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ABCinventor
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Posted on Wednesday, December 22, 2004 - 3:34 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Hi Simon, it's been a long time since I wrote to you. Well I made a hydrogen generator lately with a plastic container, some wires and 3 solar panels. The thing work well in the sunny conditions in my country, Singapore. My question is if I join 3 solar panels in series, the voltage increases 3 times. If in parallel, the voltage remains as 1 unit and the ampere increases 3 times. The similar thing appened to 3 1.5 volt batteries. I was thinking that if I could decrease the voltage of a 9v battery while remaining the resistance the same, could I increase the current? Is there a way to do this? I hope that u can give me a reply of an explanation. Thanks Simon.
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ABCinventor
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Posted on Wednesday, December 22, 2004 - 3:37 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

By the way, Simon, there is some problems regarding this forum. For example, when posting a message, using the "enter" key. for example the next paragraph, an error message will prompt.try typing "enter" in your reply after a sentence and try posting it. you will know what i mean.
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 40
Registered: 12-2004
Posted on Wednesday, December 22, 2004 - 12:19 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Nine volt batteries are made from six cells, each having 1.5 volts.

You can disassemble one, and reassemble it in a
number of different configurations, to give 1.5 volts (all cells in parallel),
3 volts (3 pairs of parallel cells, connected in series),
or 4.5 volts (two triplets).

The current will increase as the voltage decreases,
due to the lower resistance caused by the resistances
being in parallel.
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ABCinventor
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Posted on Wednesday, December 22, 2004 - 10:58 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Thanks but wasn't it dangerous to open up a 9v battery?Oh another question, i am using 2 copper wire as anode and cathode of the hydrogen generating setup and i found some blue colour mess coming from the cathode(the + wire)after about 30 minutes of electrolysis. well i am using tap water and some salt as the electrolyte. i wounder if it is related.Anyway, thanks.
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ABCinventor
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Posted on Wednesday, December 22, 2004 - 11:04 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

well the problem is solved when i use another hydrogen producer which i made yesterday which the cathode is made from graphite from a pencil.
i think there is somthing about the copper wire..
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ABCinventor
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Posted on Friday, December 24, 2004 - 9:50 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Simon, i read in my science textbook that there is
a simplified idea of how volt, current and
resistance are related. below is the diagram.
.../.\
./_V_\
/_I|R_\
V/R=I, V/I=R and IxR=V.
Given the parallel and series combination
that claims to change either volt and/or
current by sacrificing one to increase the other.
series:R=1 V=3(increase)I=1(decrease)
parallel:R=1 V=1(decrease) I=3(increase)
note that the no. is the unit.
I think that the two idea don't match.
I hope that you can explain to me how it is
related.
Thanks Simon!!!
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 42
Registered: 12-2004
Posted on Friday, December 24, 2004 - 6:33 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Sure they match.

Suppose we measure the short circuit current
from a D battery, and we see 8 amps.
The battery puts out 1.5 volts, so the internal
resistance of the battery must be about 0.19
ohms (1.5 / 8). It is difficult for an amateur
to measure low resistances with a meter, because
the resistance of the meter and its leads and
connections can be greater than the resistance
being measured. But when you put two D cells
in parallel, the resistance is half what it was
previously, making it even more difficult. It is
just this halving of the resistance, however,
that allows the current to double.

Using a more useful circuit, suppose we have a
D cell in series with a 1,000 ohm resistor and
an LED. The LED glows dimly. Why? Because
1.5 volts / 1,000 ohms is 1.5 milliamps, just
enough to barely make the LED glow. Suppose
the current through the whole circuit is .075
milliamps. The resistance of the LED and the
resistor must be 20,000 ohms. The LED thus has
19,000 ohms of resistance.

Now lets add another D cell in series to make 3 volts.
We measure the current without the LED (just the
1,000 ohms) and we get 3 millamps.
We measure the current with the LED, and we get 1.5
milliamps. The LED is brighter now.

We expect the current to double if we double the
voltage and keep the resistance the same, and
that is what we see without the LED. With the
LED in the circuit, we measured 1.5 millamps at
3 volts, so the resistance is 2,000 ohms. Thus
the LED now has 1,000 ohms of resistance. What
happened to the other 18,000 ohms? The LED has
become much more conductive -- and we see the
results as a bright glow. Be careful when
applying Ohm's Law to things that are not
resistors -- the results can be surprising when
the resistance of a device is not fixed, but a
function of the voltage or current.
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ABCinventor
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Posted on Sunday, December 26, 2004 - 6:49 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

In your explaination, we reduce voltage, keep
the resistence the same and we get less current
too;we keep voltage the same, reduce resistence
and we got a high current.
but what about making the D cells parallel?
if we take 3 D cells and construct a series
circuit, there will be 4.5 volt and current
unit x.
when we construct it in parallel, things change,
we get only 1.5 volt but, we get a current of
unit 3x right?
this only has something to do with the resistence
as it is not described.
visit this site:
http://science.howstuffworks.com/battery8.htm
as this is where i got the idea.
anyway, thanks for explaining the relationship.
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 45
Registered: 12-2004
Posted on Sunday, December 26, 2004 - 4:11 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Don't mistake what the batteries can supply
with what they actually supply in a circuit.

The current in a circuit is a function of the
voltage and the resistance, if the power source
is capable of supplying that much current. But
small batteries can only supply a certain amount
of current, and if the resistance is too low, the
voltage will drop, and the battery will only
supply the current it can. In that case, adding
another battery in parallel will double the current.
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lysdexia
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Posted on Monday, December 27, 2004 - 11:57 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

thing works
There are some problems
keeping the resistance
two ideas
explanation
resistance

What's a battery's nominal resistance?
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 62
Registered: 12-2004
Posted on Monday, December 27, 2004 - 9:41 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

I mentioned 0.19 ohms.

The internal resistance varies with the age and
charge status of the battery, and the composition.

Nickel-cadmium batteries are about 0.15 ohms,
nickel-metal-hydride comes in at about 0.77
ohms, and litium-ion at about 0.32 ohms. But
you can measure your own batteries easily if this
is an issue for some particular design you have
in mind.
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lysdexia
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Posted on Tuesday, December 28, 2004 - 8:35 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

You said internal resistance though.
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 65
Registered: 12-2004
Posted on Wednesday, December 29, 2004 - 1:12 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Internal resistance is what I am talking about.
There is no "nominal" resistance for a battery.
Batteries are named by their voltage and ampere
hours, not by their resistance.
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ABCinventor
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Posted on Wednesday, December 29, 2004 - 3:08 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

Simon, i was wondering that if i could put
more 9 v batteries, maybe 2 or 3 into the
am transmitter in parallel circuit, i can
increase the current of the radio signal
without damaging the oscilator due to the
high voltage in a series circuit?
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 67
Registered: 12-2004
Posted on Wednesday, December 29, 2004 - 3:14 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

No.
The circuit will draw the same current with one
battery as with a dozen, since the voltage is
the same, and the circuit resistance is the same.
This is plain old Ohm's Law.

Adding batteries in parallel only increases the
amount of current that can be applied, if
the circuit is drawing too much current for one
battery to supply.
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ABCinventor
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Posted on Thursday, December 30, 2004 - 12:47 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

but then how did those radio stations produce
radio waves at high watts that can be recieved
over 10 kilo metres? that must be a strong radio
amplifier or so right?
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 70
Registered: 12-2004
Posted on Thursday, December 30, 2004 - 3:50 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

A 500,000 watt radio station in Cincinnati used 11,700
volts and 65 amperes to power its transmitter
(747,500 watts of DC power in, 500,000 watts of
radio power out).
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lysdexia
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Posted on Friday, January 7, 2005 - 9:05 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

I'm asking for the other sense of nominal, and the resistance is more than that of the insides.
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 86
Registered: 12-2004
Posted on Saturday, January 8, 2005 - 12:00 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

I'm sorry, but I still can't make any sense of
your question.

None of the other 5 definitions of nominal make
any sense in that question, nor does the concept
that a battery has a resistance that is "more
than that of the insides". Resistance is voltage
divided by current. Short the battery and
measure the current and voltage, and you have the
resistance.
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lysdexia
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Posted on Saturday, January 8, 2005 - 9:51 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

But then you're measuring the whole battery, not its insides. You said the figure was internal resistance!
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Simon Quellen Field (sfield)
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Username: sfield

Post Number: 92
Registered: 12-2004
Posted on Saturday, January 8, 2005 - 10:50 pm:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

What on earth are you talking about?

Internal resistance is the resistance of the
power source, excluding external parts such as
the leads going to the meter. It is what limits
the current that the source can supply. It is
why a short circuit does not have infinite
current going through it.

Internal resistance is measured by shorting the
battery and measuring the current, then subtracting
the resistance of the conductor doing the shorting.
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lysdexia
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Posted on Sunday, January 9, 2005 - 2:54 am:   Edit PostDelete PostView Post/Check IPPrint Post   Move Post (Moderator/Admin Only)Ban Poster IP (Moderator/Admin only)

That's a really dumb way of naming it. The resistance of the "shorter" should be called periferal, not external. The battery, or any power source, has inner and outer parts, right?! It has terminals and contacts which could be stripped to change the resistance. And the [internal] resistance would be least closest to the source of the charge interface.

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