Basic Electrical Installation Work

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10. 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. .6 900 W φ ∴ The power factor most industrial loads lagging because the machines and discharge lighting used industry are mostly inductive. Example 3 A 230V supply feeds three 1.9(b) the resultant load current has much improved power factor.9(c). capaci- tor connected parallel with the load, the capacitor current leads the applied voltage 90°. This causes an additional magnetizing current drawn from the supply, which does not produce power, but does need supplied, making supply cables larger.84kW loads with power factors 0.84 load supplied power factor 0.8 and 0. It can seen from these calculations that 1.4 would require 20A cable, while the same load unity power factor could supplied with an cable. 0. There may also the problem higher voltage drops the supply cables.8 10A For p.84 1840 and 230V.4.f.8.f. 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. POWER FACTOR CORRECTION Most installations have low bad power factor because the inductive nature the load. Calculate the current each power factor. then: I 1840 230 1 8 W V A For p. Figure 10. However, using slightly bigger capacitor, the load current can pushed until ‘in phase’ with the voltage can be seen Fig.f.Alternating current theory and electrical machines 203 For (d): P VI P cos (W) 150 0. The current given by: I P V cos φ where 1.8: I 1840 230 W V 0. When this capacitor current added the load current shown Fig 10.4: I 1840 230 4 20 W V A . the p.9(a) shows industrial load with low power factor