Electrical Engineering - Energy and Power - Discussion
Discussion Forum : Energy and Power - General Questions (Q.No. 4)
4.
A certain appliance uses 350 W. If it is allowed to run continuously for 24 days, how many kilowatt-hours of energy does it consume?
Discussion:
29 comments Page 2 of 3.
Mukeshbhai ramanbhai ganvit said:
8 years ago
An electrical iron drawing 9 A from 120 V Supply main is operated for 20 minutes, the energy consumed is?
CHAVITINA RAMAKRISHNA said:
8 years ago
@Messuli.
In the 'KILO watt-hour' asked, that is why we divided with 1000 in the calculation.
In the 'KILO watt-hour' asked, that is why we divided with 1000 in the calculation.
Messuli said:
9 years ago
Help please, where do we obtain 1000 from.
Noor Nawaz said:
10 years ago
350*24 days*24 hours/1000 for conversion into kilos.
Rathi said:
1 decade ago
@Radha.
Because 1kW = 1000 W. So 201600 Watt hours is equal to 201.6 kWh.
Because 1kW = 1000 W. So 201600 Watt hours is equal to 201.6 kWh.
Radha said:
1 decade ago
Hello how you have taken 1000? why you divided ?
Pramod shinva said:
1 decade ago
1 day = 24 hours.
24 days = 576 hours.
KW H = P*TIME IN HOURS.
= 350*576 (We are used here 350 watts not a kw).
= 201600. Watt hours.
= 201600/1000.
= 201.6 KW H. IS ANSWER.
24 days = 576 hours.
KW H = P*TIME IN HOURS.
= 350*576 (We are used here 350 watts not a kw).
= 201600. Watt hours.
= 201600/1000.
= 201.6 KW H. IS ANSWER.
(2)
Tshering said:
1 decade ago
Energy(kWh)= Power(kW) x Time(hrs)
Power=350 watt i.e 350/1000= 0.350 kW
Time(hr)= 24days, i.e 24x24= 576 hrs
Therefore,
E= 576x0.350= 201.6
hence energy consumed = 201.6 kWh
Power=350 watt i.e 350/1000= 0.350 kW
Time(hr)= 24days, i.e 24x24= 576 hrs
Therefore,
E= 576x0.350= 201.6
hence energy consumed = 201.6 kWh
N KIRAN said:
1 decade ago
350w*24days*24hours = 201.6kw all are says correct answer.
Ravi Purohit said:
1 decade ago
It is Option B because 350 watt X 24 days X 24 hours per that = 201600 watt-hour = 201. 6 kilo watt-hour.
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