The rating of an electric oven is 4400W;220V.calculate its resistance when it operates at 220V.also calculate

the energy consumed in kWh in the month of september if the oven is used daily for 5hrs at the rated voltage. write the full procedure not only answer

Power = Current*Voltage

Therefore, 4400 = I*220
I = 20 amps

Now, Voltage = Current*Resistance
So, 220 = 20*R
R = 11 ohms

Second part. September has 30 days. The oven is rated at 4400 Joules/sec (a watt is a joule/sec)
This is 264,000 Joules/min Or 15,840 kJ/hour. This is 79,200 kJ/day or 2,376 MegaJoules in September. One kilowatt-hour is 3.6 MegaJoules. Therefore this is 660 kW-h's.

Hope my math checks out. I checked it. but we will see.

Watts = Voltage^2 / Resistance
4400 = 220^2 / R
R = 220^2 / 4400
==>11 Ohms.

30 days in September for 5 hours
= 150 hrs.
kwh for Sept. = 150 x 4400
==>660 kwh.

P=4400w
V=220V
I=P/V=4400/220
I=20A
so
R=V/I=220/20
R=11ohm
in a september 30 days 150 hours
so energy consumed
=4400*150=660000=660kWH

R = V^2/P
R = 4400^2/220
energy consumed =4400* 5*30/1000