# 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:
8 years ago

Help please, where do we obtain 1000 from.

Noor Nawaz said:
9 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.

(1)

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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