Basic DC electricity question

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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.
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In parallel you are dividing the current. In series you are dividing the voltage.

There is no threshold for the incandescent lamp, after all its an analog device and light output is varied from zero to 100% continuous with respect to varying the voltage. You don't see it because you haven't enough current to heat the filament up - see if you could see it in the dark. I'm sure if you have a 6V car or motorcycle battery instead of the C cells, the 12v lamp would light up.
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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).
--
Mark Lloyd
http://notstupid.laughingsquid.com
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flashlight and you'll see.
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It should light, but dimly and more yellowish.
Try measuring the 6V supply WITH that light connected. You may find that it's loading your C cells so much it's getting a lot less than 6V.

--
Mark Lloyd
http://notstupid.laughingsquid.com
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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.
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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
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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?
2. 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
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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
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On Feb 15, 1:01 am, snipped-for-privacy@cs.ubc.ca (Dave Martindale) wrote:

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
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HK wrote:

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.
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Tony Hwang wrote:

And what differene does it make? High voltage or low voltage circuit. Basically it is all govenred by simple Ohm's law.
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