Chap4

Circuit TheoremsCircuit Theorems Eastern Mediterranean UniversityEastern Mediterranean University
11
Circuit Theorems
Mustafa Kemal Uyguroğlu

Chap. 4 Circuit TheoremsChap. 4 Circuit Theorems
Introduction
Linearity property
Superposition
Source transformations
Thevenin’s theorem
Norton’s theorem
Maximum power transfer
Circuit TheoremsCircuit Theorems 22EEasternastern MMediterraneanediterranean UUniversityniversity

4.1 Introduction4.1 Introduction
Circuit TheoremsCircuit Theorems 33
A large
complex circuits
A large
complex circuits
Simplify
circuit analysis
Simplify
circuit analysis
Circuit TheoremsCircuit Theorems
‧Thevenin’s theorem Norton theorem‧
‧Circuit linearity Superposition‧
‧source transformation max. power transfer‧
‧Thevenin’s theorem Norton theorem‧
‧Circuit linearity Superposition‧
‧source transformation max. power transfer‧
EEasternastern MMediterraneanediterranean UUniversityniversity

4.2 Linearity4.2 Linearity PPropertyroperty
Homogeneity property (Scaling)
iRvi =→
kiRkvki =→
Additivity property
Rivi 222 =→
Rivi 111 =→
21212121 )( vvRiRiRiiii +=+=+→+

A linear circuit is one whose output is linearly
related (or directly proportional) to its input
Fig. 4.1
V0
I0
i

Linear circuit consist of
● linear elements
● linear dependent sources
● independent sources

mA1mV5
A2.0V1
A2V10
=←=
=→=
=→=
iv
iv
iv
s
s
s
nonlinear
R
v
Rip :
2
2
==

Example 4.1Example 4.1
For the circuit in fig 4.2 find I0 when vs=12V and
vs=24V.

KVL
Eqs(4.1.1) and (4.1.3) we get
0412 21 =+− svii
03164 21 =−−+− sx vvii
12ivx =
becomes)2.1.4(
01610 21 =−+− svii
(4.1.1)
(4.1.2)
(4.1.3)
2121 60122 iiii −=→=+

Eq(4.1.1), we get
When
When
Showing that when the source value is doubled, I0
doubles.
76
076 22
s
s
v
ivi =⇒==−
A
76
12
20 == iI
V12=sv
A
76
24
20 == iI
V24=sv

Assume I0 = 1 A and use linearity to find the actual
value of I0 in the circuit in fig 4.4.

A,24/
V8)53(thenA,1If
11
010
==
=+==
vI
IvI
A3012 =+= III
A2
7
,V14682 2
3212 ===+=+=
V
IIVV
⇒=+= A5234 III A5=SI
A510 =→= SIAI
A15A30 =←= SII

4.3 Superposition4.3 Superposition
The superposition principle states that the voltage
across (or current through) an element in a linear
circuit is the algebraic sum of the voltages across
(or currents through) that element due to each
independent source acting alone.
Turn off, killed, inactive source:
● independent voltage source: 0 V (short circuit)
● independent current source: 0 A (open circuit)
Dependent sources are left intact.

 Steps to apply superposition principle:
1. Turn off all independent sources except one source.
Find the output (voltage or current) due to that active
source using nodal or mesh analysis.
2. Repeat step 1 for each of the other independent
sources.
3. Find the total contribution by adding algebraically all
the contributions due to the independent sources.

How to turn off independent sourcesHow to turn off independent sources
Turn off voltages sources = short voltage sources;
make it equal to zero voltage
Turn off current sources = open current sources;
make it equal to zero current

Superposition involves more work but simpler
circuits.
Superposition is not applicable to the effect on
power.

Use the superposition theorem to find in the
circuit in Fig.4.6.

Since there are two sources,
let
Voltage division to get
Current division, to get
Hence
And we find
21 VVV +=
V2)6(
84
4
1 =
+
=V
A2)3(
84
8
3 =
+
=i
V84 32 == iv
V108221 =+=+= vvv

Find I0 in the circuit in Fig.4.9 using superposition.

Fig. 4.10

4.5 Source Transformation4.5 Source Transformation
A source transformation is the process of replacing
a voltage source vs in series with a resistor R by a
current source is in parallel with a resistor R, or
vice versa

Fig. 4.15 & 4.16Fig. 4.15 & 4.16
R
v
iRiv s
sss == or

Equivalent CircuitsEquivalent Circuits
R
v
R
v
i
viRv
s
s
−=
+=
i i
++
--
vv
v
i
vs
-is

Arrow of the current source
positive terminal of voltage source
Impossible source Transformation
● ideal voltage source (R = 0)
● ideal current source (R=∞)

Use source transformation to find vo in the circuit
in Fig 4.17.

Fig 4.18

we use current division in Fig.4.18(c) to get
and
A4.0)2(
82
2
=
+
=i
V2.3)4.0(88 === ivo

Find vx in Fig.4.20 using source transformation

Applying KVL around the loop in Fig 4.21(b) gives
(4.7.1)
Appling KVL to the loop containing only the 3V
voltage source, the resistor, and vx yields
(4.7.2)
01853 =+++− xvi
Ω1
ivvi xx −=⇒=++− 3013

Substituting this into Eq.(4.7.1), we obtain
Alternatively
thus
A5.403515 −=⇒=++ ii
A5.40184 −=⇒=+++− iviv xx
V5.73 =−= ivx

4.5 Thevenin’s Theorem4.5 Thevenin’s Theorem
Thevenin’s theorem states that a linear two-
terminal circuit can be replaced by an equivalent
circuit consisting of a voltage source VTh in series
with a resistor RTh where VThis the open circuit
voltage at the terminals and RTh is the input or
equivalent resistance at the terminals when the
independent source are turn off.

Property of Linear CircuitsProperty of Linear Circuits
i
v
v
i
Any two-terminal
Linear Circuits
+
-
Vth
Isc
Slope=1/Rth

Fig. 4.23Fig. 4.23

How to Find Thevenin’s VoltageHow to Find Thevenin’s Voltage
Equivalent circuit: same voltage-current relation at
the terminals.

:Th ocvV = ba −atltagecircuit voopen

How to Find Thevenin’s ResistanceHow to Find Thevenin’s Resistance

:inTh RR =
b.a −− atcircuitdeadtheofresistanceinput
circuitedopenba −•
sourcestindependenalloffTurn•

CASE 1
 If the network has no dependent sources:
● Turn off all independent source.
● RTH: can be obtained via simplification of either parallel
or series connection seen from a-b

Fig. 4.25Fig. 4.25
CASE 2
If the network has dependent
sources
● Turn off all independent sources.
● Apply a voltage source vo at a-b
● Alternatively, apply a current
source io at a-b
o
o
i
v
R =Th
o
o
Th
i
v
R =

The Thevenin’s resistance may be negative,
indicating that the circuit has ability providing
power

Fig. 4.26Fig. 4.26
Simplified circuit
Voltage divider
L
L
RR
V
I
+
=
Th
Th
Th
Th
V
RR
R
IRV
L
L
LLL
+
==

Find the Thevenin’s equivalent circuit of the circuit
shown in Fig 4.27, to the left of the terminals a-b.
Then find the current through RL = 6,16,and 36 Ω.

Find RFind Rthth
shortsourcevoltageV32:Th →R
opensourcecurrentA2 →
Ω=+
×
=+= 41
16
124
112||4ThR

Find VFind Vthth
analysisMesh)1(
:ThV
A2,0)(12432 2211 −==−++− iiii
A5.01 =∴i
V30)0.25.0(12)(12 21Th =+=−= iiV
AnalysisNodalely,Alternativ)2(
12/24/)32( ThTh VV =+−
V30Th =∴V

Circuit TheoremsCircuit Theorems 4343Fig. 4.29
transformsourceely,Alternativ)3(
V3024396
12
2
4
32
THTHTH
THTH
=⇒=+−
=+
−
VVV
VV

:getTo Li
LL
L
RRR
V
i
+
=
+
=
4
30
Th
Th
6=LR A310/30 ==LI
16=LR A5.120/30 ==LI
A75.040/30 ==LI36=LR
→
→
→

Find the Thevenin’s equivalent of the circuit in Fig.
4.31 at terminals a-b.

(independent + dependent source case)
Fig(a):findTo ThR
0sourcetindependen →
intact→sourcedependent
,V1=ov
oo
o
ii
v
R
1
Th ==

For loop 1,
2121 or0)(22 iiviiv xx −==−+−
214But iivi x −==−
21 3ii −=∴

:3and2Loop
0)(6)(24 32122 =−+−+ iiiii
012)(6 323 =++− iii
givesequationstheseSolving
.A6/13 −=i
A
6
1
But 3 =−= iio
Ω==∴ 6
1
Th
oi
V
R

⇒=−+− 0)(22 23 iivx
51 =i
Fig(b):getTo ThV
23 iivx −=
analysisMesh
⇒=+−+− 06)(2)(4 21212 iiiii 02412 312 =−− iii
.3/102 =∴i
V206 2Th === ivV oc
xvii =− )(4But 21

Determine the Thevenin’s
equivalent circuit in
Fig.4.35(a).
Solution
)caseonlysourcedependent(
o
o
i
v
R =Th0Th =V
:anaysisNodal
4/2 oxxo viii +=+

22
0 oo
x
vv
i −=
−
=But
4424
oooo
xo
vvvv
ii −=+−=+= oo iv 4or −=
:4Thus Th Ω−==
o
o
i
v
R powerSupplying

4.64.6 Norton’s TheoremNorton’s Theorem
Norton’s theorem states that a linear two-terminal
circuit can be replaced by equivalent circuit
consisting of a current source IN in parallel with a
resistor RN where IN is the short-circuit current
through the terminals and RN is the input or
equivalent resistance at the terminals when the
independent source are turn off.

Fig. 4.37Fig. 4.37
v
i
Vth
-IN
Slope=1/RN

How to Find Norton CurrentHow to Find Norton Current
Thevenin and Norton
resistances are equal:
Short circuit current
from a to b :
ThRRN =
Th
Th
R
V
iI scN ==

Thevenin or Norton equivalent circuit :Thevenin or Norton equivalent circuit :
The open circuit voltage voc across terminals a and b
The short circuit current isc at terminals a and b
The equivalent or input resistance Rin at terminals a
and b when all independent source are turn off.
ocTh vV =
NI
Th
Th N
Th
V
R R
R
= =
= sci

Find the Norton equivalent circuit of the circuit in
Fig 4.39.

:)(40.4Fig a
Ω=
×
==
++=
4
25
520
20||5
)848(||5NR
NRfindTo

NifindTo
.andterminalscircuitshort ba−
))(40.4.Fig( b
:Mesh 0420,A2 2121 =−−= iiii
Nsc Iii === A12

NIformethodeAlternativ
Th
Th
N
R
V
I =
voltagecircuitopen: −ThV ba and
:))(40.4( cFig
:analysisMesh
012425,2 343 =−−= iiAi
A8.04 =∴i
terminalsacross
V45 4 ===∴ iVv Thoc

,Hence
A14/4 ===
Th
Th
N
R
V
I

Using Norton’s theorem, find RN and IN of the
circuit in Fig 4.43 at terminals a-b.

NRfindTo )(44.4. aFig
shortedresistor4Ω•
Parallel:2||||5 xo ivΩ
Hence, 2.05/15/ === ox vi
•
Ω===∴ 5
2.0
1
o
o
N
i
v
R

NIfindTo )(44.4. bFig
xiv 2||5||10||4 ΩΩ• Parallel:
.5A,2
4
010
=
−
=xi
A72(2.5)
5
10
2 =+=+= xxsc iii
7A=∴ NI

4.8 Maximum Power Trandfer4.8 Maximum Power Trandfer
LL R
RR
V
Rip
2
LTH
TH2






+
==
Fig 4.48

Fig. 4.49Fig. 4.49
Maximum power is transferred to the load when
the load resistance equals the Thevenin resistance
as seen the load (RL = RTH).

TH
TH
THL
LTHLLTH
LTH
LLTH
TH
LTH
LTHLLTH
TH
L
R
V
p
RR
RRRRR
RR
RRR
V
RR
RRRRR
V
dR
dp
4
)()2(0
0
)(
)2(
)(
)(2)(
2
max
3
2
4
2
2
=
=
−=−+=
=





+
−+
=






+
+−+
=

Find the value of RL for maximum power transfer
in the circuit of Fig. 4.50. Find the maximum
power.

Ω=
×
+=++= 9
18
126
512632THR

W
R
V
p
RR
VVVii
Aiii
L
TH
THL
THTHi
44.13
94
22
4
9
220)0(231612
2,121812
22
max
2
221
=
×
==
Ω==
=⇒=++++−
−=−+−

Homework ProblemsHomework Problems
Problems 6, 10, 21, 28, 33, 40, 47, 52, 71

Chap4

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