f.
. There may also the problem higher voltage drops the supply cables. capacitor has the opposite effect inductor, and
so seems reasonable add capacitor load which known have
a lower bad power factor, for example, motor. then:
I
1840
230 1
8
W
V
A
For p. Calculate the
current each power factor.
Figure 10. When this capacitor current added the
load current shown Fig 10.84 load supplied power factor 0.
Example 3
A 230V supply feeds three 1.84 1840 and 230V.4:
I
1840
230 4
20
W
V
A
.
The current given by:
I
P
V
cos φ
where 1.84kW loads with power factors 0. However, using slightly bigger capacitor, the load
current can pushed until ‘in phase’ with the voltage can be
seen Fig.9(c).4
would require 20A cable, while the same load unity power factor could supplied with
an cable. 0.6 900 W
φ
∴
The power factor most industrial loads lagging because the machines
and discharge lighting used industry are mostly inductive.f. 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.8.9(a) shows industrial load with low power factor. the p.4.
POWER FACTOR CORRECTION
Most installations have low bad power factor because the inductive
nature the load. 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.Alternating current theory and electrical machines
203
For (d):
P VI
P
cos (W)
150 0.8 and 0.f. 0. 10.8:
I
1840
230
W
V 0.8
10A
For p. This causes
an additional magnetizing current drawn from the supply, which
does not produce power, but does need supplied, making supply
cables larger.
It can seen from these calculations that 1
