# EE 212 Passive AC Circuits

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5 Οκτ 2013 (πριν από 4 χρόνια και 7 μήνες)

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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

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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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.

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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

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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

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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

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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.

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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 =

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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

=
㴠㔰〠牡搯0
=
v(t)

+

=
5W

0.02 H

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EE 212

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2011

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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 (

⤠†††
w桥h攠

=

v

and
i

Power Factor
:

cosine of the angle between the current and voltage, i.e.

p.f
. = cos

If

=

=
i

v

.

䥦I

=

-
=

=
i

lags
v

.

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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

††

=
㴠捯猠c
0

= 1

Phasor Diagram

V

I

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2011

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13

Across Inductor

Lagging p.f.

Current
lags

Voltage by 90
0

v(t) = V
m

sin

t
=

i(t) = I
m

sin (

t
-

0
)

angle between v and i,

= 90
0

=
㴠捯猠㤰
0

= 0
lagging

Phasor Diagram

V

I

Clock
-
wise
lagging

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14

Across Capacitor

Current

Voltage by 90
0

v(t) = V
m

sin

t
=

i(t) = I
m

sin (

0
)

angle between v and i,

= 90
0

=
㴠捯猠㤰
0

Phasor Diagram

V

I

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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

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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

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2011

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17

Real and Reactive Power (continued)

Source

AC generator:
P is

ve

Induction gen:
Q is +ve

Synchronous:
Q is +
or

ve

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),
Q =
-

ve

Total Power in a Circuit is Zero

2010
-
2011

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18

Complex Power

Complex Power, S

= V I*
(
conjugate of I)

S = |S|
/

=

-

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|

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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

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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

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21

Power Factor Correction

(continued)

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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2011

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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