Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
What happens to the other bulbs in a series circuit if one bulb blows off?
Using the circuit given below, state which of the following statement is correct?
(a) When S1 and S2 are closed, lamps A and B are lit.22
(b) With S1 open and S2 closed, A is lit and B is not lit.
(c) With S2 open and S1 closed A and B are lit.
(d) With S1 closed and S2 open, lamp A remains lit even if lamp B gets fused.
Name two effects produced by electric current.
What is the purpose of fuse in an electrical circuit?
At the time of short circuit, the electric current in the circuit ______.
How is direction of current related to the direction of flow of electrons?
When an electric current is passed, electrons move from:
Find the odd one out.
Loudspeaker, Bar magnet/magnet, Electric motor, Microphone.
