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Question

A surface of area 1m2 kept perpendicular to the sun rays absorbs 1.8MJ of solar energy in one hour.
(a) What is the amount of electrical energy produced when a solar panel of area 5m2 is exposed perpendicular to the sun rays for 5h. Take the efficiency of solar panel as 30%.
(b) Calculate the power generated in kW.

A
13.5 MJ; 0.75 kW
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B
1.5 MJ; 0.75 kW
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C
13.5 MJ; 1.75 kW
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D
1.5 MJ; 1.75 kW
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Solution

The correct option is A 13.5 MJ; 0.75 kW
The amount of energy absorbed by 1m2 area =1.8MJh1=1.8×106Jh1
The area of solar panel =5m2.
The total amount of energy absorbed by solar panel in 1 hour =5×1.8×106Jh1=9×106Jh1
The amount of energy absorbed in 5h=5×9×106J=45×106J
Efficiency of conversion =30%.
Amount of electric energy produced =30% [45×106]J
=30100×[45×106]J=1.5×9×106J
=13.5×106J=13.5MJ.
Power generated (p)=Et=energyunittime=13.5×106J5hour
=13.5×106J5×60×60s=13.536×5×104Js1
=13.5180×104W=13.518×103W=0.75kW
Power generated =0.75kW.

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