Topic 1 b_basic_concepts_and_theorem

1
Topic 1B
Basic Concept and Theorem

2
Equivalent Circuits – The Concept
Equivalent circuits are ways of
looking at or solving circuits. The
idea is that if we can make a
circuit simpler, we can make it
easier to solve, and easier to
understand.
The key is to use equivalent
circuits properly. After defining
equivalent circuits, we will start
with the simplest equivalent
circuits, series and parallel
combinations of resistors.

3
Equivalent Circuits:
A Definition
Imagine that we have a circuit, and a portion of the circuit can be
identified, made up of one or more parts. That portion can be replaced with
another set of components, if we do it properly. We call these portions
equivalent circuits.
Two circuits are considered
to be equivalent if they behave
the same with respect to the
things to which they are
connected. One can replace
one circuit with another circuit,
and everything else cannot
tell the difference.
We will use an analogy for equivalent circuits here. This analogy is that of jigsaw puzzle
pieces. The idea is that two different jigsaw puzzle pieces with the same shape can be thought
of as equivalent, even though they are different. The rest of the puzzle does not “notice” a
difference. This is analogous to the case with equivalent circuits.

4
Equivalent Circuits:
Defined in Terms of Terminal Properties
Two circuits are considered to be
equivalent if they behave the same with
respect to the things to which they are
connected. One can replace
one circuit with another circuit, and
everything else cannot tell the difference.
We often talk about equivalent circuits as
being equivalent in terms of terminal
properties. The properties (voltage, current,
power) within the circuit may be different.

5
Equivalent Circuits: A Caution
Two circuits are considered to be equivalent if
they behave the same with respect to the
things to which they are connected. The properties
(voltage, current, power) within the circuit may be
different.
It is important to keep this concept in mind. A
common error for beginners is to assume that
voltages or currents within a pair of equivalent
circuits are equal. They may not be. These voltages
and currents are only required to be equal if they can
be identified outside the equivalent circuit. This will
become clearer as we see the examples that follow in
the other parts of this module.

6
How do we use equivalent circuits?
• We will use these equivalents to simplify circuits,
making them easier to solve. In some cases, one
equivalent circuit is not simpler than another; rather
one of them fits the needs of the particular circuit
better. The delta-to-wye transformations that we
cover next fit in this category. In yet other cases,
we will have equivalent circuits for things that we
would not otherwise be able to solve. For example,
we will have equivalent circuits for devices such as
diodes and transistors, that allow us to solve
circuits that include these devices.
• Equivalent circuits are equivalent
only with respect to the circuit outside them.

7
Delta-to-Wye Transformations
• The transformations, or equivalent circuits,
that we cover next are called delta-to-wye, or
wye-to-delta transformations. They are also
sometimes called pi-to-tee or tee-to-pi
transformations.
• These are equivalent circuit pairs. They
apply for parts of circuits that have three
terminals. Each version of the equivalent
circuit has three resistors.

8
Three resistors in a part of a circuit with three
terminals can be replaced with another version, also
with three resistors. Note that none of these resistors is
in series with any other resistor, nor in parallel with
any other resistor. The three terminals in this example
are labeled A, B, and C.
RC
RA
RB
A
C
B
R2
R3
R1
A B
C
Rest of CircuitRest of Circuit

9
The version on the left hand side is called the delta
connection, for the Greek letter ∆. The version on the
right hand side is called the wye connection, for the letter
Y. The delta connection is also called the pi (π)
connection, and the wye interconnection is also called
the tee (T) connection. All these names come from the
shapes of the drawings.
RC
RA
RB
A
C
B
R2
R3
R1
A B
C

10
Delta-to-Wye Transformation Equations
When we perform the delta-to-wye transformation
(going from left to right) we use the equations given below.
RC
RARB
A
C
B
R2
R3
R1
A B
C
1
2
3
B C
A B C
A C
A B C
A B
A B C
R R
R
R R R
R R
R
R R R
R R
R
R R R
=
+ +
=
+ +
=
+ +

11
Wye-to-Delta Transformation
Equations
RC
RA
RB
A
C
B
R2
R3
R1
A B
C
1 2 2 3 1 3
1
1 2 2 3 1 3
2
1 2 2 3 1 3
3
A
B
C
R R R R R R
R
R
R R R R R R
R
R
R R R R R R
R
R
+ +
=
+ +
=
+ +
=
Perform the Wye-to-Delta transformation

12
Equation 1
We can calculate the equivalent resistance between terminals A and B, when
C is not connected anywhere. The two cases are shown below. This is the same as
connecting an ohmmeter, which measures resistance, between terminals A and B,
while terminal C is left disconnected.
1 2 1 2
1 2
Ohmmeter #1 reads || ( ). Ohmmeter #2 reads .
These must read the same value, so || ( ) .
EQ C A B EQ
C A B
R R R R R R R
R R R R R
= + = +
+ = +
RC
RARB
A
C
B
R2
R3
R1
A B
C
Ohmmeter #1 Ohmmeter #2

13
Equations 2 and 3
So, the equation that results from the first situation is
1 2|| ( ) .C A BR R R R R+ = +
RC
RARB
A
C
B
R2
R3
R1
A B
C
Ohmmeter #1 Ohmmeter #2
We can make this measurement two other ways, and get two more equations.
Specifically, we can measure the resistance between A and C, with B left open,
and we can measure the resistance between B and C, with A left open.

14
All Three Equations
The three equations we can obtain are
1 2
1 3
2 3
|| ( ) ,
|| ( ) , and
|| ( ) .
C A B
B A C
A B C
R R R R R
R R R R R
R R R R R
+ = +
+ = +
+ = +
This is all that we need. These three equations can be
manipulated algebraically to obtain either the set of equations
for the delta-to-wye transformation (by solving for R1, R2 , and
R3), or the set of equations for the wye-to-delta transformation
(by solving for RA, RB , and RC).

15
Why Are Delta-to-Wye
Transformations Needed?
• They are like many other aspects of circuit analysis
in that they allow us to solve circuits more quickly
and more easily. They are used in cases where the
resistors are neither in series nor parallel, so to
simplify the circuit requires something more.
• One key in applying these equivalents is to get the
proper resistors in the proper place in the
equivalents and equations. We recommend that
you name the terminals each time, on the
circuit diagrams, to help you get these
things in the right places.

16
Example
Convert the Δ network to an equivalent Y network.

17
Solution
Ω=
++
×
=
++
= 5
151025
1025
1
CBA
CB
RRR
RR
R
Ω=
++
×
=
++
= 5.7
151025
1525
2
CBA
AC
RRR
RR
R
Ω=
++
×
=
++
= 3
151025
1015
3
CBA
BA
RRR
RR
R

18
Example
Transform the wye network into the delta network.
A B
C
R1=10Ω R2=20Ω
R3=40Ω

19
Solution
Ω=
×+×+×
=
++
=
70
20
401040202010
2
313221
R
RRRRRR
RB
Ω=
×+×+×
=
++
=
140
10
401040202010
1
313221
R
RRRRRR
RA
Ω=
×+×+×
=
++
=
35
40
401040202010
3
313221
R
RRRRRR
RC

20
1. Choose a reference node (“ground”)
Look for the one with the most connections!
2. Define unknown node voltages
those which are not fixed by voltage sources
3. Write KCL at each unknown node, expressing
current in terms of the node voltages (using the I-V
relationships of branch elements)
Special cases: floating voltage sources
4. Solve the set of independent equations
N equations for N unknown node voltages
Node-Voltage Circuit Analysis Method

21
Example
Calculate the node voltages in the circuit shown.

22
Solution
• Labeling
the current.

23
Solution
• At node 1, applying KCL and Ohm’s law gives
• At node 2, applying KCL and Ohm’s law gives
• Solve the simultaneous equation gives v1=
13.33V and v2 = 20V.
21
121
321
320
2
0
4
5
vv
vvv
iii
−=
−
+
−
=⇒+=
21
221
5142
5360
6
0
510
4
vv
vvv
iiii
+−=
−
+=+
−
⇒+=+

24
A “floating” voltage source is one for which neither side is
connected to the reference node, e.g. VLL in the circuit below:
Problem: We cannot write KCL at nodes a or b because
there is no way to express the current through the voltage
source in terms of Va-Vb.
Solution: Define a “supernode” – that chunk of the circuit
containing nodes a and b. Express KCL for this supernode.
Incorporate voltage source constraint into KCL equation.
R4R2 I2
Va Vb
+-
VLL
I1
Nodal Analysis w/ “Floating Voltage Source”

25
supernode
R4R2 I2
Va Vb
+-
VLL
I1
Nodal Analysis

26
Example
For the circuit shown, find the nodes voltages.

27
Solution
• The supernode contains the 2-V source, node 1 and
2, and the 10-Ω resistor.

28
Solution
• Applying KCL to the supernode, gives
• Applying KVL
• Solve , gives v1 = -7.333 V and v2 = -5.333V
202
7
4
0
2
0
2
72
21
21
21
−=+
+
−
+
−
=
++=
vv
vv
ii
12 2 vv +=

29
Node-Voltage Method and Dependent Sources
• If a circuit contains dependent sources, what to do?
Example:
–
+
–
+
80 V
5i∆
20 Ω
10 Ω
200 Ω2.4 A
i∆

30
Node-Voltage Method and Dependent Sources
• Dependent current source: treat as independent current
source in organizing and writing node eqns, but include
(substitute) constraining dependency in terms of
defined node voltages.
• Dependent voltage source: treat as independent voltage
source in organizing and writing node eqns, but include
(substitute) constraining dependency in terms of
defined node voltages.

31
–
+
–
+
80 V
5i∆
20 Ω
10 Ω
200 Ω2.4 A
i∆
Example:

32
Mesh analysis
• A loop is a closed path with no node passed more
than once.
• A mesh is a loop that does not contain any other
loop within it.
• Mesh analysis applies KVL to find unknown
current.
• Mesh analysis apply to planar circuit.
• A planar circuit is one that can be drawn in a plane
with no branches crossing one another.

33
NODAL ANALYSIS
(“Node-Voltage Method”)
0) Choose a reference node
1) Define unknown node voltages
2) Apply KCL to each unknown
node, expressing current in
terms of the node voltages
=> N equations for
N unknown node voltages
3) Solve for node voltages
=> determine branch currents
MESH ANALYSIS
(“Mesh-Current Method”)
1) Select M independent mesh
currents such that at least one
mesh current passes through each
branch*
M = #branches - #nodes + 1
2) Apply KVL to each mesh,
expressing voltages in terms of
mesh currents
=> M equations for
M unknown mesh currents
3) Solve for mesh currents
=> determine node voltages
Formal Circuit Analysis Methods
*Simple method for planar circuits
A mesh current is not necessarily identified with a branch current.

34
1. Select M mesh currents.
2. Apply KVL to each mesh.
3. Solve for mesh currents.
Mesh Analysis:

35
Example
• For the circuit shown, find the branch
currents I1, I2 and I3 using mesh analysis.

36
Solution
For mesh 1, using KVL,
For mesh 2, using KVL,
Solve, i1 = 1A, i2 = 1A
I1= 1 A, I2=1 A, I3= 0A.
123
010)(10515
21
211
=−
=+−++−
ii
iii
12
046)(1010
21
2212
=+−
=++−+−
ii
iiii

37
Problem: We cannot write KVL for meshes a and b
because there is no way to express the voltage drop
across the current source in terms of the mesh currents.
Solution: Define a “supermesh” – a mesh which avoids the
branch containing the current source. Apply KVL for this
supermesh.
Mesh Analysis with a Current Source
ia ib

38
Eq’n 1: KVL for supermesh
Eq’n 2: Constraint due to current source:
ia ib

39
Example
For the circuit, find the current i1 and i2.

41
Solution
• Apply KVL to the supermesh,
• Apply KCL to a node in the branch where the
two meshes intersect.
• Solving, i1= -3.2 A and i2 = 2.8 A.
20146
0410620
21
221
=+
=+++−
ii
iii
612 += ii

42
Mesh Analysis with Dependent
Sources
• Exactly analogous to Node Analysis
• Dependent Voltage Source: (1) Formulate
and write KVL mesh eqns. (2) Include and
express dependency constraint in terms of
mesh currents
• Dependent Current Source: (1) Use
supermesh. (2) Include and express
dependency constraint in terms of mesh
currents

43
What did you learn from you
notes?
• That all circuits cannot be solved with
simple approaches and more formal
methods are sometimes needed
• Superposition is one such theorem
• You now can use node or mesh or
superposition to solve the circuits

44
Sample Problem
200 V 50 V
80 Ohms 40 Ohms
80 Ohms
Find the voltage across the resistor in black using Superposition

45
Kill Source 1
50 V
80 Ohms 40 Ohms
80 Ohms
Short
circuit

46
Solve for V across Rblack
• You can do this by combining the resistors
in parallel (40 Ohms) and then using the
voltage divider equation
• Answer 25 Volts

47
Kill Source 2
200 V
80 Ohms 40 Ohms
80 Ohms
Short
circuit

48
Solve for V across Rblack
• You can do this by combining the resistors
in parallel (26.6) and then using the voltage
divider equation
• Answer ≈ 50 Volts

49
Combine 2 Answers
• Using Superposition
• Answer = 25 + 50 = 75 Volts

50
Example
12 A 100 Ohms
20 V
100 Ohms
50 Ohms
Find the voltage across the resistor in black .

51
Partial Solution
• Kill each source individually
• Killing current => 10 V across it
• Killing voltage=> 3 amps flows through
100 Ohms => 300 V across it
• A total of 310 V across it

52
Superposition Example 1 (1/4)
Find I1, I2 and Vab by superposition

53
Step 1: Omit current source.
By Ohm’s law and the
voltage divider rule:
11 21 0.4I I A= =

54
Step 2: Omit voltage source.
By the current divider
rule and Ohm’s law :
12
2
1.0
20ab
I A
V V
= −
=

55
Combining steps
1 & 2, we get:
1 11 12
1 2
0.6
22ab ab ab
I I I A
V V V V
= + = −
= + =

56
16 A 6 Ω
16 Ω
16 A 10 Ω
64 V
Ix
Find Ix by superposition

57
16 A 6 Ω
16 Ω
10 Ω
Ixa
16 A
Activate only the 16 A Current source at the left. Then
use Current Divider Rule:
10 16
16 13
6 10 16
xaI A A
Ω + Ω
= =
Ω + Ω + Ω

58
6 Ω
16 Ω
16 A 10 Ω
Ixb
16 A
Activate only the 16 A Current source at the right. Then
use Current Divider Rule:
10
16 5
6 10 16
xbI A A
Ω
= − = −
Ω + Ω + Ω

59
6 Ω
16 Ω
10 Ω
64 V
Ixc
16 A16 A
Activate only the 64 V voltage source at the bottom. Then
use Ohm’s Law:
64
2
6 10 16
xc
V
I A= − = −
Ω + Ω + Ω

60
16 A 6 Ω
16 Ω
16 A 10 Ω
64 V
Ix
Sum the partial currents due to each of the sources:
13 5 2x xa xb xcI I I I A A A= + + = − − =

61
• Solve for Vab by Superposition method
• Dependent source must remain in circuit for
both steps

62
1 1 30
0 1.5
4 2
ab ab
x
V V
i
−
= − +
1 1
1 1 1 9
15 1.5
4 4 2 8
ab abV V
 − 
= × − + = ÷ 
  
1
40
3
abV V=

63
2 2 12
0 1.5
2 4
ab ab
x
V V
i
−
= − +
2 2
1 1.5 1 9
7.5
2 4 4 8
ab abV V
 
= × + + = ÷
 
2
20
3
abV =

64
Combining the solutions:
1 2
40 20
3 3
ab ab ab
ab
ab
V V V
V V V
V
= +
= +
=

65
Thevenin’s Theorem
• Any circuit with sources (dependent and/or
independent) and resistors can be replaced
by an equivalent circuit containing a single
voltage source and a single resistor.
• Thevenin’s theorem implies that we can
replace arbitrarily complicated networks
with simple networks for purposes of
analysis.

66
Implications
• We use Thevenin’s theorem to justify the
concept of input and output resistance for
amplifier circuits.
• We model transducers as equivalent sources
and resistances.
• We model stereo speakers as an equivalent
resistance.

67
Independent Sources (Thevenin)
Circuit with
independent sources
RTh
Voc
Thevenin equivalent
circuit
+
–

68
No Independent Sources
Circuit without
independent sources
RTh
Thevenin equivalent
circuit

69
Example: CE Amplifier
1kΩ
Vin
2kΩ
+10V
+
–
Vo
+
–

70
Small Signal Equivalent
1kΩ
Vin 100Ib
+
–
Vo
50Ω
Ib
2kΩ
+
–

71
Thevenin Equivalent @ Output
1kΩ
Vin 100Ib
+
-
Vo
50Ω
Ib
2kΩ
RTh
Voc
+
–
Vo
+
–
+
–

72
Computing Thevenin Equivalent
• Basic steps to determining Thevenin
equivalent are
– Find voc
– Find RTh(= voc /isc)

73
Example of Thevenin’s Theorem
Find the Thevenin equivalent of the circuit to the
left of a-b. Then find the current through RL =
6 Ω, 16Ω and 36Ω.

74
Solution
• Find RTh by turning off the 32 V voltage (replacing it
with a short circuit) and the 2A current source
(replacing it with an open circuit). The circuit
becomes what is shown.
Ω=+
×
=
+=
41
16
124
112//4ThR

75
.30)25.0(12)(12
.5.0,iforSolving
.2i.0)(12432
21
1i
2211
ViiV
Ai
Aiii
Th =+=−=
=
−==−++−
• To find VTh, consider the circuit. Applying mesh
analysis to the two loops.
Solution

76
Solution
• The Thevenin equivalent circuit is shown.
When RL = 6Ω, IL = 30/10 = 3A
When RL = 16Ω, IL = 30/20 = 1.5A
When RL = 36Ω, IL = 30/40 = 0.75A

77
Thevenin/Norton Analysis
1. Pick a good breaking point in the circuit (cannot split a
dependent source and its control variable).
2. Thevenin: Compute the open circuit voltage, VOC.
Norton: Compute the short circuit current, ISC.
For case 3(b) both VOC=0 and ISC=0 [so skip step 2]

78
3. Compute the Thevenin equivalent resistance, RTh.
(a) If there are only independent sources, then short
circuit all the voltage sources and open circuit the current
sources (just like superposition).
(b) If there are only dependent sources, then must use a
test voltage or current source in order to calculate
RTh = VTest/Itest
(c) If there are both independent and dependent sources,
then compute RTh from VOC/ISC.

79
4. Thevenin: Replace circuit with VOC in series with RTh.
Norton: Replace circuit with ISC in parallel with RTh.
Note: for 3(b) the equivalent network is merely RTh, that is,
no voltage (or current) source.
Only steps 2 & 4 differ from Thevenin & Norton!

80
Method 3: Thevenin and Norton
Equivalent Circuits
vTH= open circuit voltage at
terminal (a.k.a. port)
RTH= Resistance of the
network as seen from port
(Vm’s, In’s set to zero)
Any network of sources
and resistors will appear
to the circuit connected
to it as a voltage source
and a series resistance

81
Norton Equivalent Circuit
Any network of sources and
resistors will appear to the
circuit connected to it as a
current source and a parallel
resistance
Ed Norton – Bell Labs, 1898-1983

82
Calculation of RT and RN
• RT=RN ; same calculation (voltage and current sources set to zero)
• Remove the load.
• Set all sources to zero (‘kill’ the sources)
– Short voltage sources (replace with a wire)
– Open current sources (replace with a break)

83
Calculation of RT and RN continued
• Calculate equivalent resistance seen by the load

84
Calculation of VT
• Remove the load and calculate the open circuit voltage
SROC V
RR
R
VV
21
2
2
+
==
(Voltage Divider)

85
Example
• Use Thevenin’s theorem to calculate the current
through Resistor R6. (solution I=0.72A)

86
Exercise: Draw the Thevenin Equivalent
• To find RTH remove the load, kill the sources (short voltage
sources, break current sources) and find the equivalent
resistance.
• To find VTH Remove the load and calculate the open circuit
voltage

87
• To find RTH kill the sources (short voltage sources, break
current sources) and find the equivalent resistance.
voltage
VAB = 20 - (20Ω x 0.33amps) = 13.33V

88
• To find RTH kill the sources (short voltage sources, break
current sources) and find the equivalent resistance.
voltage

89
Calculation of IN
• Short the load and calculate the short circuit current
(R1+R2)i1 - R2iSC = vs
-R2i1 + (R2+R3)iSC = 0
(KCL at v)
(mesh analysis)
RN=RTH

91
Summary: Thevenin’s Theorem
• Any two-terminal linear circuit can be replaced with a
voltage source and a series resistor which will produce the
same effects at the terminals
• VTH is the open-circuit voltage VOC between the two terminals
of the circuit that the Thevenin generator is replacing
• RTH is the ratio of VOC to the short-circuit current ISC; In linear
circuits this is equivalent to “killing” the sources and
evaluating the resistance between the terminals. Voltage
sources are killed by shorting them, current sources are
killed by opening them.

92
Summary: Norton’s Theorem
• Any two-terminal linear circuit can be replaced
with a current source and a parallel resistor which
will produce the same effects at the terminals
• IN is the short-circuit current ISCof the circuit that
the Norton generator is replacing
• Again, RN is the ratio of VOC to the short-circuit
current ISC; In linear circuits this is equivalent to
“killing” the sources and evaluating the
resistance between the terminals. Voltage sources
are killed by shorting them, current sources are
killed by opening them.
• For a given circuit, RN= RTH

93
Maximum power transfer
• Maximum power is
transferred to the
load resistance
when the load
resistance equals
the Thevenin
resistance as seen
from the load.
P
RL
RTh0
Pmax

Topic 1 b_basic_concepts_and_theorem

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