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1.7.2 Loop Analysis.
In this section we discuss another approach to electric circuits called loop analysis. This time we construct a
system of linear equations that the currents in the lines satisfy. We illustrate the concepts by means of the sane
example as in section 1.7.1.
Example 1. (taken from Example 1.2.6 on p. 14 of Fundamentals of Matrix Computations, 3rd ed. by David
Watkins)
R2 = 1 Ω
v1
i2
R5 = 2 Ω
v3
i5
v5 = 6
i1
node #1
i4
+
E = 6 volts
R1 = 1 Ω
-
R4 = 5 Ω
line #1
R3 = 2 Ω
R6 = 1 Ω
i3
i6
i7
v6 = 0
v2
v4
A loop L is a sequence a0, a1, …, ap of nodes with the following properties.
i.
A line bj joins each pair aj-1 and aj of nodes in the sequence. The line can be traversed is either
the positive direction or the negative direction as one goes from aj-1 to aj.
ii.
The first and last nodes are the same, i.e. a0 = ap.
iii.
The only nodes that are repeated are the first and last, i.e aj  ak if j  k and 0  j < p and
0  k < p.
Let's write L ~ (a0, a1, …, ap) in this case.
Example 1 (continued). In Example 1, nodes 1, 2, 4, 3 and 1 form a loop L1. Nodes 3, 4, 6, 5 and 3 also form
a loop L2. A third loop L3 is nodes 1, 2, 4, 6, 5, 3 and 1. So we have
L1 ~ (1, 2, 4, 3, 1)
L2 ~ (3, 4, 6, 5, 3)
L3 ~ (1, 2, 4, 6, 5, 3, 1)
A loop can equally well be described by the sequence b1, …, bp of lines that join the nodes. Let's write
L  (b1, b2, …, bp) in this case. Because vertices don't repeat in a loop, neither do the lines.
1.7.2 - 1
Example 1 (continued). L1 consists of lines 1 in the positive direction followed by line 3 in the positive
direction, then line 4 in the negative direction and finally line 2 in the positive direction. L2 consists of lines 4,
6, 7 and 5 all in the positive direction. L3 consists of lines 1, 3, 6, 7, 5 and 2 all in the positive direction. So
L1  (1, 3, 4, 2)
L2  (4, 6, 7, 5)
L3  (1, 3, 6, 7, 5, 2)
 qq12 
Each loop has an associated loop vector  …  where
 qM 
 1
qj =  -1
 0
if line j is traversed in the positive direction in the loop
if line j is traversed in the positive direction in the loop
if line j is not in the loop
 qq12 
and M is the number of lines. Let's write L =  …  in this case.
 qM 
Example 1 (continued). The loop vectors in Example 1 are
 11 
-1
 00 
0
1
L1 =
00
1
11
1
0
L2 =
11
0
11
1
1
L3 =
Loops are important because of Kirchoff'sVoltage Law (KVL) which states
(KVL)
the sum of the voltage changes around a loop = 0
To be more precise let u(p, q) = v(q) - v(p) be the potential difference as one goes from p to q. Then KVL states
u(a0, a1) + u(a1, a2) + … + u(ap, a0) = 0
This is true because
(v(a1) - v(a0)) + (v(a2) - v(a1)) + … + (v(a0) - v(ap)) = 0
Another way to write this is the following. Let ub1, ub2, …, ubp be the voltage changes in lines b1, …, bp as one
traverses the lines in the positive direction, then KVL says
(1)
qb1ub1 + qb2ub2 + … + qbpubp = 0
or
q1u1 + q2u2 + … + qMuM = 0
1.7.2 - 2
Example 1 (continued). This equation becomes
u1 + u3 - u 4 + u2 = 0
for L1
u4 + u6 + u 7 + u5 = 0
for L2
u1 + u3 + u 6 + u7 + u5 + u2 = 0
for L3
Note that the third equation is the sum of the first two, so not all three equations are independent. So the third
equation doesn't add anything to the first two and we shall only use the first two. The fact that the third equation
 11 
= -1, L
 00 
0
1
is a linear combination of the first two can be expressed by saying that the loop vectors L1
2
00
= 1 and
11
1
0
11
= 0 for the three loops are linearly dependent.
11
1
1
L3
In general, we want to choose linearly independent loops. There is a theorem that says that one can choose
P = M - N + 1 linearly independent loops and no more. Here M is the number of lines and N is the number of
nodes. Let's suppose this has been done and let's denote their vectors by
(2)
L1 =
 qq1121 
…
 qM1 
L2 =
 qq1222 
…
 qM2 
...
LP =
 qq1P

2P
…
 qMP 
If the circuit can be drawn in the plane without any lines crossing, then one can choose the loops as the boundary
of the regions enclosed by the lines. Corresponding to these vectors we have the equations
q11u1 + q21u2 + … + qM1uM = 0
q12u1 + q22u2 + … + qM2uM = 0
(3)
...
q1Pu1 + q2Pu2 + … + qMPuM = 0
which we can write in matrix form as
(4)
QTu = 0
 q11
q21
where Q =  .

qM1
q12 . . q1P 
q22 . . q2P 
. . . 
qM2 . . qMP
 u1 
u2
and u =  … .
 
 uM 
Example 1 (continued). Here M = 7 and N = 6 so P = M – N + 1 = 2, so we know that L1 and L2 are the most
independent loops that one can find. The equations (3) are
1.7.2 - 3
u1 + u3 - u 4 + u2 = 0
for L1
u4 + u6 + u 7 + u5 = 0
for L2
or
u 
0  u 
u
1 u
u 
u 
u1
2
(5)
1 1 1 -1 0 0
0 0 0 1 1 1
3
4
= 0
5
6
7
 11
Here Q = - 1
 00
0
1 0
0
0
1
1
1
1

.


 qq12 
A loop vector L =  …  for a given set of lines and nodes can be regarded as a vector of currents in the lines for
 qM 
some configuation of resistances and voltage sources in the lines. This is because a loop vector satisfies
Kirchoff's current law discussed in the previous section, i.e. JTL = 0 where J is the incidence matrix discussed in
the previous section. The same theorem that gave the maximal number of independent loops as P = M - N + 1
also tells us that the current vector i for a given set of values of the resistances and voltage sources can be
written as a linear combination of the loop vectors (2). Thus
i = c1L1 + c2L2 + … + cPLP
or
(6)
i = Qc
 ii12 
 cc12 
 iM 
 cP 
where i =  …  is the vector of currents in the lines and c =  …  is the vector of as yet unknown coefficients.
 ii   11
Example 1 (continued). The equation (6) is  i  = - 1
 ii   00
i   0
i1
2
3
4
5
6
7
1 0
0
0
1
1
1
1

 ( cc ) .


1
2
The last set of equations comes from Ohm's law in a similar fashion to the case of nodal analysis in the previous
section. A typical line is shown at the right. Here ij, uj and Rj are
Rj
ij
the current, voltage change and resistance in the line. It the line
has a voltage source the ej is the voltage change as we go from the
terminal where the line enters to the terminal where the line ends.
This may be either E or – E depending on whether the line goes in
1.7.2 - 4
uj
ej
-
+
the negative or positive terminal. Applying Ohm's law gives
u1 = - R1i1 + e1
...
uj = - Rjij + ej
...
u M = - R Mi M + e M
or
R1 0 … 0   i1 
 uu12 
 ee12 
… 0
i2
0
R

2
+ …
… = -…
  … 
 uM 
 eM 
i
0 0 … RM M
or
(7)
u = - Ri + 
 ee12 
where  =  … .
 eM 
 uu  00
Example 1 (continued). This equation becomes  u  = - 0
 uu  00
 u  0
u1
2
3
4
5
6
7
1 0
1
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
5
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
  ii  00
  i  + 0.
  ii  00
  i  6
i1
0
2
3
4
5
6
7
To summarize, the various relations between the voltages changes, currents and multipliers ck in the circuit are
captured in equations (4), (6) and (7), i.e.
QTu = 0
i = Qc
u = - Ri + 
Combining the first and third gives
(8)
QTRi
= QT 
Combining this with the second gives
(9)
Ac = b
where
(10)
A = QTRQ
b = QT
1.7.2 - 5
In Example 1 one has
RQ =
00
0
00
0
1 0
1
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
5
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
  11
 - 1
  00
 0
1 0
0
0
1
1
1
1




 12
- 5
 00
0




 12
- 5
 00
0




1 0
0
0
5
2
1
0
=
1 1 1 -1 0 0 0
A = Q RQ =  0 0 0 1 1 1 1 
T
0
0  0
0
1 0
0
6
1 0
0
0
5
2
1
0
=
 9 -5
- 5 8 
0
1 1 1 -1 0 0
0 0 0 1 1 1
b = QT =
=
0
6
So the equation Ac = b becomes
 9 -5
- 5 8 
( cc )
1
2
=
0
6
or
9c1 - 5c2 = 0
- 5c1 + 8c2 = 6
The first equation gives c2 = 9c1/5. Substituting in the second equation gives (- 5 + 72/5)c1 = 6 or
c1 = 30/47  0.6383. So c2 = 54/47  1.1489.
1.7.2 - 6