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
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