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Marshall132
Nov 18, 2009, 03:52 PM
While doing some electrical work the other day I noticed that on a circuit had 120v across the neutral and the neutral buss which it was at the time disconnected from. I immediately assumed dead short but there had never been a problem with the circuit breaker tripping. I was perplexed also because if I reconnected the neutral to the buss I assumed it would energize all the neutrals in the box and create both a short to ground back at the main breaker box where neutral and ground are connected and also create a 0v difference between all the hots and neutrals allowing no current to flow and nothing to function. Obviously neither of these things happened but I left the circuit off for the night and had an electrician friend look at it the next day. He found nothing wrong and told me I was getting the 120v from neutral to neutral buss because of a motor that was on in the circuit and by connecting the two contacts with my multimeter probe I was completing the circuit. This again perplexed me as I did not think checking voltage passed any current through the multimeter and if enough did to power a motor I thought it would destroy the multimeter.

So my question is this: First off shouldn't there be a voltage drop between the hot and neutral side of a motor? I thought there should be 120v on one side and 0v on the other. If there is no voltage drop wouldn't that be a short to ground? I know induction loads do not behave the same way as resistive loads do but I thought the basics were the same. Any insight would be greatly appreciated. Thank you.

medic-dan
Nov 18, 2009, 05:11 PM
Did you have the power on when doing this? If the breaker was off you should read 0v.

Your electrician friend is right.

There is a voltage drop across the motor. If it was running, you might see a small voltage drop due to the resistance in the motor, if the meter is sensitive enough. It will probably appear to be at line voltage. Reason: everything is wired in parallel.

If the motor had an open coil, you'd read line voltage. Only on a dead short would it read 0.

If the neutral is disconnected yet the hot is connected and powered then you will read line voltage to the neutral. The meter will not pass enough current to allow the motor to function. There is such little resistance in the motors coils you'll read line voltage.

Short to ground is something else entirely. In that case there is a path for electricity to flow to the ground. It could be at any point in the current path of the motor before it reaches the neutral.

KISS
Nov 18, 2009, 09:10 PM
Try this for an explanation. Take a power source, a light bulb and a switch all in series. Now measure the voltage across the switch.

1. Nearly zero when the switch is on
2. At the power supply voltage when it's off.

Draw it and you'll see why.

If you still have trouble, measure the resistance of the lamp out of the circuit.

Marshall132
Nov 22, 2009, 03:12 PM
Thank you for your response medic-dan. I still don't understand how if there is such little resistance in a motor than what limits the current flow through the circuit? In a resistive circuit there is a 120v drop, right?

medic-dan
Nov 22, 2009, 03:38 PM
I think "drop" is what is getting you confused. If you draw out a circuit, with the switch on, as KISS suggests, you'll show 0 across the switch. When that switch is open, you read line voltage, there is such little resistance in the motors windings that the meter will read "just about" line voltage. It's not really reading a "drop."j

Actually, as the motor starts running current required goes down. I forget exactly why but that is why a stalled or locked rotor will burn out a motor. I think it's counter EMF or something, been a while.

Look at your home, all the receptacles in a home are wired in parallel. If you plug three lights into an extension cord, all will read line voltage across the bulb. All are equally bright. Think about what would happen if they read 0 - you'd have a short.

Make up a little circuit and try it with a meter.

Marshall132
Nov 22, 2009, 04:16 PM
I understand that completely. You measure 120v across a resistive load, that is a 120v drop. On one side you are reading the 120v that comes from the powerplant, and on the other side you are reading earths potential, the difference is 120v. But on a motor, i.e. and induction load that does not seem to be the case. From what I gather reading the wires comeing to and from a motor you would read close to 0v, aka little to no voltage drop, very little resistance and what one would consider a short. Or is this wrong? Once the motor starts running would you read near 120 volts across the motor?

medic-dan
Nov 22, 2009, 04:38 PM
I think you've about got it. Remember, the source of power for all this is huge, the powerplant.

That is why the lights usually dim when starting a large motor load, it is almost a short but is of such short duration while the motor gets running that it doesn't trip the breaker (or fuse).

Put a motor on that extension cord and the lights, they'll all read the same voltage across them.

EPMiller
Nov 22, 2009, 05:43 PM
When a motor is not running, a coil inside it really is just a long length of wire. The resistance could be calculated if you know the gauge and the length. This resistance will be rather low on a motor of any size because the wire gauge will be sufficiently large in relation to it's length.

When a motor is running it is not a dead short, it has a resistance to the current flow on account of counter EMF. With a common voltmeter/DVM you would measure line voltage across a running motor just like a resistive load. Now stop that rotor and the motor would become almost a dead short because it is just a coil of wire that only has the reactance of the coil to resist the flow of current.

KISS
Nov 22, 2009, 06:18 PM
Actually, the motor has inductance because it is a coil of wire. So, the load can mostly be modeled as an inductor in series with a resistance. The voltage across an inductor is L di/dt.

Another way of looking at it is that the current in an inductor cannot change instantaneously, so therefore when the motor starts it sees only the resistive part of the circuit.

When you start adding inductances and capacitances in an AC circuit the phase angle between voltage and current changes.

EPMiller
Nov 22, 2009, 06:23 PM
Actually, the motor has inductance <snip>

Yes, I know. I used the word resistance because most people are using a common meter and that only measures resistance. How technical should I get? :)

KISS
Nov 22, 2009, 06:54 PM
You did OK, but counter EMF is different too. The motor does generate a small amount of voltage because it's rotating in a magnetic field. Counter-electromotive force - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Counter-electromotive_force)

The KEY points for inductance is that the curent can't change instantaneously. The key point for capacitance is that the voltage cannot change instantaneously. The correlary for capacitance, is that it acts as a short (actually called ESR or effective series resistae) when voltage is initially applied.

The P=VI or ohms law only apples to resistive circuits.

For AC circuits composed of resistance, capacitance and inductance and the voltage is sinusolidal P = VI*cos(theta). Theta is the phase angle between voltage and current. Capacitance tends to cancel inductance to make the load more resistive and inductance tends to cancel capacitance to make the load more resistive.

AC circuits are quite messy: Electrical impedance - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Electrical_impedance)

Z=sqrt((XL-Xc)^2+R^2)

Z, here is a complex impedance. With no capacitance or incuctance, it reduces to R. With inductance or capacitance, the effective R is higher.

This means that a motor with a locked rotor will draw the LRA (Locked Rotor Amps) and the motor would be damaged if this current is sustained. Inductance is what makes this rotating motor not overheat with the wire size used.

EPMiller
Nov 22, 2009, 07:08 PM
So you can see that I am not an engineer, just a tradesman. :) I know what I should see when I put my meter on things but I guess I don't know the exact technical reasons. Sigh... Guess I better back off this kind of question.

Marshall132
Nov 22, 2009, 07:09 PM
I think I get it all. I have a question to piggyback on this one. My mother just bought a device to plug in appliences and such to see how much power they are using. It gives you volts, amps, watts, VA, KWH, and so on. It also tells you the power factor. How can it do this? I was under the impression that this was something that could not easily be known.

EPMiller
Nov 22, 2009, 07:27 PM
<snip> It gives you volts, amps, watts, VA, KWH, and so on. It also tells you the power factor. How can it do this? I was under the impression that this was something that could not easily be known.

It's just math. If you know volts and amps then watts, VA, and kWh are simple calculations. At to power factor, I will let KISS address that one before I get burned. :D My guess is that it does an approximation of that value. You have to know the phase angle of the current in relation to the voltage to get that one I believe. It can be done, but the gizmo would be a bit pricier.

Marshall132
Nov 22, 2009, 07:45 PM
Yes sorry I meant to say specifically how does it give you the power factor. And is this a function of just inductance or capacitance as well?

Marshall Frye

EPMiller
Nov 22, 2009, 07:51 PM
Power factor is a function of voltage and current. Inductance and capacitance of the DUT (device under test) will affect it.

KISS
Nov 22, 2009, 08:00 PM
You can integrate for one period of voltage rather than long term and integrate for one period of current rather than long term. The time between voltage and current zero crossings and the frequency allows you to determine phase angle. The phase angle (theta) is related to the PF as cos(theta) ; Cos(0) is 1. or when it's purely resistive, PF=1.

Power factor determination for complex loads like switching power supply currents (PC power supplies) and Fluorescent lamps (non linear currents) I'll have to think about. The current drawn from those devices is inherently non-linear, but power factor correct is employed in modern ballasts and power supplies.

Now you can do all the mathematical monkey business to create what's called true RMS voltages and currents.

You can also multiply them together, peak detect them if you want etc.

There is a lot of useful info here: Power Measurement (Root Mean Square/RMS) (http://archive.chipcenter.com/circuitcellar/january00/c10r24.htm)

At one point, I had to purchase a 3 phase power meter so we could measure single phase non-sinusoidal power of < 24 volts.
I think they were about a grand.

While it is quite specialized and Kill-watt does what it does. Helps pinpoint energy hogs, but don't expect you utility to use it to compute your bill.

There is a lot of stuff here: Power meters, watt meters, kWh meters - PowerMeterStore.com (http://www.powermeterstore.com/index.php)

On the higher end of the scale there is the power line disturbance monitor.

KISS
Nov 22, 2009, 08:16 PM
There is this wierdness theat can burn you when trying to computer power with something called a phase angle fired dimmer to a lamp.
The voltage consists of pieces of a sine wave. At some point (phase angle) the electronic device conducts. It releases when the voltage crosses 0 volts.

To get even close to the right numbers, you need the true RMS values of voltage and current. The True RMS quantity is what would an equivalent DC voltage and DC current be.

Most meters are average responding, true rms reading. The sinusoid is flipped by a rectifier giving you 120 Hz. It's filtered by a capacitor to average it and multiplied to read 120 VAC.

RMS = square root of the average of the square of the voltage. Something you can do with calculus for a periodic signal.

EPMiller
Nov 22, 2009, 08:25 PM
You can integrate for one period of voltage rather than long term and integrate for one period of current rather than long term. The time period between voltage and current zero crossings and the frequency allows you to determine phase angle. The phase angle (theta) is related to the PF as cos(theta) ; Cos(0) is 1. or when it's purely resistive, PF=1.<snip>


Thanks for that post KISS. I never got calculus in high school. Didn't apply myself enough. I regret that one now. Wish I knew more about electrical engineering.

KISS
Nov 22, 2009, 09:30 PM
Here is a pretty good SIMPLE link: http://www.eznec.com/Amateur/RMS_Power.pdf

This isn't bad either: Atlantic Quality Design, Inc., What is RMS Power? (http://www.aqdi.com/rms.htm)

Here is something with a bit more to chew:
Atlantic Quality Design, Inc., What is RMS Power? (http://www.aqdi.com/rms.htm)

The title page has the formula for rms

=sqrt( 1/T * [definite Integral from 0 to T of f(t)]^2 dt

Sorry, I can't write in that fancy vbcode yet.

When you end up plugging in all the numbers like Asin(t) for the function, where A is the 0 to peak value of 120 VAC sinewave and integrating from 0 to 2*PI, out comes the magic number sqrt(2);

So the zero to peak value (Call it A) of a sinewave is A/sqrt(2).

Thus an sine wave that varies +-170 Volts has an rms value of 120V.

Stratmando
Nov 23, 2009, 06:16 AM
I would avoid disconnecting Neutrals at panel, unless it is positively NOT a Multi circuit. You can burn out some expensive equipment/appliances.

Marshall132
Nov 23, 2009, 08:19 AM
Why is that?

KISS
Nov 23, 2009, 08:35 AM
Ok, simple one. Your TV. It has a polorized plug. It should rally have a 3 prong plug. It doesn't.

The neutral is connected to the wide prong, which in turn is connected to ground.

Because components aren't 100% isulated, there is a small amount of leahage current present, so now that leakage current and whatever weird waveform that's present is TRYING to power all the other devices connected to the neutral.

Since ground is used for communications in USB and TV coax, raising these groud potentions can evoke damage. There may be some info in the surge supression sticky in the Electrical & Lightng link on the top left of this page.

Stratmando
Nov 23, 2009, 01:34 PM
Plus, if you have a circuit drawing 10 amps, and the other multicircuit hot has 1 light bulb on it.
The load being very great compared to the bulb will cause the bulb to blow, can go as high as 240 volts.

Marshall132
Nov 23, 2009, 02:30 PM
So you are saying that equipment that creates this stray current is traveling through other equipment that are tied into ground through other means than the neutral thus making it a complete circuit despite the neutral being disconnected.

Marshall132
Nov 23, 2009, 02:34 PM
And stratmando, could you contextualize your response a little more. Where does the 240 volts come from?

Stratmando
Nov 23, 2009, 02:34 PM
No, only the 2 circuits will be affected, The total voltage of the 2 circuits will equal 240 volts, without the neutral, 1 device could see 80 volts, the other would see 160 volts.
80 + 160 equals 240.
High or low voltages are dangerous.
Nothing should be going through ground.

Marshall132
Nov 23, 2009, 06:01 PM
You have completely lost me. The voltage between two different legs could be 240, though I thought it was more like 230, or 220. But you would need a short between two circuits with different legs (or the third if you have three phases, which I believe most residential electrical breakers don't) or a two pole breaker. Right? Is this what you are referring to? Or are you referring to the previously mentioned stray current and voltage?

KISS
Nov 23, 2009, 06:33 PM
In the US voltages have been bumped up a bit. The standard single phase distribution is from a 240 volt center tapped transformer. That center tap is the ground neutral bond. Connect both ends of opposite 240 phases together, you'll get zero volts out.

The second distribution that you'd see is 208 both in single and 3 phase. You'll also see 277 as a single phase voltage. 277 is used for lighting loads.

I'm thinking on the second part of the question.

Suppose we had a stove with a 4 wire connection and the sove turned on with the neutral disconnected. That neutral would se 120 V with respect to ground because it's being powered by a say a clock motor winding. Now suppose that there were 120 V outlets that used the neutral that you disconnected, say a convienience outlet. Now that 120 outlet is 240 V and limited in current by the winding of the clock motor.

It's harder to explain the varying differences, but they are caused by RF filters which tend to connect the line to ground through various leaky devices like capacitors. This is where varying voltages come from.

Here is a schematic of a typical filter: EP/VP Series (http://www.corcom.com/Series/PowerLine/EPVP/)

Note all the interconnections. Note the leakage at 125 V is .73 mA. This low of a current may also a source with a 171K output resistance. This will skew a reading when a certain voltmeters are attached esecially those that have a Z in ohms/volt measured. A 10 Meg ohm resistor (many voltmeters) in series will act as a divider.

The 10 meg ohm input Z meter needs 12 uA to measure 125 V, so weird voltages show. They can't deliver appreciable power, but could spark and could damage connected equipment.

Stratmando
Nov 23, 2009, 06:40 PM
I'm not talking about any induced voltages, stray currents or anything, Just basically if 2 circuits sharing a neutral, and the neutral is opened and a Motor was plugged in on 1 circuit(black circuit), and you had a lamp(on red circuit/opposite phase)You can get close to Line Voltage. I see More 240 than the others.
I was also talking about 2 Pole Breaker.

Stratmando
Nov 23, 2009, 06:42 PM
Check this out:
Weird Circuit Problem at Home? (http://www.thecircuitdetective.com/twocircuit.htm)

Marshall132
Nov 23, 2009, 10:53 PM
I get it now, I did not know what you meant by multi circuit.

Marshall132
Nov 24, 2009, 09:42 AM
"In the US voltages have been bumped up a bit. The standard single phase distribution is from a 240 volt center tapped transformer. That center tap is the ground neutral bond. Connect both ends of opposite 240 phases together, you'll get zero volts out."

The last sentence is confusing to me. Each phase is 120v to neutral and 240 to each other, right? Where are you getting 0v?

tkrussell
Nov 24, 2009, 01:07 PM
The single phase is across the 240 volts, the 120 volt is only a center tap of a transformer, based on ratio and physical location in a coil of wire, to deliver the 120 volts.

I found a couple of interesting diagrams. One new and shows grounding of current systems, and another very old by the verbage and obsolete 220 volts but gets the point across, that it is still only one single phase.

Stratmando
Nov 24, 2009, 02:23 PM
You should have 0 Volts between Neutral and Ground.
TK's drawing shows the Ground and Neutral Tied Together.

KISS
Nov 25, 2009, 08:49 AM
Zero volts has been defined as the reference. A reference can be anywhere. But it makes sense to make the earth the 0 volt reference point. Earth and neutral is tied to the center tap, hence that is also the zero volt reference.

What should be confusing is that wee call this single phase, because the PRIMARY is fed by a single phase system. The secondary is sort of a two-phase system with the phases tied together. These are just 180 degrees out of phase from each other, so putting the ends of a 240 line together will result in zero volts out.

Marshall132
Nov 25, 2009, 09:11 AM
So the two phases have a 240v difference, correct? And each 120v to the neutral/ground. So what do you mean when you say "putting the ends of a 240 line together will result in zero volts out"?

tkrussell
Nov 25, 2009, 10:13 AM
The last time I put two ends of a 240 volt circuit together, there was a large flash, melted metal, and a fire, due to what is called a short circuit.

Then there was zero volts.

I am interested in an explanation also.

In addition, what is "sort of a two phase"?

It is single phase, period.

KISS
Nov 25, 2009, 10:57 AM
My friend has two phase distribution in Philadelphia.

Two-phase electric power - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Two-phase_electric_power)

My transformer thing was all wromg:

Soldermans Basic Electronics: Transformers Series and Parallel Connections. (http://solderman.dapj.com/2005/09/transformers-series-and-parallel.html)

Some electronic trabformers are supplied as dual secondaries and dual primaries. The way you iterconnect them determines the voltage of the primary and current that the primary can handle. At the lower voltage it requres twice the curret and the higher voltage half. If you so happen to connect the primary in series the wrong way, you end up with zero volts out of the secondary.

The secondaries can be connected in various ways two. Namely series and parallel. You get an extra wire to deal with when the transformer is not center-tapped.

Connecting the 240's volt legs together 120 with an out of phase 120 V) should give you nearly zero volts. The phases would cancel. One way you would get double the voltage and the other the voltages would subtract.

Making the connections live and unintentensional and not with matched parts could stress out some parts.

120 in parallel with 100 VAC is not the same as 120 in parallel with 120 V.

EPMiller
Nov 25, 2009, 07:07 PM
The last time I put two ends of a 240 volt circuit together, there was a large flash, melted metal, and a fire, due to what is called a short circuit.
Then there was zero volts. <snip>

Hahahaha. I love it. Scares the bejeebers out of you too. Best one I ever did was cut into a 20A 120v line with a linesmans pliers. Big hole burned out of the side of it.

EPMiller