Physics, asked by jrhullatti, 19 days ago

A filament bulb rated 60 W operates 8
hours/day. What is the cost of the energy to
operate it for 30 days at Rs.5 per KWH.​

Answers

Answered by nehaverma63
1

Answer:

Power of refrigerator =600W=0.6kW

Total operation hours 8×30=240h

Work done or total electrical energy (W)=0.6×240=144kWh

Total cost of energy to operate =Rs.(144×4.00)=Rs.576

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