Calculating Electricity Consumption at School: Physics Homework

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SUMMARY

The discussion centers on calculating the electricity consumption of 200 fluorescent lights, each rated at 40 W, operating for 16 hours daily at a voltage of 240 V. The total energy consumed is calculated as approximately 128,000 watt-hours or 128 kWh. The conversation highlights the importance of understanding the distinction between instantaneous power and energy consumed over time, emphasizing that the voltage information is not necessary for this specific calculation. Additionally, it notes that the problem lacks clarity regarding whether the system is AC or DC, which affects the calculation due to the power factor in AC systems.

PREREQUISITES
  • Understanding of electrical power formulas, specifically P=VI
  • Knowledge of energy consumption calculations in watt-hours
  • Familiarity with AC vs. DC systems and their implications
  • Basic grasp of the concept of power factor in AC circuits
NEXT STEPS
  • Research the differences between AC and DC electrical systems
  • Learn about the concept of power factor and its significance in AC power calculations
  • Explore advanced energy consumption calculations for multiple devices
  • Study the implications of instantaneous power vs. average power in electrical systems
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This discussion is beneficial for physics students, electrical engineering students, and educators looking to enhance their understanding of electricity consumption calculations and the underlying principles of electrical systems.

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Homework Statement


A pgysics student wishes to determine the amount of electic energy consumed in one day at his school as a result of classsrom and hallway lighting. A quick survery revealed that there were approximately 200 40 W fluroescent lights opearting under a potentional difference of 240 V for 16 hours each day.

Hey guys, this isn't a hard question, I just don't know how to do it, could you please explain how?


Homework Equations


V=IR V= deltaEq/Q


The Attempt at a Solution



I don't know where to put the watts in my equations or what to do with it.
 
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Hello Hellohi! :smile:

(have a delta: ∆ wink:)
Hellohi said:

Homework Equations


V=IR V= deltaEq/Q

W = VI (= I2R) :wink:
 
Hey, I got my question right without the help but thank you very much for replying.( Doesnt know wether to drink or eat delta):D
 
Hellohi said:
Hey, I got my question right without the help but thank you very much for replying.( Doesnt know wether to drink or eat delta):D
I know u have solved the problm but just for the better understanding of everyone else who follows this link, I'd like to add..
W=VI is but only instantaneous power.
To find energy consumed u'd have to include the factor of time..
Evergy consumed =\Sigma (power consumed per device*time in hours) in the units of watt-hour.

But electrically the question is incomplete as it does not tell the student if the system is DC or AC and expects the student to complete the assignment assuming the system to be DC. If the system is AC then it also require one additional factor called the Power factor or the relationship of lead/lag of current to voltage.
Also in the question, there is no actual need of the voltage information if the computation is to be done based on the wattage and time information, it is but a distraction to confuse the student into thinking it is relevant information.
The approximate Power consumed per day is;
= 200 x 40 x 16 watt-hour
where 200 is the number of lights
40 is the instantaneous wattage or power consumed by each of the 200 lights
16 is the number of hours the power was consumed
= ~128000 watt-hour or 128KW-hr where '~' stands for approximately as data states that the power consumed is approximately 40W. K stands for Kilo or multiplier of 1000

the average power consumed per hour is =128KW-hr/24 in unit of watt
= ~128000/24 watt
= ~5.333 KW
 
Hellohi said:
( Doesnt know wether to drink or eat delta):D

You can get delta-wings :-p

barbecue them in sauce! :biggrin:
 
LOL, thanks for the laughs and help, and I should have been more clearer with my second reply. The thing is, the textbook expects me to use the formula P=VI but they don't say that in that part of the chapter, only after it. :smile:
 
Hellohi said:
LOL, thanks for the laughs and help, a...textbook expects me to use the formula P=VI but they don't say that in that part of the chapter, only after it. :smile:
Hi, but can you please share ur answer, if not the steps u used... Coz actually don't need to use the P=VI equation. I understand that the thread is over, but just wanted to know what the book expected you to do...
 
Hey man, soz for not replying earlier. The book actually expected me to use P=VI and no other formula. If I find any other way to solve this question ill let you know.

All the best wishes. :smile:
 
Just let me know if you want the answer, thanks.

All the best wishes.
 
  • #10
yeah sure, let me know the answer as I have already posted my answer in the 4th post...
Just wanted to concur with what was required...
 

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