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Re: voltages etc.



Additional:

Part two. generally the PD between earth (ground) is different from zero for
everything not connected to ground (with low resistance, the resistance amount
will determine the difference from ground potential). However, for DC the
charge is low (capacitance and physical height from ground or the wires common
to ground) and therefore, to measure it one must use a high impedance
voltmeter. A lamp is a rather poor detector of PD in this case. Try it**.
Note that he PD between our heads and our feet is as much as 200V.

"Detecting the Earth's Electricity
---------------------
by Shawn Carlson
July, 1999
---------------------
IF ELECTRIC FIELDS WERE visible, then even the most barren spot on the earth
would provide an awesome sight. Standing on a hilltop, you would see a forest
of electric-field lines shooting out of the ground everywhere, stretching up
to the ionosphere. You could watch them sweep across the horizon to gather
under storms. In fact, the earth's electric field is far more dynamic--and, for
me, more interesting--than its magnetic counterpart."

from the CD "The Twentieth Cent. Collection" (Scientific American)


AC is another matter, as audio techs. know well, it is impossible to remove the
effect of a ground loop by using heavy wire. All it does is increase the
current through the loop.

bc


** I did; the Voltage measured with a 10 meg. DMM and the + and - terminals of
a 9 V batt. were different and started @ ~ 50 mV and dropped to ~ 15. In some
trials the - terminal started @ ~ - 20 mV and went + in a few seconds.

Bob LaMontagne wrote:

Seth T Miller wrote:


My colleague connected the positive side of a 6 volt dry cell battery to
a 1.5 volt light bulb, then connected the other side of the light bulb
to the ground. The bulb does not light. Is there a voltage difference
between the positive side of the battery and the ground and if so is it
6 volts? Why doesn't the bulb light?

Her latest theory is that as soon as one touches the positive side of
the battery, any excess charges are immediately drained off and that
side of the battery becomes equalized with the ground. The chemical
reaction is not occurring, so there is no way for a voltage difference
between the positive terminal and the ground to be maintained, thus the
bulb does not light.

Is this a correct analysis? How could it be analyzed in more detail?


This is a neat question that students bring up all the time. The 6V is only
the potential difference between the ends of the battery - it does not refer
to an actual potential at either end. By attaching the positive terminal to
ground via the light bulb wires and filament, that terminal now has "ground"
potential (0 Volts reference) after a brief transient adjustment. The
negative terminal is now at -6V relative to ground. Current will not flow
through the bulb because the ground connection and the positive terminal of
the battery are at the same potential.

Bob at PC