• The instantaneous power dissipated in a

component is a product of the instantaneous

voltage and the instantaneous current
p = vi
• In a resistive circuit the voltage and current are in
phase – calculation of p is straightforward
• In reactive circuits, there will normally be some
phase shift between v and i, and calculating the
power becomes more complicated
 Power in Resistive Components
• Suppose a voltage v = Vp sin t is applied across a
resistance R. The resultant current i will be
v VP sin t
i   IP sin t
R R
• The result power p will be
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2
• The average value of (1 - cos 2t) is 1, so
1 V I
Average Power P  VP IP  P  P  VI
2 2 2
where V and I are the r.m.s. voltage and current
 Power in Resistive Components
• Suppose a voltage v = Vp sin t is applied across a
resistance R. The resultant current i will be
v VP sin t
i   IP sin t
R R
• The result power p will be
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2
• The average value of (1 - cos 2t) is 1, so
1 V I
Average Power P  VP IP  P  P  VI
2 2 2
where V and I are the r.m.s. voltage and current
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
 Power in Resistive Components

v VP sin t
i   IP sin t
R R
1  cos 2t
p  vi  VP sin t  IP sin t  VP IP (sin t )  VP IP (
2
)
2

1 V I
Average Power P  VP IP  P  P  VI
2 2 2
• Relationship between v, i and p in a resistor
 Power in Capacitors
• From discussion of capacitors one know that the
current leads the voltage by 90. Therefore, if a voltage
v = Vp sin t is applied across a capacitance C, the
current will be given by i = Ip cos t
• Then
p  vi
 VP sin t  IP cos t
 VP IP (sin t  cos t )
sin 2t
 VP IP ( )
2
• The average power is zero
• Relationship between v, i and p in a capacitor
 Power in Inductors
From discussion of inductors , one know that the current
lags the voltage by 90. Therefore, if a voltage v = Vp sin
t is applied across an inductance L, the current will be
given by i = -Ip cos t
p  vi
 VP sin t   I P cos t
 VP I P (sin t  cos t )
sin 2t
 VP I P ( )
2
Again the average power is zero
• Relationship between v, i and p in an
inductor
Circuit with Resistance and Reactance
• When a sinusoidal voltage v = Vp sin t is
applied across a circuit with resistance and
reactance, the current will be of the general
form i = Ip sin (t - )
• Therefore, the instantaneous power, p is given
p  vi
 VP sin t  IP sin(t   )
1
 VP IP {cos   cos(2t   )}
2
1 1
p  VP IP cos   VP IP cos(2t   )
2 2
 1 1
p  VP IP cos   VP IP cos(2t   )
2 2
The expression for p has two components
The second part oscillates at 2 and has an average
value of zero over a complete cycle, this is the power
that is stored in the reactive elements and then
returned to the circuit within each cycle
The first part represents the power dissipated in
resistive components. Average power dissipation is
1 V I
P  VP IP (cos )  P  P  (cos )  VI cos 
2 2 2
 1
P  VP IP (cos  )  VI cos 
2
The average power dissipation given by
1
P  VP IP (cos  )  VI cos 
2

is termed the active power in the circuit and is

measured in watts (W)
The product of the r.m.s. voltage and current VI is
termed the apparent power, S. To avoid confusion
this is given the units of volt amperes (VA)
From the above discussion it is clear that
P  VI cos 
 S cos 
In other words, the active power is the apparent power
times the cosine of the phase angle.
This cosine is referred to as the power factor
Active power (in watts)
 Power factor
Apparent power (in volt amperes)

P
Power factor   cos 
S
 Active and Reactive Power
• When a circuit has resistive and reactive parts,
the resultant power has 2 parts:
– The first is dissipated in the resistive element. This is
the active power, P
– The second is stored and returned by the reactive
element. This is the reactive power, Q , which has
units of volt amperes reactive or var
• While reactive power is not dissipated it does
have an effect on the system
– for example, it increases the current that must be
supplied and increases losses with cables
• Consider an RL circuit
– the relationship between the various forms of power
can be illustrated using a power triangle
• Therefore
• Active Power P = VI cos  watts
• Reactive Power Q = VI sin  var
• Apparent Power S = VI VA
• S2 = P2 + Q2
 Three-Phase Systems
• So far, the discussion of AC systems has been
restricted to single-phase arrangement
– as in conventional domestic supplies
• In high-power industrial applications one
often use three-phase arrangements
– these have three supplies, differing in phase by
120 
– phases are labeled red, yellow and blue (R, Y & B)
• Relationship between the phases in a three-
phase arrangement
• Three-phase arrangements may use either 3
or 4 conductors
 Power Measurement
• When using AC, power is determined not only by the
r.m.s. values of the voltage and current, but also by the
phase angle (which determines the power factor)
– consequently, you cannot determine the power from
independent measurements of current and voltage
• In single-phase systems power is normally measured
using an electrodynamic wattmeter
– measures power directly using a single meter which
effectively multiplies instantaneous current and voltage
• In three-phase systems one need to sum the
power taken from the various phases
– in three-wire arrangements we can deduce the total
power from measurements using 2 wattmeter
– in a four-wire system it may be necessary to use 3
wattmeter
– in balanced systems (systems that take equal power
from each phase) a single wattmeter can be used, its
reading being multiplied by 3 to get the total power
 Key Points
• In resistive circuits the average power is equal to VI, where V
and I are r.m.s. values
• In a capacitor the current leads the voltage by 90 and the
average power is zero
• In an inductor the current lags the voltage by 90 and the
average power is zero
• In circuits with both resistive and reactive elements, the
average power is VI cos 
• The term cos  is called the power factor
• Power factor correction is important in high-power systems
• High-power systems often use three-phase arrangements
 POLYPHASE CIRCUITS

Three Phase Circuits

Advantages of polyphase circuits

Three Phase Connections

Basic configurations for three phase circuits

Source/Load Connections
Delta-Y connections

Power Relationships
Study power delivered by three phase circuits

Power Factor Correction

Improving power factor for three phase circuits
 THREE PHASE CIRCUITS

Van

3 phase
voltage Vbn

Vcn

0 120 240

Instantaneous Phase Voltages

van (t )  Vm cos( t )(V )
vbn (t )  Vm cos( t  120)(V ) Vm  120 2

vc (t )  Vm cos( t  240)(V )
 a a

V0 Wye Connected

_
n Source
_ _
V-240 V-120
+ +
b b
c

c
 Delta Source

a a

_ + Delta Vab = | Vab |  0

Source
+ _
Vbc = Vab  -120
b b
c Vca = Vab  -240
_ +

c
 Wye – Wye System

a A
Zl

ZL

n
N

ZL ZL
b
c B C
Zl

Zl
 Delta – Delta System

a A
Zl

_ +

ZL
ZL
+ _

b
c _ + B C
Zl ZL

Zl
 Delta – Wye System

a A
Zl

_ + ZL

+ _
ZL ZL
b
c _ + B C
Zl

Zl
 a a
A
I aA I AB I CA
+

V0
_
Z Z
n
_ _
V-240 V-120 I BC
+ +
b b B C
c Z

c
 Vcn - Vbn

Vab

30 o

Van

Vbn Vab = Van - Vbn

Vab = 3 Van 30o

 Vab  Van  Vbn
| V p | 0 | V p |   120
| V p | 1  (cos 120  j sin120) 
1 3
| V | p  | V p |   j 
2 2 
 3 | V p | 30

Vbc  3 | V p |   90
Vca  3 | V p |   210
VL  3 | V p |  Line Voltage
I CA I aA = 3 I AB -30o

I AB

I aA

I BC - I CA
 INSTANTANEOUS POWER

Instantaneous Phase Voltages Balanced Phase Currents

van (t )  Vm cos( t )(V ) ia (t )  I m cos( t   )
vbn (t )  Vm cos( t  120)(V ) ib (t )  I m cos( t    120)
vc (t )  Vm cos( t  240)(V ) ic (t )  I m cos( t    240)

Instantaneous power
p(t )  van (t )ia (t )  vbn (t )ib (t )  vcn (t )ic (t )

Theorem
For a balanced three phase circuit the instantaneous power is constant

Vm I m
p( t )  3 cos  (W )
2
REVIEW OF
Y
Transformations

R1R2
Ra 
R1  R2  R3
R2 R3
Rb 
R1  R2  R3 Ra Rb  Rb Rc  Rc Ra
R1 
R3 R1 Rb
Rc 
R1  R2  R3 Ra Rb  Rb Rc  Rc Ra
R2 
 Y Rc
Ra Rb  Rb Rc  Rc Ra
R3 
Ra
Y 
REVIEW OF Rab  R2 || ( R1  R3 )  Y
Y
Rab  Ra  Rb
Transformations Y 

Ra R1 Rb R1 Rb R2 Rb R1
R1R2   R    R 
R2 ( R1  R3 ) Ra  R1  R2  R3 Rb R3
3 2
Ra Rc R1 Rc
Ra  Rb 
R1  R2  R3 R2 R3 REPLACE IN THE THIRD AND SOLVE FOR R1
Rb 
R3 ( R1  R2 ) R1  R2  R3 Ra Rb  Rb Rc  Rc Ra
Rb  Rc  R1 
R R Rb
R1  R2  R3 Rc  3 1
R1  R2  R3 R R  Rb Rc  Rc Ra
R2  a b
R (R  R )  Y Rc
Rc  Ra  1 2 3
Ra Rb  Rb Rc  Rc Ra
R1  R2  R3 R3 
Ra
SUBTRACT THE FIRST TWO THEN ADD Y 
TO THE THIRD TO GET Ra
POWER FACTOR CORRECTION
Similar to single phase case.
Use capacitors to increase the
power factor

Balanced Keep clear about total/phase

load power, line/phase voltages
Low pf
lagging

Q  Qnew  Qold
Reactive Power to be added
To use capacitors this value
should be negative

pf  cos  f  sin  f  1  pf 2 tan   pf

f
Q  P tan  f 1  pf 2

lagging  Q  0
LEARNING EXAMPLE

Pold  18.72 MW 
  Qnew  6.8 MVA
S  P  jQ pf new  0.94 leading 
P | S | cos  f Q  6.8  15.02  21.82 MVA
Q  P tan  f Qper capacitor  7.273MVA
Q | S | sin f
pf
pf  cos  f tan  f  Y  connection  Vcapacitor 
34.5
kV rms
1  pf 2
3
lagging  Qold  0  34.5  103 
2

 7.273  10  2  60  C  
6

pf  cos  f  sin  f  1  pf 2  0.626  3 
| Qold | 15.02 MVA C  48.6  F
Pold  18.72 MW
LEARNING BY DESIGN
#4ACSR wire rated at 170 A rms

Proposed new store

S1  70036.9 S2  100060kVA S3  80025.8kVA

 560  j 420 kVA  500  j866 kVA  720  j 349 kVA
Stotal  1780  j1635 kVA  241742.57 kVA
| Stotal | 2.417  106
| I line |   101.1A rms Wire is OK
3  Vline 3  13.8  103

Pold 
  Qnew  P tan  f ( new )  758.28kVA
pf new 
Q  Qnew  Qold  876.72kVA
|Q per capacitor
|  CV 2

C
876.72 10 / 3
3
 12.2  F
2  60  13.8  10  / 3
3 2