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
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.
In parallel you are dividing the current. In series you are dividing the
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.
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
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
**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
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.
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.
On Feb 15, 1:01 am, email@example.com (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
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
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
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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