This was just posted by Mitchell Jones, and I thought that you would enjoy
reading his most interesting comments. ERS
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Newman's Capacitor
The other day Evan Soule' posted some comments from Joseph Newman
concerning the operation of his refurbished version of the motor that he
constructed in Pennsylvania. According to my interpretation of his remarks,
he has some small batteries connected in series to yield 110 volts, which
are in turn hooked into a simple circuit to drive the motor. Connected in
parallel to the motor, Newman has placed a "very large" 5000 volt
capacitor, and he claims that it takes 7 seconds to reach an overcharged
condition, whereupon it discharges via a visible arc.
When I pointed out the apparent impossibility of that result, according to
current theory, and the implication -- i.e., that Joe's motor would have to
be the source of the voltage which overcharged the capacitor and, thus,
would have to be "over unity" -- I received no response from any of the
regulars in this group. The question is: why not? ("Over unity" devices are
the main topic of interest here, right?) A couple of explanations for this
thunderous silence come to mind, including the possibility that there is a
bit of confusion out there regarding the nature of capacitance. Therefore,
to get that possibility out of the way, I am going to try again.
Capacitance is that property of an electrical circuit that opposes a change
in voltage in the circuit, and hence a capacitor is a circuit element
designed to oppose a change in voltage. Basically, a capacitor consists of
two electrically conductive regions separated by an insulating medium
(dielectric). An example would be two sheets of aluminum foil separated by
a sheet of plastic. Connect a wire lead to each sheet of aluminum, roll the
three sheets up and stuff them in a plastic tube, and you have a home-made
capacitor.
Let us therefore assume that you have done so, that you connect the wire
leads to separate terminals of a DC power supply, and that you then ramp
the voltage up to 5500 volts. At that point, an arc blasts a hole in the
plastic and destroys your capacitor. You then build another one, just like
the first, ramp the voltage up to 5000, and stop. Since there is no arc,
you give your home-made capacitor a rating of 5000 volts by simply
scribbling "5000 V" on the outside of the plastic pipe.
Next, you want to determine the "size" of your home-made capacitor. You
know, on theoretical grounds, that when you connect your capacitor across
the terminals of a 5000 volt battery, the voltage across the plates of the
capacitor will start at 0 V and will rise to 5000, while the current will
start at some peak level and fall to zero.
Thus you introduce instruments to measure the average value of the current
and the time it takes for the plate voltage to rise from 0 to 5000 V.
Suppose, for example, that you discover that it takes 6 seconds for the
plate voltage to reach 5000 volts, and that the average value of the
current over that interval is .01 Amps. Since capacitance (C, in Farads)
is defined as the charge difference between the two plates (Q, in Coulombs)
divided by the final value of the voltage (Ef, in volts) between the two
plates, we have C = Q/Ef. Since when we move an electron from one plate to
the other, we simultaneously increase the charge by one unit on the plate
from which the electron was removed, and reduce it by one unit on the plate
to which the electron was moved, it follows that the charge difference
across the plates, Q, is equal to twice the average value of current (Ia,
in Amps) times the elapsed time (T, in seconds).
Thus we have Q = 2(Ia)(T). Since Ef equals the value of the source voltage
(Es, in volts), we have Ef = Es. Substituting, we have: C = 2(Ia)(T)/Es.
Plugging in our measured values, above, we get: C = 2(.01)(6)/(5000) =
..000024 Farads = 24 microfarads. Now we know the "size" of the capacitor
that we have constructed, so we write "24 mfd" on the plastic pipe next to
"5000 v."
Next, we borrow Joe Newman's motor, hook it into a circuit with a 110 volt
DC power source, place our capacitor in parallel with the motor, and
discover that after 7 seconds of operation our capacitor overcharges, arcs
across the plastic sheet, and destroys itself.
The question is, how can this be? We are no longer using a 5000 volt power
source. Our power source is 110 volts, not 5000 volts, and there is nothing
else in the circuit but the capacitor and the Newman motor. *Where did the
electromotive force come from that overcharged the capacitor and destroyed
it?* It couldn't have come from the power source, so it must have come from
the motor. (This is a no brainer, right?)
Now let's compute the amount of power involved in charging the capacitor.
As derived above, C = 2Ia(T)/(Es). Rewriting, we have: (1/2)CEs = IaT.
Multiplying both sides by Es, we have: (1/2)C(Es)^2 = IaEsT. Since DC power
(P, in watts) equals the average value of current (Ia) times the source
voltage (Es), we have P = IaEs. Substituting, we get: (1/2)C(Es)^2 = PT.
(Note: The above derivation can also be accomplished using calculus.)
Dividing both sides by T, we get: P = [(1/2)C(Es)^2]/T. Plugging in the
above numerical values, we have:
P = [(1/2)(.000024)(5000)^2]/7 = 42.9 watts, which is the minimum size of
the power source necessary to charge the capacitor. Of that amount, only a
tiny portion [P = (1/2)(.000024)(110)^2/7 = .02 watts] could have come from
the power source.
Thus out of the 42.9 watts required to charge the capacitor, 42.88 watts
had to come from the motor. Does anybody out there have any comment about
this? (Hello? Knock, knock! Is anybody there?)
--Mitchell Jones
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Received from Mitchell Jones
Evan Soule'
www.josephnewman.com