Lecture #8
EGR 260 – Circuit Analysis
Reading Assignment: Sections 4.1-4.9 in Electric Circuits, 9th Edition by Nilsson
Node equations – Procedure
1) Label each node.
2) Select a node as the reference (or ground) node.
Note: It is generally easier to pick the ground adjacent to a voltage source.
3) If the circuit has no voltage sources, skip to step 4. Otherwise:
A) For any voltage source or group of voltage sources adjacent to the ground,
all node voltages adjacent to the sources can be determined so no KCL
equation will be required.
B) For any voltage source or group of voltage sources not adjacent to the
ground, a supernode is required (to be discussed later).
4) Write a KCL equation at each node not adjacent to a voltage source and not at
the ground node. (Also write a KCL equation for each supernode.) Express
resistor currents in terms of node voltages.
5) Solve the simultaneous KCL equations. In general, the number of equations
required is:
# Node Equations = #nodes - #voltage sources - 1
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Lecture #8
EGR 260 – Circuit Analysis
Example: Use node equations determine the current I.
I
30
10
100 V
+
_
50
_ +
50 V
_
+
20 V
40
2
Lecture #8
EGR 260 – Circuit Analysis
Circuits containing dependent sources
Key: Replace the control variable in terms of node voltages.
Example: Use node equations
determine the current I1 .
6I1
40
5
I1
12 A
10
50 V
20
3
Lecture #8
EGR 260 – Circuit Analysis
Supernodes
A supernode is a node formed by drawing a surface around a voltage source or
group of voltage sources that are not adjacent to the reference (this would be a node
if the voltage sources in the supernode were set to zero or shorted).
When supernodes are used, node equations are written as follows:
• Write a KCL equation at each node not adjacent to the reference or adjacent to a
voltage source
• Write one KCL equation for each supernode.
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Lecture #8
EGR 260 – Circuit Analysis
Illustration - Supernode:
• Note that it is not possible to
place the ground adjacent to
12 A
both voltage sources in the
circuit shown, so a
supernode is necessary.
• If the ground is placed at
node D (adjacent to the 50V
source) then a supernode is
12 A
needed around the 20V
source.
• Note that the voltage
difference between nodes A
and B is 20V, so: Supernode relationship:
A
+
B
_
C
5
20 V
+
_
20
10
50 V
D
20 V
_
+
A
Supernode
C
5
_+ 50 V
B
20
10
D
A single new node is formed
by shorting the source
VA - VB = 20
C
• Think of the supernode as
the node that would exist if 12 A
the voltage source were
replaced by a short (a wire).
5
20
10
_+
50 V
D
5
Lecture #8
EGR 260 – Circuit Analysis
Example: Analyze the circuit shown using node equations. Show that Pdel = Pabs .
A
+
B
_
5
20 V
12 A
20
10
C
+
_
50 V
D
6
EGR 260 – Circuit Analysis
Lecture #8
Example: Use node equations determine V and I. Is a supernode required?
50
75 V
+
_
+
V
_
20
40 V
25
+
_
50
I
40
100
7
Lecture #8
EGR 260 – Circuit Analysis
Mesh Equations
Mesh equations (or mesh analysis) are a set of simultaneous KVL equations.
Mesh equations have one restriction: Mesh analysis can only be used if a circuit is
planar. A circuit is planar if it could be drawn on a 2D surface with no crossovers.
Example:
Is the following circuit planar?
Example:
Is the following circuit planar?
R1
R8
R1
R2
V1
R5
R4
R6
R3
R7
V1
R5
R2
R4
R3
R9
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Lecture #8
EGR 260 – Circuit Analysis
Mesh – a minimal region in a circuit that is bounded by circuit elements. A more
informal definition is that a mesh is like a window pane in a window.
R8
R9
R10
4 window
panes
R2
R11
4 meshes
R7
R12
R13
V2
Window
Circuit
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Lecture #8
EGR 260 – Circuit Analysis
Mesh current – a current associated with a mesh. Mesh currents are generally drawn
all clockwise (CW) or all counter clockwise (CCW).
R8
R9
Example: Mesh currents IA, IB, IC, and ID
are shown in the circuit to the right.
IA
R10
R2
V2
IC
IB
R11
ID
R13
R7
R12
Component currents – note that a component current is made up of either one or two
mesh currents.
R8
R9
I3
Example: Define the component currents
I
R11
IA 1 R10 IB
I1, I2, I3, and I4 in terms of mesh currents.
R2
I1 =
I2 =
I3 =
I4 =
I2
R7
V2
IC
R12
ID
R13
I4
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Lecture #8
EGR 260 – Circuit Analysis
Expressing resistor voltages in terms of mesh currents:
Example: Define the resistor voltages
R8
_
V1, V2, and V3 in terms of mesh currents.
IA
V1 =
V2 =
V3 =
R9
R2
Vx
+
V1
_
R12
IC
R11
IB
R7
V3
+
_ V
2 +
R13
R10
ID
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Lecture #8
EGR 260 – Circuit Analysis
Mesh Equations – Procedure:
1) Be sure that the circuit is planar (redraw it if necessary).
2) Label the mesh currents (generally all CW or all CCW).
3) If the circuit contains any current sources on the outer edge, the corresponding
mesh currents are defined. If the circuit contains any internal current sources, a
supermesh is required (more information later).
4) Write a KVL equation in each mesh with no current sources and one KVL
equation around each supermesh. Express resistor voltages in terms of mesh
currents (see below).
5) Solve the equations simultaneously. In general, the number of mesh equations is:
# Mesh Equations = # meshes - # current sources
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Lecture #8
EGR 260 – Circuit Analysis
Example: Use mesh equations to determine the current I. Note that this problem was
analyzed in previous class using node equations and I was determined to be 1.92 A.
I
30
10
100 V
50
+
_
25
20
40
13
Lecture #8
EGR 260 – Circuit Analysis
Example: Use mesh equations to analyze the circuit shown below. Use the results
to determine V1, V2, I3, and the power absorbed by the 80 ohm resistor.
50
+
V2
80
_
+
_
80 V
I3
5A
+
70
V1
_
30
3A
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