Download Electrical Networks

2.8 Applications to Electrical Networks.
In this section we look at applications of linear equations to electrical networks. For simplicity we restrict our
attention to networks with voltage sources and linear resistors.
2.8.1 The Full Set of Equations for an Electrical Network.
In this section we discuss the full set of
equations for an electrical network with
R1 = 2 Ω
v1
-
voltage sources and linear resistors.
These are the set of equations (9) below.
We illustrate the concepts by means of an
example.
i6 = f = 3 a
R2= 6 Ω
R5 = 7 Ω
e3 = 5 v
+
i5
e2 = 3 v
-
-
e5 = 8 v
line #1
R6
R3 = 4 Ω
-
Applications, 4th ed. by Gilbert Strang)
Consider the network at the right. Given
v2
i1
node #1
Example 1. (based on Example 1 in
section 2.5 of Linear Algebra and Its
e1 = 4 v
+
+
i3
+
i2
the values of R1, …, R5, e1, …, e5 and f
one wants to find i1, …, i5.
v4
i6 = f = 3 a
+
i4
Like the electrical networks we will be
e4 = 6 v
R4 = 3 Ω
v3
considering, the network at the right
consists of the following three types of electrical components.
lines
These are conducting paths, such as wires, through which current can flow.
voltage sources These create electrical fields that cause charged particles in lines to move and hence
produce currents in the lines. Voltage sources include batteries, power supplies and
generators.
resistors
These impede the flow of current in a line.
We number the lines 1, 2, …, m where
m = number of lines
In the above example, there are m = 6 lines. Current is the flow of charged particles in a line, mostly electrons.
The current i in the line is the rate at which charge is crossing a cross section in the line. Positive charges
moving in the positive direction in the line count positively and those moving in the opposite direction count
negatively. For negative charges it is just the reverse. To determine the sign of the current, it is necessary to
choose a positive direction in the line. Charge is measured in coulombs and current is measured in amperes.
One ampere of current flowing in a line means there is a net flow of one coulomb of charge past any cross
section in one second. We let
2.8 - 1
ij = current in line j
In Example 1 we have
i1 = current in line 1
. . .
i6 = current in line 6
The lines are connected at nodes. We number the nodes 1, 2, …, n where
n = number of nodes
In Example 1 there are n = 4 nodes.
Voltage sources have two terminals, one is the positive terminal designated by + and the other is the negative
terminal designated by -. The voltage e of a voltage source is a measure of the strength of the voltage source.
More precisely, it is the change in electrical potential as one goes from the negative terminal to the positive
terminal. We let
ej = voltage source in line j
More generally, for each point p in space one can assign an electrical potential. Let
v(p) = electrical potential at point p in space.
vk = electrical potential at in node k
In Example 1 we have
v1 = electrical potential at node 1
. . .
v4 = electrical potential at node 4
Given any pair of points p and q in space there is a potential difference v(q) - v(p) as one goes from p to q. We
let
(p, q) = v(q) - v(p) = potential difference as one goes from p to q.
j = potential difference as one traverses line j = ve(j) - vs(j)
s(j) = node at start of line j
e(j) = node at end of line j
In Example 1 we have
2.8 - 2
(1)
1
2
3
4
5
6
=
=
=
=
=
=
v2
v3
v3
v4
v4
v4
1
  
 
  
2
3
or
4
=
5
6
-
v1
v2
v1
v3
v1
v2
0
-1
-10
0
-1 1
-1
0
0
0
-1
1
2
3
4
5
6
 
or  = Av where  =
 ,
 
0
1
1
-1
0
0
0
0
0
1
1
1




 v1 
 v2 
 v3 
 v4 
0
-1
A= 0
-1
0
-1 1
-1
0
0
0
-1
0
1
1
-1
0
0
0
0
0
1
1
1




 v1 
v2
and v =  v 
 3
 v4 
A is called the line-node incidence matrix of the circuit. Note that
Ajk =
 1
 -1
 0
if line j goes into node k
if line j goes out of node k
if line j doesn't touch node k
Typically, as p and q move along lines, the potential difference between p and q doesn’t change except when p
or q move through a circuit element such as a resistor or voltage source. Voltage is measured in volts. If V is
the potential difference between two points, then the electric field transfers an amount of energy equal to QV to a
particle with charge Q that moves between these two points.
v(p) is only defined up to an additive constant. Often one picks some point p0 in space and assigns that point the
potential value zero, i.e. v(p0) = 0. Often that point is called the ground because in many situations one assigns
the surface of the earth the potential value zero.
This is reflected by the fact that Av = 0 if and only if 1 = 2 = … = m = 0 if and only if v1 = v2 = … = vn. Thus
1
1
the null space of A consists of multiples of the vector . and, hence, is one dimensional. Thus A has rank
.
1








n - 1 in general. This is reflected in the fact that the columns of A add up to the zero vector. Each column is the
negative of the sum of the others. Any n - 1 columns is a basis for the null space of A. In Example 1 A has rank
3.
2.8 - 3
In circuit problems, such as the one above, one often chooses a node to have voltage zero and labels this as node
n. In the above example, we set
(2)
v4 = 0
We use (2) in (1). Also, we put aside the last equation 6 = v4 - v2 = v2. Then we get
1
2
3
4
5
(3)
= v2 - v1
= v3 - v2
= v3 - v1
= - v3
= - v1
 



1
or
2
3
4
5





=
0
-1
0
-1
-1 1
-1
0
0
0
0
1
1
-1
0




 v1 
 v2 
 v3 
We can write this as
^ ^v
^ = 
(4)
 

where ^ = 


1
2
3
4
5
 -10
, ^ = -1
 0
 -1
1
-1
0
0
0
0
1
1
-1
0




 v1 
^ the reduced incidence matrix.
and ^v =  v2 . We shall call 
 v3 
The next set of equations comes from one of the basic principles of circuit analysis, namely, Kirchoff's Current
Law (KCL) which states
(KCL)
 sum of currents  =  sum of currents 
going into a node
going out of the node
at each node
or
ins = outs
for short. Applying this to Example 1 this gives
(5)
i1 = i2 + i6
i2 + i3 = i4
0 = i4 + i5 + i6
i
i
 ii
i
i1
0 = i1 + i3 + i5
or
-1
1
0
0
0
-1
1
0
-1
0
1
0
0
0
-1
1
-1
0
0
1
0
-1
0
1
2
3
4
5
6
i
i
or A i = 0 where A is the line-node incidence matrix discussed above and i = i
i
i
i1
2
3
T
4
5
6
2.8 - 4

.






0
0
= 0
 
0
The equations (5) are not all independent. The last equation, 0 = i4 + i5 + i6, is equivalent to the sum of the first
three. This is because the columns in A add up to the zero vector. So we can omit the last equation in (5). Also,
we replace i6 by f. This gives
- i1
(5)
- i3
- i5 = 0
i1 - i2
i2 + i3 - i4
= f
i
i
i
i
i1
or
-1 0 -1 0 -1
 1 -1 0 0 0 
 0 1 1 -1 0 
= 0
2
3
4
5




=
0
f
0
We can write this as
(6)
^ T ^i = F

i
^ is the reduced incidence matrix discussed above, ^i =  i
where 
i
i
i1
2
3
4
5

 and F = 0f .
0


We shall only consider resistors that obey Ohm’s law. This says V = iR where V is the drop in potential as one
goes through the resistor in the direction of the current, i is the current in the line and R is a constant called the
resistance of the resistor. Resistance is measured in Ohms. A resistor with resistance R has the property that in
order to make a current of i amperes flow through the resistor it is necessary to supply a potential difference of
iR to the two ends of the resistor. If U = - V is the voltage change as one goes through the resistor in the
direction of the current, then Ohm's law can be restated as U = - iR. Applying this to the first five lines in
Example 1 one has
1 = e1 - R1i1
1
2
2 = e2 - R2i2
3 = e3 - R3i3
 



3
or
4 = e4 - R4i4
4
5




e
e
e
e
e1
=
5 = e5 - R5i5
or
(7)
^ = e - Ri^
Combining this with equation (4) gives the following equation
(8)
^ ^v + Ri^ = e

Combining with equation (6) gives
(9)
^ R   ^v 
e
 ^ T  ^  =  F 
0   i 
2.8 - 5
2
3
4
5




0 R 0 0 0   i
0 0 R 0 0   i
0 0 0 R 0  i
0 0 0 0 R   i
R1 0 0 0 0
2
-
i1
2
3
3
4
4
5
5




^ R 
 ^v 
e
,
u
=

 ^  and g =  F . In expanded form (9) becomes
^
T
0  
i
or Ku = g where K = 
(10)




- 1 1 0 R1
0 -1 1 0
-1 0 1 0
0 0 -1 0
-1 0 0 0
0 0 0 -1
0 0 0
1
0 0 0
0
0
R2
0
0
0
0
-1
1
0
0
R3
0
0
-1
0
1
0
0
0
R4
0
0
0
-1
0
0
0
0
R5
-1
0
0




v1
v2
v3
i1
i2
i3
i4
i5








v1
v2
v3
i1
i2
i3
i4
i5




=




e1
e2
e3
e4
e5
0
f
0
=




4
3
5
6
8
0
3
0




In Example 1 these equations are the following.




-1 1 0 2
0 -1 1 0
-1 0 1 0
0 0 -1 0
-1 0 0 0
0 0 0 -1
0 0 0
1
0 0 0
0
0
6
0
0
0
0
-1
1
0
0
4
0
0
-1
0
1
0
0
0
3
0
0
0
-1
0
0
0
0
7
-1
0
0




Solving these equations (see section 2.9 or 2.10) one obtains
 vv 
 ii 
 ii 
i 
v1
3
(11)
1
2
 354

201
1  - 41 
16
19
 1329 
 28 
348
2
=
3
4
5
 18.6

10.6
2.16 
 -0.84
 0.68

1.53
 1.47 
18.3

2.8.2 Nodal Analysis.
In the previous section we combined equations (6) and (8) into equation (9) and solved to get the solution (11).
If all the resistances are non-zero one solve these equations in a slightly different fashion called nodal analysis.
One multiplies equation (8) by
R-1
=










1
0 0 0 0
R1
1
0
0 0 0
R2
1
0 0
0 0
R3
1
0 0 0
0
R4
1
0 0 0 0
R5
2.8 - 6
to obtain
(12)
^ ^v + ^i
R-1
= R-1e
^ T and uses (6) to obtain
Then one multiplies by 
^ TR-1
^ ^v + F = 
^ TR-1e

or
(13)
Cv^ = b
where
(14)
^ TR-1
^
C = 
^ TR-1e - F
b = 
After solving (13) for ^v one gets ^i from (12). In Example 1 one obtains
 v1 
 18.3 
 v2  = -  18.6 
 v3 
 10.6 
See section 2.9 or 2.10. To get the currents in the lines we use (12)
^i = R-1e - R-1
^ ^v =
=




- 2.16
0.84
0.68
1.53
1.47




2.8.3 Loop Analysis.
Another approach to solving the equations (9) is to solve the equation ATi = 0 for some of the ij’s in terms of the
others. To do this we reduce AT to reduced row echelon form. In Example 1 one has
-1 0 -1 0 -1 0
A =  1 -1 0 0 0 -1 
 0 1 1 -1 0 0 
T
Add row 1 to row 2 giving
-1 0 -1 0 -1 0 
 0 -1 -1 0 -1 -1
 0 1 1 -1 0 0 
Add row 2 to row 3 giving
R =
-1 0 -1 0 -1 0 
 0 -1 -1 0 -1 -1
 0 0 0 -1 -1 -1 
2.8 - 7
It in now essentially in reduced row echelon form. We write out the equation Ri = 0.
- i1
- i3
- i5
- i2 - i3
= 0
- i5 - i6 = 0
- i4 - i5 - i6 = 0
Solve for i1, i2 and i4 in terms of i3, i5 and i6. We get
i1 = - i3 - i5
i2 = - i3 - i5 - i6
i4 =
- i5 - i6
Therefore
i
i
 ii
i
i1
2
i =
3
4
5
6




 -1 
1
= 0 ,
0
0




- i3 - i5
3
=
5
6
3
5
6
5
6
-1
The vectors 1
-i -i - i
 i
 - ii - i
 i
 -1 
0
= -1 and 
1
0
-1
2
 -1 
1+i
 00 
0
-1
= i3
5
 -1 
0+i
 -11 
0
-1
 -1 
0
 -10 
1
0
6
 -1 
0
= -1 are in the nullspace of A . They are often called loop
0
1
0
3
T
vectors because they correspond to loops in the network. Since they are in the nullspace of AT, they are
orthogonal to the column space of A. It follows that
(15)
(k)T = 0
since  is in the column space of A. We replace i6 by f and combine the first two terms on the right into one. We
get
^i =
 -1
1
0
0
-1 -1
-1
0
-1
1
 i

i

3
5
-f
 + 0

-f
0
0
or
(16)
^i = L  i3  + h
 i5 
2.8 - 8
 -1
1
where L =
0
0

 
 and h =  -0f . It follows from (15) that L ^ = 0. Using (7) we get L (e - Ri^) = 0.

-f

0
-1 -1
-1
0
-1
1
0
T
i3
We use (16) to get LT(e - R(L i  + h)) = 0 or
5
(17)
i3
LTRL i  = LT(e - Rh)
5
i3
We solve this for  i  and substitute in (16) to get ^i . See section 2.9 or 2.10.
5
2.8 - 9
T