Lightbulbs and power/resistance

[annie122]

To increase the brightness of a desk lamp a student replaces a 50 W light bulb with a 100 W light bulb. Compared to the 60 W light bulb the 100 W light bulb has?

A. Less resistance and draws more current
B. Less resistance and draws less current
C. More resistance and draws more current
D. More resistance and draws less current

Answer:A

My questions are:
1. are we assuming constant voltage?
(And I'm going to ask a really basic one: are we supplied constant voltage at our homes?)

2. brightness = k*(power)?

3. is it just a fact that higher power output devices have less resistance?

 

[phinds]

wall power (well, I actually should say voltage) is constant. As for your other questions, what do you think?

 

[OmCheeto]

My questions are:
1. are we assuming constant voltage?
(And I'm going to ask a really basic one: are we supplied constant voltage at our homes?)

As a general rule, yes.

2. brightness = k*(power)?

I had to look this up. I didn't trust the wiki answer, as it didn't quite look right.

For a supply voltage V near the rated voltage of the lamp:
Light output is approximately proportional to V^3.4

But they referenced the source, and I found the answer:

V_d = Design Voltage
V_a = Applied Voltage

Rerated Candlepower = [V_a / V_d ]^3.5 x Candlepower at Design Voltage

http://www.welchallyn.com/documents/Lighting/OEM_Halogen_Lighting/MC3544HPX_Catalog_2_11_09.pdf

I plugged in 125v & 120v and came up with 115%.
125/120 = 1.04
So a 4% increase in voltage increased luminosity by 15%

Thank you for asking this question. I didn't know that.

3. is it just a fact that higher power output devices have less resistance?

You can plug numbers into the following equations and find out.

P = I*E = I²*R = E²/R
P = Watts
I = Amps
E = Volts
R = Ohms

 

[I like Serena]

I had to look this up. I didn't trust the wiki answer, as it didn't quite look right.

But they referenced the source, and I found the answer:

http://www.welchallyn.com/documents/Lighting/OEM_Halogen_Lighting/MC3544HPX_Catalog_2_11_09.pdf

I plugged in 125v & 120v and came up with 115%.
125/120 = 1.04
So a 4% increase in voltage increased luminosity by 15%

Thank you for asking this question. I didn't know that.

Wiki is not wrong. It says the same thing.

Anyway, this formula applies only for a voltage outside of the design specification, which is not something you would normally want to do. (The lamp will break.)

The wiki article also contains a table, specifying how the perceived brightness (in Lumen) varies with varying types of incandescent light bulbs when used at their design specification.
The relation in that table is:
$\qquad Brightness = k \times Power^{1.2} \quad$ up to $\quad k \times Power^{1.4}$.
depending on the type of lamp.

The amount of radiated power is always exactly the power you put into the lamp.
It's just that by far most of it is in the infrared spectrum, that is, radiated as heat.
The efficiency of a lamp is the ratio between amount of visible light (in Lumen) and the radiated power (in Watt).

The reason that you get better efficiency from a lamp with higher power specification is because the filaments burn hotter, meaning a larger percentage of power is radiated in the visible spectrum. But burning hotter also means quicker degradation.

 

[CWatters]

1. are we assuming constant voltage?
(And I'm going to ask a really basic one: are we supplied constant voltage at our homes?)

More or less constant.

2. brightness = k*(power)?

Apply conservation of energy. Where does the incoming power go?

3. is it just a fact that higher power output devices have less resistance?

If you are talking about devices with the same design voltage then in general yes.

If the design voltage is different then it's not allways true.

You can design a light bulb to suit a car (12V) or mains (110V).

V = IR
P= IV
so
P=V^2/R

Some combinations and the resulting resistance...

12V, 50W... R = 12^2/50 = 2.88 Ohms
110V, 50W ... R = 110^2/50 = 200 ohms

12V, 100W... R = 12^2/100 = 1.44 Ohms
110V, 100W ...R = 100^2/100 = 100 Ohms

So in this case a 12V, 50W bulb has a lower resistance (2.88Ohms) that a 110V, 100W bulb (100 Ohms).