Fundamental Circuit Theory - II
EE110300
Lecture 6
Mar. 25, 2002
Po-Tai Cheng
Dept. of Electrical Engineering
National Tsing Hua University
Circuit Theory, EE110300 – p.14/23
Node Analysis
1. Select a node as reference node. Assign node voltages v 1 , v2 , ...,
vn−1 to the remaining n − 1 node.
2. Apply Kirchhoff’s Current Law (KCL) to each of the n − 1
non-reference nodes. Use Ohm’s law to express the branch currents
in terms of node voltages v1 , v2 , ..., vn−1 .
3. Solve the n − 1 simultaneous equations to obtain the node voltages
v1 , v2 , ...,vn−1 .
Circuit Theory, EE110300 – p.15/23
Node Analysis-Example 1
5A
acements
4Ω
10 A
2Ω
6Ω
Circuit Theory, EE110300 – p.16/23
Node Analysis with Voltage sources
PSfrag replacements
If a voltage source is connected between the reference node and a
non-reference node, then the voltage of this non-reference node is assigned to
the value of this voltage source.
v1
v2
v3
Case 1:
v1 = v s
vs
PSfrag replacements
ref.
If a voltage source is connected between two non-reference nodes, then a
supernode is defined by enclosing the voltage source between the two
non-reference nodes and any elements connected in parallel.
Case 2:
v1 − v 2 = v s
vs
v1
v2
v3
R3
v1
v2
v2 − v 3
+
+
=0
R1
R2
R3
R1
ref.
R2
Circuit Theory, EE110300 – p.17/23
Node Analysis-Example 2
acements
3Ω
vx
3vx
20 V
6Ω
2Ω
10 A
4Ω
1Ω
Circuit Theory, EE110300 – p.18/23
Mesh Analysis
Mesh: A loop which does not contain any other loops within it.
1. Assign mesh current i1 , i2 , ..., im to the m meshes.
2. Apply Kirchhoff’s Current Law (KCL) to each of the m meshes. Use
Ohm’s law to express the voltage relationship within each mesh in
terms of mesh currents i1 , i2 , ..., im .
3. Solve the m simultaneous equations to obtain the mesh currents i 1 ,
i2 , ...,im .
Circuit Theory, EE110300 – p.19/23
Mesh Analysis-Example 1
acements
6Ω
5Ω
10 Ω
4Ω
15 V
10 V
Circuit Theory, EE110300 – p.20/23
Mesh Analysis with Current Sources
When a current source exists in one mesh, then the current of this
mesh is equal to this current source.
Case 1:
is
is =
PSfrag
replacements
is
i1
When a current source exists between two meshes, a supermesh
is defined by excluding the current source and any components in
series connection with it.
Case 2:
PSfrag replacements
(−i1 ) + i2 = is
is
i1
i2
i1 · (...) + i2 · (...) = 0
Circuit Theory, EE110300 – p.21/23
Mesh Analysis-Example 2
acements
2Ω
5A
6Ω
2Ω
4Ω
3ix
8Ω
ix
10 V
Circuit Theory, EE110300 – p.22/23
Conclusion
Basic Quantities: Voltage, Current, Power, and Energy.
Electric Components: Passive Components; Active Components.
Definition of KCL and KVL.
Use of KCL and KVL to analyze a circuit: Node Analysis; Mesh
Analysis.
Future Extension:
Circuit Analysis (EE2220): DC circuit analysis; AC circuits analysis.
Micro Electronics (EE2250): Diode; Transistors; Amplifiers.
Signal and Systems (EE3610): Continuous-time and Discrete-time
system theories.
Power Electronics (EE4815): Switch-mode power conversion
circuits.
Circuit Theory, EE110300 – p.23/23