Discussion #14 - AC Circuit Analysis

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Oct 5, 2013 (3 years and 10 months ago)

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ECEN 301

Discussion #14


AC Circuit Analysis

1

Obedience = Happiness

Mosiah 2:41


41 And moreover, I would desire that ye should consider on the
blessed and happy state of those that keep the commandments
of God. For behold, they are blessed in all things, both temporal
and spiritual; and if they hold out faithful to the end they are
received into heaven, that thereby they may dwell with God in a
state of never
-
ending happiness. O remember, remember that
these things are true; for the Lord God hath spoken it.

ECEN 301

Discussion #14


AC Circuit Analysis

2

Lecture 14


AC Circuit Analysis

Phasors allow the use of familiar
network analysis

ECEN 301

Discussion #14


AC Circuit Analysis

3

RLC Circuits

Linear passive circuit elements
: resistors (
R
), inductors
(
L
), and capacitors (
C
) (a.k.a. RLC circuits)


Assume
RLC

circuit sources are
sinusoidal


Z
R1


Z
R2



V
s
(j
ω
)


+



I
x


Z
L



Z
C


I
a
(j
ω
)

I
b
(j
ω
)


R
1


R
2


i
x


L


C


i
a
(t)

i
b
(t)


v
s
(t)


+



~

Time domain

Frequency (phasor) domain

ECEN 301

Discussion #14


AC Circuit Analysis

4

RLC Circuits
-

Series Impedances


Series Rule
: two or more circuit elements are said to be
in
series

if the current from one element
exclusively

flows into
the next element.


Impedances in
series

add the same way resistors in
series

add


Z
EQ


Z
1

∙ ∙ ∙

∙ ∙ ∙

Z
2

Z
3

Z
n

Z
N

ECEN 301

Discussion #14


AC Circuit Analysis

5

RLC Circuits
-

Parallel Impedances


Parallel Rule
: two or more circuit elements are said to be
in
parallel

if the elements share the
same

terminal
s



Impedances in
parallel

add the same way resistors in
parallel

add

Z
1

Z
2

Z
3

Z
n

Z
N

Z
EQ

ECEN 301

Discussion #14


AC Circuit Analysis

6

RLC Circuits

AC Circuit Analysis

1.
Identify the AC sources and note the excitation
frequency (
ω
)

2.
Convert all sources to the phasor domain

3.
Represent each circuit element by its impedance

4.
Solve the resulting phasor circuit using network
analysis methods

5.
Convert from the phasor domain back to the time
domain

ECEN 301

Discussion #14


AC Circuit Analysis

7

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


R
1


R
2



C


i
s
(t)


v
s
(t)


+



~

ECEN 301

Discussion #14


AC Circuit Analysis

8

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


R
1


R
2



C


i
s
(t)


v
s
(t)


+



~

1.
Note frequencies of AC sources

Only one AC source
-

ω

= 377 rad/s


ECEN 301

Discussion #14


AC Circuit Analysis

9

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


R
1


R
2



C


i
s
(t)


v
s
(t)


+



~


Z
1

= R
1


Z
2
=R
2



Z
3
=1/j
ω
C


I
s
(j
ω
)


V
s
=10e
j0


+



~

1.
Note frequencies of AC sources

2.
Convert to phasor domain

ECEN 301

Discussion #14


AC Circuit Analysis

10

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


Z
1

= R
1


Z
2
=R
2



Z
3
=1/j
ω
C


I
s
(j
ω
)


V
s
=10e
j0


+



~

1.
Note frequencies of AC sources

2.
Convert to phasor domain

3.
Represent each element by its impedance

ECEN 301

Discussion #14


AC Circuit Analysis

11

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


Z
1

= R
1


Z
2
=R
2



Z
3
=1/j
ω
C


I
s
(j
ω
)


V
s
=10e
j0


+



~

1.
Note frequencies of AC sources

2.
Convert to phasor domain

3.
Represent each element by its impedance

4.
Solve using network analysis


Use node voltage and Ohm’s law

Node a

ECEN 301

Discussion #14


AC Circuit Analysis

12

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


Z
1

= R
1


Z
2
=R
2



Z
3
=1/j
ω
C


I
s
(j
ω
)


V
s
=10e
j0


+



~

4.
Solve using network analysis


Use node voltage and Ohm’s law

Node a

ECEN 301

Discussion #14


AC Circuit Analysis

13

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


Z
1

= R
1


Z
2
=R
2



Z
3
=1/j
ω
C


I
s
(j
ω
)


V
s
=10e
j0


+



~

4.
Solve using network analysis


Use node voltage and Ohm’s law

Node a

ECEN 301

Discussion #14


AC Circuit Analysis

14

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)

= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


Z
1

= R
1


Z
2
=R
2



Z
3
=1/j
ω
C


I
s
(j
ω
)


V
s
=10e
j0


+



~

4.
Solve using network analysis


Use node voltage and Ohm’s law

Node a

ECEN 301

Discussion #14


AC Circuit Analysis

15

RLC Circuits


Example 1
: find
i
s
(t)


v
s
(t)
= 10cos(
ω
t)
,
ω

= 377 rad/s
R
1

= 50
Ω
,
R
2

= 200
Ω
, C = 100uF


Z
1

= R
1


Z
2
=R
2



Z
3
=1/j
ω
C


I
s
(j
ω
)


V
s
=10e
j0


+



~

4.
Solve using network analysis


Use node voltage and Ohm’s law

5.
Convert to time domain

Node a

ECEN 301

Discussion #14


AC Circuit Analysis

16

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)

= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

= 0.1H,
L
2

= 20mH


R
s


R
1


i
s
(t)


v
s
(t)


+



~


L
2



R
2



L
1


i
1
(t)

i
2
(t)

ECEN 301

Discussion #14


AC Circuit Analysis

17

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)

= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

= 0.1H,
L
2

= 20mH


R
s


R
1


i
s
(t)


v
s
(t)


+



~


L
2



R
2



L
1


i
1
(t)

i
2
(t)

1.
Note frequencies of AC sources

Only one AC source
-

ω

= 377 rad/s


ECEN 301

Discussion #14


AC Circuit Analysis

18

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)

= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

= 0.1H,
L
2

= 20mH


R
s


R
1


i
s
(t)


v
s
(t)


+



~


L
2



R
2



L
1


i
1
(t)

i
2
(t)


Z
s

= R
s

Z
R1
=R
1

Z
R2
=R
2

I
s
(j
ω
)

V
s
=155e
j0

+



~

Z
L1
=j
ω
L
1

Z
L2
=j
ω
L
2

I
1
(j
ω
)

I
2
(j
ω
)

1.
Note frequencies of AC sources

2.
Convert to phasor domain

ECEN 301

Discussion #14


AC Circuit Analysis

19

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)

= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

= 0.1H,
L
2

= 20mH


Z
s

= R
s

Z
R1
=R
1

Z
R2
=R
2

I
s
(j
ω
)

V
s
=155e
j0

+



~

Z
L1
=j
ω
L
1

Z
L2
=j
ω
L
2

I
1
(j
ω
)

I
2
(j
ω
)

1.
Note frequencies of AC sources

2.
Convert to phasor domain

3.
Represent each element by its impedance

ECEN 301

Discussion #14


AC Circuit Analysis

20

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)
= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

= 0.1H,
L
2

= 20mH


Z
s

= R
s

Z
1
=Z
R1
+Z
L1

Z
2
=Z
R2
+Z
L2

I
s
(j
ω
)

V
s
=155e
j0

+



~

I
1
(j
ω
)

I
2
(j
ω
)

4.
Solve using network analysis


Ohm’s law

ECEN 301

Discussion #14


AC Circuit Analysis

21

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)
= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

= 0.1H,
L
2

= 20mH


Z
s

Z
1

Z
2

I
s
(j
ω
)

V
s
=155e
j0

+



~

I
1
(j
ω
)

I
2
(j
ω
)

4.
Solve using network analysis


KCL

V(j
ω
)

ECEN 301

Discussion #14


AC Circuit Analysis

22

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)
= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

=
0.1H,
L
2

= 20mH


Z
s

Z
1

Z
2

I
s
(j
ω
)

V
s
=155e
j0

+



~

I
1
(j
ω
)

I
2
(j
ω
)

4.
Solve using network analysis


Ohm’s Law

V(j
ω
)

ECEN 301

Discussion #14


AC Circuit Analysis

23

RLC Circuits


Example 2
: find
i
1

and
i
2


v
s
(t)
= 155cos(
ω
t)V,
ω

= 377 rads/s,
R
s

= 0.5
Ω
,
R
1

= 2
Ω
,
R
2

= 0.2
Ω
,
L
1

=
0.1H,
L
2

= 20mH


Z
s

Z
1

Z
2

I
s
(j
ω
)

V
s
=155e
j0

+



~

I
1
(j
ω
)

I
2
(j
ω
)

5.
Convert to Time domain

V(j
ω
)

ECEN 301

Discussion #14


AC Circuit Analysis

24

RLC Circuits


Example 3
: find
i
a
(t)

and
i
b
(t)


v
s
(t)
= 15cos(
1500
t)V,
R
1

= 100
Ω
,
R
2

= 75
Ω
,
L

= 0.5H,
C
= 1uF


R
1


v
s
(t)


+



~

R
2

L

C

i
a
(t)

i
b
(t)

ECEN 301

Discussion #14


AC Circuit Analysis

25

RLC Circuits


Example 3
: find
i
a
(t)

and
i
b
(t)


v
s
(t)
= 15cos(
1500
t)V,
R
1

= 100
Ω
,
R
2

= 75
Ω
,
L

= 0.5H,
C
= 1uF


R
1


v
s
(t)


+



~

R
2

L

C

i
a
(t)

i
b
(t)

1.
Note frequencies of AC sources

Only one AC source
-

ω

= 1500 rad/s


ECEN 301

Discussion #14


AC Circuit Analysis

26

RLC Circuits


Example 3
: find
i
a
(t)

and
i
b
(t)


v
s
(t)
= 15cos(
1500
t)V,
R
1

= 100
Ω
,
R
2

= 75
Ω
,
L

= 0.5H,
C
= 1uF


R
1


v
s
(t)


+



~

R
2

L

C

i
a
(t)

i
b
(t)

1.
Note frequencies of AC sources

2.
Convert to phasor domain


Z
R1

V
s
(j
ω
)

+



~

Z
R2

I
a
(j
ω
)


Z
L


Z
C

I
b
(j
ω
)

ECEN 301

Discussion #14


AC Circuit Analysis

27

RLC Circuits


Example 3
: find
i
a
(t)

and
i
b
(t)


v
s
(t)
= 15cos(
1500
t)V,
R
1

= 100
Ω
,
R
2

= 75
Ω
,
L

= 0.5H,
C
= 1uF


Z
R1

V
s
(j
ω
)

+



~

Z
R2

I
a
(j
ω
)


Z
L


Z
C

I
b
(j
ω
)

1.
Note frequencies of AC sources

2.
Convert to phasor domain

3.
Represent each element by its impedance

ECEN 301

Discussion #14


AC Circuit Analysis

28

RLC Circuits


Example 3
: find
i
a
(t)

and
i
b
(t)


v
s
(t)
= 15cos(
1500
t)V,
R
1

= 100
Ω
,
R
2

= 75
Ω
,
L

= 0.5H,
C
= 1uF


+Z
R1


V
s
(j
ω
)

+



~

+

Z
R2



I
a
(j
ω
)


+Z
L



+

Z
C




I
b
(j
ω
)

4.
Solve using network analysis


Mesh current

ECEN 301

Discussion #14


AC Circuit Analysis

29

RLC Circuits


Example 3
: find
i
a
(t)

and
i
b
(t)


v
s
(t)
= 15cos(
1500
t)V,
R
1

= 100
Ω
,
R
2

= 75
Ω
,
L

= 0.5H,
C
= 1uF


+Z
R1


V
s
(j
ω
)

+



~

+

Z
R2



I
a
(j
ω
)


+Z
L



+

Z
C




I
b
(j
ω
)

4.
Solve using network analysis


Mesh current

ECEN 301

Discussion #14


AC Circuit Analysis

30

RLC Circuits


Example 3
: find
i
a
(t)

and
i
b
(t)


v
s
(t)
= 15cos(
1500
t)V,
R
1

= 100
Ω
,
R
2

= 75
Ω
,
L

= 0.5H,
C
= 1uF


+Z
R1


V
s
(j
ω
)

+



~

+

Z
R2



I
a
(j
ω
)


+Z
L



+

Z
C




I
b
(j
ω
)

5.
Convert to Time domain

ECEN 301

Discussion #14


AC Circuit Analysis

31

AC Equivalent Circuits

Th
évenin

and
Norton

equivalent circuits apply in AC analysis


Equivalent voltage/current is
complex

and
frequency dependent

Load

+

V



I

Source

V
T
(j
ω
)

+



Z
T

Load

+


V




I

I
N
(j
ω
)

Z
N

Load

+


V




I

Norton Equivalent

Th
évenin

Equivalent

ECEN 301

Discussion #14


AC Circuit Analysis

32

AC Equivalent Circuits

Computation of Th
évenin and Norton Impedances
:

1.
Remove the load (open circuit at load terminal)

2.
Zero all independent sources


Voltage sources


short circuit (
v

= 0)


Current sources


open circuit (
i

= 0)

3.
Compute equivalent impedance
across load terminals

(with load removed)

NB
: same procedure as equivalent resistance


Z
L



Z
1


V
s
(j
ω
)


+




Z
3



Z
2



Z
4


a

b


Z
1


Z
3



Z
2



Z
4


a

b

Z
T

ECEN 301

Discussion #14


AC Circuit Analysis

33

AC Equivalent Circuits

Computing
Thévenin voltage
:

1.
Remove the load (open circuit at load terminals)

2.
Define the open
-
circuit voltage (
V
oc
) across the load terminals

3.
Chose a network analysis method to find
V
oc


node, mesh, superposition, etc.

4.
Thévenin voltage
V
T

=
V
oc


Z
1


V
s
(j
ω
)


+




Z
3



Z
2



Z
4


a

b


Z
1


V
s
(j
ω
)


+




Z
3



Z
2



Z
4


a

b

+

V
T



NB
: same procedure as equivalent resistance

ECEN 301

Discussion #14


AC Circuit Analysis

34

AC Equivalent Circuits

Computing Norton

current
:

1.
Replace the load with a short circuit

2.
Define the short
-
circuit current (
I
sc
) across the load terminals

3.
Chose a network analysis method to find
I
sc



node, mesh, superposition, etc.

4.
Norton current
I
N

=
I
sc


Z
1


V
s
(j
ω
)


+




Z
3



Z
2



Z
4


a

b


Z
1


V
s
(j
ω
)


+




Z
3



Z
2



Z
4


a

b

I
N

NB
: same procedure as equivalent resistance

ECEN 301

Discussion #14


AC Circuit Analysis

35

AC Equivalent Circuits


Example 4
: find the Th
évenin equivalent


ω

= 10
3

rads/s,
R
s

= 50
Ω
,
R
L

= 50
Ω
,
L

= 10mH,
C
= 0.1uF


R
s


v
s
(t)


+



~

R
L

L

C

+

v
L



ECEN 301

Discussion #14


AC Circuit Analysis

36

AC Equivalent Circuits


Example 4
: find the Th
évenin equivalent


ω

= 10
3

rads/s,
R
s

= 50
Ω
,
R
L

= 50
Ω
,
L

= 10mH,
C
= 0.1uF


R
s


v
s
(t)


+



~

R
L

L

C

+

v
L



1.
Note frequencies of AC sources

Only one AC source
-

ω

= 10
3

rad/s


ECEN 301

Discussion #14


AC Circuit Analysis

37

AC Equivalent Circuits


Example 4
: find the Th
évenin equivalent


ω

= 10
3

rads/s,
R
s

= 50
Ω
,
R
L

= 50
Ω
,
L

= 10mH,
C
= 0.1uF


R
s


v
s
(t)


+



~

R
L

L

C

+

v
L



1.
Note frequencies of AC sources

2.
Convert to phasor domain


Z
s

Z
LD

+



~

Z
L

Z
C

V
s
(j
ω
)

ECEN 301

Discussion #14


AC Circuit Analysis

38

AC Equivalent Circuits


Example 4
: find the Th
évenin equivalent


ω

= 10
3

rads/s,
R
s

= 50
Ω
,
R
L

= 50
Ω
,
L

= 10mH,
C
= 0.1uF

1.
Note frequencies of AC sources

2.
Convert to phasor domain

3.
Find
Z
T


Remove load & zero sources


Z
s

Z
L

Z
C

ECEN 301

Discussion #14


AC Circuit Analysis

39

AC Equivalent Circuits


Example 4
: find the Th
évenin equivalent


ω

= 10
3

rads/s,
R
s

= 50
Ω
,
R
L

= 50
Ω
,
L

= 10mH,
C
= 0.1uF

1.
Note frequencies of AC sources

2.
Convert to phasor domain

3.
Find
Z
T


Remove load & zero sources

4.
Find
V
T
(j
ω
)


Remove load


Z
s

+



~

Z
L

Z
C

V
s
(j
ω
)

+

V
T
(j
ω
)



NB
: Since no current flows in the
circuit once the load is removed:

ECEN 301

Discussion #14


AC Circuit Analysis

40

AC Equivalent Circuits


Example 4
: find the Th
évenin equivalent


ω

= 10
3

rads/s,
R
s

= 50
Ω
,
R
L

= 50
Ω
,
L

= 10mH,
C
= 0.1uF


Z
s

Z
LD

+



~

Z
L

Z
C

V
s
(j
ω
)


Z
T

+



~

V
T
(j
ω
)

Z
LD