EE 212 Passive AC Circuits
Lecture Notes 2b
2010

2011
1
EE 212
Application of Thevenin’s Theorem
Thevenin's Theorem is specially useful in analyzing power systems
and other circuits where one particular segment in the circuit (the
load) is subject to change.
Source Impedance at a Power System Bus
The source impedance value (or the network impedance at the power system
bus) can be obtained from the utility for all the sub

stations of a power grid.
This is the
Thevenin
Impedance
seen upstream from the sub

station bus. The
Thevenin
Voltage
can be measured at the bus (usually the nominal or rated
voltage at the bus).
Thevenin
equivalent at the sub

station is important to determine cable,
switchgear and equipment ratings, fault levels, and load characteristics at
different times.
2010

2011
2
EE 212
A
B
Linear
Circuit
Norton’s Theorem
Any linear two terminal network with sources can be replaced by an
equivalent current source in parallel with an equivalent impedance.
A
B
Z
I
Current source I is the current which would flow between the terminals if
they were short circuited.
Equivalent impedance Z is the impedance at the terminals (looking into the
circuit) with all the sources reduced to zero.
2010

2011
3
EE 212
~
A
B
Z
E
A
B
Z
I
Thevenin Equivalent
E = IZ
Note:
equivalence is at the terminals
with respect to the external circuit.
Norton Equivalent
I = E / Z
2010

2011
4
EE 212
If a linear circuit has 2 or more sources acting jointly, we can consider each
source acting separately (independently) and then superimpose the 2 or
more resulting effects.
Superposition Theorem
Steps:
•
Analyze the circuit considering each source separately
•
To remove sources, short circuit V sources and open circuit I sources
•
For each source, calculate the voltages and currents in the circuit
•
Sum the voltages and currents
Superposition Theorem is very useful
when analyzing a circuit
that has
2 or more sources with different frequencies.
2010

2011
5
EE 212
Non

sinusoidal Periodic Waveforms
A non

sinusoidal periodic waveform, f(t) can be expressed as a sum
of sinusoidal waveforms. This is known as
a Fourier series.
Fourier
series
is expressed as:
f(t) = a
0
+
=
⡡
n
Cos
nwt
) +
=
(
b
n
Sin
nwt
)
where
,
a
0
=
average over one period (dc component)
=
a
n
=
b
n
=
for
n > 0
2010

2011
6
EE 212
Non

sinusoidal Periodic Waveforms:
Square Waveform
a
n
=
= 0
f(t) = 1
for 0 ≤ t ≤ T/2
=

1
for T/2 ≤ t ≤ T
a
0
= average over one period = 0
b
n
=
=
f(t) = a
0
+
=
⡡
n
cos
n
ω
t
) +
=
(
b
n
sin n
ω
t
)
=
(sin
ω
t
+
sin 3
ω
t
+
sin 5
ω
t
+ …..)
http://homepages.gac.edu/~huber/fourier/index.html
2010

2011
7
EE 212
Linear AC Circuits with
Non

Sinusoidal Waveforms
A linear circuit with
non

sinusoidal periodic sources
can be
analyzed using the Superposition Theorem.
Express the
non

sinusoidal function
by its
Fourier
series.
That is, the
periodic source
will be represented as
multiple
sinusoidal sources of
different
frequencies.
Use Superposition Theorem to calculate voltages and
currents for each
element in the series.
Calculate the final voltages and currents by summing up
all the
harmonics.
2010

2011
8
EE 212
Equations for RMS Values (V, I) and Power
V
rms
= √(v
0
2
+ v
1
2
+ v
2
2
+ v
3
2
+ …) peak values
V
rms
= √V
0
2
+ V
1rms
2
+ V
2rms
2
+ V
3rms
2
+ …
I
rms
= √( i
0
2
+ i
1
2
+ i
2
2
+ i
3
2
+ …) peak values
v
1
i
1
cos
1
+ v
2
i
2
cos
2
+ ..
P = V
0
I
0
+
P = 
I
rms

2
R
v
1
i
1
sin
1
+ v
2
i
2
s
in
2
+ ..
Q =
2010

2011
9
EE 212
Example: Non

sinusoidal AC source
Find the RMS current and power supplied to the circuit elements.
The circuit is energized by a non

sinusoidal voltage v(t), where:
v(t) = 100 + 50 sin
琠⬠㈵=獩渠s
琠癯汴猬†††慮l†=†=
=
㴠㔰〠牡搯0
=
v(t)
+
=
5W
0.02 H
2010

2011
10
EE 212
2010

2011
EE 212
11
Response to a sinusoidal input is also sinusoidal.
Has the same frequency, but may have different
phase
angle.
Linear Circuit with AC Excitation
~
v
i
Input signal,
v
= V
m
sin
t
=
Response
i
= I
m
sin (
琠⬠
⤠†††
w桥h攠
=
楳=瑨攠灨慳攠慮a汥=扥瑷敥渠
v
and
i
Power Factor
:
cosine of the angle between the current and voltage, i.e.
p.f
. = cos
If
=
楳i⬠
癥
Ⱐ
=
i
leads
v
†
汥慤楮朠
瀮p
.
䥦I
=
楳i

=
癥
Ⱐ
=
i
lags
v
†
污l杩湧g
瀮p
.
2010

2011
EE 212
12
Across Resistor
–
Unity p.f.
Voltage and Current are in phase
v(t) = V
m
sin
t
=
i(t) = I
m
sin
t
=
i.e.,
angle between v and i,
= 0
0
††
瀮p⸠㴠捯猠
=
㴠捯猠c
0
= 1
Phasor Diagram
V
I
2010

2011
EE 212
13
Across Inductor
–
Lagging p.f.
Current
lags
Voltage by 90
0
v(t) = V
m
sin
t
=
i(t) = I
m
sin (
t

㤰
0
)
angle between v and i,
= 90
0
瀮p⸠㴠捯猠
=
㴠捯猠㤰
0
= 0
lagging
Phasor Diagram
V
I
Clock

wise
lagging
2010

2011
EE 212
14
Across Capacitor
–
Leading p.f.
Current
leads
Voltage by 90
0
v(t) = V
m
sin
t
=
i(t) = I
m
sin (
琫
㤰
0
)
angle between v and i,
= 90
0
瀮p⸠㴠捯猠
=
㴠捯猠㤰
0
= 0 leading
Phasor Diagram
V
I
2010

2011
EE 212
15
Instantaneous Power, p(t) = v(t)
∙
i(t)
Power
v = V
m
sin
ω
t volts i = I
m
sin(
ω
t

θ
) A
p(t) = V
m
sin
ω
t
∙
I
m
sin(
ω
t

θ
)
p(t) = cos
θ
–
cos(2
ω
t

θ
)
Real Power, P = average value of p(t)
= V
rms
∙
I
rms
∙
cos
θ
p(t) = cos
θ
(1

cos2
ω
t) + sin
θ
∙sin2
ω
t
i
v
Reactive Power, Q = peak value of power exchanged every half cycle
= V
rms
∙
I
rms
∙
sin
θ
2010

2011
EE 212
16
Real and Reactive Power
Real (Active) Power, P

useful power

measured in watts

capable of doing useful work, e.g. ,
lighting, heating, and rotating objects

hidden power

measured in VAr

related to power quality
Reactive Power, Q
Sign Convention:
Power used or consumed:
+ ve
Power generated:

ve
2010

2011
EE 212
17
Real and Reactive Power (continued)
•
Source
–
AC generator:
P is
–
ve
•
Induction gen:
Q is +ve
Synchronous:
Q is +
or
–
ve
•
Load
–
component that consumes real power,
P is + ve
•
Resistive
: e.g. heater, light bulbs, p.f.=1,
Q = 0
•
Inductive
: e.g. motor, welder, lagging p.f.,
Q = + ve
•
Capacitive
: e.g. capacitor, synchronous motor (condensor),
leading p.f.,
Q =

ve
•
Total Power in a Circuit is Zero
2010

2011
EE 212
18
Complex Power
Complex Power, S
= V I*
(
conjugate of I)
S = S
/
楮i偯污r⁆=牭
=

Power Factor Angle
S

Apparent Power
measured in VA
S = P + jQ
in Rectangular Form
P

Real Power
Q
–
Reactive Power
Power ratings of generators & transformers in
VA, kVA, MVA
S
P
jQ
Re
Im
Q
P = S cos
=
噼=䥼=捯猠
Q = S sin
=
噼=䥼=獩渠
=
=
P‽=䥼
2
R
Q = I
2
X
p.f. =
S = V∙I
2010

2011
EE 212
19
Examples: Power
1: V = 10
/10
0
V, I = 20
/5
0
A. Find P, Q
2: What is the power supplied to the combined load?
What is the load power factor?
Motor
5 hp, 0.8 p.f. lagging
100% efficiency
Heater
5 kW
Welder
4+j3
W
120
volts
@60 Hz
2010

2011
EE 212
20
Power Factor Correction
Most loads are inductive in nature, and therefore, have
lagging p.f. (i.e. current lagging behind voltage)
Typical p.f. values: induction motor (0.7
–
0.9), welders
(0.35
–
0.8), fluorescent lights (magnetic ballast 0.7
–
0.8,
electronic 0.9

0.95), etc.
Capacitance can be added to make the current more
leading.
2010

2011
EE 212
21
Power Factor Correction
(continued)
•
P.F. Correction usually involves adding capacitor (in parallel) to the load
circuit, to maximize the
p.f
. and bring it close to 1.
•
The load draws less current from the source, when
p.f
. is corrected.
•
Benefits:
•

•
Therefore,
p.f
. is a measure of how efficiently the power supply is being
utilized
2010

2011
EE 212
22
Example: P.F. Correction
What capacitor is required in parallel for p.f. correction?
Find the total current drawn before and after p.f. correction.
Motor
5 hp, 0.8 p.f. lagging
100% efficiency
Heater
5 kW
Welder
4+j3
W
120
volts
@60Hz
Comments 0
Log in to post a comment