The current given by:
I
P
V
cos φ
where 1. then:
I
1840
230 1
8
W
V
A
For p. the p.
It can seen from these calculations that 1. However, using slightly bigger capacitor, the load
current can pushed until ‘in phase’ with the voltage can be
seen Fig.84 load supplied power factor 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. This causes
an additional magnetizing current drawn from the supply, which
does not produce power, but does need supplied, making supply
cables larger.
Example 3
A 230V supply feeds three 1.
.84 1840 and 230V.8.f. 10.4.8:
I
1840
230
W
V 0. 0. 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
would require 20A cable, while the same load unity power factor could supplied with
an cable.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.6 900 W
φ
∴
The power factor most industrial loads lagging because the machines
and discharge lighting used industry are mostly inductive. There may also the problem higher voltage drops the supply cables. 0.8
10A
For p.8 and 0. When this capacitor current added the
load current shown Fig 10.
Figure 10.9(b) the resultant load current has much
improved power factor.4:
I
1840
230 4
20
W
V
A
.Alternating current theory and electrical machines
203
For (d):
P VI
P
cos (W)
150 0. Calculate the
current each power factor.9(c).84kW loads with power factors 0. capaci-
tor connected parallel with the load, the capacitor current leads
the applied voltage 90°.f.f
