, hire an electrician, because as I said, if this line is overloaded you could have a potential fire.

Oh, something else just popped to mind. The breaker (sorry, got focused on the wiring). A 120v line typically have a 15A breaker, but can use a 20A if wired appropriately. I've seen this happen - you have a 120v 15A circuit, but it keeps popping everytime something comes on...so just go to a 20A breaker, right??? Well, you could ONLY if you have the right size wiring. Otherwise you'll draw more amperage through that line than what it can handle and the breaker will sit there cool as a clam and never pop.

So, check the breaker size and the wiring size and make sure they match. If they don't you'll have to fix one or the other. If this doesn't fix it, I'd check the line itself (voltage, potential, resistance) to check for problems within it.

Again though - if you are not comfortable working with electricity, hire an electrician.

Sorry if this was a little long...

Be well,

Doc

Kevin Gerstenecker
09-06-2003, 5:14 PM
Byron, I would start by checking the Amp Draw of the circuit in question. To do this, you will need a Multi-Meter that has Amp Clamp cababilities. This Amp Clamp can be a seperate unit that connects to a multi meter, or it can be incorporated as part of the meter itself. I would turn on everything that is connected to the circuit in question, and check the Amp Draw of the hot conductor at the panel. This Amp Clamp type device is clipped around the hot conductor, and thru inductance, it will read the Amperage travelling thru the conductor. If this problem just started, I would check for arcing connections, perhaps there is a Junction Box somewhere with a corrupt connection. You never want to overload a breaker with more than 80% of the rating of the breaker. For example, a 20A breaker may carry 20A, but the safe way is not to exceed 18A. The thing to remember is that when you have a Voltage Drop, the Amperage in that circuit increases, and that is where the danger exists. Is the amperage that does the work, and also the amperage that will knock you on your butt, and start a fire. You can also try this: Turn on whatever loads are associated with the circuit in question, and if it doesn't trip the breaker after about 1/2 hour or so, check the hot conductor associated with that circuit and see if the conductor (wire) is warm or hot. Just open the panel and touch the wire where it is insulated, a few inches above the breaker terminal. If it is warm to the touch, it is carrying too much load. The safest way to deal with that is to break down that circuit. Find the Junction Box, and identify the run that goes to whatever recepticles carry the most load, most of the time, and put these on another circuit. If there is not a connection problem, I would bet that putting the outdoor recpt's and the Garage on their own circuit will solve the overload problem, it that is indeed the case. Ya know, some irons, blow dryers and curling irons have an unreal amp draw. Good Luck, BE CAREFUL, and let us know what you find. If you need any other advice, just holler. :)

Howard Norman
09-06-2003, 6:50 PM
I have a circuit that is dropping voltage (lamp dims) when any appreciable load is applied - an iron for example. I removed all outlets and checked for loose or corroded connections. This is a GFI circuit controlling bathroom, outside and garage outlets. I have not replaced the breaker because it seems to be working. I did replace the two outlets on the outside of the house. Any ideas on how to track this problem down? :confused:

Since this is a GFCI circuit I suspect that it is a 15 amp circuit. That being so it probable is wired with 14 gauge wire. Not too unusual for such a circuit to dim lights when a heavy load comes on. As others have noted the question is this a new occurance or not. If the light dimming is new then something is wrong. I would suspect the connections in the service panel myself.

Howard

Byron Trantham
09-06-2003, 7:20 PM
Since this is a GFCI circuit I suspect that it is a 15 amp circuit. That being so it probable is wired with 14 gage wire. Not too unusual for such a circuit to dim lights when a heavy load comes on. As others have noted the question is this a new occurrence or not. If the light dimming is new then something is wrong. I would suspect the connections in the service panel myself.

Howard
Truth is, I don't know if it's new or not. The house is 13 years old and we're the original owners. We have, I think, always had problems with my wife's hair dryer in the bathroom. Every once in awhile it would pop the breaker. This "new problem" is a different circuit. It too, is a 15 amp circuit. I looked at the watts on the iron she is using and it's 1200 or about 10-11 amps. That, along with a 40 watt bulb, 200 watt sewing machine [that I know of] may very well account for the dimming.

Tomorrow I plan to double check the connection, both hot and neutral, for problems. My brother-in-law has a good clamp amp meter. I am going to borrow it and start testing loads.

As a general rule, how much current would have to be drawn in a 15 amp circuit (14 ga wire) to cause voltage to drop, in my case 10 volts?

Brad Schafer
09-06-2003, 9:18 PM
i saw this happen where wiring was "daisy chained" thru receptacles. one receptacle in the chain was marginal (internal short) and caused weirdities. replaced the receptacle and life was good again.

amp clamp is probably the best approach, though. good luck,


b

Don Abele
09-06-2003, 9:19 PM
As a general rule, how much current would have to be drawn in a 15 amp circuit (14 ga wire) to cause voltage to drop, in my case 10 volts?

Voltage drop is caused by resistance in the circuit. Either through the length of the wire itself or through connections along the wiring. That's also to say that there are no breaks, shorts, etc in the wiring.

I've always been told that the furthest receptacle on a run should have no more than a 5% drop in voltage (114 volts) due to resistance of the wiring. This is why as runs get longer, the size of the wire increases.

As Kevin said, amperage increases as voltage drops. So as the resistance in your line causes the 120v/15A to drop, the components plugged into those outlets will pull more amperage to try and compensate. Let me try to explain:

A 120 volt 1200 watt hairdryer pulls 10 amps: Amps = Watts/Volts

If you drop that voltage by 5% then that same hairdryer is operating at 114 volts and is pulling 10.5 amps to produce the same 1200 watts. Drop it by 8% and it is now getting 110.4 volts and pulling 10.9 amps.

Hope this helps more than confuses.

Be well,

Doc

Ted Shrader
09-07-2003, 12:44 AM
Byron -

Is this a new problem? If so, you might also check the neutral bus bar in the breaker panel. All of the lug screws should be tight. A loose connection there can exhibit the same symptoms you checked for in all the other places.

If it has always existed, then probably undersized wiring for the loads on the circuit.

I'm guessing this is new since you went hand-over-hand through the other parts of the leg.

Be careful & good luck.

Ted

Howard Norman
09-07-2003, 2:06 AM
Voltage drop is caused by resistance in the circuit. Either through the length of the wire itself or through connections along the wiring. That's also to say that there are no breaks, shorts, etc in the wiring.

I've always been told that the furthest receptacle on a run should have no more than a 5% drop in voltage (114 volts) due to resistance of the wiring. This is why as runs get longer, the size of the wire increases.

As Kevin said, amperage increases as voltage drops. So as the resistance in your line causes the 120v/15A to drop, the components plugged into those outlets will pull more amperage to try and compensate. Let me try to explain:

A 120 volt 1200 watt hairdryer pulls 10 amps: Amps = Watts/Volts

If you drop that voltage by 5% then that same hairdryer is operating at 114 volts and is pulling 10.5 amps to produce the same 1200 watts. Drop it by 8% and it is now getting 110.4 volts and pulling 10.9 amps.

Hope this helps more than confuses.

Be well,

Doc
Sorry, it doesn't work that way. The hair dryer has a finite impedence, would be resistance if it was DC rather than AC. Using your example of 1200 watts @ 120 volts with 10 amps of current. The impedence (Z) is 12 ohms. If the voltage drops to by 8% to 110.4 volts the current will be 9.2 amps. The impedence is fixed over the normal voltages that the appliance will see. If the voltage should drop to 50 volts the impedence will change but that is difficult to calculate without knowing more about the applicance.

Howard

Howard Norman
09-07-2003, 2:14 AM
As a general rule, how much current would have to be drawn in a 15 amp circuit (14 ga wire) to cause voltage to drop, in my case 10 volts?

Byron, the truth is one needs to know the length of the wire run to make that calculation. I don't have any wire resistance right here at my fingertips. If you want I can dig it up tomorrow. I think that you are on the right track by checking the terminals at the panel. If you have a good digital voltmeter it would be interesting to check the voltage, under load, at various outlets and points on the circuit including before and after the circuit breaker.

On a side issue, was your wife's hair dryer tripping the circuit breaker or the GFCI?

Howard

Byron Trantham
09-07-2003, 8:08 AM
Byron, the truth is one needs to know the length of the wire run to make that calculation. I don't have any wire resistance right here at my fingertips. If you want I can dig it up tomorrow. I think that you are on the right track by checking the terminals at the panel. If you have a good digital voltmeter it would be interesting to check the voltage, under load, at various outlets and points on the circuit including before and after the circuit breaker.

On a side issue, was your wife's hair dryer tripping the circuit breaker or the GFCI?

Howard

Actually, I can't say explicitly that I have checked the voltage, after loading, at each recptacle but I do know I have checked at various oultets and get the same [reduced] reading under load. Today I am going to go throught the netural buss bar and make sure that all connections are tight. Same thing with the Hot lead(s). As tedious as it is, I think I will also look for any outlets that have used the "push-in connection" rather than the screw terminal and make them screw type - more contact surface. I'll keep you posted.

Byron Trantham
09-07-2003, 4:08 PM
I had to replace one of the outlets in the ailing circuits and I found out that it was connected via the push in method rather than using the screw terminals. These push in type connections can't be that great for passing current. I used the screws on the new outlet. I am now going to inspect the remaining outlets in this circuit and wire them around the screws. I will let you know what happens.

Kevin Gerstenecker
09-07-2003, 7:17 PM
Byron, good idea to replace all the Stab Lock type recepticles you have. These are known to fail, and can drive you nuts looking for a problem. (I don't think anyone has to tell you that, at this point!) :D
Do just what you are doing, use the side terminal to wrap the pigtail around, and screw it down good and tight. One step better, and easier would be the back wire recepticles. You mostly see these in commercial work, as they cost a little more, but the convienience is worth it to me. They are also a commercial grade, and they last much longer than the 69 cent Eagle Recepticles you often see in the bins at the Box Stores. Good Luck, at this point I am interested to know what the actual problem turns out to be. Keep in touch! ;)

Howard Norman
09-07-2003, 7:35 PM
Byron, being basically a lazy person I went to Goggle rather than dig a book out for wire resistance. At the following URL

http://www.cirris.com/testing/resistance/wirecalc.html

I came up with the following information: For a 15 amp load 50 feet from the voltage source assuming 14 gage wire the voltage drop will be 12.45 volts due to the resistance of the 100 feet of copper wire feeding the load, remember the current has to travel through both wires. That translates into a resistance of 0.0083 ohms per foot for 14 gage copper.

However, that doesn't agree with my standard source which is the Handbook of Chemistry and Physics, 33rd edition, 1952. The Handbook gives a resistance of 0.0028 ohms per foot @ 50 degrees C or 122 degrees F. That corresponds to a voltage drop of 4.2 volts for the same 15 amp load 50 feet from the source.

I would tend to believe the Handbook of Chemistry and Physics numbers rather that the numbers from an unknown web site.

Howard

Byron Trantham
09-08-2003, 10:20 AM
Byron, being basically a lazy person I went to Goggle rather than dig a book out for wire resistance. At the following URL

http://www.cirris.com/testing/resistance/wirecalc.html

I came up with the following information: For a 15 amp load 50 feet from the voltage source assuming 14 gage wire the voltage drop will be 12.45 volts due to the resistance of the 100 feet of copper wire feeding the load, remember the current has to travel through both wires. That translates into a resistance of 0.0083 ohms per foot for 14 gage copper.

However, that doesn't agree with my standard source which is the Handbook of Chemistry and Physics, 33rd edition, 1952. The Handbook gives a resistance of 0.0028 ohms per foot @ 50 degrees C or 122 degrees F. That corresponds to a voltage drop of 4.2 volts for the same 15 amp load 50 feet from the source.

I would tend to believe the Handbook of Chemistry and Physics numbers rather that the numbers from an unknown web site.

Howard

Howard, I agree with you, if for no other reason than it's wiser to be conservative in this instance. I think I can inprove the voltage drop by re-wiring each outlet using it's screw terminals rather than the push-in terminals that the electrician used. Much more surface contact. It sounds like this "problem" has been heres all along. BTW my house is about 60' long and this circuit travels between two floors end to end. If I had only realized when they built this house I would have had 12 ga run. All my add-on circuits are 12 ga even the wiring for the overhead lights in my shop. I did that because all my lights plug into outlets in celing and I wanted to use those outlets for tools if I needed to. I'll what my results when I have finished.