Fig 1.9: Apparent power S = VI

When a load has voltage V across it and current I through it as in figure 1.9, the power that appears to flow to it is VI. However, if the load contains both resistance and reactance, this product represents neither active power nor reactive power. Since VI appears to represent power, it is called apparent power. Apparent power is given the symbol S and units of volt-amperes (VA). Thus
 S = VI (VA)        (1.5) where V and I are the magnitude of the r.m.s voltage and current respectively.
1.4.2
Reactive Power

Consider figure 1.3. During the intervals that instantaneous power PL(t) is negative, power is being returned from the load. (This can only happen if the load contains reactive elements: L or C.) The portion of power that flows into the load then back out is called reactive power. Reactive power is given by the symbol Q and units of volt-ampere reactive (VAr). Thus
Q = VIsinθ (1.6)

1.4.3
Active power
Consider again figure 1.3. Since PL represents the power flowing to the load; its average will be the average power to the load. Denote this average by the letter P. If P has a positive value, it represents the power that is really dissipated by the load. For this reason, P is called Real power. In modern terminology, real power is also called active power. The unit of active power is in watt (W). Thus
P = VIcosθ (W) (1.7) 

1.5 SOLVED PROBLEMS ON POWER FACTOR, ACTIVE POWER,
APPARENT POWER, REACTIVE POWER AND POWER
FACTOR CORRECTIONS
Example 1.7: In figure 1.10, find the following: 

(a) The active power (b) the apparent power (c) the reactive power and (d) the power factor.


Fig 1.10

Solution
Z = {R222 + (15 – 7)2

√ + (XC – XL)} = √{8} = 11.314-I = E = 100 = 8.84A
Z 11.314

Example 1.9: A load of P = 1000KW with p.f = 0.5lagging is fed by a 5KV, 50Hz source. A capacitor is added in parallel such that the power factor is improved to 0.8. Find the value of the shunt capacitance needed to improve the power factor

Solution
The value of the shunt capacitance is given by

C = P(tan- – tan-)
122-fV2
Before improvement.
P = 1000KW, cos-= 0.5, ⇒ – = cos-1(0.5) = 600
1 1- Ptan- = 1000K x tan600 = 1732.05KVAr
1- C = 1732.05 – 750.82 – 125µF
2 x 50 x (5000)2- 

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APPARENT POWER, REACTIVE POWER AND ACTIVE POWER/POWER IN A.C CIRCUIT

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