# Basic DC electricity question

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.
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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.

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The filament won't be as hot, which leads to a change in color (it won't be as white).

--
Mark Lloyd
http://notstupid.laughingsquid.com
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It is an interesting question actually. A 15w 12v bulb, when wired to a 6v battery draws .6a. Apparently your batteries can't produce .6a, so the bulb doesn't light up. Wouldn't be very bright anyhow.

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You're on the right track. The bulb filament glows from the amount of current through it. Too much current and the bulb is overbright or blows. Not enough current and the bulb glows weak or not at all. Current is a function of voltage divided by resistance. Higher voltage bulbs have higher resistance filaments. Therefore 6v divided by higher resistance equals less current and less glow.
Bob
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And, to follow up, OP could take the bulb and his VOM and measure the resistance across the filament and see what it is to see what the current draw should be...or maybe the bulb isn't good, even? :)
Or, of course, take enough batteries in series to make 12V ...
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Since power increases in proportion to voltage, but increases in proportion to current *squared*, don't you suppose he's better off to put the batteries in parallel to increase the current?
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Regards,
Doug Miller (alphageek at milmac dot com)
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On Feb 14, 10:57 am, snipped-for-privacy@milmac.com (Doug Miller) wrote:

...
But, since by the same token, I=V/R and R is what he's given w/ the particular bulb, P ~ V^2/R so same can be said of voltage...
I was only commenting wrt to the fact OP has a 12V-rated bulb and is/ was 'spearmint-ing so was only one of many alternatives to try and observe what happens, certainly not the only one..."Best" in any sense wasn't at all a part of the suggestion.
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That won't increase the current capability. It may still be insufficient for that bulb.
What is the current capacity for new C cells?
--
Mark Lloyd
http://notstupid.laughingsquid.com
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Yes.
No.
The filament resistance is too high, and/or your 6 volt source doesn't supply enough current. Are you trying to light up an automobile light bulb with four AA batteries?

It depends on the current as well as the voltage.

Depends on the device. For something as simple as a light bulb, I wouldn't think so. Try an experiment: see if you can light up an ordinary 60W incandescent light bulb (120V) from a car battery (12V).
--
Regards,
Doug Miller (alphageek at milmac dot com)
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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.
Any suggestions on a fun followup project that will help both of us learn more electricity concepts?

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HK inspired greatness with:

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

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

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.

--
Mark Lloyd
http://notstupid.laughingsquid.com
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On Wed, 14 Feb 2007 11:28:38 -0600, Chop Suey

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.
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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
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On Thu, 15 Feb 2007 00:19:54 +0000 (UTC), snipped-for-privacy@sdf.lNoOnSePsAtMar.org (Larry) wrote:

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
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Larry inspired greatness with:

They may have a resistor packaged with it, but a LED is a *DIODE*, and it requires a resistor.
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Chop Suey <none> wrote:

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
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On Fri, 16 Feb 2007 02:30:11 +0000 (UTC), snipped-for-privacy@sdf.lNoOnSePsAtMar.org (Larry) wrote:

I considered that. "Built-in" is not equivalent to "nonexistent".
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Mark Lloyd
http://notstupid.laughingsquid.com
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