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.

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.

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.

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?

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.

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?

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.

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.

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.

Quote:

---

Originally Posted by KeepItSimpleStupid

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? :)

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

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

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.

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.

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.

Quote:

---

Originally Posted by Marshall132

<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.

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

Power factor is a function of voltage and current. Inductance and capacitance of the DUT (device under test) will affect it.

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)

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

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

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.

Quote:

---

Originally Posted by KeepItSimpleStupid

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.

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?

Here is something with a bit more to chew:
Atlantic Quality Design, Inc., What is RMS Power?

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.