Calculating Electricity Consumption at School: Physics Homework

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Homework Help Overview

The discussion revolves around calculating the electricity consumption of fluorescent lights in a school setting. The original poster seeks assistance in determining the energy consumed by approximately 200 lights operating under a potential difference of 240 V for 16 hours daily.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relevance of voltage in the context of power calculations and question whether the system is AC or DC. There are attempts to clarify the relationship between power, voltage, and energy consumption, with some participants noting the need to consider time in energy calculations.

Discussion Status

Some participants have provided insights into the calculations and assumptions involved, while others express uncertainty about the textbook's expectations regarding the use of specific formulas. The conversation indicates a mix of completed attempts and ongoing inquiries into the problem's requirements.

Contextual Notes

There is mention of potential confusion due to the inclusion of voltage information, which some participants believe may not be necessary for the calculations based solely on wattage and time. Additionally, the lack of clarity regarding whether the system is AC or DC is noted as a significant factor in the discussion.

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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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