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. However, using slightly bigger capacitor, the load
current can pushed until ‘in phase’ with the voltage can be
seen Fig.9(c).
Example 3
A 230V supply feeds three 1. the p.8
10A
For p. then:
I
1840
230 1
8
W
V
A
For p. 0.f.8.4
would require 20A cable, while the same load unity power factor could supplied with
an cable.4:
I
1840
230 4
20
W
V
A
.f.Alternating current theory and electrical machines
203
For (d):
P VI
P
cos (W)
150 0.8:
I
1840
230
W
V 0.9(b) the resultant load current has much
improved power factor.
.84 load supplied power factor 0.8 and 0.9(a) shows industrial load with low power factor.
POWER FACTOR CORRECTION
Most installations have low bad power factor because the inductive
nature the load.
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.6 900 W
φ
∴
The power factor most industrial loads lagging because the machines
and discharge lighting used industry are mostly inductive.84 1840 and 230V. capacitor has the opposite effect inductor, and
so seems reasonable add capacitor load which known have
a lower bad power factor, for example, motor. capaci-
tor connected parallel with the load, the capacitor current leads
the applied voltage 90°.f.84kW loads with power factors 0. 10. There may also the problem higher voltage drops the supply cables.
Figure 10. 0.
It can seen from these calculations that 1. Calculate the
current each power factor. When this capacitor current added the
load current shown Fig 10
