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Exported earth
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<blockquote data-quote="Nicky Tesla" data-source="post: 4421" data-attributes="member: 445"><p>[</p><p></p><p>The resistors would not be in parallel though would they!</p><p></p><p>It would be 1 resistor ie the the lighting cpc having the fault current flowing down it. Then you have to consider the said 6a flowing through Ze resistance eg 0.1a which gives V=6ax0.1; v=0.6V. So the voltage across the earth rod to earth will not be 250v at all, no-where near, but just 0.6v in this case. Now the calculation for the earth rod will be I=v/R ; Amps=0.6/100 ; =0.6ma</p><p></p><p>YOUR ALL MISSING THE POINT. THE PROBLEM IS THAT IT IS A PME SYSTEM AND THE PROBLEMS OF BROKEN NEUTRALS.</p></blockquote><p></p>
[QUOTE="Nicky Tesla, post: 4421, member: 445"] [ The resistors would not be in parallel though would they! It would be 1 resistor ie the the lighting cpc having the fault current flowing down it. Then you have to consider the said 6a flowing through Ze resistance eg 0.1a which gives V=6ax0.1; v=0.6V. So the voltage across the earth rod to earth will not be 250v at all, no-where near, but just 0.6v in this case. Now the calculation for the earth rod will be I=v/R ; Amps=0.6/100 ; =0.6ma YOUR ALL MISSING THE POINT. THE PROBLEM IS THAT IT IS A PME SYSTEM AND THE PROBLEMS OF BROKEN NEUTRALS. [/QUOTE]
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