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 1 of 3.
Tapas Khan said:
3 years ago
Kilowatt/Hour Means = ( 350 * 24 * 24)/1000.
= 201.6kwh.
= 201.6kwh.
(4)
Md Kaium said:
4 years ago
KWh= (350*24*24)/1000
=201.6kwh.
=201.6kwh.
(2)
Ali BaBa said:
6 years ago
Thanks to all for explaining the answer.
(5)
Raza said:
6 years ago
Total kWH = power * time.
= 350 (watt) *24 (hour) *24 (days),
= 201600 watt hour,
= 201.6 kWH.
= 350 (watt) *24 (hour) *24 (days),
= 201600 watt hour,
= 201.6 kWH.
(9)
Pravin gaikwad said:
7 years ago
Here, it is (use in watt * total hours)/1000.
Sstyam said:
7 years ago
You are correct, Thanks @Gopi.
Mahantesh said:
7 years ago
201.6kwh is correct.
ABHI said:
8 years ago
P=9*120=1080WATT....... FOR UNITY pf.
FOR 20 MIN,
E=(1080*20)/1000 =21.6kwh.
FOR 20 MIN,
E=(1080*20)/1000 =21.6kwh.
Padma said:
8 years ago
Yes, you both are correct @Gopi and @Rajesh.
Sham said:
8 years ago
Thank you @Pramod Shinva.
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