EENG223: CIRCUIT THEORY I
DC Circuits:
Circuit Theorems
Hasan Demirel
EENG223: CIRCUIT THEORY I
Circuit Theorems
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Introduction
Linearity Property
Superposition
Source Transformations
Thevenin’s Theorem
Norton’s Theorem
Maximum Power Transfer
EENG223: CIRCUIT THEORY I
Introduction
A large
complex circuits
Simplify
circuit analysis
Circuit Theorems
‧Thevenin’s theorem
‧Circuit linearity
‧Source transformation
‧ Norton theorem
‧ Superposition
‧ Max. power transfer
EENG223: CIRCUIT THEORY I
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
EENG223: CIRCUIT THEORY I
Linearity Property
• A linear circuit is one whose output is linearly
related (directly proportional) to its input.
i
v
I0
V0
EENG223: CIRCUIT THEORY I
Linearity Property
• A linear circuit consist of :
• linear elements (i.e. R=5 W)
• linear dependent sources (i.e. vs=6Io V)
• independent sources (i.e. vs=12 V)
vs  10V  i  2A
vs  1V  i  0.2A
vs  5mV  i 1mA
2
v
p i R 
: nonlinear
R
2
EENG223: CIRCUIT THEORY I
Linearity Property
Example 4.1: For the circuit in below find Io when vs=12V
and vs =24V.
EENG223: CIRCUIT THEORY I
Linearity Property
Example 4.1: For the circuit in below find Io when vs=12V
and vs =24V.
KVL
12i1  4i2  vs  0
 4i1  16i2  3vx  vs  0
vx  2i1
(1)
(2)
Eq(2) becomes
 10i1  16i2  vs  0
Using Eqs(1) and (3) we get
2i1  12i2  0  i1  6i2
(3)
EENG223: CIRCUIT THEORY I
Linearity Property
Example 4.1: For the circuit in below find Io when vs=12V
and vs =24V.
vs
From Eq(1) we get:  76i2  vs  0  i2 
76
when vs  12V
12
I 0  i2  A
76
when vs  24V
24
I 0  i2  A
76
Showing that when the source value is doubled, Io doubles.
EENG223: CIRCUIT THEORY I
Linearity Property
Example 4.2: Assume Io = 1A and use linearity to find the
actual value of Io in the following circuit.
EENG223: CIRCUIT THEORY I
Linearity Property
Example 4.2: Assume Io = 1A and use linearity to find the
actual value of Io in the following circuit.
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
EENG223: CIRCUIT THEORY I
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, kill, inactive source(s):
• independent voltage source: 0 V (short circuit)
• independent current source: 0 A (open circuit)
• Dependent sources are left intact.
EENG223: CIRCUIT THEORY I
Superposition
• 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.
EENG223: CIRCUIT THEORY I
Superposition
• 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
• Superposition involves more work but simpler circuits.
• Superposition is not applicable to the effect on power.
EENG223: CIRCUIT THEORY I
Superposition
• Example 4.3: Use the superposition theorem to find v
in the circuit below.
EENG223: CIRCUIT THEORY I
Superposition
• Example 4.3: Use the superposition theorem to find v
in the circuit below.
Since there are two sources,
Let
v  v1  v2
Voltage division to get
4
v1 
(6)  2V
48
Current division, to get
8
i3 
(3)  2A
48
Hence v2  4i3  8V
And we find v  v1  v2  2  8  10V
EENG223: CIRCUIT THEORY I
Superposition
• Example 4.4: Find Io in the circuit below using
superposition.
EENG223: CIRCUIT THEORY I
Superposition
• Example 4.4: Find Io in the circuit below using
superposition.
Turn off 20V voltage source:
EENG223: CIRCUIT THEORY I
Superposition
• Example 4.4: Find Io in the circuit below using
superposition.
Turn off 4A current source:
EENG223: CIRCUIT THEORY I
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
vs
vs  is R or is 
R
EENG223: CIRCUIT THEORY I
Source Transformation
• Equivalent Circuits:
i
i
+
+
v  iR  vs
v
i
v
v vs

R R
-
i
-is
vs
v
EENG223: CIRCUIT THEORY I
Source Transformation
• Source transformation: Important points
1. Arrow of the current source is directed toward the positive
terminal of the voltage source.
2. Transformation is not possible when:
♦ ideal voltage source (R = 0)
♦ ideal current source (R=)
EENG223: CIRCUIT THEORY I
Source Transformation
• Example 4.6: Use source transformation to find vo in the
circuit below.
EENG223: CIRCUIT THEORY I
Source Transformation
• Example 4.6: Use source transformation to find vo in the
circuit below.
Fig.4.18
EENG223: CIRCUIT THEORY I
Source Transformation
• Example 4.6: Use source transformation to find vo in the
circuit below.
we use current division in Fig.4.18(c) to get
2
i
(2)  0.4A
28
and
vo  8i  8(0.4)  3.2V
EENG223: CIRCUIT THEORY I
Source Transformation
• Example 4.7: Find vx in the circuit below using source
transformation.
EENG223: CIRCUIT THEORY I
Source Transformation
• Example 4.7: Find vx in the circuit below using source
transformation.
KVL around the loop in Fig (b)
 3  5i  vx  18  0
(1)
Appling KVL to the loop containing only the 3V voltage
source, the 1W resistor, and vx yields:
 3  1i  vx  0  vx  3  i
(2)
EENG223: CIRCUIT THEORY I
Source Transformation
• Example 4.7: Find vx in the circuit below using source
transformation.
Substituting Eq.(2) into Eq.(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
EENG223: CIRCUIT THEORY I
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 VTh is the open circuit voltage at the
terminals and RTh is the input or equivalent resistance
at the terminals when the independent sources are
turned off.
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Property of Linear Circuits
i
i
Any two-terminal
Linear Circuits
+
v
Slope=1/RTh
-
v
Vth
Isc
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Replacing a linear two-terminal circuit (a) by its Thevenin equivalent
circuit (b).
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• How to find Thevenin Voltage?
 Equivalent circuit: same voltage-current relation at the
terminals.
 VTh  voc : open circuit voltage at a  b
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• How to find Thevenin Resistance?
RTh  Rin : input  resistance of the dead circuit at a  b.
 a  b open circuited
 Turn off all independent sources
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• How to find Thevenin Resistance?
 There are two cases in finding Thevenin Resistance RTh.
CASE 1
• If the network has no dependent sources:
● Turn off all independent sources.
● RTh: can be obtained via simplification of either parallel or
series connection seen from a-b
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• How to find Thevenin Resistance?
 There are two cases in finding Thevenin Resistance RTh.
CASE 2
• If the network has dependent sources:
● Turn off all independent sources.
● Apply a voltage source vo at a-b
v
RTh  o
io
● Alternatively, apply a current source io
vo
at a-b
RTh 
io
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• How to find Thevenin Resistance?
 The Thevenin resistance, RTh , may be negative,
indicating that the circuit has ability providing
(supplying) power.
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• After the Thevenin Equivalent is obtained, the simplified
circuit can be used to calculate IL and VL easily.
Simplified circuit
VTh
IL 
RTh  RL
RL
VL  RL I L 
VTh
RTh  RL
Voltage divider
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.8: Find the Thevenin equivalent circuit of the circuit
shown below, to the left of the terminals a-b. Then find the current
through RL = 6,16, and 36 W.
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.8: Find the Thevenin equivalent circuit of the circuit
shown below, to the left of the terminals a-b. Then find the current
through RL = 6,16, and 36 W.
Find RTh:
RTh : 32V voltage source  short
2A current source  open
4  12
RTh  4 || 12  1 
 1  4W
16
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.8: Find the Thevenin equivalent circuit of the circuit
shown below, to the left of the terminals a-b. Then find the current
through RL = 6,16, and 36 W.
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
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.8: Find the Thevenin equivalent circuit of the circuit
shown below, to the left of the terminals a-b. Then find the current
through RL = 6,16, and 36 W.
Find VTh: (3) Alternatively, source transform
32  VTH
VTH
2
4
12
96  3VTH  24  VTH  VTH  30V
Thevenin Equivalent circuit:
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.8: Find the Thevenin equivalent circuit of the circuit
shown below, to the left of the terminals a-b. Then find the current
through RL = 6,16, and 36 W.
Calculate IL:
iL 
VTh
30

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
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.9: Find the Thevenin equivalent of the circuit below at
terminals a-b.
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.9: Find the Thevenin equivalent of the circuit below at
terminals a-b.
• (independent + dependent sources case)
Find RTh:
Use Fig (a):
independen t source  0
dependent source  intact
vo 1
vo  1V, RTh  
io io
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.9: Find the Thevenin equivalent of the circuit below at
terminals a-b.
Find RTh:
For Loop 1:
 2vx  2(i1  i2 )  0 or vx  i1  i2
But v x  4i2  i1  i2
 i1  3i2
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.9: Find the Thevenin equivalent of the circuit below at
terminals a-b.
Find RTh:
For Loops 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 
 6W
io
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.9: Find the Thevenin equivalent of the circuit below at
terminals a-b.
Find VTh: Use Fig (b) : Mesh analysis
Loop 1:
i1  5 A
Loop 2: 4(i2  i1 )  2(i2  i3 )  6i2  0
12i2  2i3  20
Loop 3:  2v x  2(i3  i2 )  0
 2(4(i1  i2 ))  2(i3  i2 )  0
6i2  2i3  40
i  60 / 18
2
VTh  voc  6i2  20V
Thevenin Equivalent circuit:
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.10: Determine the Thevenin equivalent circuit in Fig (a).
Find VTh:
VTh  0
Find RTh:
Use Fig (b):
(dependent source only case)
Apply a current source io at a-b
vo
RTh 
io
Nodal anaysis :
io  ix  2ix  vo / 4
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.10: Determine the Thevenin equivalent circuit in Fig (a).
Find RTh:
But
Use Fig (b):
0  vo
vo
ix 

2
2
vo
vo vo
vo
io  ix      
4
2 4
4
or vo  4io
vo
Thus RTh   4W (Supplying power)
io
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.10: Determine the Thevenin equivalent circuit in Fig (a).
Input circuit:
Thevenin Equivalent circuit:
•
When RTh is negative, you must evaluate (check) if the two circuits are equivalent
or not!
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.10: Determine the Thevenin equivalent circuit in Fig (a).
Input circuit:
Check with RL (9W) and voltage source (10V)
Thevenin Equivalent circuit:
Check with RL (9W) and voltage source (10V)
EENG223: CIRCUIT THEORY I
Thevenin’s Theorem
• Example 4.10: Determine the Thevenin equivalent circuit in Fig (a).
Input circuit:
8ix  4i1  2(i1  i2 )  0
 2i1  6i2  0
, ix  i2  i1
, i1  3i2 (1)
2(i2  i1 )  9i2  10  0
5i2  10  i2  2A
Check with RL (9W) and voltage source (10V)
Thevenin Equivalent circuit:
 4i  9i  10  0
5i  10  i  2A
Load current in both circuits are equal.
So the Thevenin Equivalent circuit is OK.
Check with RL (9W) and voltage source (10V)
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Norton’s theorem states that a linear two-terminal
circuit can be replaced by an equivalent circuit
consisting of a current source IN in parallel with a
resistor RN .
• 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.
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Property of Linear Circuits
i
Slope=1/RN
v
Vth
-IN
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• How to find Norton Current
 Thevenin and Norton resistances are equal:
RN  RTh
 Short circuit current from a to b
gives the Norton current:
VTh
I N  isc 
RTh
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• 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
V
all independent source are turn
off.
VTh  voc
I N  isc
RTh
VTh

 RN
IN
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.11: Find the Norton equivalent circuit of the circuit in Fig 4.39.
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.11: Find the Norton equivalent circuit of the circuit in Fig 4.39.
Find RN:
Use Fig (a):
RN  5 || (8  4  8)
20  5
 5 || 20 
 4W
25
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.11: Find the Norton equivalent circuit of the circuit in Fig 4.39.
Find IN:
Use Fig (b):
(short circuit terminals a and b)
Mesh : i1  2 A,
20i2  4i1  12  0
i2  1A  isc  IN
(ignore 5W. Because it is short circuit)
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.11: Find the Norton equivalent circuit of the circuit in Fig 4.39.
Find IN:
VTh
Use Fig (c): Alternative Method IN 
RTh
VTh : (open circuit voltage accross terminals a and b)
Mesh analysis :
i 3  2 A, 25i 4  4i 3  12  0
 i 4  0.8A
voc  VTh  5i 4  4V
Hence,
VTh
IN 
 4 / 4  1A
RTh
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.11: Find the Norton equivalent circuit of the circuit in Fig 4.39.
Input circuit:
Norton’s Equivalent circuit:
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.12: Using Norton’s theorem, find RN and IN of the circuit in Fig
4.43 at terminals a-b.
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.12: Using Norton’s theorem, find RN and IN of the circuit in Fig
4.43 at terminals a-b.
Find RN:
Use Fig (a):
 4W resistor shorted
5W || vo || 2ix : Parallel
Hence,
ix  0A,
i0 
v0
1V

 0.2A
5W 5W
vo
1
 RN  
 5W
io
0.2
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.12: Using Norton’s theorem, find RN and IN of the circuit in Fig
4.43 at terminals a-b.
Find IN:
 4W ||10v || 5W || 2ix : Parallel
10  0
ix 
 2.5A,
4
10
isc  ix  2 ix   2(2.5)  7 A
5
 I N  7A
EENG223: CIRCUIT THEORY I
Norton’s Theorem
• Example 4.12: Using Norton’s theorem, find RN and IN of the circuit in Fig
4.43 at terminals a-b.
Input circuit:
Norton’s Equivalent circuit:
EENG223: CIRCUIT THEORY I
Maximum Power Transfer
• The Thevenin equivalent is useful in finding the
maximum power a linear circuit can deliver to a load.
2
 VTH 
p  i RL  
 RL
 RTH  RL 
2
EENG223: CIRCUIT THEORY I
Maximum Power Transfer
• Maximum power is transferred to the load when the
load resistance equals the Thevenin resistance as seen
from the load (RL = RTH).
2
 VTH 
p  i RL  
 RL
 RTH  RL 
2
EENG223: CIRCUIT THEORY I
Maximum Power Transfer
dp
0
dRL
• Pmax can be obtained when:
2

dp
(
R

R
)
 2 RL ( RTH  RL ) 
2
TH
L
 VTH 
0
2 2
dRL
(( RTH  RL ) )


2

(
R

R
)
 2 RL ( RTH  RL ) 
2
2  ( RTH  RL  2 RL ) 
TH
L
 VTH 
  VTH 
0
4
3
( RTH  RL )


 ( RTH  RL )

 ( RTH  RL  2 RL )  0
 ( RTH  RL )  0
RL  RTH
pmax
2
TH
V

4 RTH
EENG223: CIRCUIT THEORY I
Maximum Power Transfer
• Example 4.13: Find the value of RL for maximum power transfer in the
circuit of Fig. 4.50. Find the maximum power.
EENG223: CIRCUIT THEORY I
Maximum Power Transfer
• Example 4.13: Find the value of RL for maximum power transfer in the
circuit of Fig. 4.50. Find the maximum power.
Find RTh:
RTH
6  12
 2  3  6 12  5 
 9W
18
EENG223: CIRCUIT THEORY I
Maximum Power Transfer
• Example 4.13: Find the value of RL for maximum power transfer in the
circuit of Fig. 4.50. Find the maximum power.
Find VTh:
KVL around the outer loop:
2
A
3
 0  VTH  22V
 12  18i1  12i2 , i2  2 A, i2  
 12  6i1  3i2  2(0)  VTH
RL  RTH  9W  pmax
VTH2
22 2


 13.44 W
4 RL 4  9