5 bulb of 100 watt lighted 6 hour daily 6 fan 50 watt used for 10 hour daily 1 fridge of a of 300 watt used for 28 hour daily 1Tv of hundred watt used for 10 hour daily 1 heater of 2 kilowatt to used for 2 hour daily what is the maximum power when all the appliance used at a time at what is electrical used at a time electrical energy consumed for 30 days of cost of 1 unit is 6 find the cost of what is a safe rating of electrical fuse
Given : 5 bulb of 100 watt lighted 6 hour daily
6 fan 50 watt used for 10 hour daily
1 fridge of a of 300 watt used for 24 hour daily
1 Tv of hundred watt used for 10 hour daily
1 heater of 2 kilowatt to used for 2 hour daily
cost of 1 unit is 6
To Find : what is the maximum power when all the appliance used at a time electrical energy consumed for 30 days
find the cost of
safe rating of electrical fuse
Solution:
5 bulb of 100 watt = 5 * 100 = 500 W in 1 day = 500 * 6 = 3000 Wh
6 fan 50 watt = 6 * 50 = 300W in a day = 300 * 10 = 3000Wh
1 fridge of a of 300 watt = 300W in a day = 300 * 24 = 7200 Wh
1 Tv of hundred watt = 100W in a day = 100 * 10 = 1000 Wh
1 heater of 2 kilowatt = 2000W in a day = 2000 * 2 = 4000Wh
maximum power when all the appliance used at a time =
500 + 300 + 300 + 100 + 2000 = 3200W = 3.2kW
Voltage rating = 220V
Current = 14.54 A
safe rating of electrical fuse = 16 A
in a day = 3000 + 3000 + 7200 + 1000 + 4000
= 18200 Wh
= 18.2 kWh
= 18.2 units
cost of 1 unit is 6
Hence total cost = 18.2 * 6 = 109.2 Rs
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Answer:
no answer OK byy good morning