Change in watts and power dissipated

[aChordate]

One of my homework questions is giving me trouble:

"In, the US, the rms voltage from power outlets (known as line voltage) is 120 V. In the United Kingdom, line voltage is 230 V. If you take a lamp with a standard 100 W incandescent light bulb from the US, how much power will it dissipate if used in the UK? Assume that the electrical resistance of the bulb is constant."


So, I know that V=IR and P=VI. V1=120 V and V2= 230 V and the P of the light bulb= 100 W. Should I plug in power of the light bulb to the formula P=IV and find the current and then use that to find the excess power. For example,

P/V=I

(100W)/(120V)=I

I=0.8A

P=(0.8A)(230 V)= 200 W - 100 W = 100 W dissipated

I have no idea if I am on the right track or not, please help.

Thanks in advance.

 

[OmCheeto]

One of my homework questions is giving me trouble:

"In, the US, the rms voltage from power outlets (known as line voltage) is 120 V. In the United Kingdom, line voltage is 230 V. If you take a lamp with a standard 100 W incandescent light bulb from the US, how much power will it dissipate if used in the UK? Assume that the electrical resistance of the bulb is constant."


So, I know that V=IR and P=VI. V1=120 V and V2= 230 V and the P of the light bulb= 100 W. Should I plug in power of the light bulb to the formula P=IV and find the current and then use that to find the excess power.

bolding mine

No. Find the resistance of the lightbulb. As stated; "Assume that the electrical resistance of the bulb is constant."

For example,

P/V=I

(100W)/(120V)=I

I=0.8A

P=(0.8A)(230 V)= 200 W - 100 W = 100 W dissipated

I have no idea if I am on the right track or not, please help.

Thanks in advance.

You are welcome.

 

[aChordate]

So, should I calculate it like this:

P/V=I

(100W)/(120V)=I

I=0.83A

R=V/I = 120V/0.83A = 145 Ω

R=100 Ω

and then would I use the formula P=I^2*R ?

 

[gneill]

There's another power relation that might prove to be helpful: P = V²/R.

 

[aChordate]

P/V=I
(100W)/(120V)=I
I=0.83A

R=V/I = 120V/0.83A = 145 Ω
R=100 Ω

Then,
P = V2/R
P=(230V)^2/100Ω=529 W

So,
529W-100W = 429W or 400W dissipated?

 

[gneill]

P/V=I
(100W)/(120V)=I
I=0.83A

R=V/I = 120V/0.83A = 145 Ω
R=100 Ω

Why do you write R = 100 Ω after calculating it to be 145 Ω ?

Then,
P = V2/R
P=(230V)^2/100Ω=529 W

So,
529W-100W = 429W or 400W dissipated?

Does the question ask you to find the actual power dissipated or the difference in power dissipated? When I read the quoted question, it looks like they just want the power dissipated by that bulb when it's energized at 230V.

 

[aChordate]

Oh, oops, I was going by sig. figs. I recalculated and got 370 W dissipated.

I thought by "dissipated" meant the extra power that isn't being used by the 100W light bulb and that is why I found the difference. Hmmm, I guess not.

 

[aChordate]

Also, thank you so much for you input!

 

[gneill]

Also, thank you so much for you input!

Your 370W value looks quite reasonable. You're welcome