magnetic field

[Pat Flannery]

bz wrote:

m.



The breakdown voltage of air _decreases_ as the air pressure goes down.

Paschen's Law Vs=f(N ds) where Vs is sparking potential, ds is sparking
distance and N is gas number density.

Okay... but other than that... :-[

Pat

 

[petrus bitbyter]

"Lorraine" <lorrainewinters80@yahoo.com> schreef in bericht
news:1116881385.586286.3060@g43g2000cwa.googlegroups.com...

Hi,

I wanna convert the 12 volts DC in my car to AC 110 volts
60 Hz so I can use my appliance inside the car, what's the
simplest circuit lying around in the net where I can build
one? Pls. mention some sites and schematics as well as the
IC. All I know is the 555 timer ic, but it doesn't up
the voltage from 12 to 110 volts.

To make the circuit simpler, can I just use 30 Hz, I
guess my appliance would still run at this frequency
right? For electric fan, would it run slower at 30 Hz
versus 60 Hz or the same? How about light bulb. Would
it flicker at this lower frequency?

Thanks.

Lorr

Lorr,

Sorry, but from your questions I can only conclude that you don't have the
skills and knowledge to build such a thing from scratch. You can find
several solutions on the net but you can't judge what's good and what's not.
The cheapest and the least dangerous solution is to buy the thing.

As for your other questions:
- 30 Hz is not simpler
- Most appliances made for 60Hz will not run on 30Hz.
- A 60Hz fan motor will most likely be fried when used at 30Hz
- A incandescent lamp will flicker at 30Hz.

petrus bitbyter

 

[bz]

Pat Flannery <flanner@daktel.com> wrote in news:1194jks10opj219
@corp.supernews.com:

I did get a electrical burn years later off of a
discharging microwave oven capacitor though. That went in the right hand
and out the left hand with an audible "bang" noise, and left a small
burn at either end.

In the mid 70s I fixed electronic equipment on ships for a living.

I was on the bridge of a ship, working on the echo sounder, removing a
nurled nut holding part of the transmitter to its mounts (without first
discharging the capacitors!) when the skin on the surface of my thumb
provided a discharge path from the capacitor to ground.

Sound like a shotgun going off. Bright flash. White path along surface of
skin that took a while to go away.

I was lucky.

I think it was about 3KV, 0.05 mfd.

I think most of the discharge went through ionized air/skin because I never
felt anything.

--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap

 

[Boris Mohar]

On 23 May 2005 15:50:39 -0700, "Lorraine" <lorrainewinters80@yahoo.com>
wrote:

I'm asking this because I plan to use a DC
regulated power supply and make the signal vibrate at 1 Mhz and
direct it to the antennae to experiment with em radiation. I don't
plan to use any resistor because I want the current to be maximum
so the em produced would be maximum. I wonder if this would damage
the source. What sort of protection (diode?) that can be used in the
source to detect and prevent shorting damage (in case this happens).

You are not allowed to do that.

You are allowed to read this.
http://www.motionmountain.com/

Read the whole thing. Twice.



Regards,

Boris Mohar

Got Knock? - see:
Viatrack Printed Circuit Designs (among other things) http://www.viatrack.ca

 

[Ken Taylor]

"Lorraine" <lorrainewinters80@yahoo.com> wrote in message
news:1116888639.722597.298790@g49g2000cwa.googlegroups.com...

Guys,

Suppose you short the wire in the battery. The current will be maximum
right? Suppose it takes 24 hours for the battery to be used up in a
10 ohm load. If you short it, would the battery be used up in say
1 hour. Is the operation of the battery in the shorting mode the same
as in normal only accelerated?

The laws of physics don't change, the load however goes to the minum (being
the internal resistance of your battery plus the presumably negligable
short-circuit).

How about AC lines. If you short the wires, it sparks and explodes. So
I assume current is maximum. However, if you use wires that can never
melt or break. What would happen to the other end in the power utility.
Would it destroy their circuit?

See above. In this case the utility company relaises there are idiots around
who short out their power lines so they put in circuit breakers.

Suppose I use a regulated dc supply and I short the + and - outputs.
Would the circuit be damaged or would the wires just heat up without
any damage to the source?

If it's a good quality supply it will be itnernally protected, either
shutting down or blowing a fuse. Maybe it'll break - how much are you going
to pay for it?

I'm asking this because I plan to use a DC
regulated power supply and make the signal vibrate at 1 Mhz and
direct it to the antennae to experiment with em radiation. I don't
plan to use any resistor because I want the current to be maximum
so the em produced would be maximum. I wonder if this would damage
the source. What sort of protection (diode?) that can be used in the
source to detect and prevent shorting damage (in case this happens).

Eh? Get a transmitter and appropriate licence or precautions.

Thx

Lorr

Ken

 

[Peter Bennett]

On Tue, 24 May 2005 08:50:12 +1200, "Ken Taylor" <ken123@xtra.co.nz>
wrote:

"siliconmike" <siliconmike@yahoo.com> wrote in message
news:1116824758.587553.68400@f14g2000cwb.googlegroups.com...

Which model Samsung TV has 2000W output??

CS-29T10PA

Man this TV rocks! Now I don't need to go to theatre. I put it 10 ft
from me and watch DVDs.. Its 29".. And cost is (in US $) $600..

"Peak Music Power" is a term you need to become familiar with.

Even if they're using Peak Music Power, India must use much smaller
watts than the rest of the world.




--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Peter Bennett]

On 23 May 2005 15:50:39 -0700, "Lorraine"
<lorrainewinters80@yahoo.com> wrote:

Guys,

Suppose you short the wire in the battery. The current will be maximum
right? Suppose it takes 24 hours for the battery to be used up in a
10 ohm load. If you short it, would the battery be used up in say
1 hour. Is the operation of the battery in the shorting mode the same
as in normal only accelerated?

The energy stored in a battery is measured in "Ampere-hours".
Theoretically, if you have a 20 Ampere-hour battery, it can deliver 20
amperes for one hour, or 10 amperes for two hours, or one ampere for
20 hours, (or any combination of time and current where time X current
= 20) before becoming fully discharged.

In practice, if you discharge a battery rapidly, you will get less
total energy out of it than if you discharge it slowly. Lead-Acid
storage batteries are normally rated at a "20 hour rate" - based on
the current required to discharge them in 20 hours. So a 20 AH
battery will deliver 1 amp for 20 hours, but may only deliver 15 amps
for 1 hour.

How about AC lines. If you short the wires, it sparks and explodes. So
I assume current is maximum. However, if you use wires that can never
melt or break. What would happen to the other end in the power utility.
Would it destroy their circuit?

There are normally several fuses or circuit breakers between you and
the power company's generators - the lowest-rated fuse or circuit
breaker will blow, and remove the power, when you short the circuit.

There is always some resistance in the wiring which will limit the
maximum current you can draw.

Suppose I use a regulated dc supply and I short the + and - outputs.
Would the circuit be damaged or would the wires just heat up without
any
damage to the source?

Depends on the power supply - many regulated supplies are protected
against excessive current drain - they will shut down, or reduce the
output voltage to keep the current within limits.

I'm asking this because I plan to use a DC
regulated power supply and make the signal vibrate at 1 Mhz and
direct it to the antennae to experiment with em radiation. I don't
plan to use any resistor because I want the current to be maximum
so the em produced would be maximum. I wonder if this would damage
the source. What sort of protection (diode?) that can be used in the
source to detect and prevent shorting damage (in case this happens).

I think you need to learn a LOT more electronics before you do this!

Consult some Amateur Radio manuals for information on radio
transmitters - and be aware that transmitting a signal without an
appropriate license is illegal.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Peter Bennett]

On 23 May 2005 13:49:45 -0700, "Lorraine"
<lorrainewinters80@yahoo.com> wrote:

Hi,

I wanna convert the 12 volts DC in my car to AC 110 volts
60 Hz so I can use my appliance inside the car, what's the
simplest circuit lying around in the net where I can build
one? Pls. mention some sites and schematics as well as the
IC. All I know is the 555 timer ic, but it doesn't up
the voltage from 12 to 110 volts.

You can buy a suitable inverter much cheaper than you can build one,
particularly given the level of electronics knowledge your questions
indicate.

To select an appropriate inverter, you will need to determine the
power required by your appliance.

To make the circuit simpler, can I just use 30 Hz, I
guess my appliance would still run at this frequency
right? For electric fan, would it run slower at 30 Hz
versus 60 Hz or the same? How about light bulb. Would
it flicker at this lower frequency?

Most devices designed for 60 Hz will not operate correctly (or maybe
at all) at 30 Hz, and an incandescent light bulb would probably have a
noticeable flicker.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Ken Taylor]

"Peter Bennett" <peterbb@somewhere.invalid> wrote in message
news:jfr491l509ea4hncff6khlu0otapb3lisc@news.supernews.com...

On Tue, 24 May 2005 08:50:12 +1200, "Ken Taylor" <ken123@xtra.co.nz
wrote:

"siliconmike" <siliconmike@yahoo.com> wrote in message
news:1116824758.587553.68400@f14g2000cwb.googlegroups.com...

Which model Samsung TV has 2000W output??

CS-29T10PA

Man this TV rocks! Now I don't need to go to theatre. I put it 10 ft
from me and watch DVDs.. Its 29".. And cost is (in US $) $600..

"Peak Music Power" is a term you need to become familiar with.

Even if they're using Peak Music Power, India must use much smaller
watts than the rest of the world.


--
Peter Bennett, VE7CEI

Gotta spread them around further.

Oddly enough when I used Samsung's site search engine I didn't find that
model, but when I used Google it popped up. Hmmm, maybe they are a bit
ashamed.....

Ken

 

[N:dlzc D:aol T:com (dlzc)]

Dear Pat Flannery:

"Pat Flannery" <flanner@daktel.com> wrote in message
news:1194kkhrgfu5633@corp.supernews.com...

bz wrote:

m.

The breakdown voltage of air _decreases_ as the
air pressure goes down.

Paschen's Law Vs=f(N ds) where Vs is sparking
potential, ds is sparking distance and N is gas
number density.

Okay... but other than that... :-[

I'm not sure you understand, so...
As the corona inception voltage goes down, so does the average
velocity of the ions leaving. If the velocity goes down, the
thrust goes down by the square of the velocity. Altogether a bad
thing, epecially when rarefied air presents a good opportunity to
reduce drag.

David A. Smith

 

[N:dlzc D:aol T:com (dlzc)]

Dear Robert Clark:

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116862693.124425.124920@g47g2000cwa.googlegroups.com...

N:dlzc D:aol T:com (dlzc) wrote:
Dear Robert Clark:

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116851138.887584.183700@o13g2000cwo.googlegroups.com...
N:dlzc D:aol T:com (dlzc) wrote:
Dear Robert Clark:

"Robert Clark" <rgregoryclark@yahoo.com> wrote in message
news:1116812266.009025.315290@z14g2000cwz.googlegroups.com...
N:dlzc D:aol T:com (dlzc) wrote:
...
The "endoatmospheric ion-propulsion" engine
is clearly the same thing as the lifter drive.

Hardly. They don't need wires to the craft.

They beam the energy to the craft using lasers
or microwaves. Same drive just different source
for the power.

Change the thread title. An enclosed MHD drive
is as much like a lifter drive, as a turbine engine
is to a house fire.

You can call it MHD if you want.

Let me do a little house cleaning:
MHD = MagnetoHydroDynamics

I'm referring to any method of
propulsion that uses either electrical or magnetic
fields to propel air that has been ionized by
electrical means or otherwise. Lifters also can
work on AC current.

MHD uses AC primarily.

Then in addition to the force produced by the
asymmetric electric field you could get a force
due to the Lorentz force arising from the
magnetic field.

David A. Smith

 

[CWatters]

This is what you need...

$27

http://www.amazon.com/exec/obidos/tg/detail/-/B00005LACD/qid=1116919113/sr=8-2/ref=pd_csp_2/103-3794380-6248660?v=glance&s=electronics&n=507846

or perhaps this much bigger unit for just a few $ more.

http://www.amazon.com/exec/obidos/tg/detail/-/B00009OYH7/qid=1116919113/sr=8-1/ref=pd_csp_1/103-3794380-6248660?v=glance&s=home-garden&n=507846

 

[CWatters]

"Lorraine" <lorrainewinters80@yahoo.com> wrote in message
news:1116851318.253016.39530@g14g2000cwa.googlegroups.com...

I see. So there is a lower limit to the resistor used
to make constant volage. For example. Voltage
is constant at 10 volts if the resistor varies from
10 ohm to 1 Gigaohm in a 1A source. Right?

Almost but forget the 1A source bit. Your statement should read..

"Voltage is constant at 10 volts if the resistor varies from 10 ohm to 1
Gigaohm.

In a perfect world a "constant voltage source" would provide a constant
voltage no matter how much current the load "sucked" out of it. In practice
voltage sources are not perfect they have some internal resistance of their
own. This means that the voltage they provide isn't quite constant. It falls
slightly as the load takes more current. The power supply (in a computer for
example) might be rated at 5V +/- 2% for currents upto 20A. This means that
when the load takes taking 20A the voltage might fall to 5V-2%= 4.9V.

And if the resistor gets lower than 10 ohm, the voltage
gets down. Now are there applications in which they
intentional make the resistance go down the
critical value to make the voltage lower??

You could look uo "potential divider circuit" but that's complicated stuff
at your level.

Also why is that in larger load, the current is lesser
in value.

Ah I see your confusion. Try and think of it slightly differently... It's
not the battery that controls the current it's the load. With no load
connected the resistance of the load is infinitly high. This means that the
current is zero because I=V/R and R is very big. Now work out what happens
as R is reduced. The current starts to increase.

How come the electrons don't just get
stuck up at the entry point to the load?

They don't for the same reason that water in a tap doesn't "get stuck up"
just before a tap when it's switched off. The pressure in the pipe stops the
water (eg current) flowing so no more water accululates at the tap.

How did
the battery knows how much current to send to
the given load? What's the feedback mechanism?

Current flowing in the resistor causes a voltage drop accross the resistor.
That voltage drop opposes the voltage of the battery reducing the current.

 

[Vidar Lřkken]

Lorraine wrote:

Voltage is said to be the difference between the electric
potential between two poles of a battery. What is the
relationship between electric field and electric
potential at the poles. Is there a difference in the
electric field between the two poles?

This might sound harsh, but this group is here to _discuss_ electronics,
not to teachs you the basic of electronics. This, google, or a book can
do. I suggest you drop by amazon.com and get a decent book on the subject.
http://www.amazon.com/exec/obidos/search-handle-form/104-8338403-6128734
That search turned up a lot of titles, and you can even have a look at
the book before buying it.
I guess that google might turn up a few thousand results for basic
electronics too, if you don't want to buy a book.

--
MVH,
Vidar

www.bitsex.net

 

[Pat Flannery]

bz wrote:

In the mid 70s I fixed electronic equipment on ships for a living.

I was on the bridge of a ship, working on the echo sounder, removing a
nurled nut holding part of the transmitter to its mounts (without first
discharging the capacitors!) when the skin on the surface of my thumb
provided a discharge path from the capacitor to ground.

Sound like a shotgun going off. Bright flash. White path along surface of
skin that took a while to go away.

I was lucky.

I think it was about 3KV, 0.05 mfd.

I think most of the discharge went through ionized air/skin because I never
felt anything.

In my case I felt like a fool because it had a huge label on it warning
you to discharge it prior to doing any work on the oven; the oven was a
first generation microwave from the mid 70's and the capacitor was a
thing the size of a small hip flask.
I really felt it; I got tossed around three feet and had spots before my
eyes.

Pat

 

[ryan wiehle]

cheian07@yahoo.com wrote:

i have an old computer mainboard and it boots to DOS fine. but there's
only one problem. It always displays memory test fail.its memory slot
supports SIMM. I tried it with other SIMM but it doesn't recognize
them. I tried its original memory to other board but it is OK. now it
only display 5 KB on the screen. this problem prevents me from
installing even Win 95. it sometimes works fine w/ Dos but i prefer
Win application now.is my mainboard defective already? Please give me
suggestions on how to deal with this kind of problem. Any help would
be greatly appreciated. ps. its bios is AWARD.

ian

go into the bios and to see where the problem is, disable all
IDE ports. If it then boots you got a problem with the IDE ports.
Try same for floppy ports

 

[Pat Flannery]

N:dlzc D:aol T:com (dlzc) wrote:

I'm not sure you understand, so...
As the corona inception voltage goes down, so does the average
velocity of the ions leaving. If the velocity goes down, the
thrust goes down by the square of the velocity. Altogether a bad
thing, epecially when rarefied air presents a good opportunity to
reduce drag.


Yeah I got your point; there's something else to take into consideration

here- what happens when the Lifter hits the ozonosphere? Isn't ozone
electrically conductive, and won't that short the whole works out?

Pat

 

[bz]

Pat Flannery <flanner@daktel.com> wrote in news:1196icu1shkmue3
@corp.supernews.com:

N:dlzc D:aol T:com (dlzc) wrote:


I'm not sure you understand, so...
As the corona inception voltage goes down, so does the average
velocity of the ions leaving. If the velocity goes down, the
thrust goes down by the square of the velocity. Altogether a bad
thing, epecially when rarefied air presents a good opportunity to
reduce drag.


Yeah I got your point; there's something else to take into consideration
here- what happens when the Lifter hits the ozonosphere? Isn't ozone
electrically conductive, and won't that short the whole works out?

Not to mention what happens when it hits a cloud, or icing conditions, or
when the metal happens to be below the dewpoint and water starts to
condense on everything.


--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap

 

[Mark Fergerson]

Lorraine wrote:

Voltage is said to be the difference between the electric
potential between two poles of a battery. What is the
relationship between electric field and electric
potential at the poles. Is there a difference in the
electric field between the two poles?

You're very confused, but apparently you already knew that.

Electrons "desire" to be as far away from each other as possible,
and express that "desire" as a force they exert against each other;
that force reaches out through space no matter how far apart two
electrons may be. One way of describing this is to call it a stress
in the space between electrons.

We call that spatial stress a "potential field" because at any
point in the space betwen them, there's a potential value of force
you could measure on an electron inserted at that point. The word
"field" is a mathematical term for the gradual change in values from
point to point, and it has singularities (maximum/minimum values) at
certain points like where the most or least electrons are which are
also called "poles".

Since the force acts on the electric charge on electrons and not
their mass, it's assigned values in volts.

So, each point in the field can be assigned a "voltage",
including two points at different distances from an isolated electron.

When you assemble a battery (chemically take electrons from
_here_ and put them over _there_), you will naturally wind up with
two poles and a field between them.

That's a fair first approximation to an answer to your questions,
but you really ought to crack a text or three for more in-depth
information.

Mark L. Fergerson

 

[Mark Fergerson]

Lorraine wrote:

Guys,

Suppose you short the wire in the battery. The current will be maximum
right? Suppose it takes 24 hours for the battery to be used up in a
10 ohm load. If you short it, would the battery be used up in say
1 hour. Is the operation of the battery in the shorting mode the same
as in normal only accelerated?

Yeah, you need to look up and understand "ampere-hours".

How about AC lines. If you short the wires, it sparks and explodes. So
I assume current is maximum. However, if you use wires that can never
melt or break. What would happen to the other end in the power utility.
Would it destroy their circuit?

It'd just draw a huge current until a breaker blew (opened the
circuit), which would deprive your load of the voltage that drives
current through it.

Suppose I use a regulated dc supply and I short the + and - outputs.
Would the circuit be damaged or would the wires just heat up without
any damage to the source?

Depends. You can have voltage regulation (can only supply up to
so much current at a specified voltage), but they usually are
designed to shut themselves down on a dead short. Current regulated
supplies will just make your shorting wire get hot.

I'm asking this because I plan to use a DC
regulated power supply and make the signal vibrate at 1 Mhz and
direct it to the antennae to experiment with em radiation. I don't
plan to use any resistor because I want the current to be maximum
so the em produced would be maximum. I wonder if this would damage
the source. What sort of protection (diode?) that can be used in the
source to detect and prevent shorting damage (in case this happens).

Whatever oscillator circuit you use will only be able to
"process" so much current per cycle depending on its architecture
and the specific devices it comprises; if it's capable of handling
more than is available from the supply the output waveform will
suffer, or it may oscillate irregularly or not at all.

You really want to design something like this backwards; first
decide how much RF power you are willing to fool with which will
narrow down your oscillator design possibilities, which will in turn
narrow your power supply options considerably.

Mark L. Fergerson