Volts leaking to ground rod

bluebinky said:

Very easy to answer. 12V DC is 12V RMS by definition. What would be interesting is to see JW do the math for calcualting RMS on a sinewave (or even write down the equation).

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A 120 volt sine wave will have a peak voltage of 169.68. Now take the sine of the angle which will be .707 and multiply the peak and we will have a RMS of 120 volts. 169.68 times .708 equals 120.

Yes this is cut short due to the fact I have to get ready to go do some inspections before the day is out.

Edited to add;
the math will equal 119.68 volts

Dang! I wish I could remember. It has been 30 years since I took electronics in school and haven't used it since. I was thinking back to what was said earlier about addition and subtraction of voltages on the bus. What is jumping out at me is the fact that when you have 2 separate legs sharing a neutral. If memory serves me right, you have to add an additional neutral when a double breaker is placed on one leg, but isn't needed when the legs are shared by one neutral? That said, wouldn't the difference of both legs with the same demand be zero on the neutral? This thread is taking me back in the "way back machine". It is quite refreshing to read all this stuff. Thanks guys!

Homeownerinburb said:

Meanwhile, does anyone have an opinion as to which way the current was flowing on that ground line?

Keeping in mind that the flow continued at the same level when the main breaker to the house was turned off.

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How much current is on the Neutral wire that feeds the breaker panel when you turn the breakers off ?


Good Luck.

ankhseeker said:

Dang! I wish I could remember. It has been 30 years since I took electronics in school and haven't used it since. I was thinking back to what was said earlier about addition and subtraction of voltages on the bus. What is jumping out at me is the fact that when you have 2 separate legs sharing a neutral. If memory serves me right, you have to add an additional neutral when a double breaker is placed on one leg, but isn't needed when the legs are shared by one neutral? That said, wouldn't the difference of both legs with the same demand be zero on the neutral? This thread is taking me back in the "way back machine". It is quite refreshing to read all this stuff. Thanks guys!

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In a multi-wired, shared neutral install, current from one leg on the neutral will cancel that from the other IF the load on both sides is equal. But, since they don't add, but cancel, the size of the neutral required will never exceed that of the hot lead, so the one neutral can be the same size.

jadnashua said:

In a multi-wired, shared neutral install, current from one leg on the neutral will cancel that from the other IF the load on both sides is equal. But, since they don't add, but cancel, the size of the neutral required will never exceed that of the hot lead, so the one neutral can be the same size.

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Does that mean that the legs are opposite phase from each other?

No, they are not opposite in phase...you need to get into vector math to understand it all.

ankhseeker said:

Does that mean that the legs are opposite phase from each other?

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Yes.

You are adding vectors.

Reach4 said:

Yes.

You are adding vectors.

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If I have a vector that is 120 inches long from point A to B and then another 120 inches from B to C the distance between A and C will be 240 inches. As long as this distance is a straight line or a line that is at 180 degrees it wouldn’t matter how I measured it, it would still be 240 inches from A to C.

From A to B plus from C to B would still be 240 inches. From B to A added to C to B will still be 240 inches from A to C. From B to C added to C to A will still be 240 inches.

Isn’t vector math fun?

Reach4 said:

Yes.

You are adding vectors.

Click to expand...

If I have a vector that is 120 inches long from point A to B and then another 120 inches from B to C the distance between A and C will be 240 inches. As long as this distance is a straight line or a line that is at 180 degrees it wouldn’t matter how I measured it, it would still be 240 inches from A to C.

From A to B plus from C to B would still be 240 inches. From B to A added to C to B will still be 240 inches from A to C. From B to C added to C to A will still be 240 inches.

Isn’t vector math fun?

jwelectric said:

If I have a vector that is 120 inches long from point A to B and then another 120 inches from B to C the distance between A and C will be 240 inches. As long as this distance is a straight line or a line that is at 180 degrees it wouldn’t matter how I measured it, it would still be 240 inches from A to C.

From A to B plus from C to B would still be 240 inches. From B to A added to C to B will still be 240 inches from A to C. From B to C added to C to A will still be 240 inches.

Isn’t vector math fun?

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That was so good, it is worth repeating, so it was.

Homeownerinburb said:

Meanwhile, does anyone have an opinion as to which way the current was flowing on that ground line?

Keeping in mind that the flow continued at the same level when the main breaker to the house was turned off.

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Unlike ideal (simplified) circuits, real conductors have non-zero impedence. The connections to ground have non zero impedance, as does the ground itself. So you have a bunch of conductors in in parallel. You may have heard that electricity follows the path of least resistance. That is misleading. Electricity follows every path, but in proportion to the reciprocal of the impedance. Impedance is the AC version of resistance, and at 60 Hz, they are usually similar.

So is that much current normal? I wouldn't think so. It might be interesting to probe other ground wires in your neighborhood with a clamp ammeter.

delete...dup

ActionDave said:

Seven tenths of an amp on the earth ground is nothing in the real world.

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I am SO inclined to agree.

DonL said:

How much current is on the Neutral wire that feeds the breaker panel when you turn the breakers off ?


Good Luck.

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Well, that would vary depending on the draw in the house and which legs are being pulled harder, eh?

The reality of the alignment in the box, getting my clamp on on the neutral is hard, and gettng it off is even worse.

But there is a slight imbalance between what we expect the two legs to balance off and lay off on the neutral.

And it is about what is happening on the ground. Current coming up the ground and going to the transformer via the service drop.

jwelectric said:

A 120 volt sine wave will have a peak voltage of 169.68. Now take the sine of the angle which will be .707 and multiply the peak and we will have a RMS of 120 volts. 169.68 times .708 equals 120.

Yes this is cut short due to the fact I have to get ready to go do some inspections before the day is out.

Edited to add;
the math will equal 119.68 volts

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Uh, ok. The actual math is take the square root of the integral of the voltage squared for one period and divide it by the period. The .707 number comes from doing the math first.

Reach4 said:

Unlike ideal (simplified) circuits, real conductors have non-zero impedence. The connections to ground have non zero impedance, as does the ground itself. So you have a bunch of conductors in in parallel. You may have heard that electricity follows the path of least resistance. That is misleading. Electricity follows every path, but in proportion to the reciprocal of the impedance. Impedance is the AC version of resistance, and at 60 Hz, they are usually similar.

So is that much current normal? I wouldn't think so. It might be interesting to probe other ground wires in your neighborhood with a clamp ammeter.

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Good point on the whole multiple path thingy. Obviously true.

I won't be running around a neighborhood in which I do not live to chase a phantom current. I am satisfied that if the DWP is happy, I should be happy.

jwelectric said:

If I have a vector that is 120 inches long from point A to B and then another 120 inches from B to C the distance between A and C will be 240 inches. As long as this distance is a straight line or a line that is at 180 degrees it wouldn’t matter how I measured it, it would still be 240 inches from A to C.

From A to B plus from C to B would still be 240 inches. From B to A added to C to B will still be 240 inches from A to C. From B to C added to C to A will still be 240 inches.

Isn’t vector math fun?

Click to expand...

OK, what happens if I measure from A to B and then A to C, according to your logic, the distance must be 360 inches. Cooool I think I'll try that with my multi-meter the next chance I get...

The problem is that for the physical thing that you are claiming to model, you must subtract vectors to do what you want. Your "vector math" only works if you arbitrarily throw minus signs in here and there. Very sloppy work. If you want to argue against something which is universally accepted wold-wide, please at least get the math right.

Are you really that stubborn, or do you think I am so stupid that I couldn't possibly teach you something? You know, the only reason I am keeping at this is that you JW are the person that people come to this sight to listen to, not me.

Reach4 said:

If he does, you could refer to him as jω or jωt rather than JW.

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Now that is funny. Unfortunately the more complete e^jωt kind of loses its punch. I think jω works best.

Another thought experiment...

I have a box which has two wires coming out of it, a flashing light, and a switch. You hook up an oscilloscope to the wires, and rig a light sensor to the trigger input. After a few adjustments, what you see is a sign wave that crosses zero going positive at the beginning of the trace.

Now, I flip the switch and the scope shows the voltage crossing going negative at the beginning of the trace.

Simple question: did the switch reverse the connection of the two wires, or change the timing of the flashing light?

bluebinky said:

OK, what happens if I measure from A to B and then A to C, according to your logic, the distance must be 360 inches. Cooool I think I'll try that with my multi-meter the next chance I get...

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no. it will still be 240 inches from A to C

bluebinky said:

The problem is that for the physical thing that you are claiming to model, you must subtract vectors to do what you want. Your "vector math" only works if you arbitrarily throw minus signs in here and there. Very sloppy work. If you want to argue against something which is universally accepted wold-wide, please at least get the math right.

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I have neither subtracted nor threw any minus signs in anywhere I simply added the two vectors.

bluebinky said:

Are you really that stubborn, or do you think I am so stupid that I couldn't possibly teach you something?

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The closest to the truth that has spoken in this thread. I don’t think I would have used the word stupid though, maybe ignorant which means not knowing.

bluebinky said:

You know, the only reason I am keeping at this is that you JW are the person that people come to this sight to listen to, not me.

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Well I must sit back and think about that for a minute. I must say it took me off guard for a second. I thank you for the comment although I think it is stretched just a little.

Take an O-scope and connect it to two batteries connected in series by connecting channel A to one end and channel B to the other end and the ground to the center and you will see on the screen two batteries that looks like they are out of phase. Are they out of phase or pushing against each other or is it just the way the meter is connected? When I connect my old analog meter to a battery I see the voltage but when I reverse the leads the pointer tries to hide. I suppose when this happens the battery is out of phase with the meter.

My scope has four channels. Connect the ground to a circuit board and the channels to four different places on the board and I can see four different signals but I suppose although that board is supplied by 24 volts that it must somehow get everything twisted around so there is four different phases in that board. But then again it could be that I am just comparing four different meters so to speak at the same time.