4:
I
1840
230 4
20
W
V
A
.9(b) the resultant load current has much
improved power factor.84kW loads with power factors 0.f.9(a) shows industrial load with low power factor.
.4
would require 20A cable, while the same load unity power factor could supplied with
an cable.8.8 and 0.
It can seen from these calculations that 1. Calculate the
current each power factor.8
10A
For p.
The current given by:
I
P
V
cos φ
where 1. This causes
an additional magnetizing current drawn from the supply, which
does not produce power, but does need supplied, making supply
cables larger.f.f.
Example 3
A 230V supply feeds three 1. When this capacitor current added the
load current shown Fig 10. There may also the problem higher voltage drops the supply cables.Alternating current theory and electrical machines
203
For (d):
P VI
P
cos (W)
150 0.84 1840 and 230V. However, using slightly bigger capacitor, the load
current can pushed until ‘in phase’ with the voltage can be
seen Fig.6 900 W
φ
∴
The power factor most industrial loads lagging because the machines
and discharge lighting used industry are mostly inductive. 0. then:
I
1840
230 1
8
W
V
A
For p. capacitor has the opposite effect inductor, and
so seems reasonable add capacitor load which known have
a lower bad power factor, for example, motor.8:
I
1840
230
W
V 0.
POWER FACTOR CORRECTION
Most installations have low bad power factor because the inductive
nature the load.9(c). the p. a
result, the supply companies encourage installation engineers improve their power factor
to value close and sometimes charge penalties the power factor falls below 0.4.
Figure 10. capaci-
tor connected parallel with the load, the capacitor current leads
the applied voltage 90°. 10. 0.84 load supplied power factor 0
