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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
7/24/2019 Lecture 5 (2) 2014 Thevenin
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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.
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3
Example 1: Find the Thevenin euivalent at terminals 1!2 o"
the circuits#
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$
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
7/24/2019 Lecture 5 (2) 2014 Thevenin
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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
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/
&olution
*Th0
To "ind *th consider the circuit in Fi.# (a)
To "ind +th e use source trans"ormation
as shon Fi.# (,) and c)
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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!
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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
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7
Solution
For R
, consider the circuit in Fig. (a).
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
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Example $: Find the Norton Euivalent
9ircuit
e find R in the sae
3a as R th
R N =!!20=4
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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!
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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
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13
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1$
Example /: Find the Thevenin Euivalent
loo-in. into terminals a!, o" the circuit#
& l ti
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15
To find R Th
, consider the circuit in Fig. (a).
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
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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 " *
7/24/2019 Lecture 5 (2) 2014 Thevenin
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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
+=
+
==
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14
Example : 9ompute the value o" *L that results in
maximum poer trans"er# Find the maximum poer#
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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;
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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
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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
"
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22
To "ind +th applin. mesh analsis to 2
loops
is the sae as the $otage across the 12 ohs
resistor = 2.5<12=%0
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23
Example 7 : ;aximum oer Trans"er
2
02
0
40
=
=−
+−
th
thth
V
V V R th= =1.4+(2)()/10=%> Ω
E l 18 =,t i th N t E i l t
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2$
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.
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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
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2/
2V
+
− IN
(b)
-
-
vab
0 .
b
a
80I
I
vab!1000
8 .
ince the 50> oh resistor is shorted,
= 0, $a!
= 0
ence, i = 2 3hich eads to = (1/4) "
= 20 mA
To .et 6N consider the circuit in Fi.# (,)#
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2
Example 11
Find the current through the ga$anoeter
R th = %?//1? +
400//*00=50+240=--0 Ω
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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.* "
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"#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
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38
&olution
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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
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32
!olution
6x062
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Example 1$: =,tain the
Th?venin and Norton
euivalent circuits o" thecircuit ith respect to
terminals a and b#
33
50
V
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3$
Example 15: Find the Norton
euivalent at terminals a!b o"
the circuit
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