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Файл:The elements of the electrical circuit theory. Tutorial
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In this circuit, there is the branch that contains the EMF source and has
E
E
R
R
R
the zero resistance. It is the node of this branch that must be grounded.
Therefore, let us equate the potential of node
.
d
d to zero:
0
In this case, both the potential of the node
e are determined automatically. The potential of the node e is also equal
node
to zero, as the potential of the node
d. The EMF source
a, and the potential of the
is directed from
1
the lower potential to the higher potential. Therefore, the potential of the node
a is greater than the potential of the node d by the value of the EMF:
,
ed
0
E .
11ad
It is necessary to write the set of the equations only for the potential of
the node
b and the potential of the node c:
⎧
⎛⎞
111 1 1 1
⎪
⎜⎟
bca
⎪⎝ ⎠
RR R R R
234 4 2 3
⎨
⎪
⎪
⎩
1111
bc k
RRRR
4456
⎛⎞
⎜⎟
⎝⎠
E
3
J
It should pay attention to the fact that, in the first equation, there is the
potential of node a, since the resistor
links the node a with the node b.
2
The potential of the node a is known. Therefore, let us transpose it to the
right-hand member of the equation:
⎧
⎛⎞
111 1 1 1
⎪
⎜⎟
bc a
⎪⎝ ⎠
RR R R R
234 4 3 2
⎨
⎪
⎪
⎩
1 111
bc k
RRRR
4 456
⎛⎞
⎜⎟
⎝⎠
E
3
J
When the set of the equations will be solved, let us find the branch currents
using Ohm’s law and Kirchhoff’s current law. In order to use Ohm’s law correctly, let us put the arrows of the voltages in the circuit diagram (Fig. 12.5).
61
R
R
Fig. 12.5
There are the following relationships between the voltages and the potentials:
U ,
ab a b
U ,
bc b c
U ,
bd b d b
U ,
ec e c c
U .
ce c e c
In accordance with Ohm’s law, the branch currents are
U
ab
I
,
2
2
UE
3
bd
I
3
I
U
,
4
,
R
3
bc
4
62
U
R
R
E
E
E
ec
I
5
I
6
5
U
ce
.
6
,
For the branches with zero resistance, Ohm’s law cannot be used.
Therefore, in order to find the current
I
and the current
1
I
, let us use
7
Kirchhoff’s current law:
II J ,
12 k
III.
756
The nodal-pair method
If there are only two nodes in the circuit, the node potential method
gives to the single equation from which the voltage between the nodes can
be at once found.
For the circuit shown in Fig. 12.6, let us write the equation of the node
potential method (let us equate the potential of node b to zero):
⎛⎞
111 1 1
⎜⎟
ak
RR R R R
123 1 2
⎝⎠
EJ
12
. (12.5)
Since the voltage between the nodes is
U , the for-
ab a b a
mula (12.5) can be written in the following form:
11
EJ
12
RR
U
ab
12
111
RR R
123
k
. (12.6)
Let us generalize the formula (12.6) so that it corresponds to any arbitrary circuit with two nodes:
gJ
∑∑
U
pk
ab
pp k
g
∑
n
n
63
. (12.7)
R
R
R
Fig. 12.6
The formula (12.7) is the fraction whose numerator is the algebraic sum
of the products of the EMF sources on the own conductance of the branch
plus the algebraic sum of the current sources. The EMF sources and the current sources are considered positive if they are directed towards the node with
the greater potential. The denominator of this fraction is the sum of all the
conductances of the branches (these are both passive and active branches).
Example 12.2 (using the nodal-pair method)
In the circuit shown in Fig. 12.7, it is necessary to find the unknown
currents by means of the nodal-pair method. In accordance with the nodalpair method, the voltage between the nodes is
11
EE
12
U
ab
12
111
123
R
.
RR
Let us find the currents using Ohm’s law:
UE
I
1
I
2
I
3
UE
.
U
ab
ab
1
1
ab
,
2
,
2
3
R
R
64
E
Fig. 12.7
13. THE THEOREM OF THE COMPENSATION
Let us separate the branch from the complicated circuit. The complicated circuit that is considered with respect to any branch is called the twoterminal circuit. If there are the energy sources in the circuit, the twoterminal circuit is called the active (Fig. 13.1). In Fig. 13.1, the letter
notes the active two-terminal circuit.
Fig. 13.1
If we insert two identical and oppositely directed EMF sources into the
branch (Fig. 13.2), the distribution of the currents will not change either in
the branch itself or in the active two-terminal circuit.
Let us write the equality according to Kirchhoff’s voltage law:
A de-
IR U E. (13.1)
ca
The EMF sources in Fig. 13.2 can be any. Let both EMF sources be
such that the following equality is true
IR . (13.2)
65
Fig. 13.2
If we insert the formula (13.2) into the formula (13.1), we find that, in
this case,
U .
0
ca
We see that the voltage between the node
a and the node c is zero, that
is, the potentials of these two nodes are equal to each other. When the potentials of the nodes are equal each other, these nodes can be connected together (Fig. 13.3).
Fig. 13.3
There is no doubt that the closed loop, which appeared in Fig. 13.3, does
not have any influence on the distribution of the current between the node
and the node
b. Therefore, this closed loop can be completely removed from
the circuit (Fig. 13.4).
a
Fig. 13.4
66
This reasoning allows us to come to the following conclusion: in any
branch of the circuit, any resistor can be replaced with the equivalent EMF
source, which is equal to the voltage drop across this resistor and is directed
opposite to the current.
This theorem can also be formulated in the following form: in any
branch of the circuit, any resistor can be replaced with the equivalent current source, which is equal to the current in this resistor and is directed in
the same direction.
14. THE EQUIVALENT GENERATOR METHOD
Let us consider the certain electric circuit, which is represented as the
active two-terminal circuit and the branch (Fig. 14.1). As well as in the previous case, we can insert two identical and oppositely directed EMF sources
into the branch (Fig. 14.2), and the distribution of the currents will not
change either in the branch or in the active two-terminal circuit.
Fig. 14.1
Fig. 14.2
67
Now, let us use the superposition method and separate the circuit into
E
E
E
two circuits: in the first circuit, there will be only one EMF source
equ
(in
this case, the two-terminal circuit will become passive); in the second circuit, there will be all the other sources. In Fig. 14.3, the first circuit is on the
right and the second circuit is on the left.
Fig. 14.3
The real existing current of the branch is the sum of two currents:
'''
III . (14.1)
The current '
I can be found using Ohm’s law:
'
UE
ab equ
'
I
. (14.2)
R
The EMF sources
in Fig. 14.3 can be any. Let the EMF sources be
equ
such that the fraction in the formula (14.2) turns to zero. This condition is
true if
''0and''
UI II . (14.3)
equ ab
The relationship (14.3) means that, in Fig. 14.3, the active two-terminal
circuit is in the no-load mode; that is, the branch with the resistor R is open
or even removed from the circuit (Fig. 14.4). The voltage '
no-load voltage and hereinafter it is denoted as
U
.
nl
U is called the
ab
Any passive two-terminal circuit can always be converted into the
equivalent resistor. In this case, the original electric circuit takes the form
shown in Fig. 14.5.
68
Fig. 14.4
R
E
R
Fig. 14.5
The current in the branch can be found using Ohm’s law:
E
equ
I
equ
(14.4)
R
where
is the EMF of the equivalent generator;
equ
is the resistance
equ
of the equivalent generator; R is the resistance of the branch with the current, which has to be found.
As we found out above, the EMF of the equivalent generator is equal to
the voltage drop across the breakage of the branch with the sought current
and is directed in the current direction. The resistance of the equivalent generator is the resistance of the electric circuit, which is devoid of all the energy sources and is converted into the equivalent resistor relative to the branch
with the sought current.
Both the EMF of the equivalent generator and the resistance of the equivalent generator can be determined both experimentally and analytically.
69
Nota bene! The following algorithm can be proposed for the experi-
s
R
mental determination of the parameters of the equivalent generator:
1.
The branch with the sought current must be broken. This mode is
called the no-load mode. Measure the voltage drop across the breakage, it is
equal to the EMF of the equivalent generator and is directed in the same
direction as the sought current.
2.
The branch with the sought current must be short-circuited. This
mode is called the short-circuit mode. Measure the current in the jumper.
This current is called the short-circuit current.
3.
The resistance of the equivalent generator is equal to the ratio of the
no-load voltage to the short-circuit current:
U
RI .
equ
nl
c
Nota bene! The following algorithm can be proposed for the analytical
determination of the parameters of the equivalent generator:
1.
The branch with the sought current must be removed from the circuit
and replace with two the output terminals. As a result, the original circuit
becomes much simpler. In the future, the output terminals of the converted
circuit in no case should be lost.
2.
The voltage between the output terminals
U (that must be directed
nl
in the same direction as the sought current) can be determined by any
known method of the analysis of the electric circuits. The EMF of the
equivalent generator is equal to this voltage.
3.
All the sources must be removed from the circuit and replaced by
their internal resistances. The resulting passive circuit must be converted
relative to the output terminals into the equivalent resistor
, which is
equ
the resistance of the equivalent generator.
The sought current must be found using Ohm’s law (see the formu-
4.
la (14.4)).
The equivalent generator method is most effective when it is necessary to
determine the current in some single branch that has the variable resistance.
Example 14.1 (using the equivalent generator method)
In the circuit shown in Fig. 14.6, it is necessary to find the unknown
current
I .
3
70
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