9V Battery Current

Qwerty Keyboard wrote:
"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:4122F7B8.2EF329A9@earthlink.net...

The short circuit current of the battery is not relevant, period.
In fact, it is *stupid* to make such a useless measurement.

WTF? The only one who is *stoopid* here is you. Short circuit testing a
battery provides valuable information about how the battery will behave
under heavy loads and short circuit fault conditions. The original poster
wasn't very specific about what kind of electromagnet was in use, but smart
money bets it was a hobbyist's homebrew electromagnet with far too few turns
with far to large wire and the 9V battery impedance provides most of the
current limiting.

You are clearly so stoopid to make such a comment you evidently belong in
the lower two quartiles of this study.

http://www.apa.org/journals/psp/psp7761121.html

Even pond scum has more potential than you.

http://www.acfnewsource.org/environment/pond_scum.html

And your momma wears combat boots.

Bitch.

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Click to expand...
The only meaningful current measurement is that of the current flowing
thru the load (electromagnet in this case).
Ignoring the electromagnet, and thus its resistance, ignores the L/R
characteristic so eloquently described previously.
Placing an ammeter across a battery is not only dangerous, but gives
no meaningful info as related to working loads.
 
andy wrote:
On Tue, 17 Aug 2004 23:11:39 -0700, Chad Waldman wrote:

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

Thank you,
Chad

I've managed to get good results with a standard 250g coil of 24 swg wire
(0.56 mm) unwound from the bobbin, split in 3 equal lengths, and then
wound back on again, with the coils connected in parallel. The coil
resistance is around 1 ohm. With a 1.2Ah 12V lead acid battery, this gives
a peak current of about 10-11 amps. (The battery voltage drops to around
10.5 volts from about 12.5)

I'm using a piece of rebar (concrete reinforcing rod) as the core - this
works the best out of the metal objects I've tried. Bolts are no good.

With this setup, it's strong enough to push a neodymium magnet off the end
of the core, and lift the core against gravity. Guessing about 50-100
grams force when you try to pull the core out of the magnet with the
current on. To make a proper push/pull solenoid, you'd need a permanently
magnetised core, I think, which would be harder to find.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML: 
 or [attachment] in the subject line.
[/QUOTE]
  Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).
Click to expand...
 
On Fri, 20 Aug 2004 07:37:44 GMT, Robert Baer
<robertbaer@earthlink.net> wrote:


Placing an ammeter across a battery is not only dangerous, but gives
no meaningful info as related to working loads.
Click to expand...
It places an upper limit on how much current the battery might
possibly be putting into the electromagnet, or into any load. And it's
interesting to me, since I now know the fault current envelope I might
see if, say, I had a battery-powered gadget and a tantalum capacitor
failed or an IC latched up or something.

And it was obvious that this particular measurement could not have
been dangerous, even before I knew the results.

Roger the combat boots.

John
 
On Fri, 20 Aug 2004 17:42:16 +0100, andy
<news4@earthsong.free-online.co.uk> wrote:

On Fri, 20 Aug 2004 07:40:21 +0000, Robert Baer wrote:

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.
Click to expand...
---
Since it's Ampere-Turns that's doing the work and the number of turns
stays the same, if the current drops to one-tenth of what it was, so
will the strength of the field, so it's ten times less.

--
John Fields
 
On Fri, 20 Aug 2004 07:40:21 +0000, Robert Baer wrote:

andy wrote:

On Tue, 17 Aug 2004 23:11:39 -0700, Chad Waldman wrote:

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

Thank you,
Chad

I've managed to get good results with a standard 250g coil of 24 swg wire
(0.56 mm) unwound from the bobbin, split in 3 equal lengths, and then
wound back on again, with the coils connected in parallel. The coil
resistance is around 1 ohm. With a 1.2Ah 12V lead acid battery, this gives
a peak current of about 10-11 amps. (The battery voltage drops to around
10.5 volts from about 12.5)

I'm using a piece of rebar (concrete reinforcing rod) as the core - this
works the best out of the metal objects I've tried. Bolts are no good.

With this setup, it's strong enough to push a neodymium magnet off the end
of the core, and lift the core against gravity. Guessing about 50-100
grams force when you try to pull the core out of the magnet with the
current on. To make a proper push/pull solenoid, you'd need a permanently
magnetised core, I think, which would be harder to find.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML: 
 or [attachment] in the subject line.

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).
[/QUOTE]
I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.

-- 
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with [html] or [attachment] in the subject line.
Click to expand...
 
On Fri, 20 Aug 2004 17:42:16 +0100, andy
<news4@earthsong.free-online.co.uk> wrote:
Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.
Click to expand...

Um, I think it's 3.3 times less.

But the big picture is that the amount of field you can sustain
longterm depends on how well you can cool the coil. So for a fixed
copper cross-section and surface area, there's some maximum number of
watts you can pump into the copper before you toast it, and those
watts set the field strength. So series/parallel or wire size doesn't
change the basic limit on field strength so long as the power supply
can be adjusted.

John
 
On Sat, 21 Aug 2004 00:53:23 +0100, andy
<news4@earthsong.free-online.co.uk> wrote:

On Fri, 20 Aug 2004 15:10:07 -0700, John Larkin wrote:

On Fri, 20 Aug 2004 17:42:16 +0100, andy
news4@earthsong.free-online.co.uk> wrote:

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.


Um, I think it's 3.3 times less.

Isn't it like this?

Rp=R/3
Rs=3R

Ip=V/(R/3)=3V/R
Is=V/3R

Ip/Is=(3V/R)/(V/3R)=9
Click to expand...
Nice equations.

But it's not the current in the battery that makes magnetic field,
it's the current in the windings. Series, battery current equals
winding current. Parallel, current in each winding = 1/3 battery
current.

Think about it this way: we have three coils. Parallel, there's 10
volts across any given coil. Series, there's 3.3 volts across each
coil. So the current in any bit of wire is 3.3 times higher in the
parallel case. And the total ampere-turns product makes magnetic flux.

The windings don't know that they're in series or in parallel; they
just see current and make flux.

John
 
On Fri, 20 Aug 2004 15:10:07 -0700, John Larkin wrote:

On Fri, 20 Aug 2004 17:42:16 +0100, andy
news4@earthsong.free-online.co.uk> wrote:

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.


Um, I think it's 3.3 times less.
Click to expand...
Isn't it like this?

Rp=R/3
Rs=3R

Ip=V/(R/3)=3V/R
Is=V/3R

Ip/Is=(3V/R)/(V/3R)=9

But the big picture is that the amount of field you can sustain
longterm depends on how well you can cool the coil. So for a fixed
copper cross-section and surface area, there's some maximum number of
watts you can pump into the copper before you toast it, and those
watts set the field strength. So series/parallel or wire size doesn't
change the basic limit on field strength so long as the power supply
can be adjusted.
Click to expand...
yes, or as long as you can split the coil into as many paralleled windings
as you want.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML: 
 or [attachment] in the subject line.
 
Think about it this way: we have three coils. Parallel, there's 10
volts across any given coil. Series, there's 3.3 volts across each
coil. So the current in any bit of wire is

3.3<<xxx s/b 3.0 times higher in the

parallel case. And the total ampere-turns product makes magnetic flux.
Click to expand...
Oops: that makes the field ratio 3:1.

John
 
andy wrote:

On Fri, 20 Aug 2004 15:10:07 -0700, John Larkin wrote:

On Fri, 20 Aug 2004 17:42:16 +0100, andy
news4@earthsong.free-online.co.uk> wrote:

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.


Um, I think it's 3.3 times less.

Isn't it like this?

Rp=R/3
Rs=3R

Ip=V/(R/3)=3V/R
Is=V/3R

Ip/Is=(3V/R)/(V/3R)=9

No, it's times the number of turns. In series there are 3 times as many
Click to expand...
turns, so it's like John Larkin said. The term "ampere-turns" springs to
mind. :)

And how do you get 9 when you divide 12 by 1.2?

Thanks,
Rich
 
andy wrote:
On Fri, 20 Aug 2004 07:40:21 +0000, Robert Baer wrote:

andy wrote:

On Tue, 17 Aug 2004 23:11:39 -0700, Chad Waldman wrote:

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

Thank you,
Chad

I've managed to get good results with a standard 250g coil of 24 swg wire
(0.56 mm) unwound from the bobbin, split in 3 equal lengths, and then
wound back on again, with the coils connected in parallel. The coil
resistance is around 1 ohm. With a 1.2Ah 12V lead acid battery, this gives
a peak current of about 10-11 amps. (The battery voltage drops to around
10.5 volts from about 12.5)

I'm using a piece of rebar (concrete reinforcing rod) as the core - this
works the best out of the metal objects I've tried. Bolts are no good.

With this setup, it's strong enough to push a neodymium magnet off the end
of the core, and lift the core against gravity. Guessing about 50-100
grams force when you try to pull the core out of the magnet with the
current on. To make a proper push/pull solenoid, you'd need a permanently
magnetised core, I think, which would be harder to find.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML: 
 or [attachment] in the subject line.

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with [html] or [attachment] in the subject line.
[/QUOTE]
  Do *not* guess!
  Your "logic" is incomplete, and thus inaccurate.
  I suggest you make the measurement; you will find the series
configuration to be a lot stronger than your "estimate".
Click to expand...
 
John Fields wrote:
On Fri, 20 Aug 2004 17:42:16 +0100, andy
news4@earthsong.free-online.co.uk> wrote:

On Fri, 20 Aug 2004 07:40:21 +0000, Robert Baer wrote:

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.

---
Since it's Ampere-Turns that's doing the work and the number of turns
stays the same, if the current drops to one-tenth of what it was, so
will the strength of the field, so it's ten times less.

--
John Fields
Click to expand...
Your logic is incomplete and inaccurate.
 
John Larkin wrote:
On Fri, 20 Aug 2004 17:42:16 +0100, andy
news4@earthsong.free-online.co.uk> wrote:

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.

Um, I think it's 3.3 times less.

But the big picture is that the amount of field you can sustain
longterm depends on how well you can cool the coil. So for a fixed
copper cross-section and surface area, there's some maximum number of
watts you can pump into the copper before you toast it, and those
watts set the field strength. So series/parallel or wire size doesn't
change the basic limit on field strength so long as the power supply
can be adjusted.

John
Click to expand...
AT LAST!
You got it correct; ampere turns.
Parallel = if each coil has 100 turns, then the parallel combination has
100 turns (looks like fatter wire, folks), and one then has 1200
AmpereTurns.
Series = The turns add up (assuming the connection polarity is correct)
for 300 turns, and one then has 360 AmpereTurns.
 
On Sat, 21 Aug 2004 09:47:50 GMT, Robert Baer
<robertbaer@earthlink.net> wrote:


AT LAST!
You got it correct; ampere turns.
Parallel = if each coil has 100 turns, then the parallel combination has
100 turns (looks like fatter wire, folks), and one then has 1200
AmpereTurns.
Series = The turns add up (assuming the connection polarity is correct)
for 300 turns, and one then has 360 AmpereTurns.
Click to expand...
---
Your logic (or your arithmetic) is incomplete and inaccurate.

1200AT/360AT = 3.333, but:



12V 12V
| |
+-----------+------------+ |
| | | [10T] 1R
[10t] 1R [10T] 1R [10T] 1R |
| | | [10T] 1R
+-----------+------------+ |
| [10T] 1R
0V |
0V



4pi N I
Since B = K ---------,
l

If we assume the same diameter and length and they're both wound over
the same core, then K, 4pi, and l drop out and we're left with


B = N I

Where N is the number of turns per winding and N is the current
flowing in the winding.

If we assume 1 ohm coils for convenience, then for each winding in the
parallel case:

E 12V
I = --- = ----- = 12 amperes
R 1R

and

Bp = N I = 10 turns * 12A = 120 AT

Since there are three windings with the same diameters and lengths and
they're wound over each other the fields will add, for a total of

Bp = 3 N I = 3 * 120AT = 360AT.


For the series case we'll still have the same length and diameter, but
the current through each of the coils will be

E 12V
I = --- = ----- = 4A
R 3R

Then, for each coil,


Bs = N I = 10 turns * 4A = 40AT

and for all the coils.

Bs = 3 N I = 3 * 40AT = 120 AT


So, since Bp = 360AT and Bs = 120AT, the magnetic induction for the
parallel case will three (not 3.333) times greater than for the
series case, just like Larkin said earlier.

--
John Fields
 
On Sat, 21 Aug 2004 09:40:10 GMT, Robert Baer
<robertbaer@earthlink.net> wrote:

John Fields wrote:

On Fri, 20 Aug 2004 17:42:16 +0100, andy
news4@earthsong.free-online.co.uk> wrote:

On Fri, 20 Aug 2004 07:40:21 +0000, Robert Baer wrote:

Try connecting the three windings in series; see if you can make a
meaningful comparison of the relative strengths (series VS parallel).

I think I can guess that one - 3 3.3 ohm coils in parallel makes one ohm,
so the current is around 12A. In series would make about 10 ohm, with a
current of around 1.2A. So the strength would be about 9 times less.

---
Since it's Ampere-Turns that's doing the work and the number of turns
stays the same, if the current drops to one-tenth of what it was, so
will the strength of the field, so it's ten times less.

--
John Fields

Your logic is incomplete and inaccurate.
Click to expand...
---
Yes, you're right. Thank you.

--
John Fields
 
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