李宏毅 Hung-yi Lee

Introduction
• Node analysis:
Current +
Voltage
• Mesh analysis:
Current +
Voltage
Voltage
Current
Mesh 1
Mesh 2 Mesh 3
Potential
Mesh Current
Mesh: closed current path that contains no closed paths within it (Textbook, P149)
Meshes are “holes” in the circuits.

Mesh analysis i

1 i
2 i

2 i
3 i
1
 i
3
Target of Mesh analysis is to find the mesh currents.

Mesh analysis
Consider KCL: i b
 i c
 i d

0
Represent i i b
 i
1 b
, i c i c
 and i d i
2
 i by mesh current
1 i d
  i
2
KCL is already fulfilled

Mesh analysis
Consider KVL: v a
 v b
 v c
 v s v a
Represent v a
, v b

R a
 i
1
 i s
 and v c by mesh current v b

R b i
1 v c

R c
 i
1
 i
2


Mesh analysis
For mesh 1: R a
( i
1
-i s
) + R b i
1
+ R c
( i
1
-i
2
) = v s
For mesh 2: R c
( i
2
-i
1
) + R d i
2
+ R e
( i
2
-i
3
) = 0
For mesh 3: R e
( i
3
-i
2
) + R f i
3
+ v s
= 0
Can we always represent voltage by mesh currents?

Mesh analysis – 3 cases
• Represent the voltage of (1) voltage sources, (2) resistor, and (3) current sources by mesh current

Mesh analysis - current sources
• 1. Current sources on the perimeter of the circuit

Mesh analysis - current sources
• 2. Interior current sources
Method 1:
Consider v s as unknown variables i s v s v d v c

 v b v e


v s v s

0 Represent the voltages v d
, v b
, v c v e
…. (except v s
) by mesh current and
Solve mesh currents and v s
One more equation: i s
 i
2
 i
1

Mesh analysis - current sources
• 2. Interior current sources
Method 2: when the current source is parallel with a resistor
Textbook:
Chapter 2.5

Mesh analysis - current sources
• 2. Interior current sources
Method 2: when the current source is parallel with a resistor
Another point of view:
The voltage of is i s v s

R s i m equal to R s
.

 i x


Mesh analysis - current sources
• 2. Interior current sources
Method 3: supermesh R b
 i
1
 i
3
 i
3
R c
 i
2
 i
3


R c
 i s
 i
1
 i
3

R d i
1 i s
 i
2
 i
1 v s i s i
2
 i s
 i
1
R e i
2

R e
 i s
 i
1


Example 4.9
10 i
1
6
 i
1

5

6( i
1
-5) + 10 i
1
+ 3( i
1
+4) = 20
→ i
1
=2A
3
 i
1

4


Exercise 4.11
Special case: current sources are parallel with resistors

Exercise 4.11
i 3 mA
1

6 k
 i 2 mA
1
 i
1
5 k
 i
1

5 k

 i
1

3 m


4 k

 i
1

2 m


6 k

0

Non-planar Circuit
• Mesh analysis cannot be applied on non-planar circuit
Cannot transform this circuit into planar circuit

Non-planar Circuit
• Some non-planar circuits can be transformed into planar circuit
A B
C
A
C
B

Non-planar Circuit
• Some non-planar circuits can be transformed into planar circuit

Non-planar Circuit
• Why mesh analysis cannot be applied on nonplanar circuit?
6 meshes?
5 meshes

Node v.s. Mesh

Node v.s. Mesh
• What is the final target?
• Node analysis is not suitable for current
• Mesh analysis is not suitable for voltage i
If node analysis is used, it is tedious to find i v
+ -
If mesh analysis is used, it is tedious to find v

Node v.s. Mesh
• Number of Equation: Number of Nodes v.s. Number of
Meshes
• Number of Voltage sources v.s. Number of Current sources
• Use both to check your results

Homework
• 4.38
• 4.42

Answer
• 4.38
• V1=24, v2=-16, v3=20
• 4.42
• I1=3, i2=3.5, i3=3, v1=0, v2=15

Acknowledgement
• 感謝 林楷恩 (b02)
• 糾正錯誤的作業答案