(a) Define 1 watt.
(b) State the commercial unit of electric energy. Express it in terms of SI unit of energy.
(c) An electric refrigerator rated as 750 W operates 8 hours per day. What is the cost of the energy to operate it for the month of June at Rs. 2.50 per kWh ?
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1
Answer:
1 watt is amount of 1 joule work done in 1 sec
(b) KWH /Joule/sec
(c) energy-power×time
=750×8×30×2.50
=450000
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36
Answer:
please refer to the attachment..
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