Question

A water heater has a resistance of 5.3 ohms and takes 43.5 amp when in operation. If itis in service on the average of 2 hr per day, how many kilowatt hours of energy will be expended during a 30 day month?"

Answer 1

We have given,

$R = 5.3 \Omega$

$I = 43.5A$

Power consumed, $P = I^2R = (43.5)^2 \times 5.3 = 10.02kW$

Energy consumption in one day,

$E = 10.02 \times 2 = 20.04kWhr$

Hence, the energy consumption in 30 days $= 20.04 \times 30 = 601.2kWhr$