Basic DC electricity question

[Bob]

Here's one you might enjoy, or at least get sone ideas from.

[Mark Lloyd]

I knew that stuff by fourth grade, but then I wasn't aliterate (can read but won't).

Cumulative resistance in the circuit. Which end, is irrelevant.

In a series circuit, current is the same through all devices. Voltage is divided according to resistance. If all lights are the same, it will be divided equally.

With a 12V supply and 2 (identical) lights in series, each light will be operating on 6V. Current for this circuit will be the same as one light on 6V.

How about using both series and parallel lighting circuits?

When I was in 3rd or 4th grade, a relative gave me a book of simple experiments. The section on electricity used both series and parallel.

BTW, I remember how hard it was to find the "#6 dry cells" that book called for. Most people hadn't heard of such things. These are large round 1.5v cells, bigger than the ones they make lantern batteries out of.

[Mark Lloyd]

An incandescent lamp gives off light because the filament is hot. The frequencies given off depend on the temperature of the filament. To see light, you need enough current to heat the filament enough that it's temperature corresponds to frequencies in the visible range.

If you get light from a lower voltage/current, the light will be redder (or yellower). These colors correspond to the low end of the visible spectrum. At full voltage you get more blue (high end frequency).

[mm]

Parallel. If you put 6 volt items in series, the required voltages are the sum of the individual voltages. Just like xmas tree lights that only need a couple volts, but times 50 and you get 110 volts and they plug into the wall directly. If you tried to use a 2.5 volt power supply with 50 xmas tree bulbs in a row, you'd get no light. Same with nmotors.

mY aNswer is going to conflict a little, maybe, with the first two, but not directly.

When a battery is running low, it still has most of the voltage it did when new. It was surprising to me, but the chem teacher did the arithmetic in front of us, and he didn't make any errors.

So when a battery is 80% dischagred it still has 80% of the original voltage. It might even have been that when 90% discharged it had 90% of the orignal voltage. So by the time a 12 volt battery is down to

6 volts, it must be 98+ discharged, and it's not going to have enoug output to actually make light from a 12 volt bulb. Frankly, I don't even think amperage is the problem, and I don't think a 12 volt large size car battery could light a 12 volt bulb when it is so discharged it puts out 6 volts, though I don't remember ever having a 12 volt battery that was so low.** Or if output ampterage is still linited, assume 100 such batteries in parallel, but still with only 6 volts.

Even a brand new 6 volt battery, even a big one like a 6 volt car battery (still used for golf-carts??) I don't think could light a 12 volt bulb.

**Well actually now I do. It's a gelcell I bouught for testing indoors, and it is down to about 5.5 volts, maybe partly because when it was 7, I connected a battery charger backwards! and drove it even lower. So in the next day or two I'll try to try it with a 6 volt bulb.

[Jack]

Should they be learning to hook up

Sure, why not? Ohm's Law or not, its still a good exposure to basic circuits.

My friend graduated from MIT at sixteen. I'll bet he was doing Laplace Transform for circuit analysis and wave equations when he was at the grade 6 age.

[Bob F]

That's one way to do it.

So teach her ohm's law.

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

"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.

A crystal radio?

Bob

[tnom]

Jacobs ladder. Then when the class is over you can make it into a super deluxe Taser.

[tnom]

It will light but just barely. Remove one cell from a two cell flashlight and you'll see.

[Bob F]

Oops - I missed the 2 C batteries. Just parallel the 4 LEDs to the 2 cell in series.

[Redbelly]

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:

A 100W, 12V lamp, when run at 6V, will require 5 or 6 Amps and run at

30 to 35 Watts. The filament temperature will be in the range 1800-2100 C, and will definately be visible. BUT you're 6V supply must be capable of 6 amps, which your battery of "C" cells is NOT.

A 10W, 12V lamp will need almost 0.5 to 0.6 A at 6V, and run at 3 to

3.5 Watts. I'm not sure if C cells are capable of this or not.

A 1W, 12V lamp will need only about 50-60 mA, and your C cells will be sufficient to power it.

Hope this helps.

Mark

[Redbelly]

p.s.: I'm assuming these are filament lamps, not LED's or other.

As a test of your existing setup:

1. What is the wattage of the filament you tried?

1. What happens to the voltage across the filament when you try to power it with the C batteries? If it drops way below 6V, then it requires a lot more current than the batteries can provide. You'll need a voltmeter for this, of course.

Regards,

Mark

[Mark Lloyd]

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.

[Larry]

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

[Mark Lloyd]

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]

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

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

[Mark Lloyd]

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

[Mark Lloyd]

And none at all with SERIES circuit.

[Redbelly]

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.

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 are:

(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

[Redbelly]

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:

Regards,

Mark

[Doug Miller]

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".