# 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
Explanation:
No answer description is available. Let's discuss.
Discussion:
29 comments Page 1 of 3.

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)

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

1 day = 24 hours,
24 days = ? hours,
(days x 1 day hours) = 24 days hours,
24x24 = 576 hours.
Now: 350x576 = 201600wh,
201600/1000 = 201.6 kwh.

1 Day = 24 Hours
24Day = 24*24
= 576 Hours

Kwh = (3950*576)/1000
= 201.6 kwh

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?

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.

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

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

CHAVITINA RAMAKRISHNA said:   8 years ago
@Messuli.

In the 'KILO watt-hour' asked, that is why we divided with 1000 in the calculation.

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