Okay, that is a little deep but I will buy but could not explain it to anyone.

Why are ground and neutral connected?

Nealtw

Just apply Ohm's law to the normal situation and you will see what is going on. The National Electric Code (NEC) requires that a driven electrode have a resistance to ground of 25 Ohms or less OR that a second ground rod be installed at least 6 feet away from the first. Once I acquired the tester to do it easily I got in the habit of measuring the resistance to ground on all the ground rods I installed and all the ones at the existing installations I worked on. It takes about 5 minutes. I learned that just about every electrician didn't bother measuring because the resistance to ground of the single rod was never, in my experience of measuring them, lower than 25 Ohms. Even with the second rod the resistance still did not go below 25 Ohms. In nearly all installations, that I measured, the resistance to ground at the house was over 50 Ohms. The other end of that return path would be the Grounding Electrode at the transformer with the earth as the conductor in between.

Power Utilities are not regulated using the NEC rules. The code that the state regulators apply is the National Electrical Safety Code (NESC). Thus the Grounding Electrodes that they use are sometimes similar to those in the NEC but often very different. One common power utility transformer grounding electrode is a spiral of bare copper wire which is stapled to the but of the pole before it is set. Another is to cover the first foot of the but of the pole with copper sheathing. Whenever I could measure the utility's electrode without tampering with the installation I did so. That wasn't very often because the Utility's Grounding Electrode Conductor is often covered by a wood or plastic molding for the first 10 feet from the ground and the Linemen staple the GEC to the pole very tightly. If you cannot get the transformer jaw of the Ground Loop Impedance Tester around the GEC you cannot take the measurement.

Impedance is the complex resistance which at 60 Hz is always quite close to the resistance alone in value and is calculated in the same way. So with 50 Ohms at the Service Disconnecting Means; read main panel; and 70 or more at the pole the total Impedance of the Ground Return Pathway is 120 Ohms. The earth is so large that the Earth's effective resistance in this pathway is Zero. If the Neutral conductor of the Service Entry Conductors is intact then the voltage over that entire pathway is very low. Lets take the worse case. An old service in which the Service Disconnecting Means consist of a fused pull out and one of the fuses has opened for whatever reason. That way the surviving fuse, which is the only source of current flow, loads the neutral with all of that current. Now all of that current is flowing on the neutral and the Ground Return Pathway,

**in proportion to their resistance**, as well as on the energized conductor with the surviving fuse. There is no opposing current from the other energized Service Entry Conductor to cancel out a portion of the current and reduce the load on the neutral.

For example if the current through the intact fuse is 120 Amperes. The resistance of say 100 feet of #2 American Wire Gauge Aluminum Clad Steel Reinforced (ACSR) overhead conductor; because that is what the utility is likely to have used; is only .519 per 1000 feet. So over my example of 100 feet would have a resistance of 0.0519. Ohms law says the Current times the Resistance will equal the voltage. Thus E = I X R. E is the Electro Motive Fore measured in Volts, I is the current in Amperes. R is the Resistance in Ohms. So in our case the Voltage Drop; which is the voltage reduction caused by the resistance of the conductor; is 0.0519 Ohms of Resistance X 12o Amperes Of current = 6.228 volts over the utilities Service Conductors between the Main Bonding Jumper; in the panel cabinet that contains the SDM, and the center tap of the utility's transformer. That is the only voltage that the Ground Loop Pathway is subjected to. So with only 6.228 volts to "push it" and 120 Ohms of resistance the Ground Loop will carry about 50 Milliamperes of current.

*Yes I know that I took some shortcuts by not calculating the parallel resistance of both pathways first but any difference is outside the measuring accuracy of the available instrumentation.*
So out of all of the 120 Amperes of current the Ground Loop pathway carries only 50 Milliamperes of current. No where near half.

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Tom Horne