EE201 Circuit Theory
Ch 3. Network Theorems (Part 3)
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
Loop Analysis
❑
A loop analysis uses KVL to determine a set of loop currents in the
circuit.
– Once these loop currents are known, Ohm’s law can be used to calculate
any voltages in the network.
– Via network topology it can be shown that, in general, there are exactly B −
N + 1 linearly independent KVL equations for any network, where B is the
number of branches in the circuit and N is the number of nodes.
* Loop: a closed path that does not go twice over any node.
3
Loop Analysis
❑
As an example, consider the circuit shown below.
– There are 8 branches and 5 nodes.
– Thus, the number of independent KVL equations necessary to determine all
currents in the network is B − N + 1 = 8 − 5 + 1 = 4.
 If we define 4 loop currents as shown in
the figure, the branch currents are then
determined as
i1(t) = iA(t)
i2(t) = iA(t) − iB(t)
i3(t) = iB(t)
i4(t) = iA(t) − iC(t)
i5(t) = iB(t) − iD(t)
i6(t) = −iC(t)
i7(t) = iC(t) − iD(t)
i8(t) = −iD(t).
4
Circuits w/ Indep. Voltage Sources
❑
Consider the network shown below.
– This network has 7 branches and 6 nodes, and thus the number of linearly
independent KVL equations necessary to determine all currents in the
circuit is B − N + 1 = 7 − 6 + 1 = 2.
– Thus, we identify 2 independent loops, A-B-E-F-A and B-C-D-E-B.
– We now define loop currents, which can be used to find the physical
currents in the circuit.
– Let us assume that i1 flows in the 1st loop and i2 flows in the 2nd loop.
– Then, the branch current flowing from B to E through R3 is i1 − i2.
5
Circuits w/ Indep. Voltage Sources
❑
Consider the network shown below.
– Applying KVL to the 1st loop yields +v1 + v3 + v2 − vS1 = 0.
– KVL applied to the 2nd loop yields +vS2 + v4 + v5 − v3 = 0.
– Substituting v1 = i1R1, v2 = i1R2, v3 = (i1 − i2)R3, v4 = i2R4, and v5 = i2R5 into the
two KVL equations produces
i1(R1 + R2 + R3) − i2(R3) = vs1, and
−i1(R3) + i2(R3 + R4 + R5) = −vs2.
6
Circuits w/ Indep. Voltage Sources
❑
Consider the network shown below.
i1(R1 + R2 + R3) − i2(R3) = vs1, and
−i1(R3) + i2(R3 + R4 + R5) = −vs2.
– Rewriting the derived mesh equations in matrix form, we obtain
R1 + R2 + R3

− R3

− R3
  i1   v S 1 
=
.
R3 + R4 + R5  i2   − v S 2 
7
Mesh Current & Mesh Analysis
❑
Consider the network shown below.
– A mesh is a special kind of loop that does not contain any loops within it.
– Therefore, as we traverse the path of a mesh, we do not encircle any circuit
elements.
– For example, the circuit contains two meshes defined by the paths A-B-E-FA and B-C-D-E-B. The path A-B-C-D-E-F-A is a loop, but not a mesh.
– A mesh current is a loop current associated with a mesh.
– A mesh analysis deals with mesh currents as unknowns.
8
Example: Mesh Analysis
❑
Find the current Io.
 Results from two simultaneous
equations:
I1 = 54 mA, I2 = 12 mA
Io = 34 mA
 Since we know the branch currents,
– Mesh equations:
−12 + 6kI1 + 6k(I1 − I2) = 0
6k(I2 − I1) + 3kI2 + 3 = 0
where Io = I1 − I2.
we can use KVL around any closed
path to check the results.
9
Example: Loop Analysis
❑
Find the current Io.
 Loop equations:
−12 + 6k(I1 + I2) + 6kI1 = 0
−12 + 6k(I1 + I2) + 3kI2 + 3 = 0
 Results from two simultaneous
equations:
– Since we want to calculate the
current Io, we could use different
loops for analysis.
– Now the loop current I1 passes
through the center leg of the
network and I1 = Io.
I1 = Io = 34 mA, I2 = 12 mA
Current in the 12-V source
= I1 + I2 = 54 mA
 These results agree with the mesh
analysis.
10
Example: Nodal Analysis
❑
Find the current Io.
 Results from the node-voltage
equation:
Vo = 92 V
Io = Vo6k = 34 mA
 In this case, we had to solve only one
equation instead of two.
– We could also use nodal analysis
to find Io.
– Node-voltage equation:
Vo − 12 Vo Vo − 3
+
+
=0
6k
6k
3k
11
Symmetry of Mesh Equations
R1 + R2 + R3

− R3

− R3
  i1   v S 1 
=



R3 + R4 + R5  i2   − v S 2 
The coefficient matrix for circuits containing resistors and independent
voltage sources shows symmetry.
❑ We can write the mesh equations by inspection.
❑
– 1st equation: coefficient of i1 =  (resistances in mesh 1);
coefficient of i2 = − (resistances common to mesh 1 and mesh 2);
right-hand side =  (voltage sources in mesh 1)
– 2nd equation: coefficient of i2 =  (resistances in mesh 2);
coefficient of i1 = − (resistances common to mesh 1 and mesh 2);
right-hand side =  (voltage sources in mesh 2)
12
Example
❑
Write the mesh equations by inspection.
 Matrix form of the equations:
0 − 6k   I1   − 6 
 10k
 0 12k − 3k  I  =  6 

  2 

 − 6k − 3k 21k  I3   0 
 Dividing by 1000 yields
– Mesh equations:
(4k + 6k)I1 − (0)I2 − (6k)I3 = −6
−(0)I1 + (9k + 3k)I2 − (3k)I3 = 6
−(6k)I1 − (3k)I2 + (3k + 6k + 12k)I3 = 0
0 − 6  I1   − 0.006
 10
 0 12 − 3 I  =  0.006 .

  2 

 − 6 − 3 21 I3   0

13
Circuits w/ Indep. Current Sources
As the presence of a voltage source simplified the nodal analysis, the
presence of a current source simplifies a loop analysis.
❑ Let us consider the circuit shown below.
❑
– Although it appears that there are two unknown mesh currents, I1 goes
directly through the current source and therefore is constrained to be 2
mA. Hence, only I2 is unknown.
– Applying KVL to the rightmost mesh gives 2k(I2 − I1) − 2 + 6kI2 = 0.
– Solving the mesh equation with knowing that I1 = 2 mA, I2 = 34 mA.
14
Example: Unshared Sources
❑
Find Vo.
 1 mesh equation:
4k(I3 − I2) + 2k(I3 − I1) + 6kI3 − 3 = 0
 By solving these three equations,
we obtain
I3 = 14 mA and
Vo = 6kI3 − 3 = −32 V.
– 3 independent equations are
required.
– 2 constraint equations:
I1 = 4 mA
I2 = −2 mA
15
Example: Shared Sources
❑
Find Io.
 The remaining loop current I3 must
pass through the circuit elements not
covered by the 2 previous equations,
and cannot pass through the current
sources.
 The path for this remaining loop
current can be found by opencircuiting the current sources.
– First, we select two loop currents
I1 and I2 such that I1 and I2 pass
directly through the 2-mA and 4mA sources, respectively.
– Therefore, I1 = 2 mA and I2 = 4
mA.
16
Example: Shared Sources
❑
Find Io.
– The KVL equation for this last loop is
−6 + 1kI3 + 2k(I2 + I3) + 2k(I3 + I2 − I1) + 1k(I3 − I1) = 0.
– Solving the equations yields
I3 = −23 mA and therefore Io = I1 − I2 − I3 = −43 mA.
17
Example: Use of Supermesh
❑
Find Io.
– 3 mesh currents are specified as shown below.
– The unknown voltage across the 4-mA current source is assumed to be Vx.
– Constraint equations: I1 = 2 mA and I2 − I3 = 4 mA.
 KVL equations for meshes 2 and 3:
2kI2 + 2k(I2 − I1) − Vx = 0
−6 + 1kI3 + Vx + 1k(I3 − I1) = 0
 Adding these two equations yields
−6 + 1kI3 + 2kI2 + 2k(I2 − I1) + 1k(I3 − I1) = 0
 The unknown voltage Vx has been eliminated.
 2 constraint equations together with this
latter equation, yield the desired result.
18
Example: Use of Supermesh
❑
Find Io.
– The supermesh approach can be used to avoid introducing the unknown
voltage Vx.
– The supermesh is created by mentally removing the 4-mA current source.
 Writing the KVL equation around the dotted
path, which defines the supermesh, yields
−6 + 1kI3 + 2kI2 + 2k(I2 − I1) + 1k(I3 − I1) = 0
 This supermesh equation is the same as
that obtained earlier by introducing the
voltage Vx.
19
Circuits w/ Dependent Sources
First, we treat the dependent source as if it were an independent
source when writing the KVL equations.
❑ Then, we write the controlling equation for the dependent source.
❑ Let us consider the circuit shown below.
❑
 Controlling equation:
Vx = 4kI1
 These equations can be combined to
produce
−2kI1 + 2kI2 = 0
– Loop-current equations:
−2Vx + 2k(I1 + I2) + 4kI1 = 0
−2Vx + 2k(I1 + I2) − 3 + 6kI2 = 0
−6kI1 + 8kI2 = 3.

In matrix form, the equations are
 − 2000 2000  I1  0
 − 6000 8000 I  = 3 .

  2  
20
Example
❑
Derive loop-current equations.
 Controlling equations:
Vx = 2k(I3 − I1)
Ix = I4 − I2
 Combining these equations yields
I1 = 4k
I1 + I2 − I3 = 0
1kI2 + 3kI3 − 2kI4 = 8
– Constraint equations:
I1 = 4k
I2 = Vx2k
– Mesh equations:
−1kIx + 2k(I3 − I1) + 1k(I3 − I4) = 0
1k(I4 − I3) + 1k(I4 − I2) + 12 = 0
1kI2 + 1kI3 − 2kI4 = 12.
 In matrix form, the equations are
0
0
0  I1   0.004 
1
1

1
−1
0  I 2   0

 =
.
0 1000 3000 − 2000  I3   8


  

0 1000 1000 − 2000 I 4  12

21
Example
❑
Find Vo in the network using nodal analysis.
 KCL equation at the supernode:
−
2 V x V x − V3 3V x − V3 3V x − 4
+
+
+
+
=0
k 1k
1k
1k
1k
 KCL equation at the node labeled V3:
−

− 4V x + 2V3 = 2
V4 = 4
→ V1 = 3Vx
Simplified equations:
8V x − 2V3 = 6
 Constraint equations:
V1 − V2 = 2 Vx where Vx = V2
2 V3 − 3V x V3 − V x
+
+
=0
k
1k
1k

Solving these equations, we obtain
Vx = 2 V, V3 = 5 V, and
Vo = 3Vx − V3 = 1 V.
22
Example
❑
Find Vo in the network using loop analysis.
 KVL equations for meshes 2 and 4:
−2Vx + 1kI2 + 1k(I2 − I3) = 0
1k(I4 + I3 − I1) − 2Vx + 1kI4 + 4 = 0
 Controlling equation:
Vx = 1k(I1 − I3 − I4)
 Substituting the equations for I1 and
I3 into the 2 KVL equations yields
 The network has 4 loops, and thus
4 linearly independent equations
are required.
 Constraint equations:
I1 = 2k
2kI2 + 2kI4 = 6
4kI4 = 8.
 Solving these equations for I2 and I4,
we obtain I4 = 2 mA and I2 = 1 mA,
leading to Vo = 1 V.
I3 = −2k
23
Nodal Analysis vs. Loop Analysis
❑
It is important to examine the circuit carefully before selecting an
analysis approach.
– One method could be much simpler than another, and a little time invested
up front may save a lot of time in the long run.
❑
One consideration in the selection of a method should be the number
of linearly independent equations that must be formulated.
– For an N-node circuit, N − 1 linearly independent equations must be
formulated to solve for N − 1 node voltages.
– An N-loop circuit requires the formulation of N linearly independent
equations.
24
Nodal Analysis vs. Loop Analysis
❑
In the previous example, the same circuit was solved using both node
analysis and loop analysis.
– The circuit has four unknown node voltages, which requires four linearly
independent equations. Because there are two voltage sources, two
constraint equations are needed.
– This same circuit has four loops, which requires four linearly independent
equations.The two current sources produce two constraint equations.
 The effort required to solve this
circuit using either nodal or loop
analysis is similar.
 However, this is not true for many
circuits.
25
Nodal Analysis vs. Loop Analysis
❑
Consider the circuit shown below.
– This circuit has 8 loops.
– Selection of the loop currents such that only 1 loop current flows through
the independent current source leaves us with 7 unknown loop currents.
– This circuit has 7 nodes.
– By selecting the bottom node as the reference node, 4 of the node voltages
are known, leaving just 2 unknown node voltages−the node voltage across
the current source and the node voltage across the 3- and 6- resistors.
– Applying KCL at these 2 nodes yields 2 equations that can be solved for 2
unknown node voltages.
26
Loop Analysis: Strategy
❑
Determine the number of independent loops in the circuit.
❑
Assign a loop current to each independent loop.
– For an N-loop circuit, there are N loop currents, requiring N linearly
independent equations to solve for the loop currents.
❑
If any current sources in the circuit, either of two techniques can be employed.
– In the first case, one loop current is selected to pass through one of the
current sources. The remaining loop currents are determined in such a way
that they don’t flow through the current source.
– In the second case, a current is assigned to each mesh in the circuit and
use super mesh approach.
27
Loop Analysis: Strategy
❑
Write a constraint equation for each current source−independent or
dependent−in terms of the assigned loop current using KCL.
– NI current sources yield NI linearly independent equations.
– For each dependent current source, express the controlling variable for
that source in terms of the loop currents.
❑
Use KVL to formulate the remaining N − NI linearly independent
equations.
❑
Treat dependent voltage sources like independent voltage sources
when formulating the KVL equations.
– For each dependent voltage source, express the controlling variable in
terms of the loop currents.
28
Summary
❑
Loop Analysis
– Basic concepts
– Planar circuits
– Circuits containing only independent voltage sources
– Mesh currents and mesh analysis
– Symmetry of mesh equations
– Circuits containing independent current sources
– Supermesh approach
– Circuits containing dependent sources
– Nodal analysis vs. loop analysis
– Strategy of loop analysis
29
Recommended Problems
❑
Problems 3.2.1, 3.2.5, 3.2.12, 3.2.14, 3.2.22, 3.2.26, 3.2.30, 3.2.35, 3.2.38
Nodal and Loop Analysis Techniques
30