# Electric Power Efficiency

### Power efficiency

Power efficiency is defined as the ratio of the output
power
divided by
the input power:

*η* = 100% ⋅ *P*_{out} / *P*_{in }

η is the efficiency in percent (%).

P_{in} is the input power consumption in
watts (W).

P_{out} is the output power or actual work in watts (W).

##### Example

Electric motor has input power consumption of 50 watts.

So The motor was activated for 60 seconds and produced work of 2970
joules.

So Find the efficiency of the motor.

Solution:

*P*_{in} = 50W

*E* = 2970J

*t* = 60s

*P*_{out} = *E* / *t*
= 2970J / 60s = 49.5W

η = 100% * *P*_{out}
/ *P*_{in} = 100 * 49.5W / 50W = 99%

**Energy efficiency**

So Energy efficiency is defined as the ratio of the output energy
divided by
the input energy.

*η* = 100% ⋅ *E*_{out} / *E*_{in }

η is the efficiency in percent (%).

E_{in} is the input energy consumed in joule (J).

E_{out} is the output energy or actual work in joule (J).

#####

##### Example

Light bulb has input power consumption of 50 watts.

So The light bulb was activated for 60 seconds and produced heat of
2400 joules.

So Find the efficiency of the light bulb.

Solution:

*P*_{in} = 50W

*E*_{heat} = 2400J

*t* = 60s

*E*_{in} = P_{in} * t =
50W * 60s = 3000J

Since the light bulb should produce light and not heat:

*E*_{out} = *E*_{in }-_{
}E_{heat }= 3000J - 2400J = 600J

η = 100 * *E*_{out}
/ *E*_{in} = 100% * 600J / 3000J = 20%

#####

## See also