lecture 1,2.pdf

Chapter 10 – ACSS Power
Objectives:
•Power concepts: instantaneous power,
average power, reactive power, complex
power, power factor
•Relationships among power concepts – the
power triangle
•Balancing power in AC circuits
•Condition for maximum power transfer to a
load in AC circuits

Instantaneous power:
t
I
V
t
I
V
I
V
t
I
t
V
t
i
t
v
t
p
i
v
m
m
i
v
m
m
i
v
m
m
i
m
v
m












2
sin
)
sin(
2
2
cos
)
cos(
2
)
cos(
2
...
...
)
cos(
)][
cos(
[
)
(
)
(
)
(










ry)
trigonomet
of
(lots
Notes:
•The first term is a constant – instantaneous
power is not symmetric about the time axis!
•The second and third terms have a
frequency that is twice the frequency of v(t)
and i(t)!

Instantaneous power

Instantaneous power, rewritten:
)
sin(
2
)
cos(
2
2
sin
2
cos
)
(
i
v
m
m
i
v
m
m
I
V
Q
I
V
P
t
Q
t
P
P
t
p













where
Notes:
•P is the average (or real) power and has the
units watts [W]
•Q is the reactive power and has the units
volt-ampere-reactive [var]



T
t
t
dt
t
p
T
P
0
0
)
(
1

Look at the expressions for P (real power) and Q
(reactive power) for the three different circuit
components – R, L, and C. The expressions
depend on the difference between the phase
angle of the voltage drop across the component
and the phase angle of the current through the
component.

A. Dependent on the source
frequency
B. Dependent on the resistance
C. Equal to the voltage phase angle
D. The negative of the voltage
phase angle

Remember – for a resistor in the presence of
an AC source,
i
i
v
i
m
v
m I
R
V
R
















0
)
)(
0
(
I
V
The voltage and current for a resistor are in-
phase! Thus, for a resistor,
t
P
P
t
p
I
V
Q
I
V
I
V
P i
v
m
m
m
m
i
v
m
m
i
v
i
v









2
cos
)
(
0
)
sin(
2
;
2
)
cos(
2
0
)
sin(
;
1
)
0
cos(
)
cos(















A. The same as the voltage phase
angle
B. 90o more than the voltage phase
angle
C. 90o less than the voltage phase
angle
D. Dependent on the inductance

How can you remember the relationship
between the voltage and current phase angles
for inductors and capacitors?
ELI the ICE man!
Huh?
•In the early days of electrical engineering,
voltage was known as “electromotive
force”, and was represented by the symbol
“e” in equations!
•Current is represented by the letter “i”, as
usual.
•L is inductance, C is capacitance.

ELI the ICE man!
ELI – voltage is “ahead of” (i.e. LEADS) current
in an inductor, by 90o.
ICE – voltage is “behind” (i.e. LAGS) current in
a capacitor, by 90o.
Where did the 90o come from? Remember
Ohm’s law for phasors:
i
v
i
m
v
m
i
v
i
m
v
m
I
C
V
C
j
I
L
V
L
j





































90
)
)(
90
1
(
)
(
90
)
)(
90
(
I
V
I
V

angle
angle
angle
D. Dependent on the inductance
ELI the ICE man!

The voltage leads the current by 90o in an
inductor (ELI!) Therefore,
t
Q
t
p
I
V
I
V
Q
I
V
P m
m
i
v
m
m
i
v
m
m
i
v
i
v









2
sin
)
(
2
)
sin(
2
;
0
)
cos(
2
1
)
90
sin(
)
sin(
;
0
)
90
cos(
)
cos(

















angle
angle
angle
D. Dependent on the capacitance
ELI the ICE man!

The voltage lags the current by 90o in a
capacitor (ICE!) Therefore,
t
Q
t
p
I
V
I
V
Q
I
V
P m
m
i
v
m
m
i
v
m
m
i
v
i
v









2
sin
)
(
2
)
sin(
2
;
0
)
cos(
2
1
)
90
sin(
)
sin(
;
0
)
90
cos(
)
cos(





















Summary:
•Resistors P > 0, Q = 0
Resistors absorb real power and have
no reactive power
•Inductors P = 0, Q > 0
Inductors absorb reactive power and
have no real power
•Capacitors P = 0, Q < 0
Capacitors generate reactive power
and have no real power

Power factor (of a single component or a collection
of components):
This is a term that appears in the definition of
average power.
•When pf = 1, the component is purely resistive.
•When pf = 0, the component is purely reactive.
•To distinguish between inductive and
capacitive reactance, we use the modifiers
“leading” and “lagging”:
•When the pf is leading, the current leads the
voltage; when pf is lagging, the current lags
the voltage.
1
pf
0
),
cos(
pf 


 i
v 


A. A capacitor
B. An inductor
C. A capacitor and a resistor
D. An inductor and a resistor
ELI the ICE man!

Example (AP 10.1)
V
)
45
(
100cos
v(t)
A
)
15
cos(
20
)
(






t
t
t
i


Find the average power, the reactive power, and
the power factor.
5
.
0
)
15
45
cos(
)
cos(
pf
var
866
)
15
45
sin(
2
)
100
)(
20
(
)
sin(
2
Q
W
500
)
15
45
cos(
2
)
100
)(
20
(
)
cos(
2
P

























i
v
i
v
m
m
i
v
m
m
I
V
I
V







A. Leading
B. Lagging
C. Can’t tell from the
information
V
)
45
(
100cos
v(t)
A
)
15
cos(
20
)
(






t
t
t
i



A. Generating P and generating
Q
B. Generating P and absorbing Q
C. Absorbing P and generating Q
D. Absorbing P and absorbing Q

Example (AP 10.1), continued
V
)
45
(
100cos
v(t)
A
)
165
cos(
20
)
(






t
t
t
i


Find the average power, the reactive power, and
the power factor.
leading
866
.
0
)
165
45
cos(
)
cos(
pf
var
500
)
165
45
sin(
2
)
100
)(
20
(
)
sin(
2
Q
W
866
)
165
45
cos(
2
)
100
)(
20
(
)
cos(
2
P


























i
v
i
v
m
m
i
v
m
m
I
V
I
V







A. A resistor and a capacitor
B. A resistor and an inductor
C. None of the above
Warning – trick question!

Example (AP 10.1), continued
V
)
45
(
100cos
v(t)
A
)
165
cos(
20
)
(






t
t
t
i


The circuit in the box is generating both
average and reactive power.
•Capacitors generate reactive power,
•Only sources generate average power,
•The pf < 1, so there must also be a resistor.
Thus, the simplest circuit in the box has a
source, a resistor, and a capacitor!
leading
866
.
0
pf
var,
500
Q
W,
866
P 





Complex power:
Notes:
•|S| is apparent power
•pf is the power factor angle
•Units for both complex power and apparent
power are volt-amperes [VA]
The power triangle brings everything together!
VA
|
S
|
Q
P
S pf
j 




Q
P
|S|
pf
pf

cos
pf 

A. Purely resistive
B. RL
C. RC
D. Purely capacitive
Q
P
|S|
pf

The complex power in an AC circuit balances!


 


 0
Q
and
0
P
0
S
Example: Find the phasor
voltage and current for
the load, and show that
the power in the circuit
balances.
A
36.87
-
5
A
)
3
4
(
26
39
13
234
26
39
V
3.18
-
234.36
V
)
13
234
(
)
0
250
(
30
40
26
39




















j
j
j
j
j
j
j
L
L
L
V
I
V

balances.
var
325
)
1
](
2
)
5
)(
26
5
(
[
Q
;
0
P
var
50
)
1
](
2
)
5
)(
4
5
(
[
Q
;
0
P
0
Q
;
W
5
.
487
)
1
](
2
)
5
)(
39
5
(
[
P
0
Q
;
W
25
)
1
](
2
)
5
)(
1
5
(
[
P
var
375
)
87
.
36
0
sin(
]
2
)
5
)(
250
(
[
Q
W
500
)
87
.
36
0
cos(
]
2
)
5
)(
250
(
[
P
A
36.87
-
5
V,
3.18
-
234.36
1
j26
1
j4
1
39
1
1
S
S

































 L
L I
V

balances.
Component P[W] Q[var] S[VA]
Source 500 375 500j375
1 12.5 0 12.5+j0
j4 0 50 0+j50
39 487.5 0 487.5+j0
j26 0 325 0+j325
Total 0 0 0

Review:
•Instantaneous power
•Average (real) power
•Reactive power
•Power factor
(leading, i(t) leads v(t), RC; lagging, i(t) lags
v(t), RL)
1
pf
0
),
cos(
pf 


 i
v 

[W]
2
sin
2
cos
)
(
)
(
)
( t
Q
t
P
P
t
i
t
v
t
p 
 



)
0
(
[W]
)
cos(
2



 C
L
i
v
m
m
P
P
I
V
P 

)
0
(
[var]
)
sin(
2


 R
i
v
m
m
Q
I
V
Q 


Review, continued:
•Complex power
•Apparent power
•The complex power in an AC circuit balances!
•The power triangle shows how the different
types of power are related:
[VA]
|
| pf
S
jQ
P
S 




[VA]
|
| 2
2
Q
P
S 



 

 0
and
0
or
0 Q
P
S
Q
P
|S|
pf
pf

cos
pf 

A. Purely resistive
B. RL
C. RC
D. Can’t tell from the
triangle
Q
P
|S|
pf
pf

cos
pf 

A. Purely resistive
B. RL
C. RC
D. Purely capacitive
P=|S|

An aside, using an example:
Find an expression for the
average power delivered to
the resistor in the circuit
shown here.
R
V
dt
t
V
T
R
dt
R
t
v
T
dt
t
p
T
P
rms
T
t
t
v
m
T
t
t
T
t
t
2
2
2
2
0
0
0
0
0
0
)
(
cos
1
1
)
(
1
)
(
1



















The rms value of a periodic function, say f(t)
with period T, is the root of the mean of the
square of that function:



T
t
t
rms dt
t
f
T
F
0
0
)
(
1 2

RMS, continued:
Find the rms value of a cosine waveform.
What use is the rms value of voltage or current in
ac circuits?
2
math)
of
(lots
)
(
cos
1
)
cos(
)
(
0
0
2
2 m
T
t
t
v
m
rms
v
m
V
dt
t
V
T
V
t
V
t
v












The rms value of any periodic voltage or current
delivers the same average power to a resistor as a dc
voltage with the same value – rms values allow us to
compare the effect of various periodic voltages to the
effect of a dc voltage, so are sometimes called effective
values!

A. 120 V
B. Greater than 120 V
C. Less than 120 V

RMS, continued:
Often, we use rms values in the equations for
real and reactive power:
)
sin(
)
sin(
2
2
)
sin(
2
pf
pf
2
2
pf
2
i
v
rms
rms
i
v
m
m
i
v
m
m
rms
rms
m
m
m
m
I
V
I
V
I
V
Q
I
V
I
V
I
V
P





 








You must pay attention to the statement of problems
involving power calculations – you pick the correct
equations for P and Q depending on whether the
magnitude of the voltage (or current) is given, or the
rms value of the voltage (or current) is given!

Example 10.4
A load requires a voltage of 240 Vrms, absorbs
8kW and has a power factor of 0.8 lagging.
Find the complex power of the load and the
load impedance.
We can find the complex power of the load by
drawing the power triangle:

A. 8 kW of power
B. Power factor is 0.8
C. Power factor is lagging

Example 10.4, continued
load impedance.
Q
8000 W
|S|
pf
VA
000
,
10
6000
8000
|
|
var
6000
87
.
36
tan
8000
87
.
36
8
.
0
cos
pf
cos
2
2
2
2
-1
1












 
Q
P
S
Q
Q
pf

VA
87
.
36
000
,
10
VA
6000
8000 



 j
S

load impedance.
•How can we find the load impedance?
Start from its definition.




























46
.
3
61
.
4
87
.
36
76
.
5
76
.
5
67
.
41
240
|
|
A
67
.
41
)
8
.
0
)(
240
(
8000
pf
pf
?
|
|
about
but what
V
240
|
|
87
.
36
|
|
|
|
|
|
|
|
)
(
|
|
|
|
rms
j
Z
I
V
Z
V
P
I
I
V
P
I
Z
L
rms
rms
L
rms
rms
rms
rms
rms
L
rms
L
L
L
pf
L
L
i
v
L
L
L
L
L
I
V
I
V
I
V
I
V
I
V




A. An inductor and a
capacitor
B. A resistor and a capacitor
C. An inductor and a
resistor

Example 10.4
load impedance.
Q
8000 W
|S|
pf
VA
000
,
10
6000
8000
|
|
var
6000
87
.
36
tan
8000
87
.
36
8
.
0
cos
pf
cos
2
2
2
2
-1
1












 
Q
P
S
Q
Q
pf

VA
87
.
36
000
,
10
VA
6000
8000 



 j
S

Example 10.6
Two industrial loads each require 2500 V(rms)
at 60 Hz. Load 1 absorbs 8 kW at 0.8 leading;
load 2 absorbs 20 kVA at 0.6 lagging. The line
that transmits the power from the source to the
load has an impedance of (0.05 + j0.5) . To
begin to analyze the circuit described, draw a
block diagram.

Two loads each require 2500 V(rms) at 60 Hz.
Load 1 absorbs 8 kW at 0.8 leading; load 2
absorbs 20 kVA at 0.6 lagging. The line that
transmits the power from the source to the load
has an impedance of (0.05 + j0.5) . To begin
to analyze the circuit described, draw a block
diagram.

Find the power supplied by the source.
lagging
6
.
0
at
kVA
20
:
leading
8
.
0
at
kW
8
:
2
1
L
L
-6
kvar
8 kW
-36.87o
Find the total power required for the loads.
+
53.13o
12 kW
16
kvar
20 kVA
=
26.57o
20 kW
10
kvar
22.36
kVA

Find the power supplied by the source.
VA
57
.
26
36
.
22 


load
S
Find the power “lost” in the line. How, since we
don’t know the source voltage, so can’t find the
voltage drop across the line impedance? Find the
line current, which is the same as the load current.
A(rms)
443
.
89
)
89
.
0
)(
250
(
000
,
20
pf
pf
load
load






load
load
load
line
load
load
load
V
P
I
I
I
V
P

VA
000
,
14
400
,
20
VA
4000
400
var
4000
)
1
)(
443
.
89
)](
443
.
89
(
5
.
0
[
W
400
)
1
)(
443
.
89
)](
443
.
89
(
05
.
0
[
A(rms)
443
.
89
j
S
S
S
j
jQ
P
S
Q
P
I
load
line
source
line
line
line
line
line
line














The power company bills for the real power it
supplies to you – this includes the real power you
actually use, and the real power lost in the
transmission line.

A. Gives the power company
more money
B. Increases the cost of products
produced by the industry
C. Warms the feet of the birds
who sit on the power lines
D. All of the above

)
1
)(
)](
(
[ line
line
line
line I
I
R
P 
How can we reduce the real power lost in the line?
We can’t do much about the resistance of the
transmission line, and usually it is pretty small. Can
we make the line current smaller?
load
pf
load
load
load
line
V
P
I
I 


A. The real power required by
the load(s)
B. The voltage drop across the
load(s)
C. The power factor of the
load(s)
load
pf
load
load
load
line
V
P
I
I 


A. Maximize the load
power factor (set it to 1)
B. Minimize the load
power factor (set it to 0)
C. Neither (A) or (B)

A. Capacitor in series
with the loads
B. Inductor in series
with the loads
C. Capacitor in parallel
with the loads
D. Inductor in parallel
with the loads
26.57o
20 kW
22.36
kVA
10
kvar

We add a capacitor in parallel with the loads
because
•We don’t want to change the voltage drop
across the loads
•We want to “cancel” the positive (inductive)
reactive power required by the loads
•We want to have no effect on the real power
required by the loads
26.57o
20 kW
22.36
kVA
10
kvar
+
-10
kvar
=
20 kW
1
pf )
load(total 

The capacitor added must generate 10 kvar of
reactive power. Its voltage drop is 250 V(rms).
Calculate the value of capacitance.
F
4
.
424
)
25
.
6
)](
60
(
2
[
1
25
.
6
1
25
.
6
40
250
Z
A(rms)
40
)
1
)(
250
(
000
,
10
)
1
(
)
1
(
C





















C
C
I
V
V
Q
I
I
V
Q
cap
cap
cap
cap
cap
cap
cap
cap

This process is called power factor correction, and
is usually required by the power companies for
large industrial loads, to conserve energy – there is
a monetary penalty associated with net power
factors less than a pre-set value, like 0.95.

Example 10.6, concluded
Did this power factor correction actually reduce the
power lost to the line? By how much?
W)
00
4
be
to
(used
W
320
)
1
)(
80
)](
80
(
05
.
0
[
A(rms))
443
.
89
be
to
(used
A(rms)
80
)
1
)(
250
(
000
,
20
pfload






line
load
load
load
line
P
V
P
I
I
Therefore, there was a 20% reduction in the real
power lost to the line. This would be even more
dramatic if the original power factor of the
combined loads was less than 0.89.

AP 10.6
In the figure below, Load 1 absorbs 15 kVA at
0.6 lagging and Load 2 absorbs 6 kVA at 0.8
leading. Find the source voltage phasor, Vs.
The plan of attack:
pf
pf
v
i
load
load
load
i
load
load
line
line
load
line
S
V
P
I
I
j
j
j

















0
and
pf
)
(
1
1
1 I
I
V
V
V
V

AP 10.6, continued
If we find the total complex power of the load,
we can find the two unknowns we need – the
total real power of the load, and the total power
factor of the load!

AP 10.6, continued
-3.6
kvar
4.8 kW
-36.87o
+
53.13o
9 kW
12
kvar
15 kVA
=
31.33o
13.8 kW
8.4
kvar
16,155.5
VA
6 kVA

AP 10.6, continued
In the figure below, Load 1 absorbs 15 kVA at 0.6
lagging and Load 2 absorbs 6 kVA at 0.8 leading.
Find the source voltage phasor, Vs.
rms
rms
rms
rms
V
9
.
15
6
.
251
)
0
200
(
)
67
.
58
78
.
80
(
V
67
.
58
78
.
80
)
33
.
31
78
.
80
(
1
1
A
33
.
31
78
.
80
33
.
31
0
0
A
78
.
80
.33)
200)cos(31
(
800
,
13
pf



































load
line
S
line
line
load
line
pf
i
load
load
load
j
j
V
P
I
V
V
V
I
V
I
I


31.33o
13.8 kW
8.4
kvar
16,155.5
VA

A. Iline
B. Iline
C. Ilineline

Maximum power transfer:
Suppose we attach a load impedance to the
Thevenin equivalent of a source circuit. For what
value of load impedance will the maximum real
power be delivered to the load impedance?

Maximum power transfer, continued
 
2
2
2
2
)
(
)
(
)
1
(
Th
L
Th
L
L
Th
L
L
L
L
L
L
L
Z
L
X
X
R
R
R
R
R
P
P R









V
I
I
I
I
V
To find the resistive and reactive values of load
impedance for maximum power, take the partial
derivative of PL with respect to RL and XL, set the
partial derivatives to zero, and solve for RL and XL.

Maximum power transfer, continued
*
Th
L Z
Z 
The text sketches the derivation for the condition
for maximum real power to the load impedance:
Note that these two impedances are complex
numbers; the “star” (*) operator represents the
complex conjugate.

A. (18 – j36) 
B. (36 – j18) 
C. (18 + j36) 

A. 6045 
B. 4560 
C. 6045 

AP 10.7
The source current is 3 cos 5000t A(rms). Find
the impedance that should be connected to the
terminals a,b for maximum real power delivered
to that impedance, and find the maximum real
power delivered.

AP 10.7, continued
Phasor-transform the circuit:
  W
36
1
|
10
20
10
20
|
67
.
53
2
53.67
pf
)
10
20
(
)
10
20
(
)
40
||
20
(
18
4
Z
V(rms)
57
.
26
67
.
53
V(rms)
)
24
48
(
)
40
||
20
)((
0
3
(
*
Th










































j
j
I
V
P
j
Z
Z
j
j
j
j
j
L
L
L
Th
L
oc
V

Objectives:
•Power concepts: instantaneous power,
average power, reactive power, complex
power, power factor
•Relationships among power concepts – the
power triangle
•Balancing power in AC circuits
•Power factor correction
•Condition for maximum power transfer to a
load in AC circuits

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