Lecture #7
OUTLINE
• Node-Voltage Method (cont’d)
– circuits with dependent sources
• Source Transformations
• Mesh-Current circuit analysis
– method
– circuit with a current source
Reading
Chapter 4.3-4.9
EECS40, Fall 2003
Lecture 7, Slide 1
Prof. King
Equivalent Circuit Concept
• A network of voltage sources, current sources,
and resistors can be replaced by an
equivalent circuit which has identical terminal
properties (I-V characteristics) without
affecting the operation of the rest of the circuit.
iA
network A
of
sources
and
resistors
iB
+
vA
_
≡
network B
of
sources
and
resistors
+
vB
_
iA(vA) = iB(vB)
EECS40, Fall 2003
Lecture 7, Slide 2
Prof. King
1
Source Combinations
• Voltage sources in series can be replaced by an
equivalent voltage source:
≡
v1+v2
–
+
–
+
v2
–
+
v1
• Current sources in parallel can be replaced by
an equivalent current source:
i1
i2
EECS40, Fall 2003
≡
i1+i2
Lecture 7, Slide 3
Prof. King
Circuit Analysis Approaches
• The Node-Voltage method can always be
used to solve a circuit, but techniques for
simplifying circuits (using “equivalent
circuits”) are useful:
– series and parallel combination reductions
– ∆-Y and Y-∆ conversions
– source transformations
– Thevenin and Norton equivalent circuits
(to be covered in Lecture 8)
EECS40, Fall 2003
Lecture 7, Slide 4
Prof. King
2
A Note of Caution
• These two resistive circuits are equivalent for
voltages and currents external to the Y and ∆
circuits. Internally, the voltages and currents
are different.
a
Rc
a
b
b
R1
R2
Ra
Rb
R3
c
R1 =
c
RbRc
R2 =
Ra + Rb + Rc
EECS40, Fall 2003
RaRc
Ra + Rb + Rc
R3 =
RaRb
Ra + Rb + Rc
Lecture 7, Slide 5
Prof. King
Circuit Simplification Example
Find the equivalent resistance Rab:
2Ω
2Ω
a
a
18Ω
b
9Ω
6Ω
12Ω
≡
4Ω
b
EECS40, Fall 2003
Lecture 7, Slide 6
9Ω
4Ω
Prof. King
3
Node-Voltage Method and Dependent Sources
• If a circuit contains dependent sources, the node-voltage
equations must be supplemented with equations describing
the constraints imposed by the dependent sources.
Example:
i∆
20 Ω
10 Ω
EECS40, Fall 2003
–
+
200 Ω
2.4 A
–
+ 5i∆
80 V
Lecture 7, Slide 7
Prof. King
Source Transformations
• Consider the following two circuits:
R
–
+
vs
a
a
+
iL
vL
RL
is
R
–
+
iL
vL
RL
–
b
b
• We can exchange the series combination of vs and R
for the parallel combination of is and R, provided that
each produces exactly the same voltage and current
for any load RL
– NOTE THE POLARITY OF SOURCES
EECS40, Fall 2003
Lecture 7, Slide 8
Prof. King
4
Derivation of Relationship between vs and is
R
–
+
vs
a
a
+
iL
vL
RL
is
R
–
iL
vL
RL
–
b
iL =
+
b
vs
R + RL
iL =
R
is
R + RL
vs = Ris
EECS40, Fall 2003
Lecture 7, Slide 9
Prof. King
Source Transformation Example
6Ω
• Find Io
EECS40, Fall 2003
–
+
60 V
3Ω
Io
3Ω
Lecture 7, Slide 10
– 15 V
+
Prof. King
5
Formal Circuit Analysis Methods
MESH* ANALYSIS
(“Mesh-Current Method”)
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
1) Select M 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
*A mesh is a loop that does not enclose any other loops.
A mesh current is not necessarily identified with a branch current.
EECS40, Fall 2003
Lecture 7, Slide 11
Prof. King
Mesh Analysis: Example #1
1. Select M mesh currents.
2. Apply KVL to each mesh.
3. Solve for mesh currents.
EECS40, Fall 2003
Lecture 7, Slide 12
Prof. King
6
Mesh Analysis with a Current Source
ia
ib
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.
EECS40, Fall 2003
Lecture 7, Slide 13
Prof. King
Mesh Analysis: Example #2
ia
ib
Eq’n 1: KVL for supermesh
Eq’n 2: Constraint due to current source:
EECS40, Fall 2003
Lecture 7, Slide 14
Prof. King
7