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Why and when fast decouple method is udes?

Answer»

Hey ur method is below

Dear
Here we will use the method of Power = voltage2resistance method .please follow the TOTAL solution you will understand the problem  as I trying do it with simple words and simple method to solve this problem .
The rating of the BULB is 200 V -100 W .
So the RESISTANCE of the bulb is , R =V2P=(200)2100=400 Ω
Now the bulb burn for 4 hours /DAY .
So , the energy CONSUMED by the bulb is , E =V2R×t=100×4=400 Wh=4001000=0·4 BOT unit
Now the cost of one unit energy is rupees 4·60  then the cost of energy consumed by the bulb is ,4·60×0·4=1·84 rupees
Regards



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