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Файл:The elements of the electrical circuit theory. Tutorial
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R
R
R
R
Fig. 10.2
Let us convert two active branches into one equivalent branch, but the
branch with current
I will remain unconverted, and therefore the current in
3
this branch will not change (Fig. 10.3). The parameters of the equivalent
branch are
ggg
equ
12
R
equ
E
equ
111
⇒ ,
12
11
EE
12
12
11
RR
12
equ
R
12
R
RR
12
,
R
.
Fig. 10.3
51
There is no doubt that, in the converted circuit (see Fig. 10.3), the cur-
R
I can be easily found using Ohm’s law:
rent
3
E
equ
I
3
equ
.
R
3
11. THE TRANSFERENCE OF THE SOURCES
OUT OF THE BRANCH
There is, in the complicated circuit, the branch with the ideal EMF
source and the zero resistance (Fig. 11.1). It is necessary to convert the circuit so that there is no the zero resistance branch in it.
Fig. 11.1
If we remove the ideal EMF source from the branch, then the potential
of the node d will be equal to the potential of the node 0, these nodes can
52
be joined together and there will be no the zero resistance branch in the
circuit.
The ideal EMF source (that is denoted as E) is located between the node
d and the node 0 and it is directed towards the node 0. Let us add exactly the
same ideal EMF sources E to all the branches that are joined with the node
0, but these ideal EMF sources E will be directed not towards the node 0,
but from the node 0 (Fig. 11.2). In this case, the potential of the node 0 will
change, but the potentials of the nodes a, b, c and d will remain unchanged.
Fig. 11.2
Now, there are two equal and oppositely directed ideal EMF sources in
the branch. These ideal EMF sources compensate each other, so they can be
removed from the branch (Fig. 11.3).
The nodes, between which there is the zero resistance branch and there
is no the EMF source, can be joined together. The new node has the same
potential as the potential of the node d before converting. Thus, both the
zero resistance branch and the node 0 are removed from the circuit
(Fig. 11.4).
53
Fig. 11.3
Fig. 11.4
Let us call this converting to the transference of the EMF source out of
the branch. In this converting, instead of the original EMF source, exactly
the same EMF sources appear in the other branches, which are joined with
54
the node. However, these new EMF sources are directed oppositely (if the
J
J
J
J
J
R
R
original EMF source was directed towards the node, then, in the equivalent
circuit, the EMF sources are directed from the node, and vice versa).
It is necessary to note that the transference of the EMF source out of the
branch is applicable to any branches, and not only to the zero resistance
branches.
For the transference of the current source out of the branch, it takes only
to connect exactly the same current sources in parallel to the other branches,
but so that the distribution of the currents in the circuit does not change.
Example 11.1 (the transference of the current source out of the branch)
In the circuit shown in Fig. 11.5, it is necessary to transfer the current
source out of the branch.
Fig. 11.5
The current of the source
node d. Therefore, in the converted circuit, the current
exits from the node c and enters into the
k
must also exit
k
from the node c and enter into the node d (Fig. 11.6).
However, let us suppose that the current
the node a; and, at the same time, the current
enters, for example, into
k
exits from the node a. In
k
this case, the distribution of the currents in the circuit will not change
(Fig. 11.7). This reasoning allows us to connect two current sources
parallel to the resistors
and
1
(Fig. 11.8).
3
55
k
in
J
R
R
Fig. 11.6
Fig. 11.7
This reasoning allows us to connect two current sources
to the resistors
and
1
(Fig. 11.8).
3
in parallel
k
Fig. 11.8
56
12. THE NODE POTENTIAL METHOD
For the analysis of the currents in the circuit, it takes only to know the
potentials of all the nodes. Then all the currents can be found using Ohm’s
law.
The electric circuit is the system of the branches that are connected to
each other by the nodes. There is no doubt that any branch is connected by
its ends to two nodes. However, let us suppose that the branch connects any
two nodes to each other. If there is no the branch between two nodes, we
can always suppose that the branch with the infinite resistance connects
these nodes. This reasoning will allow us to consider the node potentials of
the electric circuit in the most general form.
Let many branches are joined with the node n, and each branch, at its
opposite end, is also joined with the node (Fig. 12.1).
Fig. 12.1
In the electric circuit, all the branches can be divided into three groups.
The first group is the branches, which contain the EMF source and have
the finite conductance. It is necessary to note that this group can also include the branches with the finite conductance, but without the EMF source.
Let us further denote such branches with the letter i (Fig. 12.2 a).
57
The second group is the branches with the current source. These branch-
b
es have the infinite resistance (i.e. their conductance is zero). It is necessary
to note that this group can also include the branches with the infinite resistance, but without the current source. Let us further denote such branches
with the letter k (Fig. 12.2 b).
a)
)
Fig. 12.2
The third group is the branches, which contain the EMF source and have
the zero resistance (i.e. their conductance is infinity). As we found out
above, in such branches, we can always transfer the EMF source out of the
branch. Therefore, let us further consider only the circuit in which there are
no such branches, which would contain the EMF source and have the zero
resistance.
Let the branches i and the branches k are joined with the node n. Let us
denote the nodes that are opposite to the node n with the letters i and k
(Fig. 12.3). It goes without saying, the number of the branches can be any.
Fig. 12.3
Let us write Kirchhoff’s current law for the node n:
IJ
∑∑
ik
ik
58
0
. (12.1)
However, on the other hand, Ohm’s law for the active branch gives the
g
following expression:
Let us insert the expression (12.2) into the expression (10.1):
Let us rearrange the terms in the expression (12.3):
Here we have obtained the equation for the potential of an arbitrary
node. It is obvious that, in this equation, there are the parameters of the
branches, which are connected to this node, and the potentials of the nodes,
with which these branches touch.
Kirchhoff’s current law allows us to write the independent equations for
all the nodes except one node. At the same time, the potential can only be
determined up to a certain constant. Therefore, the potential of one node can
be equated, for example, to zero. In this case, the set of the equations, which
has the unique solution, can be obtained.
Nota bene! Using the general expression (12.4), let us formulate the al-
gorithm for the analysis of the electric circuit by means of the node potential
method.
1.
An arbitrary node of the circuit must be grounded, that is, the poten-
tial of this node must be equated to zero. To shorten the equations, it is reasonable to ground the node that is connected to the largest number of the
branches. If there is, in the circuit, one branch that contains the EMF source
and has the zero resistance, it is the node of this branch that must be
grounded. In this case, the potential of the second node of this branch is determined automatically. If there are several such branches in the circuit, it is
necessary to remove them by means of the transference of the EMF source
out of the branch.
For other nodes, write the set of the equations using, for each equa-
2.
tion, the following rules:
The node potential, which is on the principal diagonal of the set of the
equations, multiply by the sum of the conductances of the branches that are
connected to the node.
IEg . (12.2)
iinii
∑∑
inii k
ik
∑∑ ∑ ∑
ni ii ii k
ii i k
Eg J
gEgJ
0
. (12.3)
. (12.4)
59
From the resulting product, subtract the potentials of all the other
nodes multiplied by the conductances of the linking branches.
The right-hand member of the equation is the algebraic sum of the
products of the EMF sources in all the branches, which are connected to the
node, on the own conductance of the branch plus the algebraic sum of the
current sources, which are connected to the node. The EMF sources and the
current sources are considered positive if they are directed towards the node.
The EMF sources and the current sources are considered negative if they are
directed from the node.
When the node potentials have been obtained, find the currents of the
3.
branches using Ohm’s law and Kirchhoff’s current law. For the branch with
zero resistance, Ohm’s law cannot be used. In this case, Kirchhoff’s current
law must be used.
In the properly written set of the equations, the principal determinant
4.
is symmetric about the principal diagonal. The dimensionality of the equations is Ampere.
Example 12.1 (using the node potential method)
In the circuit shown in Fig. 12.4, it is necessary to find the unknown
currents by means of the node potential method.
Fig. 12.4
60
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