Advertisements
Advertisements
Question
An electrical appliance is rated at 1000 KVA, 220V. If the appliance is operated for 2 hours, calculate the energy consumed by the appliance in: (i) kWh (ii) joule.
Advertisements
Solution
Given: V = 220 volt, 1000 KVA
Time = 2 hours
(i) In kWh Energy consumed = Pt = 2000 kWh
(ii) In joule
We know, 1kWh = 3.6 × 106 J
So, 2000 kWh = `(2000 xx 3.6 xx 10^6)/1`
= 7.2 × 109 Joule
= 7.2 × 109 J
APPEARS IN
RELATED QUESTIONS
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor?
Name two effects produced by electric current.
Define the following:
Super conductors
My mother was operating washing machine on Sunday morning. Suddenly, she saw a spark coming out of the electric board and the electric current in the house failed. An electrician was called to look into the matter. What should he have noticed.
State expression for Cells connected in series.
What determines the direction of flow of electrons between two charged conductors kept in contact?
An electric bulb is marked 100 W, 250 V. How much current will the bulb draw if connected to a 250 V supply?
One unit of electrical energy consumed is equal to 1000 kilowatt-hour.
Problems for practice:
If 3A current flows through a circuit, then convert the current in terms of a milliampere.
