Download Nodal Analysis

1.7 Applications to Electrical Networks.
In section 1.1 we talked about applications of systems of equations to getting approximate solutions of boundary
value problems. In this section we look at applications of linear equations to electrical networks with voltage
sources and linear resistors.
1.7.1 Nodal Analysis.
In this section we discuss one approach called nodal analysis. In nodal analysis we construct a system of linear
equations that the voltages at the nodes satisfy. We illustrate the concepts by means of an example.
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
The electrical networks we consider consist 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 = 7 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
ij = current in line j
1.7.1 - 1
In Example 1 we have
i1 = current in line 1
. . .
i7 = current in line 7
The lines are connected at nodes. We number the nodes 1, 2, …, N where
N = number of nodes
In Example 1 there are N = 6 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. 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 in node 1
. . .
v6 = electrical potential at in node 6
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
u(p, q) = v(q) - v(p) = potential difference as one goes from p to q.
uj = potential difference as one traveses 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
u 1 = v2 - v 1
u 2 = v1 - v 3
u 3 = v4 - v 2
(1)
u 4 = v4 - v 3
u 5 = v3 - v 5
u 6 = v6 - v 4
or




u1
u2
u3
u4
u5
u6
u7




=
u 7 = v5 - v 6
1.7.1 - 2




-1 1 0 0 0 0
1 0 -1 0 0 0
0 -1 0 1 0 0
0 0 -1 1 0 0
0 0 1 0 -1 0
0 0 0 -1 0 1
0 0 0 0 1 -1
  vv 
  vv 
  vv 

1
2
3
4
5
6


or u = Jv where u =


u1
u2
u3
u4
u5
u6
u7

,




J=


-1 1 0 0 0 0
1 0 -1 0 0 0
0 -1 0 1 0 0
0 0 -1 1 0 0
0 0 1 0 -1 0
0 0 0 -1 0 1
0 0 0 0 1 -1
v

v 
 and v =  vv 
v 

v 

1
2
3
4
5
6
J is called the line-node incidence matrix of the circuit. Note that
Jjk
 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 0, i.e. v(p0) = 0. Often that point is called the ground because in many problems one assigns the
surface of the earth the potential value 0.
Many circuit problems, such as the one above, have a single voltage souce. In such cases one often labels the
negative terminal of the voltage source as node N and chooses this node to have voltage 0. In the above
example, we set
(2)
v6 = 0
Then the positive terminal of the voltage souces has voltage E where
E = potential difference between the positive and negative terminal of the voltage source
In such cases one often labels the positive terminal of the voltage source as node N-1. This node has voltage E.
In the above example, one has
(3)
v5 = 6
Using (2) and (3) in (1) we get
u 2 = v1 - v 3
u 3 = v4 - v 2
u 4 = v4 - v 3
u
u
 uu
u
u1
u 1 = v2 - v 1
or
u 5 = v3 - 6
u 6 = - v4
and u7 = 6. We can write this as
1.7.1 - 3
2
3
4
5
6




=
1
0
 00
0
-1 1
0
-1
0
0
0
0
-1
0
-1
1
0
0
0
1
1
0
-1




0
0
06
0
0
 v1 
 v2   v3 
 v4 
u^ = K ^v – e
(4)
u
u
where u^ = u
u
u
u1
2
3
4
5
6

1
 , K = 0

 00

0
-1 1
0
-1
0
0
0
0
-1
0
-1
1
0
0
0
1
1
0
-1




0
0. We shall call K the reduced incidence
06
0
0
 v1 
^v =  v2  and e =
 v3 
 v4 
matrix. We also set
n = N – 2 = remaining number of nodes whose voltage values are to be found.
In the above example, one has
n = 4
The next set of equations comes from one of the basic priciples 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
i2 = i1
i1 = i3
i5 = i2 + i4
(5)
or
i3 + i4 = i6
i7 = i5




-1
1
0
0
0
0
1
0
-1
0
0
0
0
-1
0
1
0
0
0
0
-1
1
0
0
0
0
1
0
-1
0
0
0
0
-1
0
1
0
0
0
0
1
-1








i1
i2
i3
i4
i5
i6
i7




0
0
00
0
0
=
i6 = i7


or J i = 0 where J is the line-node incidence matrix discssed above and i =


T
i1
i2
i3
i4
i5
i6
i7

.


The equations (5) are not all independent. In fact the last equation, i6 = i7, is equivalent to the sum of the first
six. So we can omit the last equation. Also, we shall wait to use the next to last equation, i7 = i5, until later.
This leaves us with the first four, i.e.
1.7.1 - 4
i2 = i1
i
i
 ii
i
i1
i1 = i3
i5 = i2 + i4
-1
1
0
0
or
i3 + i4 = i6
1
0
-1
0
0
-1
0
1
0
0
-1
1
0
0
0
-1
0
0
1
0
2
3
4
5
6




0
0
= 0
 
0
We can write this as
(6)
KT ^i = 0
i
i
where K is the reduced incidence matrix discussed above and ^i = i
i
i
i1
2
3
4
5
6

.


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. The conductance G of a resistor is the
1
reciprocal of the resistance, i.e. G = . With this definition Ohm's law becomes i = - GV. Applying this to the
R
first six lines in Example 1 one has
i1 = - u1
i2 = - u2
1
i3 = - 2 u3
1
i4 = - 5 u4
1
i5 = - 2 u5
or




i1
i2
i3
i4
i5
i6




1
2
=
i6 = - u6
or
(7)
0 0 0 0 0 


- 0 0 0 0 0
0 0 0 0 0 
0 0 0 0 0 1 
1 0 0 0 0 0
0 1 0 0 0 0
^i = - G u^
1.7.1 - 5
1
5
1
2
u
u
 uu
u
u1
2
3
4
5
6




0 0 0 0 0 


where G = 0 0 0 0 0 is the diagonal matrix with the conductances along the diagonal.
0 0 0 0 0 
0 0 0 0 0 1 
1 0 0 0 0 0
0 1 0 0 0 0
1
2
1
5
1
2
To summarize, the various relations between the potentials, voltages changes and currents in the circuit are
captured in equations (4), (6) and (7), i.e.
u^ = K ^v – e
KT ^i = 0
^i = - G u^
Combining the first and third gives
(8)
^i = G(e - Kv^)
Combining this with the second gives
(9)
Av^ = b
where
(10)
A = KTGK
b = KTGe
In Example 1 one has
00 10 0 00 00 00 -11 10 -10 00 
0 0 0 0 0   00 -10 -10 11 
0 0 0 0 0   00 00 10 -10 
0 0 0 0 0 1 
 10
0
0
0
1 0 0 0 0 0
-1
1
2
GK =
1
5
=
1
2
0
0
0
0
-1
1
-1
1
A = KTGK =  0

0
1
0
-1
0
0
-1
0
1
0
0
-1
1
0
0
1
0
0
0
0
-1
1.7.1 - 6
1 0
0 -1
-
1
2
0
1
0 -5
0
1
2
0
0
1 0
0 -1
1
-2
0
1
0 -5
0
1
2
0
0



0
- 1
0
0
1
2
1
5



0
- 1
0
0
1
2
1
5
- 1

0
0
2 -1 -1
=
3
2
17
10
1 1
2 5
0
-
0
0
1
-2
1
-5
17
10




0 0 0 0 0  00
0 0 0 0 0  00
0 0 0 0 0  60
0 0 0 0 0 1 
1 0 0 0 0 0
0 1 0 0 0 0
=
0
0
03
0
0
0
0
-1
0
0
03
0
0
1
2
Ge =
1
5
1
2
0
-1
1
T
b = K Ge =  0

0
1
0
-1
0
0
-1
0
1
0
0
-1
1
0
0
1
0
0
0
= 3
 
0
So the equation Av^ = b becomes
- 1

- 1
0
2 -1 -1
3
2
17
10
1 1
2 5
0
-
0
0
1
-2
1
-5
17
10
v
v
  vv


0
 = 0
3
3
0
4
1
2
Solving gives
 v1 
3.0638
 v2  = 2.4255
 v3 
3.7021
 v4 
1.1489
See section 1.7.3. To get the currents in the lines we use (8)
1
 0
-
 0
0

0
-1
^i = Ge - GKv^ =
0
0
0
0
3
0
1 0
0 -1
1
-2
0
1
0 -5
0
1
2
0
0
See section 1.7.3
1.7.1 - 7
 3.0638
 2.4255

3.7021
 1.1489
0
- 1
0
0
1
2
1
5
0.6383
0.6383
0.5106

1.1489
1.1489
0.6383
=