Chapter 10
Sinusoidal Steady–State
Power Calculations
10.1
10.2
10.3
10.4
10.5
10.6
Instantaneous power
Average power & reactive power
The rms value and power calculations
Complex power
Power calculations
Maximum power transfer
1
Overview

Nearly all electric energy is supplied in the form
of sinusoidal voltages and currents (i.e. AC,
alternating currents), because
1.
Generators generate AC naturally.
2.
Transformers must operate with AC.
3.
Transmission relies on AC.
4.
It is expensive to transform from DC to AC.
2
Key points

How to decompose a sinusoidal instantaneous
power into the average power and reactive
power components? What are the meanings?

How to decompose a sinusoidal instantaneous
power into the in-phase and quadrature
components?

Why and how to do power factor correction?

For a specific circuit, how to maximize the
average power delivered to a load?
3
Section 10.1
Instantaneous Power
4
Definition

“Instantaneous” power is the product of the
instantaneous terminal voltage and current, or
p(t )   v (t )  i (t ).

Positive sign is used if
the passive sign
convention is satisfied
(current is in the direction
of voltage drop).
5
Sinusoidal power formula
Absolute timing is
unimportant
v (t )  Vm cos(t   ),
v (t )  Vm cos(t   v ),

 i (t )  I m cos(t ),

i (t )  I m cos(t  i ),
     ;
v
i

cos    cos   
By cos  cos  

,
2
2
 p(t )  Vm I m cos(t   ) cos(t )
Vm I m
Vm I m

cos  
cos(2t   ).
2
2
Constant, Pavg Oscillating at frequency 2
6
Relation among i(t), v(t), p(t)
7
Section 10.2
Average and Reactive
Power
1.
2.
3.
Decompose the instantaneous power
in different ways
Instantaneous powers of resistive,
inductive, and capacitive loads
Power factor and reactive factor
8
Definitions
Vm I m
Vm I m
cos  
cos(2t   )
p (t ) 
2
2
Vm I m
cos   cos 2t cos   sin 2t sin  

2
 P 1  cos 2t   Q sin 2t ,
where
In-phase
quadrature
Vm I m
Vm I m
cos  , Q 
sin .
P
2
2
Average power Reactive power (volt(watt, or W)
ampere reactive, or VAR)
9
Relation among components of power
10
What is average power ?

The power transformed from electric to nonelectric energy or vice versa.

The average of instantaneous power p(t).

The average of P(1+cos2t), which is a power
component in-phase with the current i(t).

The circuit dissipates (delivers) electric energy if
P > 0 (P < 0).
11
What is reactive power ?

The power exchanged among (1) the magnetic
field in an inductor, (2) the electric field in a
capacitor, and (3) the electric source.

The magnitude of -Q(sin2t) (while its average
equals zero), which is a power component in
quadrature with i(t) (leading or lagging i(t) by
90 or T/4 in time).

Reactive power cannot do work.
12
Power for resistive loads
v (t )  Ri (t ),    0, v (t ) and i (t ) are in - phase.
Vm I m Vm I m
Vm I m
1  cos 2t   0 .

cos 2t 
p (t ) 
2
2
2

p(t) > 0 at all times,  P is maximized, Q = 0.
13
Power for inductive loads
v (t )  Li(t ),    90 , i (t ) lags v (t ) by T 4 .
Vm I m
Vm I m

p (t )  0 
cos(2t  90 )  0 
sin 2t .
2
2

p(t) is halved by 0-level, P = 0, Q > 0 is maximized.
14
Power for capacitive loads
i (t )  Cv(t ),    90 , i (t ) leads v (t ) by T 4 .
Vm I m
Vm I m

p (t )  0 
cos(2t  90 )  0 
sin 2t .
2
2

p(t) is halved by 0-level,  P = 0, Q < 0.
15
Power factor & reactive factor

The above examples show that the relative
phase  between v(t) and i(t) determines
whether the electric power is delivered to the
load or simply exchanges between EM fields.

Power factor and reactive factor, defined as:
pf  cos  , rf  sin  ,
quantitatively describe the impact of  on power
delivery.
1.
Lagging pf: inductive, Q > 0, 0 <  < 180;
2.
Leading pf: capacitive, Q < 0, -180 <  < 0.
16
Section 10.3
The rms Value and Power
Calculations
17
Definition of root-mean-square (rms) value

The rms value of any periodic (not necessarily
sinusoidal) function y(t) of period T is the
“square root of the mean value of the squared
function” (Section 9.1).
Yrms

1

T

t0  T
t0
2
y (t )dt
The function square makes it suitable in
describing the concept of power.
18
rms value of sinusoidal functions

Consider a sinusoidal voltage: v(t) = Vm cos(t + )
Vrms
1

T
Vm2

T


t0  T
t0
t0  T
t0
V cos (t   )dt
2
m
2
Vm2 T Vm
1  cos(2t  2 )
dt 

T 2
2
2

rms value is 1 2 times of the amplitude,
independent of frequency  and phase .

The ratio of rms value to the function amplitude
changes with the functional shape.
19
Power formulas in terms of rms values

The average power P and reactive power Q
due to sinusoidal voltage and current are:
Vm I m
P
cos   Vrms I rms cos  ,
2
Vm I m
Q
sin   Vrms I rms sin .
2

A sinusoidal voltage source of rms value Vrms
and a dc voltage source of constant voltage Vs
deliver the same average power to a load
resistance R if Vrms = Vs.
20
rms values in daily life

Voltage rating of residential electric wiring 220
V/110 V are given in terms of rms values.

E.g. A lamp rated by {120 V, 100 W} has:
1.
resistance R = (Vrms)2/P = 144 ;
2.
rms current Irms = Vrms/R = 0.83 A;
3.
peak current Im = 2Irms = 1.18 A, which is critical
in safety.
21
Section 10.4
Complex Power
22
Definition

The complex power S (volt-amps, VA) is:
S  P  jQ
|S|: apparent power
(VA)

Q: reactive power
(volt-amp-reactive,
VAR)
P: average power
(watts, W)
23
Example 10.4

Q: An electric load has Vrms = 240 V, P = 8 kW, pf =
0.8 (lagging);  (1) S = ? (2) Irms = ? (3) ZL = ?
(1) Lagging pf, Q > 0, (2) P > 0,  0 < < 90; (3)
cos = 0.8;  = 36.87.
P = |S|cos, 8 kW = |S|(0.8),  |S| = 10 kVA.
S = P + jQ = |S| =1036.87 = (8+j6) kVA.
P = VrmsIrmscos, Irms= (8 kW)/[(240 V)(0.8)] = 41.67 A.
ZL = VL/IL = (Vrms/Irms) = (5.7636.87) .
24
Section 10.5
Power Calculations
1.
2.
Complex powers in a circuit
Power factor correction
25
Power calculations by voltage & current phasors
Vm I m
Vm I m
S  P  jQ 
cos   j
sin 
2
2
Vm I m
Vm I m
1 *
cos   j sin   

  VI ,
2
2
2
where V  Vm v , I  I mi ,    v  i .
Beides, S  Vrms I*rms ,
where Vrms
Vm
Im

 v , I rms 
i .
2
2
26
Power calculations by impedance
 Vrms  I rms Z , Z  R  jX ,
 S  I rms Z I
I
2
rms
R 
PI

2
rms
*
rms
2
 I rms Z
jX   P  jQ,
R, Q  I
2
rms
Z = R+jX
= |Z|e j
X.
A power consumer has to suppress its load
reactance (power factor correction, making X  0)
such that a smaller apparent power |S| is
sufficient to deliver the specified average power.
27
Example 10.5 (1)

Q: The complex powers delivered to the load,
line, and source.
ZLN
ZL
2500
Vs

 5  36.87  A (rms).
IL 
Z LN  Z L 40  j 30
VL  I L Z L  5  36.87 39  j 26   234.4  3.18  V (rms).
28
Example 10.5 (2)
S L  VL I*L  234.4  3.18 5  36.87   975  j 650  VA.
rms values
S LN  I L Z LN  52 1  j 4   25  j100 VA.
2
S s   Vs I*L   2500 5  36.87    1000  j 750 VA.
power delivered to the source,
passive sign convention
S L  S LN  S s
 975  j 650  25  j100   1000  j 750 VA  0,
 the complex powers are conserved.
29
Example 10.6: PF correction (1)

Q: Let {P1 = 8 kW, pf1 = 0.8 (leading)}, {|S2| = 20 kVA,
pf2 = 0.6 (lagging)}.  (1) How to make the total pf
=1? (2) What are the powers lost in the line PLN
before and after the pf correction?
ZLN
( f = 60 Hz)
(fixed VL)
30
Example 10.6: (2)
pf1 = 0.8 (leading)

pf2 = 0.6 (lagging)
pftot = 0.89 (lagging)
To make pftot= 1, one has to add a parallel
capacitance C that gives {PC = 0, QC = -10 kVAR}.
2
2
2
V
V
V
2
 Z C  jX C ,  QC  I rms
X C  rms2 X C  rms  L ,
XC
XC XC
2502
VL2
 XC 

 6.25 ,
QC  10 kVAR
1
1
1

 424.4 μF.
XC  
,C  
C
X C
2 (60)( 6.25)
31
Example 10.6 (3)

The average power lost in the line is:
2
PLN  I s RLN ,
where the line current phasor (rms) is:
I s  S L VL  ,  S L  VLI*s .

 PLN 
SL
VL
2
2
2
RLN
SL

(0.05 )
2
( 250 V)
400 W, before pf correction, S L  22.36 kVA,

320 W, after pf correction, S L  20 kVA.
32
Section 10.6
Maximum Power Transfer
1.
2.
Unrestricted optimal load impedance
Restricted optimal load impedance
33
Conclusion

For a general circuit with 2 output terminals a, b,
the optimal load impedance that will consume
*
the maximum average power is Z L  Z Th
, where
ZTh is the Thévenin impedance of the circuit.
34
Proof (1)

The rms current phasor I through the load is:
VTh
VTh
I

,
Z Th  Z L RTh  RL   j  X Th  X L 
where ZTh  RTh  jX Th , Z L  RL  jX L .

The average power P delivered to the load is:
2
2
P ( RL , X L )  I RL 
VTh RL
RTh  RL    X Th  X L 
2
2
.
35
Proof (2)

The maximum average power occurs when the
two partial derivatives are zero:
 2 VTh RL  X Th  X L 
P

X L R  R 2   X  X 2
Th
L
Th
L
2

 P

 RL


2
 0,  X L   X Th  (1)
 V 2 R 2  R 2 
Th
L
  Th
 0,  RL  RTh  ( 2)
4

RTh  RL 
X L   X Th 
*
 Z L  ZTh
.
36
Example 10.8: Max power transfer w/n restriction (1)

Q: Without restrictions on ZL, determine (1) ZL
that results in the maximum average power Pmax
transferred to ZL, (2) the value of Pmax.
(not rms)

Apply source transformation to {200 V, 5 , 20
}, we got a simplified circuit:
37
Example 10.8 (2)

Apply source transformation to {160 V, 4+j3 ,
-j6 }, we got the Thévenin circuit:
Z Th  4  j 3 //(  j 6)  5.76  j1.68 .
160
I Nt 
A,  VTh  I Nt Z Th  19.2  53.13  V.
4  j3
38
Example 10.8 (3)
VTh
ZTh
Max power
occurs if
ZL = (ZTh)*
VTh
19.2  53.13
I

 1.67  53.13  A,
Z Th  Z L
2(5.76)
 I rms  I
2  1.18 A,
2
 Pmax  I rms
RL  (1.18) 2 (5.76)  8 W.
39
Example 10.9: Max power transfer with restriction (1)

Q: What are the optimal load impedance ZL that
lead to the maximum average power if 0  RL 4
k, -2 k  XL  0 are required?
ZTh
ZL
40
Example 10.9 (2)

Without restriction, ZL = (ZTh)*,  RL = RTh = 3 k,
XL = -XTh = -4 k, respectively.

The result can be
verified by
calculating average
power P for possible
combinations of (RL,
XL).
41
Example 10.9 (3)

Since XL = -4 k is outside the permitted range
of -2 k XL  0, one first sets XL as close to the
boundary as possible,  XL = -2 k.

In this case, one has to modify the optimal RL
formula when XL  -XTh:
 P

 RL



 V 2 R 2  R 2   X  X 2
L
Th
L
  Th Th
 0,
4

RTh  RL 
X L   X Th 
 RL  R   X Th  X L   RTh .
2
Th
2
42
Example 10.9 (4)
 RL  R   X Th  X L  
2
Th
2
3k   4k  2k 
2
2
 3.6 k  RTh .
43
Key points

How to decompose a sinusoidal instantaneous
power into the average power and reactive
power components? What are the meanings?

How to decompose a sinusoidal instantaneous
power into the in-phase and quadrature
components?

Why and how to do power factor correction?

For a specific circuit, how to maximize the
average power delivered to a load?
44
Practical Perspective
Hair Dryer
45
Physics

A resistor heated by the sinusoidal current, and a
fan that blows the warm air out.

Heater tube is made of coiled nichrome (80%
nickel, 20% chromium) wire, because of

High resistivity:  (= RA/L)
 10-6 m, while other
metals have   10-8 m.

No oxidation when
heated red hot (longer
life time).
46
Circuit

A connection partway divides the heater tube coil
into two pieces.

The position of a four-position switch controls the
heat setting.
47
Low switch setting

The four-position switch makes R1 and R2 in
series, giving the lowest output power.
48
Medium switch setting

The four-position switch makes R1 in vain, giving
the medium output power.
49
High switch setting

The four-position switch makes R1 and R2 in
parallel, giving the highest output power.
50