Active and Reactive Components of Circuit Current I
VOLT-AMPERES AND REACTIVE VOLT-AMPERES
Active component is that which is in phase with the applied voltage V i.e. I cos φ. It is also known as
‘wattful’ component.
Reactive component is that which in quadrature with V i.e. I sin φ. It is also known as ‘wattless’ or ‘idle’
component.
Pa = apparent power abbreviated va in volt ampere (va)
= EI
Pr = reactive power abrbreviated R-va in reactive volt-ampere (var) = EIsinθ
P = real power in watts = EI cosθ
The Power Triangle
The real, reactive, and apparent powers supplied to a load are related by the power triangle.
The adjacent side of this triangle is the real power P supplied to the load, the opposite side ofthe
triangle is the reactive power Q supplied to the load, and the hypotenuse of the triangle is the apparent
power S of the load.
The quantity cos θ is usually known as the power factor of a load. The power factor is defined as the
fraction of the apparent power S that is actually supplying real power to a load.
where θ is the impedance angle of the load.
Where θ is the impedance angle of the load. Note that cos θ = cos (—θ), so the power factor produced
by an impedance angle of +30° is exactly the same as the power factor produced by an impedance angle
of —30°. Because we can’t tell whether a load is inductive or capacitive from the power factor alone, it
is customary to state whether the current is leading or lagging the voltage whenever a power factor is
quoted.
Reactive volt amperes, abbreviated R-va , units for these quantities are var ( volt-amperes
reactive ) and kvar . The reactive volt-amperes by EI gives the reactive fator , abbreviated RF.
Active, Reactive and Apparent Power
EXAMPLE: A load 0f 250 kva, operating at a power factor of 0.86 lagging, is connected to a 2, 300 volt
ac source. Calculate a) power, b) current c) reactive kilovolt-amperes d) reactive factor.
GIVEN:
Pa = 250 kva , pf = 0.86 lagging , E = 2,300 volt
SOLUTION:
NOTE: Pa = EI = 250 KVA OR 250,000 VA
COSθ = pf = 0.86
Θ=arccos 0.86 = 30.6834
A) Power = EI COSθ = 250 ( 0.86 ) = 215 kw ( kilowatt ) or 215,000 watts
B) Pa = EI , I = Pa / E , I = 250,000 / 2,300 = 108.6956 AMPERE
C) REACTIVE KILOVOLT AMPERE
COSθ = 0.86 , θ = arccos 0.86 , θ = 30. 6834
Pr = R-KVA = kva sinθ = Pa sinθ = 250 SINθ = 250 SIN 30. 6834 = 127.5734 kvar
D) REACTIVE FACTOR = RF , SINθ = SIN 30. 6834 = 0.5102
EXAMPLE: The power input to an electric motor is 8.8 kw when operating normally from a 230 volt ac
source. If the current is 45 ampere. Under this condition, calculate the a) power factor b) reactive
factor, and c) reactive volt ampere ( kvar).
GIVEN: P = EICOSθ = 8.8 KW OR 8,800 WATTS
E= 230 VOLT , I = 45 AMPERE,
SOLUTION:
A) PF = COSθ
P= EI COSθ
COSθ = P / EI = 8,800 / (230)(45) = 0.8502
COSθ = PF ( POWER FACTOR) = 0.8502
Θ = arccos 0.8502 = 31.76650
B) Reactive factor = RF = Sinθ = sin 31.76650 = 0.5264
C) KVAR (Reactive kilo volt ampere ) = Pa sinθ = KVA sinθ = EI SINθ = 230 (45) SIN 31.7665
PR = kvar = 5.4488 kvar
EXAMPLE: A single phase, 7.46 kW motor is supplied from a 400-V, 50-Hz a.c. mains. If its efficiency is
85% and power factor 0.8 lagging, calculate (a) the kVA input (b) the reactive components of input
current and (c) kVAR.
A) KVA INPUT POWER
Pa = VI = 10,970.5882 voltampere = 10.97 kva
b) Input current I = Pa / E = 10,970.5882 / 400 = 27.4264 ampere
active component of current = I cosθ = 27.4264 ( 0.8 0 ) = 21.9411 amp
cos θ = 0.8
θ = arccos 0.8 = 36.8698
RF = SINθ = SIN 36.8698 = 0.5999
reactive component of current = I sinθ = 27.4264 ( sin36.8698 ) = 16.4558 amp
B)
PR = Pa sinθ = 10,970.5882 ( 0.5999 ) = 6581.2558 var
PROBLEM 3:
An ac voltage source supplying power to a load with impedance Z = 20<-30° Ω. Calculate the current I
supplied to the load, the power factor of the load, and the real, reactive, apparent, and complex power
supplied to the load.
The current supplied to this load is
The power factor of the load is
(Note that this is a capacitive load, so the impedance angle θ is negative, and the current leads the
voltage.)
The real power supplied to the load is
P = 120 ( 6) COS 300 = 623. 5382 WATTS
The apparent power supplied to the load is
Pa = EI = 120 ( 6 ) = 720 VA ( VOLT -AMPERE )
The complex power supplied to the load is
= 623.5382 -j360 VA
SEATWORK:
1.An AC CIRCUIT TAKES A LOAD OF 160 KVA AT A LAGGING POWER FACTOR OF 0.75 WHEN CONNECTED
TO A 460 VOLT SOURCE. Calculate the current, pf , power , kvar and reactive power.