ECN Forum Forums General Topics of the Trade General Discussion Area Megger Death

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