quick question about watts.

if i pluged your average 100 watt bulb into an average outlet, it would
draw just under 1 amp, right? well, if i pluged this same bulb into a
10 volt 10 amp power supply (100 watts) would i get the same result as
the regular outlet?

 

[John Popelish]

ngdbud@hotmail.com wrote:

if i pluged your average 100 watt bulb into an average outlet, it would
draw just under 1 amp, right? well, if i pluged this same bulb into a
10 volt 10 amp power supply (100 watts) would i get the same result as
the regular outlet?

No. The lamp allows 1 amp to pass through when 100 volts is impressed
across it, because ut has approximately 100 ohms of resistance (at
normal operating temperature). If you lower the voltage, the current
will go down, and the total power consumed by the lamp will go down,
too. To complicate things a bit, the resistance also drops as the
filament temperature drops, so the current does not stay proportional
to the voltage. But the current still is lower with a lower voltage,
just not as low as it would have been if the filament held a constant
resistance. Remember that ohms is just a word that means volts per
ampere.

--
John Popelish

 

[Don Klipstein]

In article <1105321050.516507.286010@c13g2000cwb.googlegroups.com>,
ngdbud@hotmail.com wrote:

if i pluged your average 100 watt bulb into an average outlet, it would
draw just under 1 amp, right? well, if i pluged this same bulb into a
10 volt 10 amp power supply (100 watts) would i get the same result as
the regular outlet?

No.

The voltage applied to the lightbulb would be 10 volts. Ideally, that
is. Ideally, a power supply delivers its rated voltage to the load even
if the load when supplied this voltage draws an amount of current
different from what the power supply is rated for.

Having a load drawing less current than the power supply is rated for is
OK, while drawing more current than the supply is rated for can result in
voltage being less than the rating of the power supply and/or overheating
of the power supply.

Power supplies that lack regulators will produce slightly to somewhat
higher voltage when you draw less than rated current. Possibly
significantly more, but I have yet to see or hear about twice as much.

So what happens when you apply 10 volts to a 100 watt lightbulb:

If the lightbulb is an ideal resistor, it would draw 10/120 as much
current at 10 volts as it would at 120 volts.

At 120 volts, 100 watts means .833 amp. 10/120 of that is .0694 amp.

But incandescent lamps are famously not linear resistors, since the
resistance of most metals varies roughly proportionately with temperature
in degrees K. (Very roughly - tungsten at 2800 Kelvin has about 15 times
as much resistance as it does at 300 K, not all that close to proportional!)

Roughly very roughly and there are variations, but the resistance of an
incandescent lamp is usually roughly proportional to the square root of
the applied voltage (within the range of voltages at which the lamp
glows). Likewise the current is roughly proportional to the square root
of the applied voltage. this means that a 100 watt 120V lightbulb at 10
volts would draw about .24 amp.

(I just tried this and got .24 amp at 9.3 volts and about .255 amp at
11.2 volts, interpolating to about .245-.246 amp at 10 volts - plus or
minus meter reading tolerances. BEWARE - some lightbulbs will not follow
my above "rule of thumb" that closely, with deviations becoming more
severe with greater degree of undervoltage.)

Go ahead and conect your 120V 100 watt lightbulb to your 10 volt 10 amp
power supply. Expect the bulb to glow a dim orange or reddish orange,
possibly not visible in normal room light and fair to good chance not if
the bulb is frosted or "soft white". Even at 12 volts a 120V 100 watt
lightbulb is a little dim for use as a night light.

- Don Klipstein (don@misty.com)

 

[Jamie]

ngdbud@hotmail.com wrote:

if i pluged your average 100 watt bulb into an average outlet, it would
draw just under 1 amp, right? well, if i pluged this same bulb into a
10 volt 10 amp power supply (100 watts) would i get the same result as
the regular outlet?

nope

 

Think of the current rating as meaning the supply can produce up to 10
amps.

 

let me see if i got it, as the voltage goes down so does the heat, and
the cooler it is the less resistance it has, so it pulls less amps.
this decrease in power makes for dimmer light. Do i have it figured
out?

 

[John Popelish]

ngdbud@hotmail.com wrote:

let me see if i got it, as the voltage goes down so does the heat, and
the cooler it is the less resistance it has, so it pulls less amps.
this decrease in power makes for dimmer light. Do i have it figured
out?

Almost.

As the voltage goes down, it pushes less current through the
resistance of the filament. Watts (heat) is the product of volts and
current, so with both the volts and current going down, the heat does
down.

But the lower temperature lowers the filament's resistance, so that
the lower voltage pushes more current through the now lower resistance
than it would have, if the resistance had remained constant, but still
less that it did at the higher voltage and temperature.

So the current falls as the voltage falls, but not in proportion to
the voltage. (cut the voltage in half and the current is something
like 3/4ths as large) This keeps the filament warmer than if the
resistance had remained the same.

Figure 1 in this page shows that for a +-20% change in lamp voltage,
the current varies +-10%
http://www.gilway.com/html/appl-tungsten.html

The difficulty in capturing such situations accurately in words is why
differential equations were invented.

--
John Popelish

 

[Don Klipstein]

In article <1105901390.874463.213490@z14g2000cwz.googlegroups.com>,
ngdbud@hotmail.com wrote:

let me see if i got it, as the voltage goes down so does the heat, and
the cooler it is the less resistance it has, so it pulls less amps.
this decrease in power makes for dimmer light. Do i have it figured out?

Actually, pulling less amps is from less voltage.

If resistance is constant, then amps pulled is volts divided by the
resistance.

But with a lightbulb, resistance decreases as voltage decreases, so as
voltage decreases the amps go down less than they would if the resistance
did not change.

A 100 watt 120V lightbulb has a resistance of 144 ohms at 120V.
According to Ohm's Law:

I=E/R, I=120/144, which is .833 amp

But apply various voltages to a hypothetical ideal 144 ohm resistor and
an actual 100 watt lightbulb:

(Figures for the lightbulb at voltages less than 120V I am predicting
based on what I know about them, and are mostly interpolations along with
two extrapolations and are not actual measurements except 9.3V and 11.2V
which are actual measurements.)
(ALSO please note that a 144 ohm 100 watt resistor whose resistance is 144
ohms both cold and at full power operating temperature is a hypothetical
item - I have yet to see this in any electronics parts supplier's
catalog! Please also note that 144 ohms is not even a standard resistor
value but 100, 120 and 150 ohms are standard resistor values, and
resistors of wattage rating more than 10 watts are less common and less
standardized than ones rated 10 watts or less!)


VOLTAGE RESISTANCE of RESISTANCE of CURRENT thru CURRENT thru
144 ohm resistor 100W 120V lamp 144 ohm res. 120V 100W lamp
--------------------------------------------------------------------------

120 V 144 ohms 144 ohms .833 amp .833 amp
100 V 144 ohms 131.5 ohms .6944 amp .76 amp
80 V 144 ohms 117.6 ohms .5556 amp .68 amp
60 V 144 ohms 102 ohms .41667 amp .59 amp
50 V 144 ohms 93 ohms .347 amp .54 amp
40 V 144 ohms 83 ohms .2778 amp .481 amp
30 V 144 ohms 71.9 ohms .20833 amp .417 amp
24 V 144 ohms 64.2 ohms .1667 amp .374 amp
20 V 144 ohms 58.5 ohms .1389 amp .342 amp
15 V 144 ohms 50.5 ohms .1041557 amp .297 amp
12 V 144 ohms 45.3 ohms .0833 amp .265 amp

11.2 V 144 ohms 43.9 ohms .0777 amp .255 amp
10 V 144 ohms 40.8 ohms .069444 amp .245 amp
9.3 V 144 ohms 38.8 ohms .0646 amp .24 amp

6 V 144 ohms 31 ohms .041667 amp .193 amp

..1 V 144 ohms 9.7 ohms .6944 mA 10.3 mA

Hope this helps,

- Don Klipstein (don@misty.com)