Question

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​

Answer 1

Answer:

no answer OK byy good morning

Answer 2

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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