How much power does a light bulb connected to a 120 V outlet use if it draws .5 A of?

In summary, a light bulb using 120 V will use .5 A of current. To expend one kilowatt hour of energy, you would have to leave the light bulb on for 3,600,000 seconds (16 hours).
  • #1
graphicer89
55
0

Homework Statement




A) How much power does a light bulb connected to a 120 V outlet use if it draws .5 A of current? Show your calculations


B) One kilowatt hour is a measure of energy which is equal to 3,600,000 J. How many hours would you have to leave on the light bulb from part A (60 W) in order to expend one kilowatt hour of energy (Remember that 1 W= 1 J/s)

C) If one kilowatt hour costs $.10 how long would you have to leave the light bulb from Part A) on in order to spend $1.00 on electricity? Show your calculations



Homework Equations



A) Current x voltage = energy

B) ?

The Attempt at a Solution



A) 120 V x .5 A = 60 W

B) I know that 60 w = 60 joules...so would it be 3,600,000 joules / 60 joules = 60,000 hours? I honest do not think that makes sense ..please help me out...

C) If i got the right answer from B...then 60,000 hours x .10 = 6,000 hours?...please help me out i am so close to finishing physics...and i just need help with this...
 
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  • #2


60w = 60J per second.
So it's 60,000 seconds, for one kilowatt hour

for part C, how many kilowatt hours are you looking at?
 
  • #3


graphicer89 said:
B) I know that 60 w = 60 joules...so would it be 3,600,000 joules / 60 joules = 60,000 hours? I honest do not think that makes sense ..please help me out...
You should have divided by 60 watts (not joules) there, and then you would have gotten 60000 seconds. :wink:
 
  • #4


graphicer89 said:
A) Current x voltage = energy

Not quite, that should be

Current x voltage = power

As the others hinted, power and energy are different, but related, concepts.
 
  • #5


oh ok ...so A is right then...
B...if its 60,000 seconds...divide that by 60 (60 seconds in minute)and i get 1,000 hours which makes a little bit more sense but still i don't know if I am off or I am on the answer...i feel that i have to keep dividing...so i think its something 16 hours?
C...hm well 60,000 seconds divided by .10 = 6,000 then divide again by 60 (seconds in 1 minute) gives me 100 so its 100 minutes or 1 1/2...same thing with this i really feel off...please help me out...this is basically easy math but i don't know why i can't capture the point...
 
  • #6


graphicer89 said:
oh ok ...so A is right then...
Yes, the power drawn by this bulb is 60 W.
B...if its 60,000 seconds...divide that by 60 (60 seconds in minute)and i get 1,000 hours which makes a little bit more sense but still i don't know if I am off or I am on the answer...i feel that i have to keep dividing...so i think its something 16 hours?
60,000 seconds divided by 60 gives you 1000 minutes, not hours. Divide again by 60 to get hours.

(16 hours is about right.)
C...hm well 60,000 seconds divided by .10 = 6,000 then divide again by 60 (seconds in 1 minute) gives me 100 so its 100 minutes or 1 1/2...same thing with this i really feel off...please help me out...this is basically easy math but i don't know why i can't capture the point...
First figure out how many kilowatt-hours you need to spend $1.00. And you already know how long it takes to burn one kilowatt-hour from part B. Multiply!
 
  • #7


Doc Al said:
Yes, the power drawn by this bulb is 60 W.

60,000 seconds divided by 60 gives you 1000 minutes, not hours. Divide again by 60 to get hours.

(16 hours is about right.)

First figure out how many kilowatt-hours you need to spend $1.00. And you already know how long it takes to burn one kilowatt-hour from part B. Multiply!

On C...so if it takes let's say 16 hours for one KW hour... $.10 x 10= $1.00 so i multiply 16 hours x 10= 160 hours for it to reach $1.00...is this right?
 
  • #8


graphicer89 said:
On C...so if it takes let's say 16 hours for one KW hour... $.10 x 10= $1.00 so i multiply 16 hours x 10= 160 hours for it to reach $1.00...is this right?
Exactly. (But 16 hours is only roughly right for B. Figure that part out more carefully.)
 
  • #9


Doc Al said:
Exactly. (But 16 hours is only roughly right for B. Figure that part out more carefully.)

Well i got 16.6666667 but basically rounded up = 17 hours...so then for C its 170 hours...am i on or off?
 
  • #10


graphicer89 said:
Well i got 16.6666667 but basically rounded up = 17 hours...so then for C its 170 hours...am i on or off?
I say you are smack on... bullseye!
 
  • #11


Doc Al said:
I say you are smack on... bullseye!

Oh wow i actually smack into it...but seriously thanks for the help...
 

Related to How much power does a light bulb connected to a 120 V outlet use if it draws .5 A of?

1. What is the formula for calculating power?

The formula for calculating power is P = VI, where P is power in watts, V is voltage in volts, and I is current in amperes.

2. How do I calculate the power of a light bulb connected to a 120 V outlet?

To calculate the power of a light bulb connected to a 120 V outlet, you can use the formula P = VI. Simply plug in 120 V for V and the given current value in amperes for I to calculate the power in watts.

3. Can I use a different unit for voltage and current in the power formula?

Yes, you can use different units for voltage and current as long as they are compatible. For example, you can use millivolts for V and milliamperes for I, as long as they are both in the same unit of measurement.

4. What is the power consumption of a light bulb drawing .5 A from a 120 V outlet?

The power consumption of a light bulb drawing .5 A from a 120 V outlet can be calculated by using the formula P = VI. Simply plug in 120 V for V and .5 A for I to calculate the power consumption in watts.

5. How does changing the current or voltage affect the power consumption of a light bulb?

Changing the current or voltage can directly affect the power consumption of a light bulb. According to the power formula P = VI, increasing the current or voltage will result in a higher power consumption, while decreasing the current or voltage will result in a lower power consumption.

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