Nothing To Do With Cars, But I Need Help (ill erase it when im done)

[GreyGoose006]

so my physics homework is due tonight and i have a question about the relationship between volts and watts.

the question says that a european lightbulb rated at 100w with 220v power is brought to the US. how many watts does it put out on 120w.

i was thinking that it would be simply (120/220)*100=54w, but that isnt right.
anyone know?

for some reason i cant wrap my head around this whole electricity thing lol.

[dave92cherokee]

If I calculated it right it should be 29.75. What you do to get that is you know it is 220V and uses 100W. Next figure out what the resistance of the bulb is which is 220V squared divided by the 100W of power used. That gives you 484 ohms of resistance. Then you need to figure out how much current is going to be used on the 120V system so you take the 120V and divide it by the 484 ohms and that gives you .2479 amps. Take that measurement and multiply it times the 120V and that gives you the 29.75W. Check out this site it should help you with doing any electrical equations like that.

http://www.sengpielaudio.com/calculator-ohm.htm

[Selectron]

Yep, just confirming what Dave said - I was typing my reply at the same time so I'll post it anyway and it may help clarify the problem for you.

You start out knowing the European voltage and wattage - from that can be determined the current, and from that can be determined the resistance.

Calculate the current first, using I = P / V, where I is the current (amps), P is the power (watts), and V is volts, so:

I = P / V
I = 100 / 220
I = 0.454545 Amps

You can now calculate the filament resistance, using R = V / I, where R is resistance, so:

R = V / I
R = 220 / 0.454545
R = 484.0 ohms


So how many watts will the bulb dissipate with 120 volts applied? Well firstly, calculate the current, using I = V / R, so:

I = V / R
I = 120 / 484.0
I = 0.248 amps

You can now calculate the wattage, using P = V * A

P = V * A
P = 120 * 0.248
P = 29.76 watts

And that's the answer.

========================
Once you're familiar with the relationships between voltage, current, resistance, and power, you can skip steps and tackle the problem thus:

R = V * V / P
R = 220 * 220 / 100
R = 484.0 ohms

then:

P = V * V / R
P = 120 * 120 / 484.0
P = 29.75 watts

The discrepancy of 0.01 watts between the two answers is because I only used six decimal places for the current, at step one in the first calculation, so retain as many decimal places as possible throughout, to achieve greater accuracy.

[bearcat58]

I just a little confused on a couple of things. I know that “Power in is Power out!”
Watts (Power) is the product of Amps (current) x Volts.

In a resistive load like a light bulb, the resistance does not change significantly (well maybe a little with the difference in heat). We can assume that it does not, so this parameter does not even enter into the equation.

Power (watts) = Volts x Amps.


100W = 220V x ?A
Or
?A = 100W/220V.

A = 0.4545

Complete- 100W = 220V x 0.4545A
A 100W light bulb will draw 0.4545 amps of current at 220 volts of electrical potential.

NOW let us change the Voltage.

100W = 120V x ?A

?A = 100W/120V

A = 0.8333

Complete- 100W = 120V x 0.833A
A 100W light bulb will draw 0.833 amps of current at 120 volts of electrical potential.

The Power (Watts) Does not change.
If you change the voltage, and the Resistance of the load does not change, the Amp draw is what is changing. NOT THE WATTS.

It will still be a 100W light bulb.

[bearcat58]

My Bust…………..nix that last post.

I just looked up my old electrical engineering text (early 80’s)

Power in is Power out has to do with Inductive loads like transformers and Motors.

Light bulb have a fixed filament resistance and it’s watt rating is based off of the amount of voltage applied. Resistance is defined as a resistance of current flow. So this 100W bulb at 220V will only allow 0.4545 amps of current to flow. The resistance will still remain the same.

If we drop the volts and maintain the current……..then 120V x 0.4545A will yield 54.54W of Power.

[curtis73]

Good catch... 100w is 100w regardless. Halving the voltage should double the amperage.

[534BC]

As the voltage drops, so does the current (the power doesn't remain constant). The bulb is no longer a "100 watt bulb" at the new voltage.

[KiwiBacon]

Originally Posted by Selectron

P = V * V / R
P = 120 * 120 / 484.0
P = 29.75 watts

Agreed.
A lightbulb is simply a resistive load and we can assume (for a homework problem) that the resistance stays the same.

In reality it's a non ohmic device and will run cooler (and dimmer) on lower voltage so will have less resistance.
This will draw more current and power than predicted above, but that's far beyond the scope of a normal homework problem.

[2.2 Straight six]

tell your professor that europe as a whole runs mostly on 120v, the UK runs 230-240v. so his/her question sucks.

[KiwiBacon]

Originally Posted by 2.2 Straight six

tell your professor that europe as a whole runs mostly on 120v, the UK runs 230-240v. so his/her question sucks.

Sure about that?
This page: http://www.kropla.com/electric2.htm
Shows germany, belgium, switzerland etc all on 230v.

[2.2 Straight six]

wow, didn't know that. never been to belgium, been to germany twice and switzerland a few time. don't really care as long as stuff works when i plug it in.

[KiwiBacon]

Originally Posted by 2.2 Straight six

wow, didn't know that. never been to belgium, been to germany twice and switzerland a few time. don't really care as long as stuff works when i plug it in.

Modern multi-voltage power supplies are great. Plug anything in anywhere and it won't blow up.

[GreyGoose006]

ok, so the voltage is squared...
that makes sense when i look at the formulas, but 100w to 29ish watts seemed rediculously low.
well thank you everyone.

and yes kiwi, the question does suck.
i doubt the damn thing would even screw in. lol.

thanks for the link btw.



should i delete this or are we having fun talking about electricity?

[GreyGoose006]

Originally Posted by bearcat58

Light bulb have a fixed filament resistance and it’s watt rating is based off of the amount of voltage applied. Resistance is defined as a resistance of current flow. So this 100W bulb at 220V will only allow 0.4545 amps of current to flow. The resistance will still remain the same.

If we drop the volts and maintain the current……..then 120V x 0.4545A will yield 54.54W of Power.

thats what i thought too, but apparently not according to the book

[KiwiBacon]

Originally Posted by GreyGoose006

ok, so the voltage is squared...
that makes sense when i look at the formulas, but 100w to 29ish watts seemed rediculously low.
well thank you everyone.

Not only that, but the 29 watts is the power consumption, the light output will be much less than 29% of original. It'll put out proportionally more heat and the ligt will be more yellow than white.