Neural wire shows HOT !!! can you suggest what the issue might be?
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Neural wire shows HOT !!! can you suggest what the issue might be?
So, my wife was using the hairdryer in the bath room, connected to the GFCI plug when the power to the plug goes out. I toggled the circuit breaker on/off, still not power. Suspecting the GFCI outlet is bad, I replaced it with a new one, and still no power to the plug and all the other outlets that it feeds.
Did some basic troubleshooting using a light tester and a volt meter and noticed that there is no volatge detected between the Hot (black) and Neutral (white) wire. however, there is 110 voltage between the Hot and ground, and surprisingly 110 voltage between Neutral and ground.
So jut to make sure I have my facts straight. In a good circuit, if I place a light bulb between Hot and Neutral, the light should come on. The light bulb should also come on if I place it between the HOT and ground. The light should NOT come on if I place it between Neutral white wire and the ground.
Do I have my fact straight? if the Neutral and ground turns my test light on, does this mean I have a short somewhere? If there is a short, wouldn't the circuit breaker go off?
Is there some kind of device that can help in troubleshooting where the short is?
Thanks.
Did some basic troubleshooting using a light tester and a volt meter and noticed that there is no volatge detected between the Hot (black) and Neutral (white) wire. however, there is 110 voltage between the Hot and ground, and surprisingly 110 voltage between Neutral and ground.
So jut to make sure I have my facts straight. In a good circuit, if I place a light bulb between Hot and Neutral, the light should come on. The light bulb should also come on if I place it between the HOT and ground. The light should NOT come on if I place it between Neutral white wire and the ground.
Do I have my fact straight? if the Neutral and ground turns my test light on, does this mean I have a short somewhere? If there is a short, wouldn't the circuit breaker go off?
Is there some kind of device that can help in troubleshooting where the short is?
Thanks.
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Can you please expand on "open neutral" meaning? is the neutral being Hot called open neutral ? i thought open neutral means when neutral line is disconnected ..Thanks
#4
You have two wires feeding a receptacle or circuit. Hot and neutral. If the neutral opens.... any
load connected will allow voltage to be measured on the neutral side because the power goes thru the load and then on to the neutral.. The neutral should always be at 0v with respect to ground.
load connected will allow voltage to be measured on the neutral side because the power goes thru the load and then on to the neutral.. The neutral should always be at 0v with respect to ground.
#6
At the panel..... the neutral/white wire is connected to the neutral bar which is at 0v measured to ground. This white wire carries neutral to your receptacle. If that wire opens..... it is no longer at 0v potential. The load on the circuit is connected to both white and black. So if there is no neutral..... the voltage travels thru the load and onto the neutral wire.
It can't cause a short as the neutral is open.
You have to identify everything on the affected circuit..... both working and non working. The problem is going to be at the last working device on that circuit or at the first non working device.
It can't cause a short as the neutral is open.
You have to identify everything on the affected circuit..... both working and non working. The problem is going to be at the last working device on that circuit or at the first non working device.
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Ok.. I will allow myself couple more questions and a few more hours of troubleshooting before I call an electricians ......
This is the part i am not fully understanding. So we know there is voltage on both Black and White. In your diagram, if the neutral white is cut (open), how and where does voltage travel to it? My limited knowledge was telling me that that can only happen if the hot wire is somehow touching the neutral wire .. so once again, if you dont mind, how/where is the white getting the power?
This is the part i am not fully understanding. So we know there is voltage on both Black and White. In your diagram, if the neutral white is cut (open), how and where does voltage travel to it? My limited knowledge was telling me that that can only happen if the hot wire is somehow touching the neutral wire .. so once again, if you dont mind, how/where is the white getting the power?
#9
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To prove there is an open neutral, put a load (e.g. a light bulb) between the hot and the ground. If the bulb lights up then it is confirmed.
I don't know if the GFCI would interfere with this test, so bypass the GFCI if possible.
Once you confirmed there is an open neutral, you just have to work your way up/down the "outlet chain" to find the open neutral. It could be a loose wire in a wire nut.
I don't know if the GFCI would interfere with this test, so bypass the GFCI if possible.
Once you confirmed there is an open neutral, you just have to work your way up/down the "outlet chain" to find the open neutral. It could be a loose wire in a wire nut.
#10
This is the part i am not fully understanding. So we know there is voltage on both Black and White. In your diagram, if the neutral white is cut (open), how and where does voltage travel to it? My limited knowledge was telling me that that can only happen if the hot wire is somehow touching the neutral wire .. so once again, if you dont mind, how/where is the white getting the power?
#11
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That includes disconnecting the GFCI outlet, because it is also a load (a tiny "appliance").
#12
Scenario 1. The light bulb is turned on and lighting up normally and there is no open (no broken) neutral. Using some real world numbers although still hypothetical, possibly exaggerated, and/or possibly rounded off, we might have 120 volts hot to neutral at the panel, one volt dropped (lost; consumed) in the black wire (say with a resistance of one ohm) going to the light, one volt dropped in the white wire (one ohm) going back to the panel, leaving about 118 volts for the light bulb (say, 100 ohms when glowing brightly) to use. With the black wire, the light bulb, and the white wire all in sequence as one current path, the voltage dropped in each section is proportional to the resistance in that section and the current is the same through each section. The maximum possible and conceivable current is about 1.2 amps since given Ohm's Law, the current is the voltage (120) divided by the resistance (100 ohms). The estimated voltage drop (voltage measured) of almost zero between the neutral terminal of the light and the neutral in the panel comes from about 1.2 amps going back times the resistance of the length of white wire, say, one ohm.
Scenario 2. (why you measured nearly 120 volts neutral to ground at the light fixture box now that you have an open neutral) The partial stretch of white wire starting at the light fixture is not part of the current path because the far end is not connected to anything (was broken off somewhere, somehow). The items in sequence in the single current path are: black wire from panel to light (say about one ohm), the light bulb (100 ohms, actually less because the bulb is cooler and not glowing), your voltmeter (typically 10000 ohms for analog meter with needle), the ground wire or conduit shell back to the panel (say one ohm), and the interconnection (bond) between ground and neutral in the panel (a fraction of an ohm). Maximum conceivable current flow is (120 volts divided by 10102 ohms) or about 12 milliamps). The share of volts dropped in the light bulb is (approx. 12 ma times 100 ohms) is approx. 1.2 volts. The share of volts dropped in the meter and showing on the meter dial is (approx 12 ma times 10000 ohms or) nearly the full 120 volts.
Ohm's Law, simplified: At all times in any portion of any circuit, the voltage measured between the start and end of the portion in question is equal to the current flowing through times the resistance of that portion. Or, the current drawn by any portion of any circuit equals the voltage applied divided by the resistance of that portion.
Most of the time, science teachers, students, electricians, even electrical engineers round off numbers representing electrical quantities to no more than two significant figures before going on to the next calculation. So the 10102 ohms above would be rounded to 10000 ohms almost immediately after it was seen in a preceding calculation.
Now that you took all the time to read all this, here is the booby surprise. I have not given you any help in finding exactly where the open neutral is.
Scenario 2. (why you measured nearly 120 volts neutral to ground at the light fixture box now that you have an open neutral) The partial stretch of white wire starting at the light fixture is not part of the current path because the far end is not connected to anything (was broken off somewhere, somehow). The items in sequence in the single current path are: black wire from panel to light (say about one ohm), the light bulb (100 ohms, actually less because the bulb is cooler and not glowing), your voltmeter (typically 10000 ohms for analog meter with needle), the ground wire or conduit shell back to the panel (say one ohm), and the interconnection (bond) between ground and neutral in the panel (a fraction of an ohm). Maximum conceivable current flow is (120 volts divided by 10102 ohms) or about 12 milliamps). The share of volts dropped in the light bulb is (approx. 12 ma times 100 ohms) is approx. 1.2 volts. The share of volts dropped in the meter and showing on the meter dial is (approx 12 ma times 10000 ohms or) nearly the full 120 volts.
Ohm's Law, simplified: At all times in any portion of any circuit, the voltage measured between the start and end of the portion in question is equal to the current flowing through times the resistance of that portion. Or, the current drawn by any portion of any circuit equals the voltage applied divided by the resistance of that portion.
Most of the time, science teachers, students, electricians, even electrical engineers round off numbers representing electrical quantities to no more than two significant figures before going on to the next calculation. So the 10102 ohms above would be rounded to 10000 ohms almost immediately after it was seen in a preceding calculation.
Now that you took all the time to read all this, here is the booby surprise. I have not given you any help in finding exactly where the open neutral is.
Last edited by AllanJ; 05-02-18 at 08:19 AM.