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?
20.16 kWh
201.6 kWh
2.01 kWh
8.4 kWh
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
29 comments Page 1 of 3.

Raza said:   6 years ago
Total kWH = power * time.

= 350 (watt) *24 (hour) *24 (days),
= 201600 watt hour,
= 201.6 kWH.
(5)

Ali BaBa said:   5 years ago
Thanks to all for explaining the answer.
(4)

Tapas Khan said:   2 years ago
Kilowatt/Hour Means = ( 350 * 24 * 24)/1000.
= 201.6kwh.
(3)

Md Kaium said:   3 years ago
KWh= (350*24*24)/1000
=201.6kwh.
(1)

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.
(1)

Rathi said:   1 decade ago
@Radha.

Because 1kW = 1000 W. So 201600 Watt hours is equal to 201.6 kWh.

Pravin gaikwad said:   6 years ago
Here, it is (use in watt * total hours)/1000.

Sstyam said:   6 years ago
You are correct, Thanks @Gopi.

Mahantesh said:   7 years ago
201.6kwh is correct.

ABHI said:   7 years ago
P=9*120=1080WATT....... FOR UNITY pf.
FOR 20 MIN,
E=(1080*20)/1000 =21.6kwh.


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