D11.1 AC mesh&nodal

7/24/2019 D11.1 AC mesh&nodal

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1

AC Nodal and Mesh Analysis

Discussion D11.1

Chapter 44/10/2006


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2

AC Nodal Analysis


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How did you write nodal euations

!y inspection"

1v

2v

3v

1i

3i #

i


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4

Writing the Nodal Equations by Inspection

1 2 2 1

2 2 3 4 3 2

3 3 # 3

0 2

0

0 s

G G G v

G G G G G v

G G G v i

+ + + =

+

1v

2v

3v

1i 3i #i

$%he &atri' Gis sy&&etric( g)*+ g*)and all o, the o,,-diaonal ter&s

are neatie or ero.

%he i)the kthco&ponent o, the ector i + the ale!raic su& o, the

independent currents connected to node k( with currents enterin the

node ta)en as positie.

%he g)*ter&s are the neatie su& o, the conductances connected to

%H node kand node j.

%he g)) ter&s are the su& o, all conductances connected to node k.

=Gv i


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#

1

2

3

1 1 2 1 0 2

1 1 1 4 1 3 1 3 0

0 1 3 1 3 1 2 1

v

v

v

+ + + = +

v1 v2 v3

1

2

3

1.# 1 0 2

1 1.#53 0.333 0

0 0.333 0.533 1

v

v

v

=

Example with resistors


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or steady-state AC circuits we can

use the sa&e &ethod o, writin nodal

euations !y inspection i, we replace

resistances with i&pedances andconductanceswith admittances.

7et8s loo) at an e'a&ple.


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Problem 4.! in text

Chane i&pedances to ad&ittances


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5

1

2

1 / 2 / 2 2

/ 2 / 2 4

j j

j j =

"

"

1

2

1 0.# 0.# 2

0.# 0.# 4

j j

j j

=

"

"

1

2

4.243 4#

#.531 121

=

"

"

o

o


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:

#atlab $olution 1

2

1 0.# 0.# 2

0.# 0.# 4

j j

j j

=

"

"

1

2

4.243 4#

#.531 121

=

"

"

o

o


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Nodal %nalysis &or 'ircuits 'ontaining "oltage $ources

(hat 'an)t be (rans&ormed to 'urrent $ources

$ ;, a oltae source is connected !etween

two nodes( assu&e te&porarily that the

currentthrouh the oltae source is)nown and write the euations !y

inspection.


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Problem 4. in text Note< "2+ 10

1

2 2

6 4#1/ 2 / 4 / 4

/ 4 / 4

j j

j j

+ =

"

" I

assu&e I2

1

2

6 4#1/ 2 / 4 / 4

/ 4 / 4 10

j j

j j

+ =

"

I


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12

1

2

6 4#1/ 2 / 4 / 4

/ 4 / 4 10

j j

j j

+ =

"

I

( ) 11/ 2 / 4 2.# 6 4#j j+ = "

( ) 1 2/ 4 2.#j j = " I

1

2

1/ 2 / 4 0 6 4# 2.#

/ 4 1 2.#

j j

j j

+ + =

"

I

1

2

0.# 0.2# 0 4.243 6.943

0.2# 1 2.#

j j

j j

+ + =

"

I

1

2

14.2# 31.2#

#.546 105.4

=

"

I

%* + ,

1* + % ,


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13

Problem 4. in text Note< "2+ 10

assu&e I2

1

2

14.2# 31.2#

#.546 105.4

=

"

I

10 9.12# 31.2#

2= = "I


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#atlab $olution

1

2

14.2# 31.2#

#.546 105.4 =

"

I

1

2

1/ 2 / 4 0 6 4# 2.#

/ 4 1 2.#

j j

j j

+ + =

"

I


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1#

AC Mesh Analysis


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How did you write &esh euations

!y inspection"


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$%he &atri' -is sy&&etric( r)*+ r*)and all o, the o,,-diaonal ter&s

are neatie or ero.

Writing the #esh Equations by Inspection

2

1

1

1 # 9 9 # 1

9 2 6 9 6 2

# 3 # 3

6 4 6 5 4

0

00

0 0

0 0

s

s

s

VR R R R R i

R R R R R i

VR R R i

R R R R i V

+ + + + = + + +

%he v)the kthco&ponent o, the ector v + the ale!raic su& o, the

independent oltaes in &esh k( with oltae rises ta)en as positie.

%he r)*ter&s are the neatie su& o, the resistances co&&on to

%H &esh kand &esh j.

%he r)) ter&s are the su& o, all resistances in &esh k.

-i + v


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Example with resistors

1

2

3

4

2 4 1 4 1 0 4

4 3 2 4 0 2 0

1 0 3 1 0 2

0 2 0 2 4 1 2

i

i

i

i

+ +

+ + = +

+ +

-i + v


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1:

or steady-state AC circuits we can

use the sa&e &ethod o, writin &esh

euations !y inspection i, we replace

resistanceswith impedancesandconductances with ad&ittances.

7et8s loo) at an e'a&ple.


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Problem 4. in text/ 0ind I!and I1

1

2

2 1 1 1 12

1 1 1 0 6

j j

j j + = + +

I

I

1

2

3.9:# 15.43

2.653 116.6

=

I

I

o

o


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#atlab $olution 12

2 1 1 1 12

1 1 1 0 6

j j

j j

+ = + +

I

I

1

2

3.9:# 15.43

2.653 116.6 =

I

I

o

o


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What happens i& we have independent

current sources in the circuit2

1. Assu&e te&porarily that the oltae across eachcurrent source is )nown and write the &esheuations in the sa&e way we did ,or circuits

with only independent oltae sources.2. ='press the current o, each independent current

source in ter&s o, the &esh currents and replaceone o, the &esh currents in the euations.

3. >ewrite the euations with all un)nown &eshcurrents and oltaes on the le,t hand side o, theeuality and all )nown oltaes on the r.h.s o,the euality.


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Problem 4.43 in text/ 0ind I3

Assu&e you )now "2

1

2 2

63 1 2 2

2 2 2 2

j j

j j

+ = +

I

I "

1

2

63 1 2 2

2 2 2 2 2

j j

j j

+ = +

I

"

Note I2+ -2


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1

2

1.414 51.59

9.396 12.#3

= =

I*

"

o

o

1

2

63 1 2 2

2 2 2 2 2

j j

j j

+ = +

I

"

( ) 13 1 4 4 6j j+ + + =I

( ) 1 22 2 4 4j j = I "

1

2

3 1 0 2 4

2 2 1 4 4

j j

j j

+ = +

I

"

%* + , 1

* + % ,


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2#

#atlab $olution

1

2

1.414 51.59

9.396 12.#3 =

I

"

o

o

1

2

3 1 0 2 4

2 2 1 4 4

j j

j j

+ = +

I

"


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Problem 4.43 in text/ 0ind I3

Assu&e you )now "2

Note I2+ -2

1

2

1.414 51.59

9.396 12.#3

=

I

"

o

o

0 1 2 1 2= = +I I I I

0 1.414 51.59 2 0 2.605 32.49= + = I o

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D11.1 AC mesh&nodal