*pf).*The power factor is only applied to alternating current and sometimes confusing.

Efficiency is the ratio between
the output power to input power

h = output power / input power

Efficiency value is ranging from
0 to 1. In the real world the efficiency value is typically
ranged from 0.6 to 0.8, the bigger the better. Efficiency
applies to direct current (dc), also applies to alternating
current (ac). Power lost at the output is
usually converted into heat, occurs in: coils, resistors, condensers,
transformers, transistors, ICs, etc.. Power can also lost due to
friction, usually occurs in: electric motors, generators, alternators, pumps,
compressors, air conditioning (ac), etc..

Especially on
alternating current, there is other thing that affect the power output and make it less than the input power, it is the power factor (

*pf*):*pf*= real power / apparent power = watt / voltampere

In direct current, power in watt unit is always equal to the power unit voltampere (va). Because power factor in
direct current is always close to or equal to 1 (one).

In alternating current, a
resistive load which is produced by
resistor
component, has a power factor close to 1, because the wave of voltage and the wave of current are always synchronous. So, if the equipment contains only resistors such as heater,
soldering iron, incandescent lamps, etc., the
power in watts can be considered equal to the power in voltamperes unit.

For inductive load, then the alternating current wave is not synchronized with the wave of voltage, current phase lags behind the voltage phase. Therefore, the
voltage multiplied by amperage is not the same as watt. So power
factor is lower than 1 (one).

Likewise, if
the alternating current gets capacitive load, the voltage
lags behind the current phase. So power factor is lower than 1 (one).

Power factor can also be
calculated by the difference degree in phase of voltage and amperage:

*pf*= cos q

The
picture below shows charts of voltage,
amperage, and power, with a phase difference of 30 degrees or power factor = cos q = cos 30 = 0.87, this is a very good power factor.

The picture below shows charts of voltage, amperage, and power, with a phase difference of 90 degrees or power factor = cos q = cos 90 = 0, very low power factor. We can see power curve is going down below zero line or negative, and positive part of power chart is about the same as negative part of power chart, so effective power, or power in watt is close to zero.

One way to
improve power factor is by installing parallel
capacitors
and located as close as possible to inductive loads (coils,
motors, etc.). The greater the load the greater capacitor,
therefore cost consideration will be needed. Some power loss and converted to heat and current leaks in capacitor will
reduce efficiency. For electric motors, always apply load close to the load rate, where the motor is designed to work most efficiently. Avoid equipments operate
with voltage above rating.

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