Circuit Theorems
Mustafa Kemal Uyguroğlu
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Chap. 4 Circuit Theorems
 Introduction
 Linearity property
 Superposition
Source transformations
 Thevenin’s theorem
 Norton’s theorem
 Maximum power transfer
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4.1 Introduction
A large
complex circuits
Simplify
circuit analysis
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‧Thevenin’s theorem
‧Circuit linearity
‧source transformation
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‧ Norton theorem
‧ Superposition
‧ max. power transfer
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4.2 Linearity Property
Homogeneity property (Scaling)
i  v  iR
ki  kv  kiR
Additivity property
i1  v1  i1 R
i2  v2  i2 R
i1  i2  (i1  i2 ) R  i1 R  i2 R  v1  v2
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 A linear circuit is one whose output is linearly
related (or directly proportional) to its input
 Fig. 4.1
i
I0
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V0
5
 Linear circuit consist of
● linear elements
● linear dependent sources
● independent sources

vs  10V  i  2A
vs  1V  i  0.2A
vs  5mV  i  1mA
2
v
p i R 
: nonlinear
R
2
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Example 4.1
 For the circuit in fig 4.2 find I0 when vs=12V and
vs=24V.
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Example 4.1
 KVL
12i1  4i2  vs  0
 4i1  16i2  3vx  vs  0
vx  2i1
(4.1.1)
(4.1.2)
(4.1.2) becomes
 10i1  16i2  vs  0
(4.1.3)
Eqs(4.1.1) and (4.1.3) we get
2i1  12i2  0  i1  6i2
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Example 4.1
Eq(4.1.1), we get
vs
 76i2  vs  0  i2 
76
When vs  12V
12
I 0  i2  A
76
vs  24V
24
When
I 0  i2  A
76
Showing that when the source value is doubled, I0
doubles.
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Example 4.2
 Assume I0 = 1 A and use linearity to find the
actual value of I0 in the circuit in fig 4.4.
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Example 4.2
If I 0  1A, then v1  (3  5) I 0  8V
I1  v1 / 4  2A, I 2  I1  I 0  3A
V2
V2  V1  2 I 2  8  6  14V, I 3   2A
7
I 4  I 3  I 2  5A  I S  5A
I 0  1A  I S  5A
I 0  3A  I S  15A
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4.3 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.
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 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.
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How 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
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 Superposition involves more work but simpler
circuits.
 Superposition is not applicable to the effect on
power.
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Example 4.3
 Use the superposition theorem to find in the
circuit in Fig.4.6.
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Example 4.3
Since there are two sources,
let
V  V1  V2
Voltage division to get
4
V1 
(6)  2V
48
Current division, to get
Hence
8
i3 
(3)  2A
48
v2  4i3  8V
And we find
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v  v1  v2  2  8  10V
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Example 4.4
 Find I0 in the circuit in Fig.4.9 using superposition.
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Example 4.4
Fig. 4.10
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Example 4.4
Fig. 4.10
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4.5 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
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Fig. 4.15 & 4.16
vs
vs  is R or is 
R
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Equivalent Circuits
i
i
+
+
v  iR  vs
v
i
v vs

R R
-
v
-
i
v
-is
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vs
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 Arrow of the current source
positive terminal of voltage source
 Impossible source Transformation
● ideal voltage source (R = 0)
● ideal current source (R=)
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Example 4.6
 Use source transformation to find vo in the circuit
in Fig 4.17.
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Example 4.6
Fig 4.18
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Example 4.6
we use current division in Fig.4.18(c) to get
2
i
(2)  0.4A
28
and
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vo  8i  8(0.4)  3.2V
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Example 4.7
 Find vx in Fig.4.20 using source transformation
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Example 4.7
Applying KVL around the loop in Fig 4.21(b) gives
 3  5i  vx  18  0
(4.7.1)
Appling KVL to the loop containing only the 3V
voltage source, the1 resistor, and vx yields
 3  1i  vx  0  vx  3  i
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(4.7.2)
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Example 4.7
Substituting this into Eq.(4.7.1), we obtain
15  5i  3  0  i  4.5A
Alternatively
 vx  4i  vx  18  0  i  4.5A
thus
vx  3  i  7.5V
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4.5 Thevenin’s Theorem
 Thevenin’s theorem states that a linear twoterminal circuit can be replaced by an equivalent
circuit consisting of a voltage source VTh in series
with a resistor RTh where VTh is 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.
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Property of Linear Circuits
i
i
Any two-terminal
Linear Circuits
+
v
Slope=1/Rth
-
v
Vth
Isc
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Fig. 4.23
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How to Find Thevenin’s Voltage
 Equivalent circuit: same voltage-current relation
at the terminals.

VTh  voc : open circuit voltage at a  b
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How to Find Thevenin’s Resistance

RTh  Rin :
input  resistance of the dead circuit at a  b.
 a  b open circuited
 Turn off all independen t sources
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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
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Fig. 4.25
CASE 2
 If the network has dependent
sources
● Turn off all independent sources.
● Apply a voltage source vo at a-b
vo
RTh 
io
● Alternatively, apply a current
source io at a-b R  vo
Th
io
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 The Thevenin’s resistance may be negative,
indicating that the circuit has ability providing
power
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Fig. 4.26
Simplified circuit
VTh
IL 
RTh  RL
RL
VL  RL I L 
VTh
RTh  RL
Voltage divider
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Example 4.8
 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 .
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Find Rth
RTh : 32V voltage source  short
2A current source  open
4  12
RTh  4 || 12  1 
 1  4
16
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Find Vth
VTh :
(1) Mesh analysis
 32  4i1  12(i1  i2 )  0 , i2  2A
i1  0.5A
VTh  12(i1  i2 )  12(0.5  2.0)  30V
(2) Alternatively, Nodal Analysis
(32  VTh ) / 4  2  VTh / 12
VTh  30V
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Example 4.8
(3) Alternatively, source transform
32  VTH
VTH
2
4
12
96  3VTH  24  VTH  VTH  30V
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Fig.
4.29
43
Example 4.8
To get iL :
VTh
30
iL 

RTh  RL 4  RL
RL  6  I L  30 / 10  3A
RL  16 I L  30 / 20  1.5A
RL  36 I L  30 / 40  0.75A
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Example 4.9
 Find the Thevenin’s equivalent of the circuit in Fig.
4.31 at terminals a-b.
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Example 4.9
 (independent + dependent source case)
To find RTh : Fig(a)
independen t source  0
dependent source  intact
vo 1
vo  1V, RTh  
io io
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Example 4.9
 For loop 1,
 2vx  2(i1  i2 )  0 or vx  i1  i2
But  4i  vx  i1  i2
 i1  3i2
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Example 4.9
Loop 2 and 3 :
4i2  2(i2  i1 )  6(i2  i3 )  0
6(i3  i2 )  2i3  1  0
Solving these equations gives
i3  1/ 6A.
1
But io  i3  A
6
1V
 RTh 
 6
io
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Example 4.9
To get VTh : Fig(b) Mesh analysis
i1  5
 2vx  2(i3  i2 )  0  vx  i3  i2
4(i2  i1 )  2(i2  i1 )  6i2  0  12i2  4i1  2i3  0
But 4(i1  i2 )  vx
i2  10 / 3.
VTh  voc  6i2  20V
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Example 4.10
 Determine the Thevenin’s
equivalent circuit in
Fig.4.35(a).
 Solution
(dependent source only case)
vo
RTh 
VTh  0
io
Nodal anaysis :
io  ix  2ix  vo / 4
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Example 4.10
0

v
v
o
But ix 
 o
2
2
vo
vo vo
vo
io  ix       or vo  4io
4
2 4
4
vo
Thus RTh   4 : Supplying power
io
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Example 4.10
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Example 4.10
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4.6 Norton’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.
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Fig. 4.37
i
Slope=1/RN
v
Vth
-IN
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How to Find Norton Current
Thevenin and Norton
resistances are equal:
RN  RTh
Short circuit current
from a to b :
VTh
I N  isc 
RTh
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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.
VTh  voc
IN  isc
RTh
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VTh

 RN
RTh
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Example 4.11
 Find the Norton equivalent circuit of the circuit in
Fig 4.39.
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Example 4.11
To find RN Fig 4.40(a) :
RN  5 || (8  4  8)
20  5
 5 || 20 
 4
25
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Example 4.11
To find iN (Fig.4.40(b))
short  circuit terminals a and b .
Mesh : i1  2A, 20i2  4i1  i2  0
i2  1A  isc  IN
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Example 4.11
Alternativ e method for IN
VTh
IN 
RTh
VTh : open  circuit voltage across terminals a and b
( Fig 4.40(c)) :
Mesh analysis :
i 3  2 A, 25i 4  4i 3  12  0
 i 4  0.8A
voc  VTh  5i 4  4V
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Example 4.11
Hence ,
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VTh
IN 
 4 / 4  1A
RTh
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Example 4.12
 Using Norton’s theorem, find RN and IN of the
circuit in Fig 4.43 at terminals a-b.
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Example 4.12
To find RN Fig .4.44(a)
 4 resistor shorted
 5 || vo || 2ix : Parallel
Hence, ix  vo / 5  1 / 5  0.2
vo
1
 RN  
 5
io
0.2
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Example 4.12
To find I N
Fig .4.44(b)
 4 ||10v || 5|| 2ix : Parallel
10  0
ix 
 2.5A,
4
10
isc  ix  2 ix   2(2.5)  7 A
5
 I N  7A
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4.8 Maximum Power Trandfer
2
 VTH 
p  i RL  
 RL
 RTH  RL 
2
Fig 4.48
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Fig. 4.49
 Maximum power is transferred to the load when
the load resistance equals the Thevenin resistance
as seen the load (RL = RTH).
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dp
2  ( RTH  RL )  2 RL ( RTH  RL ) 
 VTH 
4

dRL
(
R

R
)


TH
L
2
 ( RTH  RL  2 RL ) 
V 
0
3

 ( RTH  RL )

0  ( RTH  RL  2 RL )  ( RTH  RL )
2
TH
RL  RTH
pmax
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VTH2

4 RTH
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Example 4.13
 Find the value of RL for maximum power transfer
in the circuit of Fig. 4.50. Find the maximum
power.
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Example 4.13
RTH
6  12
 2  3  6 12  5 
 9
18
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Example 4.13
 12  18i1  12i2 , i2  2 A
 12  6ii 1  3i2  2(0)  VTH  0  VTH  22V
RL  RTH  9
pmax
2
TH
2
V
22


 13.44W
4 RL 4  9
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Homework Problems
 Problems 6, 10, 21, 28, 33, 40, 47, 52, 71
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