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ECE 3144 Lecture 14
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Loop analysis
• In loop analysis, the unknown parameters
are currents and KVL is employed to
determine the unknowns.
• If there are N independent loops, N
independent equations are needed to
describe the network.
2
Nodal analysis vs. loop analysis
We would like to consider nodal analysis and loop analysis as mirror
scenarios:
• In nodal analysis, the
unknown parameters are
nodal voltages and KCL is
used to solve the problem.
• If there are N nodes in the
network, N-1 equations
are required to solve the
problem.
• KCL is applied to each
nonreference node.
• In loop analysis, the
unknown parameters are
current and KVL is
employed to determine the
unknowns.
• If there are N independent
loops, N independent
equations are needed to
describe the network.
• KVL is applied to each
independent loop.
3
Loop analysis case 1: containing independent voltage source only
vS1
a
2 kW
b vS2
+
d
+
12 V
2 kW
v1
c
2 kW
v2
-
I1
4V
I2
e
•Let us first identify how many independent loops here:
+
Loop 1 and loop 2
v3
•The unknown variables are loop currents: I1 and I2
- •Now we determine the physical currents through
each branch.
f
Branch d->a: I1
Branch a->b: I1
Branch b->c: I2
Branch c->f: I2
Branch f->e: I2
Branch b->e: I1-I2
Applying KVL to loop 1:
v1 – vS1 + v2 = 0
=>
2000I1 + 2000(I1-I2) -12 = 0
=>
4000I1 - 2000I2 = 12
=>
-2000 (I1-I2) + 4 + 2000I2 = 0
(1)
Applying KVL to loop 2:
-v2 + vS2 + v3= 0
=> -2000I1 + 4000I2 = -4
(2)
There are two equations and two unknowns I1 and I2. You can solve them.
I1 =10/3 mA and I2 =2/3 mA
4
Loop analysis case 1: cont’d
The two equations are put into the matrix format:
 4000
 2000

2000   I1  12 
 



4000   I 2   4 
Again, we notice that the coefficient matrix for this type of circuits is symmetrical. We
will explain later why this coefficient matrix is symmetrical and how to write the mesh
equations by inspection.
An important definition: what is mesh: A mesh is a special kind of loop that does not
contain any loops within it. So when you traverse the path of a mesh, you do not
encircle any circuit elements. The majority of our loop analysis will involve writing
KVL equations for meshes, thus loop analysis usually is called mesh analysis.
The following example derives the general interpretation on the symmetry of the
coefficient matrix.
5
Loop analysis case 1: independent voltage source only
There are two independent loops (meshes) in
this circuit.
Applying KVL to loop 1
v1 + v3 + v2 – vS1 = 0
=>
i1R1 + (i1-i2)R3 + i1R2 - vS1 = 0
i1(R1 + R2 + R3) - i2R3 =vS1
=>
(1)
Applying KVL to loop 2:
v4 + v5 –v3 + vS2 = 0
=>
i2R4 + i2R5 – (i1-i2)R3 + vS2 = 0
-i1R3 +i2(R3 + R4 + R5) =-vS2
In the matrix format
=>
(2)
 R3
 R1  R 2  R3
  i1  vs1 

 i 2   vs2

R
3
R
3

R
4

R
5

  

It is again the Ohm’s law in matrix format: RI = V.
6
Remember in nodal analysis case 1 we have Ohms' law in matrix format GV = I
Loop analysis case 1: independent voltage source only
•
•
•
•
the R matrix is symmetrical.
In the first equation, the coefficient of i1 is the sum of all resistors through
which mesh current i1 flows; the coefficient of i2 is the negative of the sum of
the resistances common to mesh current 1 and mesh current 2. The right-hand
side of the equation is the algebraic sum of the voltages sources in mesh 1.
The sign of the voltage source is positive if it aids the assumed direction of
current flow 1 and negative if it apposes the assumed flow direction.
In the second equation, the coefficient of i2 is the sum of all resistors through
which mesh current i2 flows; the coefficient of i1 is the negative of the sum of
the resistances common to mesh current 1 and mesh current 2. The right-hand
side of the equation is the algebraic sum of the voltages sources in mesh 2.
The sign of the voltage source is positive if it aids the assumed direction of
current flow 2 and negative if it apposes the assumed flow direction.
In general, we assume all of the mesh currents to be in the same direction
(clockwise or counterclockwise). If KVL is applied to mesh j with mesh
current ij, the coefficient of ij is the sum of all resistors in mesh j; the
coefficients for other mesh currents ik (kj) are the negative sum of the
resistors common to mesh k and mesh j.The right-hand side of jth equation is
equal to the algebraic sum of the voltage sources in mesh j. These voltage
sources have a positive sign if they aid the current flow ij and a negative sum if
they appose it.
7
Loop analysis case 2: with dependent voltage source
We deal with circuits with dependent voltage source the same way as the independent
voltage source except that we need to write the controlling equations for the dependent
voltage source. However the symmetry of the R matrix may no longer exist.
In loop 1, apply KVL:
2 kW
2000 Ix
I1
2000 I x   2 kW I1   4 kW I1  I 2   0
+
6V
4 kW
I2
2 kW
Ix
Vo

(1)
In loop 2, apply KVL
  6 V   2 kW I 2   4 kW I 2  I1   0
(2)
Write the controlling equation for the dependent voltage source
I x  I2
(3)
There are three equations for three unknowns I1, I2, Ix.
I1 =3 mA I2 = 3 mA
Ix = 3 mA
8
Homework for lecture 14
• Problem 3.43, 3.45, 3.48, 3.51
• Due Feb 18
9