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simple capacitor test circuit

G

Greegor

Jan 1, 1970
0
G > What made you choose water?

MAT > He's all wet. ;-)

Even if it had mineral oil dialectric it's not much plate area.

John Larkin > I think it's one of those fringey free-energy hydrogen
generators.

I don't think stainless fits that idea.
And they usually get both hydrogen and oxygen though
they could intend to throw away (vent) the oxygen.

I always thought it was neat how you can take water
and break it into hydrogen and oxygen but then
if you combine those two gases and ignite they
can be quite energetic.

Is the energy from that process worth the cost
of the electricity used to break up the hydrogen
and the oxygen?

I took it for granted that it was a lossy energy conversion.

When they reconsidered this stuff for automobile use,
what exactly killed that idea off?

The explosive hazard? Conversion costs? What else?
 
J

Jeroen Belleman

Jan 1, 1970
0
[...]
I always thought it was neat how you can take water
and break it into hydrogen and oxygen but then
if you combine those two gases and ignite they
can be quite energetic.

Is the energy from that process worth the cost
of the electricity used to break up the hydrogen
and the oxygen?

Of course not. You can't win. You can't break even.
I'm surprised by the very question. What are you?
I took it for granted that it was a lossy energy conversion.

When they reconsidered this stuff for automobile use,
what exactly killed that idea off?

Nothing beats the convenience of a tank full of liquid fuel.

Jeroen Belleman
 
W

whit3rd

Jan 1, 1970
0
perhaps build ac bridge and compare them with a known capacitor.

That's the most accurate way; known C and unknown form half the bridge,
and two resistors (really, a potentiometer ) form the other half.
The capacitor ratio sets the voltage division on the capacitance
half, and resistor ratio sets it on the resistor half. An AC source
drives both (I'd use a wall-tumor transformer to start with), and an
AC meter determines a null if they're balanced. Adjust the potentiometer
for best null. C1/C2 = R1/R2 when the output is minimum.

The AC meter is unnecessary, really; a speaker would do (you just listen
to the signal).

Less accurate, but also effective, you could oscillate with a known resistor, in
a NE555 (or otheer '555) astable circuit, and use charge-pump
circuitry (two diodes and a capacitor or two) into the milliamp inputs
on your multimeter. If the frequency is low enough, replace the charge
pump with a flashing light or ticking speaker and use a stopwatch.
 
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