7. In a house 3 bulbs of 100 watt each lighted for 5 hours daily, 2 fans of 50 watt each used
for 10 hours daily and an electric heater of 1.00 kW is used for an hour daily. Calculate the
tota/energy consumed in a month of 31 days and its cost at the rate of 3.60 per kWh.
Answers
Answer:
Energy due to three bulbs
Energy = kWh
= 300/1000 × 5 × 31
= 93/2
= 46.5 units.
Energy due to two fans,
Energy = kWh
= 100/1000 × 10 × 31
= 31 units.
Energy due to electric heater,
Energy = kWh
= 1000/1000 × 1 × 31
= 31 units.
So total energy used in 31 days is 46.5 + 31 + 31 = 108.5 units.
Now rate is rs.3.60/units,
So,
Cost = 3.60 × 108.5
= Rs. 390.6
Hope it helped.
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Answer:
Total Energy consumed is 108.5 Kwh and Cost is Rs 390.6
Explanation:
We Have :-
3 bulbs of 100 watt each lighted for 5 hours daily
2 fans of 50 watt each used for 10 hours daily
Electric heater of 1.00 kW used for an hour daily
Rate of 3.60 per Kwh
To Find :-
Total Energy consumed in a month of 31 days and Cost
Formula Used :-
Energy consumed = Power × time
Solution :-
Energy consumed by Bulbs :-
Power = 3 * 100 watt = 300 watt = 0.3 Kw
Time = 5 hours in one day
= 5 * 31 in a month
= 155 hours
Energy = 0.3 * 155
= 46.5 Kwh
Energy consumed by Fans :-
Power = 2 * 50 = 100 watt = 0.1 Kw
Time = 10 hours in one day
= 10 * 31 in a month
= 310 hours
Energy = 0.1 * 310
= 31 Kwh
Energy consumed by Heater :-
Power = 1 Kw
Time = 1 hours in one day
= 1 * 31 in a month
= 31 hours
Energy = 1 * 31
= 31 Kwh
Total Energy consumed = 46.5 + 31 + 31
= 108.5 Kwh
Cost :-
1 Kwh = Rs 3.60
108.5 Kwh = Rs 3.60 * 108.5
= Rs 390.6