I'm wondering just how deadly meggers are, if anybody here has ever heard of anybody being killed or shocked hard using a megger and if the low amperage isn't dangerous while the high voltage is trying to kill him. Also, I've read online that a long cable can hold megger-induced voltage for a long time and that you have to ground it for "a few minutes" after a megger test. That seems excessive to me - a few minutes of grounding?

I'm thinking about buying a megger (not sure whether to get 500v or 1000v). Anything you can tell me about megger dangers or accidents involving them will be appreciated. I like to learn from other people's mistakes!

So...exactly what will happen to me (and in what order) if I grab both leads and someone cranks 1000v through me? Is the duration too short to kill me? If it can kill me, what body part fails first that causes death? Has anybody here got experience being zapped by one of these things?

Thanks in advance for sharing your wisdom and experience.

I believe the megger is set to give off about a milliamp.

I once asked a similar question as a reply in this thread .

I'm sure the others here can add a whole lot more from personal experiences, etc.

If you have wire connectors on the ends, do they hold all the voltage in?

Spark Master,
HOW do you call yourself "Spark Master" if you've never 'experienced' a megger?

They are set for milliamps. An old teacher of mine used to have the class hold hands and the student at the front and rear of the line had to grab a lead......... fun. I'm pretty sure they would not allow this anymore. Not deadly (Personal Opinion) unless you have a health condition, such as a pacemaker, but not something to play with arbitrarily.

Most 1000 volt meggers have a 500 volt switch nowadays, so buy the 1000 and get both. DC meggers are better for longer runs.

Because of capacitance in conductors (remember, a capacitor is a conductive material, wrapped in an insulative material, wrapped in a conductive material - can you say cable in conduit boys and girls?) so it WILL hold a charge and WILL require grounding out. You only have to ground it for a brief period, or wait a few minutes (up to an hour) for the charge to dissipate on it's own.

Yup, I can tell embarrassing stories about this, but won't

You will find a megger to be one of the most useful tools you'll ever own. Fluke makes a great one that is volt meter, ohm meter, capacitance checker all in one, a bit pricey, but great. I disdain that they taught this in my apprenticeship but seem to not teach it anymore, real pity.

Also, if you're a geez (THANK you veddy much) you may want to look at the old amprobe wind ups, no batteries to go bad. I don't trust that 'lectricity, got's ta have my windup. OK, I own both.

Of course, NOBODY beats Biddle for the meggers, AND the training. Both a bit pricey, but I recommend them highly.

Good luck, and........... wear gloves.

[This message has been edited by George Corron (edited 07-30-2003).]

how many joules is the average megger ? i wonder if one could defib (in a pinch) with a one?

Steve (aka... any port in a storm) sparky

[This message has been edited by sparky (edited 07-30-2003).]

Her's where I show my ignoarance, but aren't joules a measurement of stored/generated then discharged energy? I know all our strobe power supplies and things like defibrilators and "thumper" partial discharge cable fault locators, ( http://www.hipotronics.com/ ) by example are devices expressed in joules. But my megger can apply a fairly steady voltage into a cable under test, but does so at a maximum 5 milliamp output.

If you are talking about the voltage that capactively builds up and can lay stored on a long cable run, I guess that would be applicable, but would vary with a number of factors including the length of the run and the amount of energy wou were able to dump into the cable in the first place.


[This message has been edited by US Coreman (edited 07-30-2003).]

A joule {pronounced 'jewel'} is equal to 1 watt·second of energy, or 1/3600 of a watt∙hour [NOT kilowatthour.] CF Dalziel and his noted Dalziel's Equation are pertinent in electric-shock response.

[Small aside — there are portable DC insulation-test sets that produce 160kV.]

People tell me "It's the amp load that will kill you, not the voltage".

If I say, "Okay then, grab onto the hot lead of 4160", they say, "Well...at high voltages, the voltage alone will kill you."

Can somebody explain this?

When I hear these two statements, there seems to be a big gray area of how much voltage and how much amperage it takes to hurt me. How much voltage/amperage does it take to hurt somebody? 480 at 1 amp? 120 at 5 amps? 4160 at 0 amps?

Thanks again

Most of the modern electronic meggers will deliver a maximum of a few milliamps into a short-circuit. The current likely to flow through your body if you grab hold of the probes will obviously be less due to your body's resistance.


It's not the high voltage in itself which would kill you, but rather the high current flow which results from that high voltage.

It is the current which does the damage during an electric shock. How much harm is done depends upon the intensity of the current, which parts of the body it flows through, and for how long. As the current and the duration are increased, so there is more chance of burns. Current passing through the vital organs of the body (e.g. the heart) is far more dangerous than that passing between, say, two fingers on the same hand.

Ohm's Law applies to the human body just as to metallic conductors. Thus at higher voltages you will get more current for any given value of body resistance.

I [i]knew[/i] the subject of this topic sounded familiar...

The megger is a great tool if you do a lot of fishing.

Thanks, pauluk...so you're saying that when the megger runs into a dead short, the dead short is the path of least resistance, and I won't soak up much current?

And you're saying that my body's resistance creates an amp load which burns me?

This means that even if there is no amp load downhill from me, if I grab a hot lead my body will then pull an amp load which would vary depending on how much meat and bone it was going through?

In other words, my body would be like a light bulb, consuming energy?

To ThinkGood: har har re. Megadeth

To LK:
------------------------------------------
"The megger is a great tool if you do a lot of fishing."
------------------------------------------
DO TELL! Is it like that joke with the game warden? Makes me wonder...what is the depth range of a 1000v megger with a DeWalt 18v drill (3rd gear) driving it? I guess you could just motor around the lake with the drill running the megger, dragging a net behind...

As far as a power line is concerned, yes. If you connect yourself across that line, then current flows through your body at a rate dependent upon your body's resistance (I = E / R). Your body will dissipate heat the same as any other load (P = I x E).

The power line coming into your house has a very low source impedance, comprised of just the internal resistance of the transformer and that of the connecting cables. This source impedance will be just a fraction of an ohm. That's why you can connect a heavy load yet the voltage stays reasonably constant. It's also why you get a massive current flow if you short-circuit the supply (e.g. if the source impedance was 0.1 ohm, then a short-circuit at 120V would result in I = 120 / 0.1 = 1200 amps).

A megger, on the other hand, has a much higher source resistance. A 500V megger which can deliver a maximum of, let's say 5mA, would have a source resistance of 100,000 ohms. It's that high internal resistance which limits the current.

If you touched the probes in such a way that the resistance through your body was also 100,000 ohms, then the 500V source would be divided equally between the megger's internal resistance and that of your body. You would experience a shock of 2.5mA at 250V.

Thanks a lot, pauluk. I appreciate the work you put into that answer.

So a 500v megger has resistors built in that might total 100,000 ohms so the operator doesn't get shocked?

And without those resistors in the megger, it would put out a lot more current?

As a side note: When the voltage goes up, the current tends to flow closer to your surface in your body. This was indirectly used by performers who hooked themselves up to a very high voltage sources with a very low current to create cool effects like sparks.

(I remember in occasion in school when the teacher hooked up one of the girls to a 100kV source so that her hair stood out... )

Hey C-H,

I saw a couple of those shows (when I was in the Philippines) where they sent "1,000,000 volts" through a guy who was standing on top of the device holding a 2x4 in his hands with a piece of tin foil. When they turned it on, the 2x4 burst into flames.

Does anybody know if that was AC or DC?

There were lightning bolts 6 feet long cracking off the device in all directions.

Can somebody explain what happened there? How does the 2x4 burn and the person doesn't?

Spark Master,

BTW, I think LK meant that a Megger would be good to drive worms out of the ground. (to use for bait when fishing)

Bill

George,
You're right about the Megger being a very handy piece of equipment!.
I've found heaps of faults, that I wouldn't have normally found, had I not been able to test Insulation Resistance to Earth.
The one I use has 250V, 500V and 1000V ranges.

Bill, that's pretty funny.

The fish & game warden noticed that his next door neighbor was coming home with his trailered boat filled to the brim with fish, day after day. He asked the fisherman how he gets so lucky. The fisherman said, "Well, come out fishing with me tomorrow and I'll show you."

The next day they were out on the lake. The fisherman lit a stick of dynamite, threw it in the lake, it detonated and a bunch of fish floated to the surface. He used his net to scoop them up and put them in the boat.

The warden said, "You can't use dynamite to catch fish like this, it's illegal."

The fisherman lit a stick of dynamite, handed it to the warden and asked, "Are you going to talk or fish?"

Seems like a drill-operated megger-type device would be so much quieter...

Well, the internal resistance isn't comprised entirely of resistors deliberately inserted in series.

Part of the internal resistance is the windings of the transformer which generates the high voltage (I'm talking about a modern, electronic type megger here).

Maybe another example would help: Look at a 12V automobile battery versus a 12V transistor radio battery. Both have an EMF, or voltage source, of 12 volts.

The difference is that the chemical structure of the lead-acid car battery gives it a much lower internal resistance than that of the equivalent-EMF radio battery.

The latter might have an internal resistance (caused by its chemical composition) of around 10 ohms. That means that even if you put a dead short on the terminals, the current is limited by that 10-ohm internal resistance (I = 12 / 10 = 1.2A).

But the car battery has a much lower internal resistance, maybe in the region of 0.01 ohms. Short the terminals of that and the current flow is 1200 amps!

I own a small 500/1000 volt megger made by AVO It is a model BM2222. It is powered by six AA batterys.It has a built in powered low resistance meter. It works excellent for troubleshooting motors and such. I would reccomend having one on every electricians truck.

One more thing, A few years ago I was testing a Large AC motor 6000 Hp. I had it apart and in a rewind shop. The reccomended Hi pot voltage was twice the voltage plus a 1000...I hipotted the motor to 10k volts and got distracted and went to lunch. Needless to say when I did discharge it packed quite a punch. Meggers are an excellent tool and just need to be used properly,and discharged after to ground after.

And:


Sounds like a Van De Graaff (sp???) Generator.

This device creates high potential Static Charges, which may easily exceed 100K Volts between Charged body and opposite potential (typically Earth).


If done with a Van De Graaff Genny, then it's neither AC or DC - only Static.
If done with a Tesla coil, or similar plasma tossing Induction Animal, then this would be AC.


This sounds more like a Tesla coil, but a Van De Graaff Genny could do the same, especially if the air was dusty.


It's a trick!

Plasmas from a Tesla coil will flow around and sometimes through wood, but rarely would this only cause the wood to burn and not ignite or shokken ze sh*@zenhouzen out of the person holding the wood!

Static charge transfers would unlikely carry enough energy to ignite wood alone, but if the wood was soaked with a highly flammable material - such as Gasoline - a few sparks will get the flames going!

Man, I would hate to have that guy's job!

Scott35

Pauluk, thanks again, your answers are great. I sat and thought about that until I rose to a new level, referring back and forth from your explanation to the cover of my Ugly's Electrical References book until I got it. I know how to figure volts, amps and watts calculations, but haven't done much with ohms yet. This thread is taking me headlong into the land of Ohm. Thanks!!! It's going to be great to take a resistance reading and calculate how many amps I'm looking at. That was a hole in my knowledge that had to be filled.

Can I can test the resistance through a battery?

Is there a way to test resistance while a wire is hot, or does it always have to be dead?

Scott35, I believe it was a Tesla coil. Thanks to you, too, for the enlightening answer!

What do people normally do with a Tesla coil or a Van De Graaf generator?

Here's a cheap new megger I found, has anybody used one of these?
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=2548791771&category=4678#ebayphotohosting

$55 isn't too bad for a new megger, is it worth it? Seems like a megger's job is only to crank out some voltage to see if there's a short...

I tried to copy and paste the picture alone, it didn't work, so I had to post the url.

Are new meggers better than the old antique ones? I like the looks of the old ones. They have that Ben Franklin look to them. Maybe there's something I need to know about them that somebody here can tell me. Thanks for the help, guys, this message board is a godsend.

mlk682, how long did the AC motor hold the charge before it discharged? Was that a 4160v motor?

Here's a shot of the Van de Graf generator(s) at the Museum of Science in Boston. They give a great show there. A bit loud (and scary) for small children when the lightning bolts start flying around!



(The person you see is in something resembling a large Bird Cage)

Bill

[This message has been edited by Bill Addiss (edited 08-02-2003).]

Amazing.

The coil I saw was a pedestal about 5 feet tall, around 18" to 24" in diameter. The guy stood on top of it, holding the 2x4 with foil wrapped around one end for a handle. That kept the gasoline off his hands, as Scott35 told us.

It must have been scary to be the first one to stand on that thing and say, "Switch."

I wouldn't want to grab the leads of a megger and get surprised by some physics law I'm not aware of.

So a megger can't shock me very badly, right? It's not going to knock me out, stop my heart or anything, even if I'm standing in water? I haven't tested the feeling of various amounts of amperage through my body, so I have no idea of how much it would hurt to get hit by a megger, even with 5mA. Maybe somebody can tell me, like George Corron who did it in class, or somebody who accidentally came in contact with one. I want to know what a megger can do to me, worst case scenario. Is it dangerous if used improperly, or does all the built-in resistance make it as safe as a GFI that won't allow me to take a lethal shock? Will I just feel a little tickle? If the unit gets wet, is there a danger of getting more than I bargained for if the water causes the power to get re-routed around the resistors?

[This message has been edited by Spark Master Flash (edited 08-02-2003).]

Don’t forget that with insulation-resistance testing you are charging the natural capacitor whose two “plates” are the normally grounded and normally ungrounded {er, “phase” conductors.} That stored energy can be discharged through you, and should not be dismissed as always harmless.

Another mode of stored energy might be the inductance in a winding should one mistakenly do a test where the instrument is paralleled not with a capacitor, but an inductor.

What you have to always keep in mind is that volts, amps, watts, and ohms are so interlinked that you can't change one without changing at least one of the others. When you get into the realm of AC, then reactance also plays a big part, but I think you'll find it best to get a thorough understanding of the DC basics before trying to tackle that.

Not directly with an ohmmeter, because the meter supplies power from its own batteries and the EMF from the battery you're trying to test would interfere with the reading (that's the best case scenario; the worst is that you'd burn out your meter!).

You can calculate the internal resistance of the battery indirectly, however. What you need to do is measure the open-circuit voltage at the battery terminals using a voltmeter of fairly high resistance.

Next, you connect a load across the battery (e.g. a lamp), and measure both the current flowing and the voltage now appearing at the battery's terminals. You can then use Ohm's Law to calculate the internal resistance of the battery.

An example:
You get an off-load voltage of 12.6V. When you connect a load which draws 2A the voltage across the battery drops to 11.8V.

The voltage being lost across the internal resistance of the battery is therefore

12.6 - 11.8 = 0.8V

You know that 2A is flowing through the battery, so the internal resistance by Ohm's Law is then

R = E / I = 0.8 / 2 = 0.4 ohm.


Again, you can't use your ohmmeter because the circuit is energized, but you can apply Ohm's Law again if you know the current and voltage.

Measure the voltage between the ends of the wire you want to test, check the current flowing through it, then the resistance is R = E / I.

That's for DC or non-reactive AC. It gets more complex when inductance and capacitance are involved.

Just be a little careful how you take that assumption.

A GFI certainly provides a very high level of protection, but a shock at just below 6mA could still be a little risky in some situations.

The other point is that the GFI trips only if the shock you receive is from line to ground. If you get yourself across hot and neutral (and there is no appreciable current flowing to ground at the same time), then it won't help you one bit.



[This message has been edited by pauluk (edited 08-03-2003).]

Pauluk, a million thanks.

So...in the case of the DC battery showing 12.8 volts, can I assume that the battery is capable of providing up to 31.5 amps? I=E/R = 12.8/.4 = 31.5 amps.

Maybe the voltage will drop and change all the numbers as the amp load is applied. Do I work from the full, unladen voltage to calculate amps and ohms, or do I have to let the amp load bring the voltage down before I'll have good numbers to work with?

--------------------------------------------
"The voltage being lost across the internal resistance of the battery is therefore

12.6 - 11.8 = 0.8V

You know that 2A is flowing through the battery, so the internal resistance by Ohm's Law is then

R = E / I = 0.8 / 2 = 0.4 ohm."
-------------------------------------------

So it looks like you're saying that we're not working off the full EMF when we calculate the resistance through a battery, but instead, we work off the voltage drop and the full amp reading. Seems like the formula would be this:

R=EMF lost/I

It would be easy to make the mistake of working off the full voltage divided by the amps drawn, which would be 11.8 volts/2 amps = 5.9 ohms.

Please correct me if I'm wrong on any of this, I appreciate it.


[This message has been edited by Spark Master Flash (edited 08-03-2003).]

Correct -- Although as you need a dead short across the terminals to achieve that much current, you couldn't put it to any productive use. There's also the point that such a small battery wouldn't supply this much power (12.8V x 31.5A = 403.2W) for more than a very brief period. The battery would be fully exhausted very quickly and the EMF (and therefore the current also) would quickly taper off.

Ohm's Law applies equally to the entire circuit, or to any portion of the circuit. You just have to make sure you match up the values correctly.

Let's take the last part of your post and apply it to the example in hand:

What you've done there is to calculate the resistance of the load. We know that 2A is flowing through the whole circuit, and 11.8V is the voltage which appears across the load. That's applying Ohm's Law to the load portion of the circuit.

Let's extend it to the whole circuit. We know from the previous measurement that the actual EMF of the battery is 12.6V. Now, applying Ohm's Law to the entire circuit we get:

R = E / I = 12.6 / 2 = 6.3 ohms.

That's the resistance of the complete circuit, including the internal resistance of the battery.

We've already calculated the internal resistance of the battery as 0.4 ohm. Add that to the 5.9 ohms of the load and you get the same result for total circuit resistance of 6.3 ohms. It checks out.

That's right. It's appying Ohm's Law to the appropriate part of the circuit:

1. To calculate the resistance of the entire circuit (battery+load), you need the full EMF which is powering that circuit - 12.6V.

2. To calculate the resistance of the load, you need the voltage which appears across the load, i.e. 11.8V.

3. To calculate the internal resistance of the battery, you use the voltage which is dropped internally by the battery's resistance, i.e. 0.8V.

Pauluk, I love this stuff. Electrical sure is a thinking man's game.

So it appears to me that resistance goes up as the amperage increases. Kind of like forcing water through a sponge, as more water is forced through the sponge per second, the sponge resists the flow more, to where it takes more force to push the water through the sponge. Am I right? Eventually, the sponge must act like a wall, where it would take tremendous force to push the water through it.

You've certainly opened my eyes to some things I needed to know. Thanks to all of you, especially Pauluk, you really broke it down for me and it came at a good time. I've used my Fluke to check voltage, amps and dead shorts. Now I'll be using it to check resistance, and it sure opens up a whole new world.

Pauluk, you said "Your body will dissipate heat the same as any other load (P = I x E)."

Are watts basically heat given off by a device (or by my body if it's being electrocuted)?

The way I understand it, it's very similar to "Fluid hydraulics" (i.e. pumping fire apparatus). Can you get X amount of water (Gallons Per Minute) through Y hose that is Z feet long? How much energy (Friction Loss in psi) will it take to accomplish this?

Example - 175 GPM through 50' of 2.5" hose takes 3 psi FL to do this, which must be included when calc'ing your pump pressure.

To push twice as much water through (350 GPM), it takes four times as much energy (12 Psi).

In fire engineering, it's not the pressure, it's the GPM. Just like wiring - it's not the voltage, it's the amperage.

Can you get 15A through 14ga AWG CU? Sure.

Can you get 30A through 14ga AWG CU? Sure, but the wire will heat up beyond the design specifications of the insulation (safe operating temp), and it creates a hazard. And, just like hose, the more you try to push through it (GPM/A), the more if will fight the flow (FL/R).

The water analogy is very commonly used in textbooks. Basically, if you have a certain amount of resistance in a hose, then to increase the flow (GPM) you need greater pressure (PSI).

A copper conductor has a certain amount of resistance. To increase the flow of current (amps), you also need a higher pressure (volts).

Here's a quick diagram which might help you visualize the internal resistance angle a little better (apologies for the poor quality -- I just drew it "freehand" very quickly):

The part within the dotted blue box represents the battery with its positive and negative terminals. It's shown represented as the actual source of EMF and a series resistance.

Assume that the source of EMF is the constant 12.6V we've been using in the example above. Whenever we draw current from the battery there has to be a voltage dropped across that internal resistance. In our example, that internal resistance is 0.4 ohm, so if you draw 2 amps of current you "lose" 0.8V across that internal resistance.

The EMF remains the same, but because we can only get to the two terminals of the battery, we can get only 11.8 volts to the load.

Does that make it any clearer?

In reality, the EMF and internal resistance cannot be separated in this way -- The resistance is an integral part of the chemical composition of the battery. This representation just makes it a little easier to understand what's happening inside the battery.


[This message has been edited by pauluk (edited 08-05-2003).]

That's good stuff...thanks for the diagram.

Now if I have a 20 amp circuit, 120 volts, resistance of .01 ohms, then if I short it to ground it will deliver 1200 amps.

Then, for about 5 cycles, it's still delivering 1200 amps - until the 5 cycles are over, at which time the circuit breaker pops? Please correct me if I'm wrong.

So the circuit breaker waits 5 cycles before doing anything...and during the brief time it takes for those 5 cycles to occur, the 1200 amp load goes through the 20 amp breaker, but the amp load doesn't have time to hurt me (short duration)?

Is this correct?

What's the best method? I'm thinking connect one lead to each of my 2 downrigger cables and let her rip. I should probably make sure I am not touching the side of my aluminum boat.

[This message has been edited by jdevlin (edited 08-05-2003).]

I think the megger "fishfinder"/worm harvester is somewhat of a yarn.

A short on a 120V circuit with an overall loop resistance of 0.01 ohm will actually result in 12,000 amps!

The idea is that the breaker opens before the wiring has time to become hot enough to cause a problem

My bad, I meant to say .1 ohms instead of .01 ohms.

You can calculate the amount of energy which will be "let through" before the circuit is broken so long as you know how quickly the breaker will trip at the specific current level involved.

To use your example, if the C/B will open in 5 cycles at 1200A, then we can work out how much energy is dissipated:

Trip time = 5 / 60Hz = 0.08333 sec.

Power dissipated = 120V x 1200A = 144,000W.

One watt is equal to one joule per second, so:

Energy = 144,000 x 0.08333 = 12,000 joules.

Is 60 Hz the standard frequency for 120v?

Is it the same for 240, 208, 277 and 480?

Thanks

The transformers for various supplies (120/240V 1-ph, 120/208, 277/480 etc.) are all connected to the same HV distribution network, so the frequency will always be the same.

60Hz is the standard frequency for public supplies in the U.S.A., Canada, much of Latin America, and parts of the Caribbean. Most of the rest of the world (e.g. Europe, Australia, most of Africa) uses 50Hz.

Verrry Innnteresting.