EE 212 Passive AC Circuits
Lecture Notes 2b
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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.
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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
A
Linear
Circuit

B
I
Z
B
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.
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Z
~
E
A
A

B
Thevenin Equivalent
E = IZ
I
Z
B
Norton Equivalent
I=E/Z
Note:
equivalence is at the terminals
with respect to the external circuit.
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Superposition Theorem
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.
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.
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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) = a0 +  (an Cos nwt) +  (bn Sin nwt)
where,
a0 = average over one period (dc component) =
2 T
an =
f(t)  cos(n t)  dt

0
T
2
bn =
T
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
T
0
1 T
f(t).dt

0
T
for n > 0
f(t)  sin(n t)  dt
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Non-sinusoidal Periodic Waveforms: Square Waveform
2
f(t) = 1 for
0 ≤ t ≤ T/2
= -1 for T/2 ≤ t ≤ T
Volts
1
-T
-T/2
0
T/2
T
1.5T
-1
a0 = average over one period = 0
time
-2
2 T
an =  f(t)  cos(n t )  dt = 0
T 0
2
2 T
(1  cos n )
bn =  f(t)  sin(n t)  dt =
nπ
T 0
f(t) = a0 +  (an cos nωt) +  (bn sin nωt)
=
1
1
4
(sin ωt + sin 3ωt + sin 5ωt + …..)
3
5

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http://homepages.gac.edu/~huber/fourier/index.html
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.
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Equations for RMS Values (V, I) and Power
Vrms = √V02 + V1rms2 + V2rms2 + V3rms2 + …
Vrms = √(v0
1 2 1 2 1 2
+ v1 + v2 + v3 + …) peak values
2
2
2
2
Irms = √( i0
P = V 0 I0 +
2
1 2 1 2 1 2
+ i1 + i2 + i3 + …) peak values
2
2
2
1
2
1
v1 i1 cos 1 + v2 i2 cos 2 + ..
2
P = |Irms| 2 R
1
v
i
sin

+
v2 i2 sin 2 + ..
Q=
1 1
1
2
2
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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 t + 25 sin 3t volts,
+
and
 = 500 rad/s
5W
0.02 H
v(t)

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Linear Circuit with AC Excitation
i
Input signal, v = Vm sin t v ~
Response to a sinusoidal input is also sinusoidal.
Has the same frequency, but may have different phase angle.
Response i = Im sin (t + )
where  is the phase angle between v and i
Power Factor:
cosine of the angle between the current and voltage, i.e. p.f. = cos 
If  is + ve,
i leads v  leading p.f.
If  is - ve,
i lags v  lagging p.f.
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Across Resistor – Unity p.f.
R
i(t)
Voltage and Current are in phase
v(t) = Vm sin t
v(t)
~
i(t) = Im sin t
i.e., angle between v and i,  = 00
 p.f. = cos  = cos 00 = 1
Phasor Diagram
I
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Across Inductor – Lagging p.f.
L
Current lags Voltage by 900
i(t)
v(t) = Vm sin t
v(t)
~
i(t) = Im sin (t-900)
angle between v and i,  = 900
p.f. = cos  = cos 900 = 0 lagging
Phasor Diagram
V
Clock-wise
lagging
I
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Across Capacitor – Leading p.f.
C
Current leads Voltage by 900
i(t)
v(t) = Vm sin t
v(t)
i(t) = Im sin (t+900)
~
angle between v and i,  = 900
p.f. = cos  = cos 900 = 0 leading
Phasor Diagram
I
V
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Power
v = Vm sin ωt volts
i = Im sin(ωt - θ) A
i
v
Instantaneous Power, p(t) = v(t) · i(t)
p(t) = Vm sin ωt · Im sin(ωt - θ)
p(t) =
Vm I m
V I
cos θ – m m cos(2ωt-θ)
2
2
Real Power, P = average value of p(t)
= Vrms·Irms·cos θ
Vm I m
p(t) =
cosθ(1-cos2ωt) +
2
Vm I m
sinθ·sin2ωt
2
Reactive Power, Q = peak value of power exchanged every half cycle
= Vrms·Irms·sin θ EE 212
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Real and Reactive Power
Reactive Power, Q
Real (Active) Power, P
- useful power
- measured in watts
- capable of doing useful work, e.g. ,
lighting, heating, and rotating objects
Sign Convention:
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- hidden power
- measured in VAr
- related to power quality
Power used or consumed:
Power generated:
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+ ve
- ve
16
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
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Complex Power
Complex Power,
S = |S| /
S = V I* (conjugate of I)
S = P + jQ in Rectangular Form
in Polar Form
 - Power Factor Angle
P - Real Power
Q – Reactive Power
|S| - Apparent Power
measured in VA
P = |S| cos  = |V| |I| cos 
Q = |S| sin  = |V| |I| sin 
|S| = |V|·|I|
Power ratings of generators & transformers in
VA, kVA, MVA
Im
P
p.f. =
|S|
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jQ
|S|
P = |I|2 R
Q = |I|2 X
Q
Re

P
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Examples: Power
1: V = 10/100V, I = 20/50A. 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
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Heater
5 kW
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Welder
4+j3 W
120 volts
@60 Hz
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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.
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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
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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
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Heater
5 kW
Welder
4+j3 W
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120 volts
@60Hz
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