Watts are calculated by the formula:
Watts = Voltage * Current
(current is in amps for this formula)
You say you have 30 volts and .002 amps (ie; 2 milliamps) so you multiply these together and get .06
watts. So, you can basically use any resistor you can get your hands on since the smallest axial
lead resistors are generally .25 watts (ie; 1/4 watt). In other words, use a 1/4 or 1/2 watt resistor.
Resistors don't have any polarity requirement. They can be connected either
way.
Using Ohm's law (current=voltage/resistance), with 30 volts and 20,000 ohms you will have 1.5
milliamps (ie; .0015 amps). Actually, you will have less current since there will be the added
resistance of the distilled water in the cell. (Resistances in series add together)
I might use a 10,000 ohm resistor myself but there is no problem with the larger resistor, it will
just take longer to brew.
Since you have DC volts, you don't need a diode unless you are talking about the so-called "current
regulating diode" to regulate the current. If you are using the "current regulating diode" you
don't have to use the resistor since the "diode" will change it's resistance to keep the current at
the preset level. It will work better than the resistor in that the brewing process will be faster
than in a cell regulated by a resistor. The proper polarity must be used with the "current
regulating diode."
Also, the silver MUST be connected to the positive (+) battery terminal. The other electrode can be
stainless steel which should be connected to the negative (-) battery terminal. Everything should
be connected in series (ie; in single file, so to speak...). A current meter would also be
connected in series. A voltmeter would be connected in parallel, across any part to read the
voltage drop across that part. If you connect a voltmeter (set for reading volts) in series, it
will block the current.
Dan
-------------------------------------------
* From: * Peter M. Stellas[SMTP:[email protected]]
* Sent: * Wednesday, November 22, 2006 8:48:51 PM
* To: * [email protected]
* Subject: * RE: CS>technical question
Dan,
Thanks for that response. What I am about to construct is the following
experimental generator. The cathode will be a piece of stainless steel
tubing that will sit on a nylon base inside the glass container. This
base will center the tube in the glass jar and support it at three
points, in an elevated position so that water can circulate under it.
The glass jar will sit on the inverted can/night light heater described
elsewhere. Thus the heated water in the center will flow up the immersed
s/s tube and flow over it, flowing downward against the colder glass
surface.
At a distance of about 1.5 “ away from the inside or outside surface of
the s/s tube, I will position four, long, U-shaped 12 gauge silver
electrodes, at positions 12, 3, 6, and 9 o’clock with respect to the
circumference of the s/s tube as viewed from the top. This will provide
a lot of silver surface for the anode and a lot of s/s surface for the
cathode. From here I need some help from the experts on how to select
diodes, resistors, etc, and how to mount them in the circuitry to
provide regulated current at the suggested 0.002 amps. Ken suggested a
20KOhm resistor, but did not mention the watt rating. I am assuming that
the watt rating of the resistor is the peak wattage that the dc
generator can put out, i.e 12 watts for a 30 vdc and 0.400 amp output.
In this case I can find a 50 watt, 20K Ohm resistor for my application.
Does it make any difference how the resistor is hooked up, meaning which
side the power enters and which side it exits? I am sure that it makes a
difference in diodes. Amazingly, no one at the electronics supply store
that I visited knew anything about circuitry. Real professionals!
Peter
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