phillipreis
Member
I am a bit rusty on theory. Two questions, how does an open neutral cause 220v to occur on a 120v circuit, and why does voltage drop cause heat from increased resistance or another way. If this is too remedial I apologize.
open neutrals and voltage drop
- Thread starter phillipreis
- Start date
- Status
Re: open neutrals and voltage drop
With an open neutral you often get the equivalent of two (or more) 120-volt loads hooked in series on a 240-volt circuit. If the loads are perfectly balanced this is not a problem, each load would see 1/2 the voltage or 120-volts. If the loads are not equal then one load would see more than 120-volts and the other less. The more they are uneven the larger the voltage difference.
Voltage Drop:
With an inductive load (such as a motor) reduced voltage (due to voltage drop) will result in increased amperage. Increased amperage leads to increased voltage drop which results in increased amperage... well you get the picture. The voltage that is "dropped" is actually being used to heat the conductor, so a conductor that is experiencing voltage drop will often be operating at a higher than normal temperature. The higher temperature increases the resistance of the wire, which causes additional voltage drop and again we get a cycle started. That is why it is important to limit voltage drop. A purely resistive load such as incandescent lighting will just consume or produce less wattage at reduced voltage, so it is not quite as severe a problem with resistive loads.
Hope that helps,
With an open neutral you often get the equivalent of two (or more) 120-volt loads hooked in series on a 240-volt circuit. If the loads are perfectly balanced this is not a problem, each load would see 1/2 the voltage or 120-volts. If the loads are not equal then one load would see more than 120-volts and the other less. The more they are uneven the larger the voltage difference.
Voltage Drop:
With an inductive load (such as a motor) reduced voltage (due to voltage drop) will result in increased amperage. Increased amperage leads to increased voltage drop which results in increased amperage... well you get the picture. The voltage that is "dropped" is actually being used to heat the conductor, so a conductor that is experiencing voltage drop will often be operating at a higher than normal temperature. The higher temperature increases the resistance of the wire, which causes additional voltage drop and again we get a cycle started. That is why it is important to limit voltage drop. A purely resistive load such as incandescent lighting will just consume or produce less wattage at reduced voltage, so it is not quite as severe a problem with resistive loads.
Hope that helps,
dave_asdf
Member
- Location
- tampa florida
Re: open neutrals and voltage drop
------)(-----------B------------------>
`````)(``````````````> 120V```````> load 1
``|``)(```````````````````````````>
240V )(-----------N------------------|
```|`)(`````````````> 120V````````>
`````)(```````````````````````````> load 2
------)(-----------A------------------>
try to ignore all the ``` marks, had to space out the "drawing"
Concidering the above transformer is a 1:1 ratio, you can see how an open in the neutral would give you 240V across both of the loads.
in a resistive component heat is the power that component uses. for example, lets say you have a 10 amps of 120V electricity flowing through a resistort. the ammount of power or heat you are producing is 1200w (P=IV). the resistor is heating up to whatever temperature it has to to pump out 1200 watts of energy (less temp if it's under water or air blowing on, or it's a large resistor). resistors can act one of two ways. as temperature goes up, so does resistance -or- as temperature goes up resitance goes down. these are positive and negative temperature coefficients. pretty much all conductors have a positive coefficient (temp up, resistance up) and have a "temperature curve" associated with their resistance. this is how RTDs (electronic temp detectors) work.
------)(-----------B------------------>
`````)(``````````````> 120V```````> load 1
``|``)(```````````````````````````>
240V )(-----------N------------------|
```|`)(`````````````> 120V````````>
`````)(```````````````````````````> load 2
------)(-----------A------------------>
try to ignore all the ``` marks, had to space out the "drawing"
Concidering the above transformer is a 1:1 ratio, you can see how an open in the neutral would give you 240V across both of the loads.
in a resistive component heat is the power that component uses. for example, lets say you have a 10 amps of 120V electricity flowing through a resistort. the ammount of power or heat you are producing is 1200w (P=IV). the resistor is heating up to whatever temperature it has to to pump out 1200 watts of energy (less temp if it's under water or air blowing on, or it's a large resistor). resistors can act one of two ways. as temperature goes up, so does resistance -or- as temperature goes up resitance goes down. these are positive and negative temperature coefficients. pretty much all conductors have a positive coefficient (temp up, resistance up) and have a "temperature curve" associated with their resistance. this is how RTDs (electronic temp detectors) work.
ryan_618
Senior Member
- Location
- Salt Lake City, Utah
Re: open neutrals and voltage drop
Dave, my compliements on your ingenuity of drawing a transformer and circuit. Well done!!!
Dave, my compliements on your ingenuity of drawing a transformer and circuit. Well done!!!
Re: open neutrals and voltage drop
Sorry Fellas
with a balanced load would only drop 120 v across each resistor from a 240 v center tapped winding.
And with any imbalance on a 240 v Edison circuit 240 volt could never drop across any one load due to a lost or broken neutral.
Roger
Sorry Fellas
that's impossible, a lost neutral
with a balanced load would only drop 120 v across each resistor from a 240 v center tapped winding.
And with any imbalance on a 240 v Edison circuit 240 volt could never drop across any one load due to a lost or broken neutral.
Roger
Re: open neutrals and voltage drop
Roger:
If Load 1 is 1 ohm and Load 2 239 ohms then Load 1 will have 1 volt and load 2 239V. So, it is not impossible.
Ok, that ratio load 1 / load 2 is not impossible but is is high unlikely. But if load 1 is 50 VA and load 2 is 1 KVA, the voltage in load 1 will be 12V while load 2 will recieve 228V; and this ratio is more likely to happen.
Joe.-
Roger:
If Load 1 is 1 ohm and Load 2 239 ohms then Load 1 will have 1 volt and load 2 239V. So, it is not impossible.
Ok, that ratio load 1 / load 2 is not impossible but is is high unlikely. But if load 1 is 50 VA and load 2 is 1 KVA, the voltage in load 1 will be 12V while load 2 will recieve 228V; and this ratio is more likely to happen.
Joe.-
Re: open neutrals and voltage drop
Joe, although 239 is close, it is not 240. If this were a test question your answer (or the students you may have teaching) would be wrong.
What may seem insignificant may not be.
So, it is impossible to have full system voltage dropped across any one load (unless there is only one load)in a series circuit.
Roger
Joe, although 239 is close, it is not 240. If this were a test question your answer (or the students you may have teaching) would be wrong.
What may seem insignificant may not be.
So, it is impossible to have full system voltage dropped across any one load (unless there is only one load)in a series circuit.
Roger
phillipreis
Member
Re: open neutrals and voltage drop
Thank you to all those replying to my message. These questions resulted from explaining to a resident why they cannot use a standard dimmer on a fluorescent lamp, and from my early days before taking some classes when I removed a neutral from a wire nutted group of neutrals to make room for the neutral to the hanging lamp I was installing in a bar. The cash registers went to high speed and the satellite system fried along with the receiver and coffee makers, all 120V.
I am more enlightened now and apologize for any arguments that resulted.
Thank you to all those replying to my message. These questions resulted from explaining to a resident why they cannot use a standard dimmer on a fluorescent lamp, and from my early days before taking some classes when I removed a neutral from a wire nutted group of neutrals to make room for the neutral to the hanging lamp I was installing in a bar. The cash registers went to high speed and the satellite system fried along with the receiver and coffee makers, all 120V.
I am more enlightened now and apologize for any arguments that resulted.
Re: open neutrals and voltage drop
Roger, the quote you made from dave_asdf's answer tells " ....would give you 240V across both of the loads", an let me say it again: "both loads".
I didn't pass the test because the answer was tricky ("And with any imbalance on a 240 v Edison circuit 240 volt could never drop across any one load due to a lost or broken neutral" -you said).
So, at least give me half point in that answer and I promise to give more attention to the questions next time.
Roger, the quote you made from dave_asdf's answer tells " ....would give you 240V across both of the loads", an let me say it again: "both loads".
I didn't pass the test because the answer was tricky ("And with any imbalance on a 240 v Edison circuit 240 volt could never drop across any one load due to a lost or broken neutral" -you said).
So, at least give me half point in that answer and I promise to give more attention to the questions next time.
- Status