4:
I
1840
230 4
20
W
V
A
. 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. Calculate the
current each power factor. capacitor has the opposite effect inductor, and
so seems reasonable add capacitor load which known have
a lower bad power factor, for example, motor.
Example 3
A 230V supply feeds three 1.4.84 1840 and 230V. 0.8
10A
For p.f.8:
I
1840
230
W
V 0.
It can seen from these calculations that 1.84 load supplied power factor 0. This causes
an additional magnetizing current drawn from the supply, which
does not produce power, but does need supplied, making supply
cables larger.8 and 0. capaci-
tor connected parallel with the load, the capacitor current leads
the applied voltage 90°.9(b) the resultant load current has much
improved power factor.9(c). the p. 0.6 900 W
φ
∴
The power factor most industrial loads lagging because the machines
and discharge lighting used industry are mostly inductive. When this capacitor current added the
load current shown Fig 10.Alternating current theory and electrical machines
203
For (d):
P VI
P
cos (W)
150 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.f. then:
I
1840
230 1
8
W
V
A
For p. However, using slightly bigger capacitor, the load
current can pushed until ‘in phase’ with the voltage can be
seen Fig. There may also the problem higher voltage drops the supply cables.
.84kW loads with power factors 0.4
would require 20A cable, while the same load unity power factor could supplied with
an cable. 10.8.f.
The current given by:
I
P
V
cos φ
where 1.
Figure 10
