Lecture 5 (2) 2014 Thevenin

7/24/2019 Lecture 5 (2) 2014 Thevenin

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1

Lecture No 5

Circuit Analysis - Part 4

•Nodal Voltage Method •

Mesh-Current Method •Superposition and Source Transformation

•Thevenin and Norton EquivalentCircuits



2

Thevenin’s Theorem (1)

It states that a linear two-terminalcircuit (Fig. a) can be replaced by anequivalent circuit (Fig. b) consistingof a voltage source V

TH in series with

a resistor RTH

,

where

• VTH is the open-circuit voltage at the

terminals.

• RTH is the input or equivalent resistance atthe terminals when the independentsources are turned off.



3

Example 1: Find the Thevenin euivalent at terminals 1!2 o"

the circuits#



$

To find R Th

, consider the circuit in Fig. (a).

R Th

= 20 + 10||40 = 20 + 400/50 = 28 ohms

To find Th

, consider the circuit in Fig. (!).

"t node 1, (40 # $1)/10 = % + &($1 # $2)/20' + $1/40, 40 = $1 # 2$2 (1)

"t node 2, % + ($1 $2)/20 = 0, or $1 = $2 # *0 (2)

o$ing (1) and (2), $1 = %2 , $

2 = -2 , and

Th = $

2 = 92 V

(a)

10

40

20

R Th 40V

%

−

v1

3 A

10

40

20

VTh

%

v2

(b)

&olution



5

'o to "ind Thevenin’s Theorem

arameters *th and +thTo find R th

Short the voltage sources,open the current sourcesand find the equivalentresistance, this will be th

!"# $ %!, "# $ &Ω, i $ '.

6

4

(a)

R Th

6

2A

6

4

(b)

6 2A

+VTh

To find Vth

*eturn the sources ,ac- and

"ind the volta.e at the open

circuit ,ranch (or at the parallel

,ranch

Example 2



/

&olution

*Th0

To "ind *th consider the circuit in Fi.# (a)

To "ind +th e use source trans"ormation

as shon Fi.# (,) and c)



Norton’s Theorem (1)

It states that a linear two-terminal circuitcan be replaced by an equivalent circuitof a current source I

N in parallel with a

resistor RN ,

*here • I

N is the short circuit current through

the terminals.

• RN is the input or equivalent resistance

at the terminals when the independent sources are turned off.

The Thevenin’s and Norton equivalent circuits are

related by a source transforation!



4

Example 3:Find the Norton euivalent o" the

circuit#

To "ind the Norton arameters:

*N0*th the same a as *th

To "ind 6N06 short circuit# The short circuit current is the 6N



7

Solution

For R


R

= (* + *)||4 = % ohs

For

, consider the circuit in Fig. (!). The 4oh resistor is shorted so that 4" current is eua di$ided !et3een the t3o *

oh resistors. ence,

= 4/2 = 2 A

(b)

4

6

6

IN

4A

(a)

4

6

6 R N



18

Example $: Find the Norton Euivalent

9ircuit

e find R in the sae

3a as R th

R N =!!20=4



11

Example $ cont#

"66ing

eshanasis

i1=2"

i = i2

20 2 4 i

1 12=0

2=1"

Find the current in the short circuit terinas a!



12

Norton’s Theorem (ependant &ources)

"#a$le %

Find the +orton equivalentcircuit of the circuit shown

below.

RN $ 'Ω, I

N $ '.

2

(a)

6

2v#

−

+v#

+v#

&V−i#

i

2

(b)

6 &' A

2v#

−

+v#

sc



13



1$

Example /: Find the Thevenin Euivalent

loo-in. into terminals a!, o" the circuit#

& l ti



15

To find R Th


R Th = 10||10 + 5 = 10 ohms

To find Th

, consider the circuit in Fig. (!).

$ ! = 175 = 5 , $

a = 20/2 = 10

8ut, $a +

Th + $

! = 0, or

Th = $

a # $

! = 105=5

(a)

10

R Th

a b

10

(b)

10

20V

%

−

1A10

VTh

%

ba

&olution



1/

;aximum oer Trans"er (1)

L

ThTH L

R

V P R R

4

2

a7 =⇒=

If the entire circuit is replaced byits "hevenin equivalent ecept forthe load, the power delivered tothe load is/

"he power transfer profile withdifferent 0

For maimum power dissipated

in 0, 1ma, for a given "#, and !"#,

L

LTh

Th

L R

R R

V Ri P

2

2

+

==

d li d t th l d " ti " *



1

oer delivered to the load as a "unction o" *Load#

" function is a7. 3hen the deri$ati$e =0

'4

&

0

0222

0)22()(

0)(

)22()(

0))(

(

0

2

a7

22

222

222

4

222

9

2

2

th

th

Lth

Lth

L Lth Lth Lth

Lthth L Lthth

Lth

Lthth L Lthth

L

Lth

th L

L

R

V P

R R R R

R R R R R R Rif

R RV R R RifV

R R

R RV R R RV

dR

dP

R R

V R

dR

dP

P

=

=

=−

=−−++

=+−+

=+

+−+

=

=+

=

2

2

22

)(

)(

Lth

th L

L

Lth

th

L

R R

V R

R R R

V Ri P

+=

+

==



14

Example : 9ompute the value o" *L that results in

maximum poer trans"er# Find the maximum poer#



17

&olution

To "ind *th consider the circuit inFi.# (a)

To "ind +th use source trans"ormationas in the circuit

:se otage di$ider rue;



28

The Thevenin’s Euivalent circuit

For maximum poer trans"er*L0*th 03Ω

<nd the maximum poer is:

Ω=== 5.*

%<4

-'

4

&22

a7

th

th

R

V P

E l 4 "i d th Th i E i l t " th



21

Example 4: "ind the Thevenin Euivalent o" the

9ircuit# Find 6L i" *L0/Ω "# $in% R

th b&

tu'nin o$$ th#

32V sou'*#

('#+la*in it ,ith

a sho't *i'*uit)

an% tu'nin o$$

th# 2 A sou'*#

('#+la*in it ,ith

an o+#n *i'*uit)

R th=4!!12 -1=4

"



22

To "ind +th applin. mesh analsis to 2

loops

is the sae as the $otage across the 12 ohs

resistor = 2.5<12=%0



23

Example 7 : ;aximum oer Trans"er

2

02

0

40

=

=−

+−

th

thth

V

V V R th= =1.4+(2)()/10=%> Ω

E l 18 =,t i th N t E i l t



2$

Example 18: =,tain the Norton Euivalent

at terminals a!, o" the circuit

To .et * appl a 1 m< source at the terminals a and , as shon in Fi.



25

To .et *N appl a 1 m< source at the terminals a and , as shon in Fi.#

(a)#

>e assume all resistances are in - ohms all currents in m< and all

volta.es #

(a)

-

-vab

8 . 0 .

b

a

80I

I

vab!1000

1mA

"t node a, ($a!

/50) + 0 = 1 (1)

"so,

= ($a!

/1000), or = $a!

/000 (2)

Fro (1) and (2), ($a!

/50) # (0$a!

/000) = 1, or $a!

= 100

R

= $a!

/1 = 100 . ohms



2/

2V

+

− IN

(b)

-

-

vab

0 .

b

a

80I

I

vab!1000

8 .

ince the 50> oh resistor is shorted,

= 0, $a!

= 0

ence, i = 2 3hich eads to = (1/4) "

= 20 mA

To .et 6N consider the circuit in Fi.# (,)#



2

Example 11

Find the current through the ga$anoeter

R th = %?//1? +

400//*00=50+240=--0 Ω



24

Example 11 cont#

1=(220/%? Ω+1? Ω)<1? Ω=55

2=(220/400Ω+*00Ω)<*00=1%2

"66ing ?@ around oo6 a!

- 1 + th + 2 =0

th =

"nd the current 3i !eg=(/(--0+40)=4.* "



27

"#a$le &2

2etermine the value of 0 that will

draw the maimum power fromthe rest of the circuit shown below.

3alculate the maimum power.

2

4

& V

−

(a)

&

v#

+

i

v' −v#

* V

−

io

& +VTh

+

v#

2

−v# 4

(b)

Fig. a

$4 "o determine "#

Fig. b$4 "o determine !"#

RL $ 5.66Ω, P m $ 6.7'*

;aximum oer Trans"er (2)

& l ti



38

&olution



31

Thevenin’s Theorem : < circuit ith a

dependant current source

"#a$le &

Find the "hevenin equivalentcircuit of the circuit shown

below to the left of theterminals.

V TH

$ .&&!, RTH

$ .55Ω

6 V

% #

4

−

(a)

&!%#

i&

i2

i& i2

o

+VTh

b

a

&!%# & V−

'!%

#

%

(b)

a

b

4

# i



32

!olution

6x062



Example 1$: =,tain the

Th?venin and Norton

euivalent circuits o" thecircuit ith respect to

terminals a and b#

33

50

V



3$

Example 15: Find the Norton

euivalent at terminals a!b o"

the circuit

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Lecture 5 (2) 2014 Thevenin