Question

An electrical appliance is rated at 1000 kVA, 220 V. If the appliance is operated for 2 hours, calculate the energy consumed by the appliance in:
A. kWh.
B. joule.

Answer 1

Hint: The electrical energy is numerically equal to the product of power and time. One kilowatt-hour is equal to the product of 1000 W or $1 Js^{-1}$ and 1 hour i.e., 3600 s.

Complete step by step answer:
 When an electrical appliance of power P watt is used for time t second, the electrical energy consumed is W and Work done or electrical energy W = P×t.
The electrical energy is generally measured in bigger units i.e., watt-hour (Wh) and kilowatt-hour(kWh). Watt-hour and kilowatt-hour are the commercial units of electrical energy.
One kilowatt-hour is the electrical energy consumed by an appliance of power 1 kW when it is used for 1 hour and it is used as the commercial unit of electrical energy supplied to our houses by the electric company and the electrical power consumed is said to be 1 kilowatt if a potential difference of 1 kV causes a current of 1 A to flow through it.
Electrical energy = P × t
W = 1000 kVA × 2 hours
W = 2000 kWh
1 kWh = 1 kW × 1h
$\implies$ 1 kWh = 1000 W × (60×60) s [1 hour = 60 min and 1 min = 60 s]
$\implies$ 1 kWh = $1000 Js^{-1}$ × 3600 s [1W = $1Js^{-1}$]
$\implies$ 1 kWh = 3600000 J
$\therefore$ 1kWh = $3.6×10^6$ J
Hence, 2000 kWh in joule is $2000 × 3.6×10^6$ J
Energy in joule = $7200 × 10^6$ J
Energy in joule = $7.2 × 10^9$ J

Note:
In a house, all the electrical appliances are rated at a potential difference of 220V.
The energy in kWh means the unit of power should be kW (kVA) and the unit of time is hour.