It can seen from these calculations that 1.8.8:
I
1840
230
W
V 0.
. 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.f. 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. 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. capaci-
tor connected parallel with the load, the capacitor current leads
the applied voltage 90°.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
.84kW loads with power factors 0.9(a) shows industrial load with low power factor.8
10A
For p.
The current given by:
I
P
V
cos φ
where 1. When this capacitor current added the
load current shown Fig 10.
Figure 10.Alternating current theory and electrical machines
203
For (d):
P VI
P
cos (W)
150 0.
POWER FACTOR CORRECTION
Most installations have low bad power factor because the inductive
nature the load. There may also the problem higher voltage drops the supply cables.6 900 W
φ
∴
The power factor most industrial loads lagging because the machines
and discharge lighting used industry are mostly inductive. 0. capacitor has the opposite effect inductor, and
so seems reasonable add capacitor load which known have
a lower bad power factor, for example, motor. 0. the p.84 load supplied power factor 0.84 1840 and 230V. 10.9(b) the resultant load current has much
improved power factor.9(c).8 and 0.4. Calculate the
current each power factor.
Example 3
A 230V supply feeds three 1
