Active power
 The fact that power is always positive reveals that it always
flows from the generator to the resistor.
 This is one of the basic properties of active power: although it
pulsates between zero and max. it never changes direction.
 The average power is clearly midway between 2P and zero, and
so its value is P = 2EI/2 = EI watts.
 That is precisely the power indicated by the wattmeter.
 The generator is an active source and the resistor is an active
load.
 The symbol for active power is P and the unit is the watt (W).
The kilowatt (kW) and megawatt (MW) are frequently used.
Lecture 27
Electro Mechanical System
1
Reactive power
 The resistor is replaced by a reactor X . As a result, current I
lags 90° behind the voltage E
 we can draw the waveforms of E and I & by multiplying their
instantaneous values, we obtain curve of instantaneous power
 Positive waves correspond to instantaneous power delivered
by the generator to the reactor and the negative waves
represent power delivered from the reactor to the generator
 Duration of each wave corresponds to one quarter of a cycle
 Frequency of power wave is therefore twice the line frequency
 Power that surges back and forth is called reactive power (Q)
 To distinguish this power from active power, another unit is
used the var. Its multiples are kilovar (kvar) and megavar (Mvar)
Lecture 27
Electro Mechanical System
2
Reactive load & reactive source
 Reactive power involves real power that
oscillates back and forth between two devices
over a transmission line.
 Consequently, it is impossible to say whether
the power originates at one end of the line or
the other.
 It is useful to assume that some devices
generate reactive power while others absorb it.
 In other words, some devices behave like
reactive sources and others like reactive loads.
 A reactor is considered to be a reactive load
that absorbs reactive power.
Lecture 27
Electro Mechanical System
3
Example
A reactor having an inductive reactance of 4 Ω is connected to the
terminals of a 120 V ac generator.
a) Calculate the value of the current in the reactor
IL = E / XL = 120V / 4 Ω = 30 A
b) Calculate the power associated with the reactor
Q = EI = 120 X 30 = 3600 var = 3.6 kvar
c) Calculate the power associated with the ac generator
Because the reactor absorbs 3.6 kvar of reactance power, the ac
generator must be supplying it. Consequently, the generator is a
source of reactive power: it delivers 3.6 kvar.
d) Draw the phasor diagram for the circuit
The phasor diagram show that current I, lags 90° behind voltage E
Lecture 27
Electro Mechanical System
4
Capacitor and reactive power
 If we add capacitor having a reactance of 4 Ω to the circuit.
 capacitor current IC = 120 V/4Ω = 30A & leads the voltage by 90°
 The vector sum of IL and IC is zero and so the ac generator is
no longer supplying any power at all to the circuit.
 The current in the reactor has not changed; consequently, it
continues to absorb 30 A X 120 V = 3.6 kvar of reactive power.
 Where is this reactive power coming from? It can only come
from the capacitor, which acts as a source of reactive power.
 Q = Elc = 120 V X 30 A = 3600 var = 3.6 kvar
 It means a capacitor is a source of reactive power. It acts as a
reactive power source whenever it is part of a sine-wave-based,
steady-state circuit.
Lecture 27
Electro Mechanical System
5
Capacitor and reactive power






Remove the reactor from the circuit. Capacitor is now alone.
It still carries a current of 30 A, leading the voltage E by 90°
Capacitor acts as a source of reactive power, delivering 3.6 kvar
Where does this power go? Capacitor delivers it to the generator
How, can a passive device(capacitor) produce any power?
Reactive power really represents energy, like a pendulum, swings
back and forth without ever doing any useful work
 The capacitor acts as a temporary energy-storing device
 Instead of storing magnetic energy, it stores electrostatic energy
 If we connect a varmeter, it will give a negative reading of
EI = – 3600 var, showing that reactive power is indeed flowing
from the capacitor to the generator.
Lecture 27
Electro Mechanical System
6
Example
An ac generator is connected to a group of R, L, C circuit elements
carrying, the currents shown. Calculate the active and reactive power
associated with the generator.
The two resistors absorb active power given by
P = I2R =(16.122 X 2) + (142 X 4) = 784 + 520 = 1304 W
The 3 Ω reactor absorbs reactive power:
QL = I2XL = 142 X 3 = 588 var
The 3.5 Ω capacitor generates reactive power:
QC=I2Xc = 202 X 3.5 = 1400 var
R, L, C circuit generates a net reactive power of 1400 – 588 = 812 var
In conclusion, the ac generator is a source of active power (1304 W)
and a receiver of reactive power (812 var).
Lecture 27
Electro Mechanical System
7
Distinction between active and reactive power
 Active and reactive power, are separate quantities and one
cannot be converted into the other.
 Both place a burden on the transmission line.
 Active power eventually produces a tangible result (heat,
mechanical power, light, etc.)
 Reactive power represents power that oscillates back and forth.
 All ac inductive devices such as magnets, transformers, ballasts,
and induction motors, absorb reactive power because one
component of the current they draw lags 90° behind the voltage.
 The reactive power plays a very important role because it
produces the ac magnetic field in these devices.
 A building, shopping center, or city may be considered to be a
huge active/reactive load connected to an electric utility system.
 Load centers contain thousands of electromagnetic devices that
draw both reactive power (to sustain their magnetic fields) and
active power (to do the useful work).
Lecture 27
Electro Mechanical System
8
Combined active & reactive loads: apparent power
 Loads that absorb active power P and reactive power Q is made
up of a resistance and an inductive reactance.
 Resistor draws a current Ip, while the reactor draws a current Iq
 Consequently, Ip is in phase with E while Iq lags 90° behind.
 Resultant line current I lags behind E by an angle . I  I p2  I q2
 If we connect a wattmeter and a varmeter , the readings will both
be positive, indicating P = EIp watts and Q = EIq vars
 An ammeter will indicate a current of I amperes, so the power
supplied to the load should be equal to EI watts.
 This is incorrect because the power is composed of an active
component (watts) and a reactive component (vars). For this
reason the product EI is called apparent power, symbol is S.
 Apparent power is expressed in voltamperes. (kVA) & (MVA).
Lecture 27
Electro Mechanical System
9