That statement was:
"If the voltage is insufficient, no power supply, no matter what current it
can deliver, will light the bulb."

I think they were referring to:

I=E/R

Assuming R is constant as in many cases then when E goes to zero than I must
follow. You just cannot pump any current into the circuit when E goes to
zero. This is true except when R goes to zero than I goes to infinity or
both R and E go to zero and this equation becomes "undefined". Funny things
happen in absolute zero temperature conditions - sorry I didn't past school
this far to explain things like anti-gravity, infinite current and so on.

On Wed, 14 Feb 2007 12:50:18 -0800, "Bob F"
wrote:


"Doug Miller" wrote in message
t...
In article W6HAh.3741$Aa5.925@trnddc01, "Pop`"
wrote:
w_tom wrote:
Light bulbs work at may voltages. For example, a 120 volt
incandescant bulb will work at 60 volts AND last for maybe 100 years
continuous. As voltage drops, bulb life expectancy increases
exponentially (about a factor of 12). As voltage drops, bulb
intensity also decreases exponentially. Sure, the 12 volt bulb will
work at 6 volts. But its light output will be massively diminished
and its efficiency is decreases (less light per amp of electricity).

So yes, a 12 volt lamp will work on 6 volts (if power supply can
provide sufficient current). Just not work very well.

NO, it will not. There will be insufficient heat in the element to
provide
any light or even a glow with a 12V bulb at 6V DC. A 12VDC bulb will
begin
to dim substantially at 9 VDC and may not even be visible in the light of
a
room.

I guess you didn't understand the part where he said "if power supply can
provide sufficient current."


If the voltage is insufficient, no power supply, no matter what current it
can deliver, will light the bulb.


However 6V probably IS sufficient to light a 12V bulb somewhat.
Measuring the voltage while it is actually connected to the bulb could
tell you something. It doesn't STAY at 6V no matter what.


Do not profess knowledge you do not have.

That's good advice, Poop. Maybe you should take it, too.

You say?

Bob

--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Wed, 14 Feb 2007 11:38:55 -0600, Chris Friesen
wrote:

Chop Suey wrote:

You do know that LED's must be wired in series with a resistor to make
them glow, right?

I agree that is the proper way to do it.

However, they work just fine by themselves if the voltage supply is low
enough, or if you parallel enough LEDs together.


Essentially false. However if you use a small enough battery you may
be able to use its internal resistance (it cannot supply enough
current to destroy the LED without dropping below the LEDs forward
voltage). You may find this in small, cheap flashlights.

Chris

--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

The LEDs being produced today for illumination do not require any resistor.
They are connected directly to the battery or other power source.

--
Contentment makes poor men rich. Discontent makes rich men poor.
--Benjamin Franklin

Larry Wasserman - Baltimore Maryland - lwasserm(a)sdf.lonestar.org

On Wed, 14 Feb 2007 13:17:25 -0500, Goedjn wrote:

On Wed, 14 Feb 2007 11:28:38 -0600, Chop Suey
wrote:

HK inspired greatness with:

FYI, we're building a small airboat. There are four 3v LED lights powered
by 2 "C" batteries and a single 6v motor powered by 4 "C" batteries.


You do know that LED's must be wired in series with a resistor to make
them glow, right?


Actually, they don't. They have to be wired in series with a resistor
to keep them from eating your battery whole, but as long as you have
enough voltage, they'll light.

You can make the worlds smallest flashlight by just taping
the leads of an LED to either side of a watch battery.

A LED acts as a voltage regulator (regulating the voltage across it to
about 2V). It will draw as much current as needed to do this. If the
LED is across a 12V battery it will "happily" destroy itself trying to
maintain 2V across that 12V battery. I made that mistake once.
Immediately, there was a POP and half the LED disappeared.
If the supply voltage is below that, you don't get any light at all.

The flashlights that use LEDs without resistors use batteries that
cannot supply more current than about 20mA. The nearly-infinite load
from the LED lowers the battery output voltage until it's below that
which the LED will conduct. Then the battery recovers and the cycle
begins again.

That is, you need a battery with limited current capacity. You're
using the battery's internal resistance (Rth, if you care) as the LED
series resistor.
--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On 14 Feb 2007 09:32:25 -0800, "Bob" wrote:

On Feb 14, 11:04 am, "HK" wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are supposed
to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed a
12v power supply. Yikes.

Nothing wrong with that as long as they are in series.


Is Grade 6 too early to teach Ohm's Law? Should they be learning to hook up
basic circuits without learning Ohm's Law and the concepts of voltage,
current and resistance?

No. a good mix of math and science.



Some of this stuff is coming back to me. Am I correct in assuming that the
problem with wiring the lights in series is there is too much cumulative
resistance at the far end of the circuit to light the lights?

Go back to Ohms law. 2X resistance = 1/2 current


2x resistance * 2x voltage = 1/1 current (as in 2 6V bulbs in series
on 12V).


FYI, we're building a small airboat. There are four 3v LED lights powered
by 2 "C" batteries and a single 6v motor powered by 4 "C" batteries.

Aahhh... Remember that we've been talking incandescent light bulbs.
A LED is a diode and not a bulb even though it gives off light.
Although the same principles apply since LED's use a dropping resistor
to limit current, similar to items in parallel.


LEDs can be connected in series as long as the voltage is high enough.
Then, only one resistor is needed.



Any suggestions on a fun followup project that will help both of us learn
more electricity concepts?

Google or your local library.
--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Wed, 14 Feb 2007 18:29:08 GMT, (Doug Miller)
wrote:

In article , "HK" wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are supposed
to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed a
12v power supply. Yikes.

So what's the problem? Connect them in series.

Or in parallel if the power supply is big enough. C cells may not be.


Is Grade 6 too early to teach Ohm's Law? Should they be learning to hook up
basic circuits without learning Ohm's Law and the concepts of voltage,
current and resistance?

Some of this stuff is coming back to me. Am I correct in assuming that the
problem with wiring the lights in series is there is too much cumulative
resistance at the far end of the circuit to light the lights?

Too much cumulative resistance at some point, anyway -- not necessarily all
the way at the end.

FYI, we're building a small airboat. There are four 3v LED lights powered
by 2 "C" batteries and a single 6v motor powered by 4 "C" batteries.

Any suggestions on a fun followup project that will help both of us learn
more electricity concepts?

Browse around at the web site for Edmund Scientific
//http://scientificsonline.com/
Lots of good stuff there.
--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Wed, 14 Feb 2007 13:06:33 -0800, "Bob F"
wrote:


"HK" wrote in message
. ..
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are
supposed
to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed
a
12v power supply. Yikes.

That's one way to do it.


Is Grade 6 too early to teach Ohm's Law? Should they be learning to hook
up
basic circuits without learning Ohm's Law and the concepts of voltage,
current and resistance?

So teach her ohm's law.


Some of this stuff is coming back to me. Am I correct in assuming that
the
problem with wiring the lights in series is there is too much cumulative
resistance at the far end of the circuit to light the lights?

There is no "far end" of a circuit unless you have long small
wires creating resistance.

And none at all with SERIES circuit.


FYI, we're building a small airboat. There are four 3v LED lights powered
by 2 "C" batteries and a single 6v motor powered by 4 "C" batteries.

"3v LED" suggests LEDs with built in resisters. Remember LEDs
are polarized - if you connect them backwards, they won't light.
2 3v LEDs in series should work fine off of 6V. Do that twice for
4 bulbs.


Any suggestions on a fun followup project that will help both of us learn
more electricity concepts?

A crystal radio?
http://sci-toys.com/scitoys/scitoys/...ade_radio.html

Bob


--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On 14 Feb 2007 15:45:38 -0800, "Redbelly"
wrote:

On Feb 14, 6:25 pm, wrote:
If the voltage is insufficient, no power supply, no matter what current it
can deliver, will light the bulb.

Oh, really? Let's see the explanation of that.

He's correct. It's simple electricity 101.

OK, fine -- explain it.

A power supplies current draw depends on what the load is
and what the voltage is.

The current draw is not determined by the rating of the supply.

FYI, that's only true if the required current is less than the rating
of the supply.

If the required current is more than the supply's rating, then actual
current IS determined by the supply.

Yes and no. When you draw more current than rated three things can
happen.

1. The supply goes over current until something burns up
2. The supply shuts down electronically (the voltage goes to nil)
3. The voltage diminishes slowly as resistive losses in the supply
use a higher and higher percentage of the voltage drop.

In any case the supply limits the current by limiting the voltage at
the output.

On Feb 14, 12:04 pm, "HK" wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are supposed
to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed a
12v power supply. Yikes.

Is Grade 6 too early to teach Ohm's Law? Should they be learning to hook up
basic circuits without learning Ohm's Law and the concepts of voltage,
current and resistance?

Some of this stuff is coming back to me. Am I correct in assuming that the
problem with wiring the lights in series is there is too much cumulative
resistance at the far end of the circuit to light the lights?

FYI, we're building a small airboat. There are four 3v LED lights powered
by 2 "C" batteries and a single 6v motor powered by 4 "C" batteries.

Aha! LED's, and not filaments. Please DISREGARD my post elsewhere in
this thread, where I ramble on about what happens when a filament is
supplied with 6V instead of the designed-for 12V.

I'm puzzled, in your original post you refer to a "12 V light". Is
that referring to these four LED's, wired in series? But above here
you are saying they are actually powered by 3V, indicating that they
are in parallel. Without a clear picture of what is going on, I'm not
sure what advice to give you.

Any suggestions on a fun followup project that will help both of us learn
more electricity concepts?

You could buy some resistors and red or yellow LED's from Radio Shack
(if you're in the U.S.). They'll also sell battery holders, and some
wires with alligator clips. You can experiment with calculating the
LED current for different resistors, and seeing how the LED dims/
brightens when you try different resistors. For a 2V red LED driven
by 3V, some resistors to try a

(3V - 2V)/(0.030 A) = 33 ohms
(3V - 2V)/(0.010 A) = 100 ohms
(3V - 2V)/(0.003 A) = 330 ohms

(Even if the LED says it is rated for 20 mA, driving it at 30 mA for 5
or 10 seconds will be okay.)

Do you have a voltmeter? You could show that the voltages across the
resistor and LED always add up to equal the voltage of the battery
(which you should also measure, don't just assume it's 1.5)

I admit these suggestions aren't as exciting as a model boat that
lights up and has a running motor, but it is a starting point to start
a 6th grader on learning some of the basics of electricity.

Regards,

Mark

On Feb 14, 12:04 pm, "HK" wrote:

Any suggestions on a fun followup project that will help both of us learn
more electricity concepts?


I should also mention, there is a book by Forrest M. Mims called
"Getting Started in Electronics". Many of the regulars in
sci.electronics.basics recommend it for beginners, though I myself am
not familiar with it. It's available at Amazon:
http://www.amazon.com/Getting-Starte...e=UTF8&s=books

Regards,

Mark

In article , wrote:
On 14 Feb 2007 15:45:38 -0800, "Redbelly"
wrote:

On Feb 14, 6:25 pm, wrote:
If the voltage is insufficient, no power supply, no matter what current
it
can deliver, will light the bulb.

Oh, really? Let's see the explanation of that.

He's correct. It's simple electricity 101.

OK, fine -- explain it.

A power supplies current draw depends on what the load is
and what the voltage is.

The current draw is not determined by the rating of the supply.

FYI, that's only true if the required current is less than the rating
of the supply.

If the required current is more than the supply's rating, then actual
current IS determined by the supply.

Yes and no. When you draw more current than rated three things can
happen.

1. The supply goes over current until something burns up
2. The supply shuts down electronically (the voltage goes to nil)
3. The voltage diminishes slowly as resistive losses in the supply
use a higher and higher percentage of the voltage drop.

In any case the supply limits the current by limiting the voltage at
the output.

IOW -- the current *is* determined by the supply. :-)

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

In article , Mark Lloyd wrote:
On Wed, 14 Feb 2007 18:29:08 GMT, (Doug Miller)
wrote:

In article , "HK"
wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are supposed
to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed a
12v power supply. Yikes.

So what's the problem? Connect them in series.

Or in parallel if the power supply is big enough. C cells may not be.

If you connect two 6V lamps in parallel to a 12V supply, you're quite likely
to see the two lamps doing an excellent imitation of fuses -- especially "if
the power supply is big enough".

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

IOW -- the current *is* determined by the supply. :-)

If you hooked a 10 amp supply to a toothpick would it draw 10 amps?

Multiple choice....

Assume a fixed resistance of 1 ohm, and a 10 volt power supply
rated at 5 amps. What will best determine the current draw?

A. The current draw will be the measured voltage.

B. The supplies rating of 5 amps

HK wrote:

I'm trying to help my daughter with a school project and while I know my way
around house wiring, I'm a relative newbie when it comes to low voltage.

Let's say I have a 6 volt DC power supply (4 "C" batteries). If I want to
drive a small motor and some lights, I just find 6 volt motors and lights
and wire them in parallel or series.

Why is it that I can't light up a 12v light with a 6 volt power supply? I
always test my batteries using a multi-meter and as the battery ages, the
voltage drops. With low voltage, the device (flashlight, etc.) still works
but the light is weak. So, wouldn't a 12v light just be weak if I use a 6v
power supply?

Is there some kind of voltage threshold at which a device won't work?

Like I said, basic question.


Hi,
It depends on the lamp's rating. If you lower the voltage to half you
have to double the current to get same Wattage. Can the battery do that
with the lamp you have? Automotive type bulb draws quite bit of curent.

On Thu, 15 Feb 2007 01:35:28 GMT, (Doug Miller)
wrote:

In article , Mark Lloyd wrote:
On Wed, 14 Feb 2007 18:29:08 GMT, (Doug Miller)
wrote:

In article , "HK"
wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are supposed
to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed a
12v power supply. Yikes.

So what's the problem? Connect them in series.

Or in parallel if the power supply is big enough. C cells may not be.

If you connect two 6V lamps in parallel to a 12V supply, you're quite likely
to see the two lamps doing an excellent imitation of fuses -- especially "if
the power supply is big enough".

I have done that before. They usually do not burn out immediately, but
glow brighter (and whiter) for awhile.

Of course that is not what I was talking about before. Maybe you
missed that what I said was an ALTERNATIVE to using 12V.
--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

In article , Mark Lloyd wrote:
On Thu, 15 Feb 2007 01:35:28 GMT, (Doug Miller)
wrote:

In article , Mark Lloyd
wrote:
On Wed, 14 Feb 2007 18:29:08 GMT, (Doug Miller)
wrote:

In article , "HK"
wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are supposed

to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed a

12v power supply. Yikes.

So what's the problem? Connect them in series.

Or in parallel if the power supply is big enough. C cells may not be.

If you connect two 6V lamps in parallel to a 12V supply, you're quite likely
to see the two lamps doing an excellent imitation of fuses -- especially "if
the power supply is big enough".

I have done that before. They usually do not burn out immediately, but
glow brighter (and whiter) for awhile.

Of course that is not what I was talking about before. Maybe you
missed that what I said was an ALTERNATIVE to using 12V.

?

I was responding to your suggestion, visible above, that "if the power supply
was big enough" he could connect two 6V lamps in parallel to a 12V supply.

That won't work for very long.

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

Tony Hwang wrote:
HK wrote:

I'm trying to help my daughter with a school project and while I know
my way around house wiring, I'm a relative newbie when it comes to low
voltage.

Let's say I have a 6 volt DC power supply (4 "C" batteries). If I
want to drive a small motor and some lights, I just find 6 volt motors
and lights and wire them in parallel or series.

Why is it that I can't light up a 12v light with a 6 volt power
supply? I always test my batteries using a multi-meter and as the
battery ages, the voltage drops. With low voltage, the device
(flashlight, etc.) still works but the light is weak. So, wouldn't a
12v light just be weak if I use a 6v power supply?

Is there some kind of voltage threshold at which a device won't work?

Like I said, basic question.

Hi,
It depends on the lamp's rating. If you lower the voltage to half you
have to double the current to get same Wattage. Can the battery do that
with the lamp you have? Automotive type bulb draws quite bit of curent.
Hi,
And what differene does it make? High voltage or low voltage circuit.
Basically it is all govenred by simple Ohm's law.

"Pop`" writes:

So yes, a 12 volt lamp will work on 6 volts (if power supply can
provide sufficient current). Just not work very well.

NO, it will not. There will be insufficient heat in the element to provide
any light or even a glow with a 12V bulb at 6V DC. A 12VDC bulb will begin
to dim substantially at 9 VDC and may not even be visible in the light of a
room.

A counterexample: I just connected a 12 V automobile headlight lamp
(the high beam filament of an H4 lamp, rated 60 W at 12 V) to a bench
power supply. The power supply current limits and can't raise the
voltage above 4.8 V. At that voltage, the 12 V bulb is definitely
glowing. It's reddish rather than white light, but it's unmistakably
glowing. If I reduce the voltage further, I can still see a dull glow
at 1.5 V, one eighth of the rated 12 V.

At 6 V, it would be whiter and brighter, but 4.8 V is enough.

Do not profess knowledge you do not have.

So, what bulb did you test? Which 12 V rated bulb does not light up at
all at 6 V? Surely you tried it before writing?

Dave

"Redbelly" writes:

Lots of factors are at play here. Changing the supply voltage will
change the filament temperature, and hence the filament resistance.
So cutting the voltage in half will not cut the current in half.

I made a spreadsheet which accounts for the changes in temperature and
resistance, and here's what I found:

Where did you get the formulas for these? As you say, an incandescent
lamp filament is far from a fixed resistor, but how does resistance and
temperature change with voltage?

Dave

On Feb 14, 12:04 pm, "HK" wrote:
Any suggestions on a fun followup project that will help both of us learn
more electricity concepts?

Put a speed controller on the motor. A simple rheostat (variable
resistor) demonstrates how motor speeds can be changed.

Get various resistors. Demonstrate how the different color bands
correspond to ohm numbers - and then measure them with the meter.
Radio Shack once sold little color wheels. Dial in the color and read
te number.

Reading the values and then measuring with the meter should teach
something about how the real world 'varies' from what is should be.

Meanwhile some numbers for that light bulb. A 12 volt light
operating at 6 volts will output about 1/10 the lumens and consume 33%
of watts as compared to 12 volt power consumption. Obviously the bulb
is less efficient at those lower voltages. At 6 volts, the bulb will
last about 8200 times longer.

A 12 volt bulb at 9 volts will output 40% the light at 12 volts and
consume about 60% of the wattage. Unlike Pop', numbers can be
provided with replies.

On Feb 14, 12:04 pm, "HK" wrote:
Any suggestions on a fun followup project that will help both of us learn
more electricity concepts?

Put a speed controller on the motor. A simple rheostat (variable
resistor) demonstrates how motor speeds can be changed.

Get various resistors. Demonstrate how the different color bands
correspond to ohm numbers - and then measure them with the meter.
Radio Shack once sold little color wheels. Dial in the color and read
te number.

Reading the values and then measuring with the meter should teach
something about how the real world 'varies' from what is should be.

Meanwhile some numbers for that light bulb. A 12 volt light
operating at 6 volts will output about 1/10 the lumens and consume 33%
of watts as compared to 12 volt power consumption. Obviously the bulb
is less efficient at those lower voltages. At 6 volts, the bulb will
last about 8200 times longer.

A 12 volt bulb at 9 volts will output 40% the light at 12 volts and
consume about 60% of the wattage. Unlike Pop', numbers can be
provided with replies.

FYI, we're building a small airboat. There are four 3v LED lights
powered
by 2 "C" batteries and a single 6v motor powered by 4 "C" batteries.


You do know that LED's must be wired in series with a resistor to make
them glow, right?

I hooked them up in parallel and they seem to work just fine. Incredibly
bright too.

On Feb 15, 1:01 am, (Dave Martindale) wrote:
"Redbelly" writes:
Lots of factors are at play here. Changing the supply voltage will
change the filament temperature, and hence the filament resistance.
So cutting the voltage in half will not cut the current in half.
I made a spreadsheet which accounts for the changes in temperature and
resistance, and here's what I found:

Where did you get the formulas for these? As you say, an incandescent
lamp filament is far from a fixed resistor, but how does resistance and
temperature change with voltage?

Dave

A decent approximation is that the resistance is proportional to the
absolute temperature. A more accurate relationship is given at the
end of this post.

For my calculation, I used an iterative approach which goes as follows
(somewhat lengthy description, skip if not interested. NOT FOR
MATHOPHOBES!):

Step "0". the design values:

First, assume three conditions at the filament's designed operating
point, eg. power, voltage, and filament temperature (3000 Kelvin is a
reasonable number). Current and resistance are then determined from
power and voltage.

Step 1. 1st iteration for new operating point:
Next, change the voltage but do the calculation for the same
resistance, as if it were fixed. This gives a new power and current,
which will be wrong, but our iterative approach will converge to the
correct values. We also need to calculate a new temperature using the
blackbody radiation law:

Power is proportional to T^4
or in other words
T is proportional to P^0.25

The way we REALLY use this relation is:

T_1 = T_0 * (P_1 / P_0)^0.25
where "0" refers to the original operating point, and "1" refers to
the updated temperature and power.

Step 2. 2nd iteration of values:
Recalculate the resistance, R_2, using:

R_2 = R_0 * (T_1 / T_0)

Since voltage is fixed, we can recalculate power and current from

I_2 = V / R_2
P_2 = V * I_2

And then a new temperature from
T_2 = T_0 * (P_2 / P_0)^0.25

Step 3.
Recalculate the resistance:
R_3 = R_0 * (T_2 / T_0)

Then recalculate I_3, P_3, and T_3 as in Step 2.

Steps 4, 5, etc.
Keep repeating the process until the power, temperature, and
resistance converge to steady values that don't change significantly.

That's it, really. A somewhat more accurate description (for
tungsten, anyway) is that resistance is proportional to T raised to
the power of 1.2. Accounting for this did not change the results by
very much.

Regards,

Mark

On Wed, 14 Feb 2007 21:06:29 -0500, wrote:


IOW -- the current *is* determined by the supply. :-)

If you hooked a 10 amp supply to a toothpick would it draw 10 amps?


If it's a 1-ohm toothpick.

Multiple choice....

Assume a fixed resistance of 1 ohm, and a 10 volt power supply
rated at 5 amps. What will best determine the current draw?

A. The current draw will be the measured voltage.

B. The supplies rating of 5 amps

On Thu, 15 Feb 2007 00:19:54 +0000 (UTC),
(Larry) wrote:

The LEDs being produced today for illumination do not require any resistor.
They are connected directly to the battery or other power source.

That seems unlikely. There's probably a resistor somewhere. You may
have a LED module with built-in resistor (it's still a resistor, where
ever it is). Those modules are often being sold for use in flashlights
now (just replace the bulb). It could be the battery. Little batteries
(button cells) have significant internal resistance.

I actually tested that on a new LED (they're not THAT expensive). That
LED isn't working any more.
--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Thu, 15 Feb 2007 09:19:26 -0500, "HK" wrote:



FYI, we're building a small airboat. There are four 3v LED lights
powered
by 2 "C" batteries and a single 6v motor powered by 4 "C" batteries.


You do know that LED's must be wired in series with a resistor to make
them glow, right?

I hooked them up in parallel and they seem to work just fine. Incredibly
bright too.


LEDs in parallel with each other? It's unlikely that all would light,
since the threshold voltages would be slightly different, and the one
that's lowest would prevent the others from lighting.

As someone said earlier, your LEDs probably are actually LED modules,
and come with built-in resistors. You have a separate resistor in
series with each LED.

Modern LEDs can appear very bright. I noticed that with the holiday
lights I had this year (yes, I know that's "last year", but it is
still less than 2 months ago). Those LEDs look brighter than the
miniature incandescent's.
--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Thu, 15 Feb 2007 02:52:43 GMT, (Doug Miller)
wrote:

In article , Mark Lloyd wrote:
On Thu, 15 Feb 2007 01:35:28 GMT, (Doug Miller)
wrote:

In article , Mark Lloyd
wrote:
On Wed, 14 Feb 2007 18:29:08 GMT, (Doug Miller)
wrote:

In article , "HK"
wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are supposed

to build a basic circuit and working model.

For instance, my daughter assumed that to power two 6v lights, she needed a

12v power supply. Yikes.

So what's the problem? Connect them in series.

Or in parallel if the power supply is big enough. C cells may not be.

If you connect two 6V lamps in parallel to a 12V supply, you're quite likely
to see the two lamps doing an excellent imitation of fuses -- especially "if
the power supply is big enough".

I have done that before. They usually do not burn out immediately, but
glow brighter (and whiter) for awhile.

Of course that is not what I was talking about before. Maybe you
missed that what I said was an ALTERNATIVE to using 12V.

?

I was responding to your suggestion, visible above, that "if the power supply
was big enough" he could connect two 6V lamps in parallel to a 12V supply.


That just might have worked IF I had forgotten what I said, and you
had edited my quote to say that. "12V" does not appear in that quote
at all.

That won't work for very long.

OK, I left something out (saying to use 6V, or clarifying what "big
enough" means). That's no excuse to stick in something that doesn't
belong.
--
Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

In article , Mark Lloyd wrote:
On Thu, 15 Feb 2007 02:52:43 GMT, (Doug Miller)
wrote:

In article , Mark Lloyd
wrote:
On Thu, 15 Feb 2007 01:35:28 GMT, (Doug Miller)
wrote:

In article , Mark Lloyd
wrote:
On Wed, 14 Feb 2007 18:29:08 GMT, (Doug Miller)
wrote:

In article , "HK"
wrote:
Thanks everyone. Good information here.

This is a Grade 6 project and I was a little distressed to find that even
the basic concepts of electricity haven't been taught yet they are
supposed
to build a basic circuit and working model.
For instance, my daughter assumed that to power two 6v lights, she needed a
12v power supply. Yikes.

So what's the problem? Connect them in series.

Or in parallel if the power supply is big enough. C cells may not be.

If you connect two 6V lamps in parallel to a 12V supply, you're quite likely
to see the two lamps doing an excellent imitation of fuses -- especially "if
the power supply is big enough".

I have done that before. They usually do not burn out immediately, but
glow brighter (and whiter) for awhile.

Of course that is not what I was talking about before. Maybe you
missed that what I said was an ALTERNATIVE to using 12V.

?

I was responding to your suggestion, visible above, that "if the power supply
was big enough" he could connect two 6V lamps in parallel to a 12V supply.


That just might have worked IF I had forgotten what I said, and you
had edited my quote to say that. "12V" does not appear in that quote
at all.

Apparently you *did* forget what you said, or at least you forgot what you
were responding to -- which was the girl's belief that, to light two 6v
lamps, she should connect them to a 12v supply.

*I* said that would work fine if they were connected in series -- which is
true, as each lamp would see 6v.

*You* said "or connect them in parallel if the power supply is big enough".

That won't work for very long.

OK, I left something out (saying to use 6V, or clarifying what "big
enough" means). That's no excuse to stick in something that doesn't
belong.

I didn't stick *anything* in there, and you know it.

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

Mark Lloyd writes:

I hooked them up in parallel and they seem to work just fine. Incredibly
bright too.

LEDs in parallel with each other? It's unlikely that all would light,
since the threshold voltages would be slightly different, and the one
that's lowest would prevent the others from lighting.

If they're all from the same batch, they're probably pretty well matched
in forward voltage, and it only takes a little bit of series resistance
within each LED to approximately balance the current between LEDs.
This isn't a *good* way of connecting multiple LEDs, but it's not
automatically doomed to failure.

Modern LEDs can appear very bright. I noticed that with the holiday
lights I had this year (yes, I know that's "last year", but it is
still less than 2 months ago). Those LEDs look brighter than the
miniature incandescent's.

The little 1 W and 3 W LEDs appearing in flashlights are now brighter
than anything I ever saw from any incandescent bulb in the same size of
flashlight (2 AA battery).

Dave

"Doug Miller" wrote in message
. net...
In article , "Bob F"
wrote:

If the voltage is insufficient, no power supply, no matter what current
it
can deliver, will light the bulb.

Oh, really? Let's see the explanation of that.


There are hundreds of amps available from a car battery. Ever notice you
don't get electrocuted when you touch both terminals? It's the same thing.

Bob

"Doug Miller" wrote in message news:%KOAh.6000

Yes and no. When you draw more current than rated three things can
happen.

1. The supply goes over current until something burns up
2. The supply shuts down electronically (the voltage goes to nil)
3. The voltage diminishes slowly as resistive losses in the supply
use a higher and higher percentage of the voltage drop.

In any case the supply limits the current by limiting the voltage at
the output.

IOW -- the current *is* determined by the supply. :-)

Wrong - the current is LIMITED by the supply. If the supply
can supply the needed current, a bigger supply at the same voltage
will NOT increase the current.

Bob

"Mark Lloyd" wrote in message

LEDs in parallel with each other? It's unlikely that all would light,
since the threshold voltages would be slightly different, and the one
that's lowest would prevent the others from lighting.

As someone said earlier, your LEDs probably are actually LED modules,
and come with built-in resistors. You have a separate resistor in
series with each LED.

He said 3V LEDs in one post - that implies the internal resister.

Bob

In article , "Bob F" wrote:

There are hundreds of amps available from a car battery. Ever notice you
don't get electrocuted when you touch both terminals? It's the same thing.

Mostly, you don't get electrocuted from a car battery because it's DC.

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

Larry inspired greatness with:

The LEDs being produced today for illumination do not require any resistor.
They are connected directly to the battery or other power source.


They may have a resistor packaged with it, but a LED is a *DIODE*, and
it requires a resistor.

"Doug Miller" wrote in message
et...
In article , "Bob F"
wrote:

There are hundreds of amps available from a car battery. Ever notice you
don't get electrocuted when you touch both terminals? It's the same
thing.

Mostly, you don't get electrocuted from a car battery because it's DC.

You have a very strange concept of electricity.
I wonder where you get these ideas.

You don't get electrocuted from a car battery because it's
only 12 volts, which isn't enough in most conditions to push
the needed few milliamperes through your body to electrocute
you.

Bob

(Doug Miller) writes:

Mostly, you don't get electrocuted from a car battery because it's DC.

Oh? Try grabbing the main battery terminals of a hybrid gas-electric
car. It's DC, at several hundred volts. You won't like it.

Actually, don't try it. You might not be able to let go, and you might
die.

Dave

In article ,
Chop Suey none wrote:
Larry inspired greatness with:

The LEDs being produced today for illumination do not require any resistor.
They are connected directly to the battery or other power source.


They may have a resistor packaged with it, but a LED is a *DIODE*, and
it requires a resistor.


Well, yes that is true. However they do not require an external
resistor in order to make a complete lighting circuit. Whatever resistance
is needed in built into the unit itself.

--
Make it as simple as possible, but no simpler.

Larry Wasserman - Baltimore Maryland - lwasserm(a)sdf.lonestar.org

On Thu, 15 Feb 2007 13:14:53 -0800, "Bob F"
wrote:


"Doug Miller" wrote in message news:%KOAh.6000

Yes and no. When you draw more current than rated three things can
happen.

1. The supply goes over current until something burns up
2. The supply shuts down electronically (the voltage goes to nil)
3. The voltage diminishes slowly as resistive losses in the supply
use a higher and higher percentage of the voltage drop.

In any case the supply limits the current by limiting the voltage at
the output.

IOW -- the current *is* determined by the supply. :-)

Wrong - the current is LIMITED by the supply. If the supply
can supply the needed current, a bigger supply at the same voltage
will NOT increase the current.

Bob

You can have a current limiting supply but most are not.

A simple supply is rated at a certain current at a certain voltage,
much like a amplifier is rated at a certain power at a certain
distortion level.

You can turn your amp up higher than its rating. The result is more
distortion along with more power. In a simple supply you can draw
more current than it is rated for. The result will be more current at
a lower voltage. How far you can go beyond the rating is dependent
on the design.

Bob F inspired greatness with:

Mostly, you don't get electrocuted from a car battery because it's DC.

You have a very strange concept of electricity.
I wonder where you get these ideas.

You don't get electrocuted from a car battery because it's
only 12 volts, which isn't enough in most conditions to push
the needed few milliamperes through your body to electrocute
you.


Heh... :-))


--

You ****ing gormless ****rag. How is he supposed to do that if the ****ing
machine bluescreens when it ****ing boots? **** me dead, there's some dumb
****s around.

--Kadaitcha Man
Message-ID: