Power question

Aprilshowers

Question:
How much power does a light bulb connected to a 120-V outlet use if it draws .5A of current?
P=1.V
=.5 x 120
=60Watts
Does that look correct?

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Andrew Mason

Homework Helper
Aprilshowers said:
Question:
How much power does a light bulb connected to a 120-V outlet use if it draws .5A of current?
P=1.V
=.5 x 120
=60Watts
Does that look correct?
Yes. I think you meant P = IV

AM

mezarashi

Homework Helper
As your output uses Alternating Current. Just make sure that the values for voltage and current you have there are RMS (root mean square). The definition for real power is the integral of the dot product of voltage and current over time. RMS simplifies this however. By dividing the peak values by square root two, you can do simple multiplication and the equation P = IV still applies.

Nonetheless, I am happy to say that the 120-V rating in most European/North American homes is the RMS value.

Last edited:

dextercioby

Homework Helper
It doesn't say about alternating current.

Daniel.

FredGarvin

dextercioby said:
It doesn't say about alternating current.

Daniel.
I can't say I know of many 120VDC outlets around...

NewScientist

If you are dealing with alternating currents the equation becomes:

P = I.V.cos(h)

Where h is the phase shift between the graph of voltage and of current.

-NewScientist

Andrew Mason

Homework Helper
NewScientist said:
If you are dealing with alternating currents the equation becomes:

P = I.V.cos(h)

Where h is the phase shift between the graph of voltage and of current.

-NewScientist
There is a 0 phase shift between voltage and the current passing through a light bulb, which is resistance with virtually 0 reactance.

AM

NewScientist

Exactly, but there is not always 0 phase shift, that is why in that case cos (h)=1 as h = 0. Therefore the result can be simplified to IV.

In some cases cos(h) is not 1 as the phase shift is not 0 and so this ter mis useful.

-NewScientist

Andrew Mason

Homework Helper
NewScientist said:
Exactly, but there is not always 0 phase shift, that is why in that case cos (h)=1 as h = 0. Therefore the result can be simplified to IV.

In some cases cos(h) is not 1 as the phase shift is not 0 and so this ter mis useful.

-NewScientist
If the light bulb, which is a pure resistor, is drawing .5 amps (RMS) with 120 volts (RMS) across it, it is consuming 60 watts of power (RMS). There is always a 0 phase shift if the load, like a light bulb, is a pure resistor.

AM

NewScientist

I didn't read the specific first post and I saw a couple of bumbling post on P = IV in regards to AC, and so I posted the formula. I was speaking of other examples when I said that the cos(phase shift) was an important term.

Oh and if it were a flourescent lamp it wouldn't be in phase.

-NewScientist

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