- Power can be defined as the rate of doing work.
- It is the work done in unit time.
- It decides how quickly work could be done.
- It can be given by the formula:

Or,

$\mathrm{p}=\frac{\mathrm{W}}{\mathrm{t}}$- The work, power and energy are all scalar quantities.
- When work is done, an equal amount of energy is consumed. Power as the rate at which energy consumed can be given by.

or

$\mathrm{p}=\frac{\mathrm{E}}{\mathrm{t}}$

The SI unit of power is watt.

**Watt**

The power of an agent is one watt if it does work at the rate of 1 joule per second or if it consumes energy at the rate of 1 joule per second.

It can be expressed as:

$1\mathrm{Watt}=\frac{1\mathrm{Joule}}{1\mathrm{Second}}$or

$1\mathrm{W}=1\mathrm{J}{\mathrm{s}}^{\u20131}$$1\mathrm{kilowatt}=1000\mathrm{watt}$

or

$1\mathrm{kW}=1000\mathrm{W}$1 megawatt = 1000,000 watts or 1 MW =1000,000 W or

$1\mathrm{MW}={10}^{6}\mathrm{W}$

Power of motor vehicles and other machines are given in terms of Horsepower (hp).

1 Horse power = 746 watts or 1 H.P. = 746 W

The power at any given instant of time is called instantaneous power.

Average-power is defined as the total energy consumed divided by the total time taken.

Power is always dependent on work done.

If a person does work at different rates his power also differs at different times. In such cases the concept of average-power is used.

Consider Bulb A of 60 watt bulb is switched ON 24 hours a day and there is another Bulb B of 60 watt turned ON for only 12 hours. Given the context, calculate the energy consumed by both the bulbs in one day.

For the first 12 hours both bulb A and B are turned ON therefore,

$\mathrm{Power}=60+60=120\mathrm{watts}$$\mathrm{Energy}=\mathrm{Power}\mathrm{x}\mathrm{Time}$

$=120\times 12=1.44\mathrm{kWh}(\mathrm{kilo}\mathrm{watt}\mathrm{hour})\phantom{\rule{0ex}{0ex}}$

Now for the next 12 hours only bulb A would remain ON:

power = 60 watt

energy = power x time

= 120 x 12 = 1.44 kWh (kilo watt hour)

$\mathrm{Now}\mathrm{for}\mathrm{the}\mathrm{next}12\mathrm{hours}\mathrm{only}\mathrm{bulb}\mathrm{A}\mathrm{would}\mathrm{remain}\mathrm{ON}:$$\mathrm{power}=60\mathrm{watts}$

$\mathrm{energy}=60\times 12=0.72\mathrm{kWh}$

The power consumed during the whole day varies because one bulb is turned ON for only 12 hours so we need to calculate average power,

$\mathrm{Average}\mathrm{power}=\frac{\mathrm{Total}\mathrm{energy}\mathrm{consumed}}{\mathrm{Total}\mathrm{time}\mathrm{taken}}$

$=\frac{(1.44+0.7224)}{2}$

$=0.092\mathrm{kW}$

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