Terminology (1/4)
Fig. 4.3: A circuit illustrating nodes, branches, meshes, paths
and loops
R1
a
Principles of Electrical
Engineering - I (14:332:221)
v1
+
−
b
i1
R2
c
Chapter 4 Notes (Part 1)
v2
R5
R3
d
e
i3
+
−
R7
i6
I
i4
R6
R4
f
i2
g
i3
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Terminology (3/4)
Terminology (2/4)
• Node: A point where two or more circuit elements join, e.g., node a
• Essential node: A node where three or more circuit elements join
e.g., node b
• Path: A trace of adjoining basic elements with no elements included
more than once
e.g., (v1, R1, R5, R6)
R1
a
• Branch: A path that connects two nodes
e.g., R1
• Essential branch: A path which connects two essential nodes
without passing through an essential node
e.g., (v1, R1)
• Loop: A path whose last node is the same as the starting node
e.g., (v1, R1, R5, R6, R4, v2)
b
R1
a
b
i1
+
−
v1
2
R2
R3
d
c
R5
e
i3
+
v2 −
i2
R6
R4
v1
R7
i6
I
c
i4
v2 −
g
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3
Terminology (4/4)
R1
a
R2
d
+
−
v2
R3
R5
e
R4
I
R4
R6
R7
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4
• Solving a circuit with respect to be essential
branches is easier
– be ≤ b
– If there are ne essential nodes in the circuit, ne−1
equations may be derived from KCL
– The remaining be−(ne−1) equations must be derived
from KVL
I
R6
f
14:332:221, Spring 2004
– If there are n nodes in the circuit, n−1 equations may
be derived from KCL
– The remaining b−(n−1) equations must be derived
from KVL
R5
e
g
• A circuit with b branches requires b independent
equations to solve
b
R3
R7
Simultaneous Equations
• Mesh: A loop that does not enclose any other loops
e.g., (v1, R1, R5, R3, R2)
• Planar circuit: A circuit that can be drawn on a a plane with no
crossing branches
e.g., Fig. 4.3 is a planar circuit while Fig. 4.2 is a non-planar
circuit
c
d
f
i3
v1
R2
+
f
+
−
+
−
g
5
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6
1
Node-Voltage Method (2/6)
Node-Voltage Method (1/6)
4 essential nodes (b, c, e, f)
Select node f as the reference node
Must solve for vb, vc and ve
• A technique to simplify analysis, i.e., reduces the
number of simultaneous equations
– Analysis is done using essential nodes
– Given ne essential nodes, node-voltage method
requires ne−1 equations to describe the circuit
R1
a
v1
• One of the essential nodes is chosen as the
reference node (usually the node that connects
the most branches)
• Once reference node is selected, node
voltages are determined
R2
c
v2
b
i1
+
−
R4
R3
d
e
i3
+
−
i2
R6
f
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7
Node-Voltage Method (3/6)
14:332:221, Spring 2004
vb vb − ve vb − v1 − v2
+
+
−I =0
R6
R4
R1
v
v − v2
KCL at node e : 3 + e
=0
R5 R2 + R3
KCL at node b :
– (e.g., node c: vc = v2)
– Reduces circuit to only two unknowns (vb and ve)
R1
R1
b
+
−
c
v2
b
i1
R2
R3
R4
e
i3
+
−
R5
v1
i2
i5
R6
vc = v2
I
c
v2
i4
+
−
i1
R2
R3
R4
−
R5
9
Node-Voltage Method (5/6)
+
40 Ω
−
1
50 V
+
−
2
5Ω
40 Ω
iφ
I
i4
− +
3
100 Ω
50 Ω
2
4A
3
50 Ω
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14:332:221, Spring 2004
• KCL still holds at the supernode (2,3)
• Formation of the supernode allows v3 to be expressed in
terms of v2 → Reduces circuit to a single unknown (v2)
v3 = v 2 + 10iφ
v − 50 v 2 v 2 + 10iφ
+
+
−4=0
KCL at supernode (2,3) :
5
50
100
v −v
v − 50
where iφ = 2 1 = 2
5
5
1
100 Ω
10
Node-Voltage Method (6/6)
• When a voltage source is the only element between two
essential nodes, these nodes can be combined to form a
supernode
10iφ
5Ω
iφ
i5
R6
f
14:332:221, Spring 2004
1
i2
e
i3
+
f
50 V
8
Node-Voltage Method (4/6)
• If a voltage source is the only element on a branch
connecting the reference node and a non-reference
node, then this greatly simplifies the circuit analysis
v1
I
i4
R5
A node voltage is defined as the voltage rise from
the reference node to a non-reference node
i5
4A
50 V
11
+
−
5Ω
40 Ω
iφ
2
3
50 Ω
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100 Ω
4A
12
2
Mesh Current Method (1/2)
Mesh Current Method (2/2)
R1
• Mesh-current method allows a circuit to be solved
with be−(ne−1) independent equations
– be ≡ Number of essential branches
– ne ≡ Number of essential nodes
v1
i1
+
−
• Mesh-current method applies only to planar
circuits
• A mesh current is the current that exists only in
the perimeter of a mesh
Effectively, a virtual current that may not actually exist
in any a particular circuit branch (e.g., i2 in Fig. 4.18)
i3
v1
+
−
R3
ia
+
−
ia
5Ω
ic
14
⇒ 50 = 9i a − 5ib + 6ic
ib
3Ω
100 V
50 V
ia
+
−
ic
5Ω
50 V
4Ω
15
Node-Voltage v.s. Mesh-Current (1/4)
Combine with KVL
around mesh b
2Ω
and ic = 5 + ia
+
−
4Ω
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i1 = ia

⇒ i2 = ib
i = i − i
2 b a
− 100 + 3(ia − ib ) + 2(ic − ib ) + 50 + 4ic + 6ia = 0
ic = 5 + i a
+
−
v2
Supermeshes (2/2)
but v is unknown.
5A
−
ib
10 Ω
+
v
+
−
• To create a supermesh, remove the current source by
avoiding its branch when writing mesh current equations
10 Ω
100 V
− v 2 = −ia R3 + ib (R2 + R3 )
R2
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KVL around mesh b : 0 = 10ib + 2(ib − ic ) + 3(i − ia )
2Ω
v2
i1 = ?

i2 = ?
i = ?
2
v1 = ia (R1 + R3 ) − ib R3
Supermeshes (1/2)
ib
+
−
R3
13
• When a branch contains a current source, an unknown
voltage is introduced to the mesh equations
3Ω
i2
R1
• Apply KVL around be−(ne−1) meshes
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R2
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16
Node-Voltage v.s. Mesh-Current (2/4)
300 Ω
Which method to use depends on a number of
circuit issues:
150 Ω
– Does one method result in fewer simultaneous
equations?
– Does the circuit contain supernodes?
– Does the circuit contain supermeshes?
– Will solving some portion of the circuit give the
requested solution?
– Is the circuit non-planar?
14:332:221, Spring 2004
Fig. 4.29
i∆
256 V
+
−
1
100 Ω
200 Ω
2
250 Ω
+ 50i
∆
−
3
400 Ω
+ 128 V
−
Mesh-current: Must solve 5 loop equations (plus constraint for i∆)
Node-voltage: Need only 2 node equations plus 1 constraint
equation for v2:
50 ⋅ (v3 − v1 ) v3 − v1
v2 = 50i ∆ =
=
300
6
17
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18
3
Node-Voltage v.s. Mesh-Current (3/4)
4Ω
193 V
+ v −
∆
+
+
−
6Ω
v0
0.4v∆
−
15 Ω
+ 0.8v
θ
−
0.5 A
7.5 Ω
Mesh-current or node-voltage?
Fig. 4.33
2Ω
2.5 Ω
Node-Voltage v.s. Mesh-Current (4/4)
8Ω
10 Ω
+
20 V
+ v −
θ
Mesh-current: Current sources reduces problem to a single
mesh equation and 3 constraint equations
4Ω
256 V
14:332:221, Spring 2004
ix
+
1 mesh-current equation
and 1 constraint equation
R
is
R
vs
R
vs
RL
a
+
−
RL
b
Resistor in parallel with vs does not change voltage va,b
ia
R
b
ia
a
RL
a
R
is
RL
b
Resistor in series with is does not change current ia
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21
Source Transformations (3/4)
b
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22
Source Transformations (4/4)
Exercise 4.19 (continued)
1.6 Ω
b
20
a
Rp
is
b
b
+
−
Rs
RL
iL
Reduces to 2 mesh-current
equations and 2 constraint
equations
+ 50i
x
−
8Ω
Source Transformations (2/4)
a
a
RL
4
3 node-voltage equations
and 1 constraint equation
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• A voltage source in series with a resistor may be
replaced by a current source in parallel with the
same resistor, or, visa versa
vs
R
v
iL =
and iL =
is
⇒ is = s
R + RL
R + RL
R
Exercise 4.19
25 V
3
6Ω
−
19
Source Transformations (1/4)
iL
3Ω
2
2Ω
+
−
−
4A
1
vs
+
2A
−
Node-voltage: Requires 3 node-voltage equations and 2
constraint equations (due to voltage-controlled sources)
R
1 node-voltage equation
1.6 Ω
b
+
− 60 V
20 Ω
120 V
+
+
36 A
+
6Ω
5Ω
−
6A
v
20 Ω
5Ω
6Ω
v
8Ω
−
1.6 Ω
b
b
+
v
8Ω
−
30 A
2.4 Ω
a
a
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6Ω
a
+
36 A
36 A
5Ω
8Ω
1.6 Ω
b
12 A
12 A
−
a
6A
20 Ω
v
8Ω
30 A
−
1.92 Ω
a
⇒ vb,a = (30)(1.92) = 57.6 V ⇒ v = 48 V
23
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24
4