Answer Posted / sachin

Power factor Basics
In all industrial electrical distribution systems, the
major loads are resistive and inductive.
Resistive loads are incandescent lighting and resistance
heating. In case of pure resistive loads,
the voltage (V), current (I), resistance (R) relations are
linearly related, i.e.
V = I x R and Power (kW) = V x I
Typical inductive loads are A.C. Motors, induction
furnaces, transformers and ballast-type
lighting. Inductive loads require two kinds of power: a)
active (or working) power to perform
the work and b) reactive power to create and maintain
electro-magnetic fields.
Active power is measured in kW (Kilo Watts). Reactive power
is measured in kVAr (Kilo
Volt-Amperes Reactive).
The vector sum of the active power and reactive power make
up the total (or apparent)
power used. This is the power generated by the SEBs for the
user to perform a given amount of
work. Total Power is measured in kVA (Kilo Volts-Amperes)

The active power (shaft power required or true power
required) in kW and the reactive
power required (kVAr) are 90° apart vectorically in a pure
inductive circuit i.e., reactive power
kVAr lagging the active kW. The vector sum of the two is
called the apparent power or kVA, as
illustrated above and the kVA reflects the actual
electrical load on distribution system.
The ratio of kW to kVA is called the power factor, which is
always less than or equal to
unity. Theoretically, when electric utilities supply power,
if all loads have unity power factor,
maximum power can be transferred for the same distribution
system capacity. However, as the
loads are inductive in nature, with the power factor
ranging from 0.2 to 0.9, the electrical dis distribution
network is stressed for capacity at low power factors. So
COS PHI= KW/KVA

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