Circuits
Lecture 4: Superposition
李宏毅 Hung-yi Lee
Outline
• Matrix Equation for Node and Mesh analysis
• Chapter 4.1, 4.2
• Superposition
• Chapter 2.4
Node Analysis
v1:
v2:
v3:
 v1

Ra
v1  v 2
 v1
Rb

Rc
v2  v3
Re

v 2  v1
Rc
 v2

v3  v2
Rd

 v3
Rf
  is 
Re
 is
0
vs
Ra
v s  v1

Ra
v1  v 2

v 2  v1
Rb

Rc
v2  v3
 v1
 v2
Rc

v3  v2
Rd

Re
 1
1



 R
Rb
a

 1
1
v1   
Rc
 Rc
 1
1
v2   
 R
Re
e

 v3
Rf
 is  0
0
Re
 is
vs
1 
1
 v1 
v


i

2
s
R c 
Rc
Ra
1
1 
1
v2 


v3  0

Rd
Re 
Re
1 
v  i

3
s
R f 
You can directly write the
matrix equation below.
(textbook, P139)
Node Analysis
 1
1
1




Rb Rc
 Ra
1


Rc


0


1

1
Rc

Rc
1
Rd
1

1
Re
Re
 R v   s 
Resistance
Node
Sources
potentials

vs 
0


i 
  v1   s R 
a
1

  
v2  
0

 
Re

  v 3  
is
1
1 






R e R f 
 1
1



 R
Rb
a

 1
1
v1   
Rc
 Rc
 1
1
v2   
 R
Re
e

vs
1 
1
 v1 
v


i

2
s
R c 
Rc
Ra
1
1 
1
v2 


v3  0

Rd
Re 
Re
1 
v  i

3
s
R f 
Node Analysis
 R v   s 
v    R  s 
1
 1
1
1



Rb Rc
 Ra
1


Rc


0


1

 v1   a 11
  
v  a
 2   21
 v 3   a 31
1
Rc

Rc
1

vs 



i

  v1   s R 
a
1

  
v2  
0


 
Re


 v 
is
1
1  3  





R e R f 
0

Rd
1
Re
a 12
a 22
a 32
1
Re
vs 

i 
a 13   s R 
a


a 23 
0



a 33  
is




v1, v2, v3 is the weighted sum of is and vs
Node potential is the weighted sum of the values
of sources
Voltage (potential difference) is the weighted
sum of the values of sources
Mesh Analysis
For mesh 1
Ra(i1-is)+Rbi1+Rc(i1-i2)-vs=0
For mesh 2
Rc(i2-i1)+Rdi2+Re(i2-i3)=0
For mesh 3
Re(i3-i2)+Rfi3+vs=0
 Ra  Rb  Rc

 Rc


0

 Rc
Rc  Rd  Re
 Re
You can directly write the
matrix equation below.
  i1   R a i s  v s 
  

 R e  i2 
0
  

R e  R f   i3    v s 
0
(textbook, P153)
Mesh Analysis
Mesh
Current
Resistance
Sources
 R i   s 
 Ra  Rb  Rc

 Rc


0

 Rc
Rc  Rd  Re
 Re
  i1   R a i s  v s 
  

 R e  i2 
0
  

R e  R f   i3    v s 
0
Mesh Analysis
 Ra  Rb  Rc

 Rc


0

 Rc
Rc  Rd  Re
 Re
  i1   R a i s  v s 
  

 R e  i2 
0
  


R e  R f   i3    v s 
0
 R i   s 
i    R  s 
1
 i1   b11
  
i  b
 2   21
 i3   b 31
b12
b 22
b 32
b13   R a i s  v s 


b 23
0


b 33    v s 
i1, i2, i3 is the weighted sum of is and vs
Mesh currents are the weighted sum of the values
of sources
Currents of the braches are the weighted sum of
the values of sources
Linearity
Based on node and mesh analysis:
y
ax
i
i
i
• y: any current or voltage for an element
• xi: current of current sources or voltage of
voltage sources
Any current (or voltage) for an element is the
weighted sum of the voltage (or current) of the
sources.
Linearity - Example
i1  a1 x1  a 2 x 2  a 3 x 3
x1
x2
x3
Any current (or voltage) for an element is the
weighted sum of the voltage (or current) of the
sources.
Not apply on Power
• xi: current of independent current sources or voltage of
independent voltage sources
Voltage:
v
ax
i
Current:
i
p
bx
i
i
i
i
Power:
i
cx
i
i
i
Power:



p  vi    a i x i    b i x i 
 i
 i

Proportionality Principle – One
Independent Sources
Find i1 and v1
when vs is 9V,
72V and 0.9V
Complex
Circuit
i1  a 1 v s
v1  a 2 v s
v s  9V
i1  1 A
v1  10 V
v s  72 V
i1  8 A
v1  80 V
v s  0 . 9V
i1  0 . 1 A
v1  1V
Superposition Principle – Multiple
Independent Sources
• Example 2.10
• Find i1
x2
i1  a 1 x1  a 2 x 2  a 3 x 3
 i1 1  i1  2  i1  3
We can find i1-1, i1-2, i1-3
separately.
x1
x3
When x2=0 and x3=0,
The current through 2Ω is i1-1.
Superposition Principle – Multiple
Independent Sources
x2  0
• Example 2.10
• Find i1
i1  a 1 x1  a 2 x 2  a 3 x 3
 i1 1  i1  2  i1  3
We can find i1-1, i1-2, i1-3
separately.
Current of current
source set to be zero.
x1
x3  0
Open
Circuit
i1 1  30 / 6  4  2   2 . 5 A
Superposition Principle – Multiple
Independent Sources
• Example 2.10
• Find i1
x2
i1  a 1 x1  a 2 x 2  a 3 x 3
 i1 1  i1  2  i1  3
We can find i1-1, i1-2, i1-3
separately.
x1
x3
To find i1-2, we set x1=0 and x3=0.
Now the current through 2Ω is i1-2.
Superposition Principle – Multiple
Independent Sources
x2
• Example 2.10
• Find i1
i1  a 1 x1  a 2 x 2  a 3 x 3
 i1 1  i1  2  i1  3
We can find i1-1, i1-2, i1-3
separately.
x1  0
Voltage of voltage
source set to be zero.
x3  0
Short
Circuit
i1  2  3  4 /  2  6   4   1 A
Superposition Principle – Multiple
Independent Sources
x 0
2
• Example 2.10
• Find i1
i1  a 1 x1  a 2 x 2  a 3 x 3
 i1 1  i1  2  i1  3
We can find i1-1, i1-2, i1-3
separately.
i1  3    8 
x1  0
x3
6
 4 A
6  4  2 
Superposition Principle – Multiple
Independent Sources
• Example 2.10
• Find i1
x2
i1  a 1 x1  a 2 x 2  a 3 x 3
 i1 1  i1  2  i1  3
We can find i1-1, i1-2, i1-3
separately.
set x2=0 and x3=0
set x1=0 and x3=0
set x1=0 and x2=0
x1
x3
i1 1  2 . 5 A
i1  2  1 A
i1  3   4 A
i1   0 . 5 A
Superposition Principle – Multiple
Independent Sources
• Steps to apply Superposition Principle:
• If the circuit has multiple sources, to find a voltage or
current for an element
• For each source
• Keep the source unchanged
• All the other sources set to zero
• Voltage source’s voltage set to 0 = Short circuit
• Current source’s current set to 0 = open circuit
• Find the voltage or current for the element
• Add all the voltages or currents obtain by individual
sources
Remind
• Always using superposition when there are
multiple sources?
One circuit (3 sources)
Three circuits (1 source)
v.s.
Concluding Remarks
y
ax
i
i
This equation only for circuits
with sources and resistors.
i
•
•
y: any current or voltage for an element
xi: current of current sources or voltage of voltage sources
Proportionality Principle, Superposition Principle
Can be used in any circuit in this course
Linearity
• A circuit is a multiple-input multiple-output (MIMO)
system
• Input: current of current sources or voltage of voltage
sources
• Output: the current or voltage for the elements
input
Circuit
(System)
+
v
-
i output
All circuits in this
courses are
linear circuits.
Linearity
• All linear circuits are linear system
• Linear Circuit:
All circuits in this
• Sources
courses are
• Linear Elements:
linear
systems.
• Resistor, Capacitor, Inductor
i
v
R
Linearity
• Linear System:
• Property 1:
Input: g1(t), g2(t), g3(t), ……
output: h1(t), h2(t), h3(t), ……
Input: Kg1(t), Kg2(t), Kg3(t), ……
output: Kh1(t), Kh2(t), Kh3(t), ……
Proportionality Principle
Linearity
• Linear System:
• Property 2:
Input: a1(t), a2(t), a3(t), ……
output: x1(t), x2(t), x3(t), ……
Input: b1(t), b2(t), b3(t), ……
output: y1(t), y2(t), y3(t), ……
Input: a1(t)+ b1(t), a2(t)+ b2(t), a3(t)+ b3(t), ……
output: x1(t)+y1(t), x2(t)+y2(t), x3(t)+y3(t), ……
Superposition Principle
Linearity
• Linear System:
• Property 2:
Input: a1(t), a2(t), a3(t), ……
output: x1(t), x2(t), x3(t), ……
Input: b1(t), b2(t), b3(t), ……
output: y1(t), y2(t), y3(t), ……
Input: a1(t)+ b1(t), a2(t)+ b2(t), a3(t)+ b3(t), ……
output: x1(t)+y1(t), x2(t)+y2(t), x3(t)+y3(t), ……
Superposition Principle
v t 
Linearity
v t 
i t 
g 1 t 
0
g t 
g t   g 1 t   g 2 t 
0
Superposition Principle can be i t 
applied on any circuit in this
course (Textbook: Chapter 6.5).
g 2 t 
Homework
• 2.50
Given vs
and R3,
find vb
Homework
• 2.52
Given is, find vs such that v4= 36V
Thank you!
Answer
• 2.50
• -12V
• 2.52
• 60V