Physics, asked by Silverstar007, 1 year ago

Gow to find the power rating of the machine if the efficiency is 40 percent

Answers

Answered by vijayjain41
1
3. Explaining Power



Suppose I have a simple 1 kW electric heater:

When the heater is running it uses 1 kW of electricity and it puts out 1 kW of heat.

If I turn it on for ten seconds, the power it uses at any time during those ten seconds is 1 kW. If I turn it on for one hour, the power it uses at any time in that hour is 1 kW. If I leave it on for a year, the power it uses at any time during that year is 1 kW. It is a 1 kW heater and no matter how long it is on or off, it remains a 1 kW heater.

If I turn on that same heater for five minutes and turn it off for 15 minutes, when it is on, the heater uses 1 kW and when it is off it uses 0 kW. If I am going to power this from a generator (or solar panel) I need to be able to provide a maximum power of 1 kW. If I repeat this cycle (on for a quarter of the time), then in an hour the heater would be on 15 minutes and off for 45 minutes. When the heater is on it uses 1 kW and when it is off it uses 0 kW. The average power will be 0.25 kW.

A heater that is 2 kW is more powerful. It has twice the power of a 1 kW heater. If I turn on the 2 kW heater, it will heat the room twice as fast as if I had used the 1 kW heater. The electrical circuit that feeds my 2 kW heater must be capable of supplying more power than a circuit that feeds a 1 kW heater. 5

In any given period of time, the 2 kW heater will use twice the energy as the 1 kW heater.


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