Motivation
For the following circuit, how do we determine
(1) the voltage across each resistor, (2) current
through each resistor, (3) power generated by
each current source, etc.
Methods of Analysis
EEE110 Electric Circuits
Anawach Sangswang
Dept. of Electrical Engineering
KMUTT
What do we need to know to determine the
answer?
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Motivation
Nodal Analysis
We need to keep in mind that in electric
circuit, everything obeys the basic laws
A general procedure for analyzing circuits
using node voltages as the circuit variables
Kirchhoff’s current laws (KCL)
Kirchhoff’s voltage laws (KVL)
Ohm’s law
Example 1
How do we apply the laws?
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Nodal Analysis: Steps
Nodal Analysis: Steps
S1: Select a node as the reference node.
S2: Assign voltages v1, v2,…, vn-1 to the
remaining n-1 nodes. The voltages are
referenced with respect to the reference node.
S3: Apply KCL to each of the n-1 non-reference
nodes. Use Ohm’s law to express the branch
currents in terms of node voltages.
S4: Solve the resulting simultaneous equations
to obtain the unknown node voltages.
Determine v1 and v2
Step 1: Pick the reference node
Step 2: Assign voltages to the non-reference
nodes v1 and v2 (v1 > v2: arbitrary)
Step3: KCL
I1 = I 2 + i1 + i2
@Node1:
@Node2:
I 2 + i2 = i3
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Nodal Analysis: Steps
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Example 3.1
Calculate the node voltages
Still in step 3: We apply Ohm’s law to express
the unknown currents i1, i2, and i3 in terms of
node voltages
Assigning the voltages and currents
Node1:
v −0
v −v
v −0
i1 = 1
, i2 = 1 2 , i3 = 2
R1
R2
R3
5=
The system equations become
I1 = I 2 +
I2 +
v1 v1 − v2
+
R1
R2
v1 − v2 v2
=
R2
R3
 1
1 
1
 +  v1 − v2 = I1 − I 2
R2
 R1 R2 
 1
1
1 
− v1 +  +  v2 = I 2
R2
 R2 R3 
v1 − v2 v1
+
4
2
Node2:
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3v1 − v2 = 20
i2 + i4 = i1 + i5
v1 − v2
v
+ 10 = 5 + 2
4
6
−3v1 + 5v2 = 60
Step4 : Solving for v1 and v2
i1 = i2 + i3
v1 = 20V, v2 =
40
V
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Dependent Current Source
Node1: 3 = i1 + ix
Node2: ix = i2 + i3
3=
v1 − v2 v2 − v3 v2
=
+
2
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4
Node3: i1 + i2 = 2ix
Voltage Source
v1 − v3 v1 − v2
+
4
2
v1 = 4.8V , v2 = 2.4V , v3 = −2.4V
v1 − v3 v2 − v3
v −v
+
=2 1 2
4
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A super-node is formed
by enclosing a dependent
or independent voltage
source connected
between two nonreference nodes and any
elements connected in parallel with it.
Note: We analyze a circuit with super-nodes using
the same steps mentioned before except that the
super-nodes are treated differently.
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Voltage Source
KCL@supernode
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Example 3.3
i1 + i4 = i2 + i3
Find the node voltage
v1 − v2 v1 − v3 v2 v3
+
= +
2
4
8 6
KCL @supernode
KVL@supernode
−v2 + 5 + v3 = 0
v2 − v3 = 5
Note
2 = i1 + i2 + 7
The voltage source in
the supernode provides a constraint equation
needed to solve for the node voltage
A supernode requires both KCL and KVL
v1 v2
+ = −5
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KVL @supernode
−v1 − 2 + v2 = 0
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v2 − v1 = 2
v1 = −7.333V
v2 = −5.333V
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Mesh (Loop) Analysis
Mesh Analysis
Mesh analysis provides another general
procedure for analyzing circuits using mesh
currents as the circuit variables
Nodal analysis applies KCL to find unknown
voltages in a given circuit, while mesh analysis
applies KVL to find unknown currents
Recall: A loop is a closed path with no node
passed more than once
Definition: A mesh is a loop which does not
contain any other loops within it
Is the following a mesh?
abefa
bcdeb
abcdefa
Steps
S1:Assign mesh currents i1, i2, …, in
S2: Apply KVL to each mesh and use Ohm’s law to
express the voltages in terms of the mesh currents
S3: Solve the resulting n simultaneous equations
to get the mesh currents
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Mesh Analysis
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Example
S1: Assign i1, i2
S2: KVL @ each mesh
Find the currents I1, I2, I3
Mesh1
−15 + 5i1 + 10(i1 − i2 ) + 10 = 0
Mesh1
−V1 + i1 R1 + (i1 − i2 ) R3 = 0
or 3i1 − 2i2 = 1
or ( R1 + R3 )i1 − R3i2 = V1
Mesh2
Mesh2
(i2 − i1 ) R3 + i2 R2 + V2 = 0 or
−10 + 10(i2 − i1 ) + 6i2 + 4i2 = 0
− R3i1 + ( R2 + R3 )i2 = −V2
or
i1 = 2i2 − 1
Solving the above equations gives
S3: Just solving for i1 and i2
The branch (measurable) and mesh currents
i1 = i2 = 1A
Therefore
I1 = i1 , I 2 = i2 , I 3 = i1 − i2
I1 = i1 = 1A, I 2 = i2 = 1A, I 3 = i1 − i2 = 0
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Mesh Analysis: Current Source
Mesh Analysis: Current Source
A current source exists only in one mesh
−10 + 4i1 + 6(i1 − i2 ) = 0
i2 = −5
The current source in the supermesh provides the
constraint equation necessary to solve the mesh
currents
A supermesh has no current of its own
A supermesh requires the application of both KVL
and KCL
A current source exists between 2 meshes
Supermesh
−20 + 6i1 + 10i2 + 4i2 = 0
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Example
Find i1 - i4
Supermesh
6i2 + 2i1 + 4i3 + 8(i3 − i4 ) = 0
KCL@node P: i2 = i1 + 5
KCL@node Q: i3 + 3I o = i2 but
Apply KCL to a node in the branch where the
meshes intersect i1 + 6 = i2
Solving both equations we get i1 = −3.2 A, i2 = 2.8 A
Note
I o = −i4
i2 = i3 − 3i4
Mesh4: 8(i4 − i3 ) + 2i4 + 10 = 0
The above equations give
i1 = −7.5 A, i2 = −2.5 A, i3 = 3.93 A, i4 = 2.143 A
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