EE201 Circuit Theory
Ch 3. Network Theorems (Part 1)
Bongjin Kim
Associate Professor
Electrical Engineering, KAIST
*Lecture materials are modified version of the contents provided by Prof. Minkyu Je (KAIST)
Outline
❑
Nodal Analysis
– Circuits Containing Only Independent Current Sources
– Circuits Containing Dependent Current Sources
– Circuits Containing Independent Voltage Sources
– Circuits Containing Dependent Voltage Sources
❑
Loop Analysis
– Circuits Containing Only Independent Voltage Sources
– Circuits Containing Independent Current Sources
– Circuits Containing Dependent Sources
2
Nodal Analysis
❑
In a nodal analysis, the variables in the circuit are selected to be the
node voltages.
– The node voltages are defined with respect to a common point in the
circuit.
– One node is selected as the reference node, and all other node voltages are
defined with respect to that node.
– Quite often, this node is the one to which the largest number of branches
are connected.
– It is commonly called ground because it is said to be at ground-zero
potential, and it sometimes represents the chassis or ground line in a
practical circuit.
3
Nodal Analysis
❑
Value of knowing all the node voltages in a network
 The bottom node is selected
as the reference and labeled
with the ground symbol.
 VS, Va, Vb and Vc are all
measured with respect to the
bottom node.
 Once these node voltages are
known, we know the voltage
across every element in the
network.
 Thus, we can immediately
calculate any branch current
or the power supplied or
absorbed by any element.
Nodal and Loop Analysis Techniques
4
Nodal Analysis
❑
Value of knowing all the node voltages in a network
V1 = VS − Va = 12 − 3 = 9 V
3 3
= V
2 2
3 3 9
V5 = Vb − Vc = − = V
2 8 8
V3 = Va − Vb = 3 −
I1 =
VS − Va
9
=
= 1 mA
9k
9k
I3 =
Va − Vb 3 2 1
=
= mA
3k
3k 2
I5 =
Vb − Vc 9 8 1
=
= mA
9k
9k 8
I2 =
Va − 0 3
1
=
= mA
6k
6k 2
I4 =
Vb − 0 3 2 3
=
= mA
4k
4k 8
Nodal and Loop Analysis Techniques
5
Nodal Analysis
❑
As a general rule, if we know the node voltages in the circuit, we can
calculate the current through any resistive element using Ohm’s law:
I=
(Vm − Vn )
.
R
– Note carefully that both voltages Vm and Vn are measured with respect to
the same point, that is, ground.
– In a nodal analysis, this concept is central to the manner in which we write
the equations necessary to determine all the node voltages.
Nodal and Loop Analysis Techniques
6
Nodal Analysis
❑
In a nodal analysis, we employ KCL equations in such a way that the
variables contained in these equations are the unknown node voltages
of the network.
– One of the nodes in an N-node circuit is selected as the reference node,
and the voltages at all the remaining N − 1 nonreferernce nodes are
measured with respect to this reference node.
– Exactly N − 1 linearly independent KCL equations are required to determine
the N − 1 unknown node voltages.
– Therefore, once one of the nodes in an N-node circuit has been selected as
the reference node, our task is reduced to identifying the remaining N − 1
nonreference nodes and writing one KCL equation at each of them.
Nodal and Loop Analysis Techniques
7
Circuits w/ Indep. Current Sources
❑
Consider the network shown below.
– This network contains three nodes, and thus N − 1 = 3 − 1 = 2 linearly
independent KCL equations are required to determine the N − 1 = 2
unknown node voltages.
– We select the bottom node as the reference node, and then the voltage at
the two remaining nodes labeled v1 and v2 are measured with respect to
this node.
Nodal and Loop Analysis Techniques
8
Circuits w/ Indep. Current Sources
❑
Consider the network shown below.
– Applying KCL at node 1 yields
− i A + i 1 + i 2 = 0.
 Applying KCL at node 2 yields
− i2 + iB + i3 = 0, or
Using Ohm’s law,
− G2 (v 1 − v 2 ) + iB + G3 (v 2 − 0) = 0
− i A + G1(v 1 − 0) + G2 (v 1 − v 2 ) = 0, or
which can be expressed as
(G1 + G2 )v 1 − G2v 2 = i A
Nodal and Loop Analysis Techniques
− G2v 1 + (G2 + G3 )v 2 = − iB .
9
Circuits w/ Indep. Current Sources
❑
Consider the network shown below.
– Therefore, the following two equations need to be solved for the two
unknown node voltages v1 and v2.
(G1 + G2 )v 1 − G2v 2 = i A
− G2v 1 + (G2 + G3 )v 2 = − iB
– Three techniques will be demonstrated for solving linearly independent
simultaneous equations: Gaussian elimination, matrix analysis, and
MATLAB mathematical software package.
Nodal and Loop Analysis Techniques
10
Example
❑
Let us determine all node voltages and branch currents.
6 k
4 mA
12 k
6 k
1 mA
Using the parameter values, the equations become
1
 1
1
V1 
+  − V2   = 1m and
 12k 6k 
 6k 
1
1
1
− V1   + V2  +  = −4m.
 6k 
 6k 6k 
The equations can be written as
V1 V2
−
= 1m and
4k 6k
Nodal and Loop Analysis Techniques
−
V1 V2
+
= −4m.
6k 3k
11
Example
❑
Let’s use Gaussian elimination to solve
V1 V2
V V
−
= 1m and − 1 + 2 = −4m.
4k 6k
6k 3k
We solve the first equation for V1 in terms of V2:
2
V1 = V2 + 4.
3
This value is then substituted into the second equation to yield
−
1 2
 V
 V2 + 4  + 2 = −4m or V2 = −15 V.
6k  3
 3k
This value for V2 is substituted back into the equation for V1 in terms of V2:
2
V1 = V2 + 4 = −6 V.
3
Nodal and Loop Analysis Techniques
12
Example
❑
Let’s use matrix analysis to solve
V1 V2
V V
−
= 1m and − 1 + 2 = −4m.
4k 6k
6k 3k
Using matrix analysis, we can write GV = I, where in this case
 1

G =  4k
1
−
 6k
−
1
6k  , V =  V1  , and I =  1m .
V 
 − 4m
1




2

3k 
The solution to the matrix equation is V = G−1I and therefore,
 1
 V1   4k
V  =  1
 2  −
 6k
−1
1
−   1m
6k
.
1   − 4m

3k 
To calculate the inverse of G, we need the adjoint and the determinant.
Nodal and Loop Analysis Techniques
13
Example
❑
Let’s use matrix analysis to solve
V1 V2
V V
−
= 1m and − 1 + 2 = −4m.
4k 6k
6k 3k
The adjoint is
 1

Adj G =  3k
1

 6k
1
6k  ,
1

4k 
and the determinant is
1
 1  1   1  1 
|G | =     − −
−
=
.


2
 4k   3k   6k   6k  18k
Therefore,
 1
 V1 
2  3k
=
18
k
 1
V 
 2

 6k
Nodal and Loop Analysis Techniques
4 
 1
1
−
6k   1m = 18k 2  3k 2 6k 2  =  − 6 .
 1
1   − 4m
1   − 15 

− 2

2
4k 
 6k
k 
14
Example
❑
Let’s use MATLAB to solve
V1 V2
V V
−
= 1m and − 1 + 2 = −4m.
4k 6k
6k 3k
Multiplying both equations by 12k yields
3V1 − 2V2 = 12 and
− 2V1 + 4V2 = −48.
In matrix form, the equation is
 3 − 2  V1   12
− 2
 V  =  − 48 .
4

  2 

Then the data entries and solution using MATLAB are as follows:
>> G = [ 3
G =
3
-2
-2;
-2
-2
4
Nodal and Loop Analysis Techniques
4 ]
>> I = [ 12;
I =
12
-48
-48 ]
>> V = inv(G)*I
V =
-6.0000
-15.0000
15
Example
❑
Knowing the node voltages, we can determine all the currents using
Ohm’s law.
6 k
12 k
1 mA
I1 =
V1 − 6
1
=
= − mA
R1 12k
2
I2 =
V1 − V2 − 6 − (−15) 3
=
= mA
R2
6k
2
I3 =
V2 − 15
5
=
= − mA
R3
6k
2
Nodal and Loop Analysis Techniques
4 mA
6 k
16
Circuits w/ Indep. Current Sources
❑
Let’s examine the network shown below.
– This network has four nodes.
– The node at the bottom of the circuit is selected as the reference node and
labeled with the ground symbol.
– Since N = 4, N − 1 = 3 linearly independent KCL equations will be required
to determine the three unknown nonreference node voltages labeled v1, v2,
and v3.
Nodal and Loop Analysis Techniques
17
Circuits w/ Indep. Current Sources
❑
Let’s examine the network shown below.
 Node 2:
− i 2 + i 4 − i 5 = 0.
v 2 − v1 v 2 − 0 v 2 − v 3
+
+
=0
R2
R4
R5
 1
1
1
1 
1
 − v 3
+ v 2 
+
+
=0
R2
R
R
R
R
 2
4
5
5
– Node 1:
− v1
– Node 1:
 Node 3:
i 1 − i A + i 2 − i 3 = 0.
i 3 + i 5 + i B = 0.
v1 − 0
v − v 2 v1 − v 3
− iA + 1
+
=0
R1
R2
R3
v 3 − v1 v 3 − v 2
+
+ iB = 0
R3
R5
 1
1
1 
1
1
 − v 2
v 1
+
+
− v3
= iA
R
R
R
R
R
 1
2
3
2
3
− v1
Nodal and Loop Analysis Techniques
 1
1
1
1 
 = − iB
− v2
+ v 3 
+
R3
R5
R
R
 3
5
18
Circuits w/ Indep. Current Sources
❑
Grouping the node equations together, we obtain
 1
1
1 
1
1
 − v 2
v 1
+
+
− v3
= iA
R2
R3
 R1 R2 R3 
− v1
 1
1
1
1 
1
 − v 3
+ v 2 
+
+
=0
R2
R
R
R
R
 2
4
5
5
− v1
❑
 1
1
1
1 
 = − iB .
− v2
+ v 3 
+
R3
R5
R
R
 3
5
The equations can also be written in matrix form as
 1
1
1
+
+
R R
R3
2
 1
1

−

R2

1

−
R3

Nodal and Loop Analysis Techniques
1
R2
1
1
1
+
+
R2 R4 R5
1
−
R5
−
1
R3
1
−
R5
1
1
+
R3 R5
−


 v 1   i A 
 v  =  0  .

  2 
 v 3   − iB 


19
Circuits w/ Indep. Current Sources
❑
Symmetrical G matrix
– The node equations for networks containing only resistors and
independent current sources can always be written in symmetrical form.
1
1
+
R R
2
 1
 − 1
R2

1 
R2 

1
1
+
R21 R3 
−
1
1
1
+
+
R R
R3
2
 1
1

−

R2

1

−
R3

Nodal and Loop Analysis Techniques
1
R2
1
1
1
+
+
R2 R4 R5
1
−
R5
−
1 
R3 

1 
−
R5 
1
1
+

R3 R5 
−
20
Circuits w/ Indep. Current Sources
❑
Symmetrical G matrix
– We can take advantage of this fact and learn to write the equations by
inspection.
1
1
+
R R
2
 1
 − 1
R2

1 
R2   v 1   i A 
=

1
1  v 2   − iB 
+
R12 R3 
−
– 1st equation: The coefficient of v1 is the sum of all the conductances
connected to node 1, the coefficient of v2 is the negative of the
conductance connected between node 1 and node 2, and the right hand
side is the sum of the currents entering node 1 through current sources.
– 2nd equation: The coefficient of v2 is the sum of all the conductances
connected to node 2, the coefficient of v1 is the negative of the
conductance connected between node 2 and node 1, and the right-hand
side is the sum of the currents entering node 2 through current sources.
Nodal and Loop Analysis Techniques
21
Circuits w/ Indep. Current Sources
❑
Symmetrical G matrix
 1st equation: The coefficient of v1 is
the sum of all the conductances
connected to node 1, the coefficient
of v2 is the negative of the
conductance connected between
node 1 and node 2, the coefficient of
v3 is the negative of the conductance
connected between node 1 and node
3, and the right hand side is the sum
of the currents entering node 1
through current sources.
1
1
1
+
+
R R
R3
2
 1
1

−

R2

1

−
R3

1
R2
1
1
1
+
+
R2 R4 R5
1
−
R5
−
Nodal and Loop Analysis Techniques
1 
R3   v   i 
 1
A
1   
−
v2 = 0 

R5    
1
1  v 3   − iB 
+

R3 R5 
−
 The other two
equations are obtained
in a similar manner.
22
Circuits w/ Indep. Current Sources
❑
For the node j with node voltage vj ,
– Coefficient of vj: sum of all the conductances connected to node j
– Coefficients of the other node voltages (e.g., vj-1 , vj+1): negative of the sum
of the conductances connected directly between these nodes and node j
– Right-hand side of the equation: sum of the currents entering the node via
current sources
❑
The left-hand side of the equation represents the sum of the currents
leaving node j and the right-hand side represents the currents entering
node j.
Nodal and Loop Analysis Techniques
23
Example
❑
Determine the node voltages.
4 mA
2 k
 4
 0

 − 2
 1
 1
1 
 − v 2 (0) − v 3   = − i A
v 1
+
 R1 R2 
 R1 
2 k
4 k
4 k
2 mA
1 k
0 − 2  v 1   − 16
2 − 1 v 2  =  8
  

−1
7 v 3   0
Nodal and Loop Analysis Techniques
 1
 1 
1 
 − v 3 
 = i A − iB
− v 1(0) + v 2 
+
R
R
R
 4
 3
4
− v 1(
 1
 1 
1
1
1 
 = 0
 + v 3 
) − v 2 
+
+
R1
R
R
R
R
 4
 1
4
5
1
1
+
 2k 2k

 0

 − 1

2k
0
1
1
+
4k 4k
1
−
4k
1

  v   − 4m
2k
  1 
1

−
 v 2  =  2m 
4k

1
1
1  v 3   0 
+
+
2k 4k 1k 
−
24
Example
❑
Let’s use Gaussian elimination to solve the equations.
 4
 0

 − 2
0 − 2  v 1   − 16
2 − 1 v 2  =  8
  

−1
7 v 3   0
0 − 2  v 1   − 16
 4
 0
2 − 1 v 2  =  8

  

 0 − 2 12 v 3   − 16
 4
 0

 0
0 − 2  v 1   − 16
22
0 v 2  =  80
  

0 11 v 3   − 8
 4
 0

 0
22 0
 0 22

 0 0
0  v 1   − 96
0 v 2  =  80
  

11 v 3   − 8
Nodal and Loop Analysis Techniques
0 − 2  v 1   − 16
2 − 1 v 2  =  8
  

0 11 v 3   − 8
 1 0 0  v 1   − 4.3636 
0 1 0 v  =  3.6364 

  2 

0 0 1 v 3   − 0.7273 
25
Example
❑
Let’s use matrix analysis to solve the equations.
We can write GV = I, where in this case
 4
G= 0

 − 2
0 − 2
v 1 
 − 16
2 − 1 , V = v 2  , and I =  8 .
 



v 3 
−1
7
 0
The solution to the matrix equation is V = G−1I and therefore,
v 1   4
v  =  0
 2 
v 3   − 2
Nodal and Loop Analysis Techniques
0 − 2
2 − 1

−1
7
−1
 − 16
 8


 0
26
Example
❑
Let’s use matrix analysis to solve the equations.
To calculate the inverse of G, we use the following formula.
𝑎
A−1 = 𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓
𝑖
−1
𝑒𝑖 − 𝑓ℎ
1
𝑐ℎ − 𝑏𝑖
=
|A|
𝑏𝑓 − 𝑐𝑒
𝑓𝑔 − 𝑑𝑖
𝑎𝑖 − 𝑐𝑔
𝑐𝑑 − 𝑎𝑓
𝑑ℎ − 𝑒𝑔
𝑏𝑔 − 𝑎ℎ
𝑎𝑒 − 𝑏𝑑
| A | = a(ei − fh) + b(fg − di ) + c (dh − eg )
Therefore,
| G | = 4  (2)(7) − (−1)(−1) + 0  (−1)(−2) − (0)(7) − 2  (0)(−1) − (2)(−2) = 44
v 1 
13 2
v  = 1  2 24
 2  44 
v 3 
 4 4
Nodal and Loop Analysis Techniques
4   − 16
 − 192  − 4.3636 
1 
4   8 =
160 =  3.6363 

 44 
 

8  0
 − 32  − 0.7273 
27
Example
❑
Let’s use MATLAB to solve the equations.
 4
 0

 − 2
0 − 2  v 1   − 16
2 − 1 v 2  =  8
  

−1
7 v 3   0
>> G = [ 4
G =
4
0
-2
0
>> I = [ -16;
I =
-16
8
0
-2;
0
2
-1
8;
0
2
-1;
-2
-1
7 ]
-2
-1
7
0 ]
>> V = inv(G)*I
V =
-4.3636
3.6364
-0.7273
Nodal and Loop Analysis Techniques
28
Summary
❑
Nodal Analysis
– Value of knowing all the node voltages in a network
– Circuits containing only independent current sources
– Techniques to write node-voltage equations
– Symmetry of node-voltage equations
– Various examples
– Techniques to solve node-voltage equations
Nodal and Loop Analysis Techniques
29
Recommended Problems
❑
Problems 3.1.1, 3.1.2, 3.1.4, 3.1.6, 3.1.9 (in textbook)
30