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Файл:The elements of the electrical circuit theory. Tutorial
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Let us calculate the real existing currents (see Fig. 7.2):
111
222
333
61122
72233
0.634 AII
0.224 AII
1.51 AII
0.634 0.224 0.858 AII I
0.224 1.51 1.734 AII I
;
;
;
;
.
Let us now check the analysis using the power balance. When calculating the power, we must direct the currents and the voltages so that
the supposed power is positive, i.e. the currents in the branches with
the EMF sources must coincide with the EMF sources in direction
(see Fig. 7.2).
As a result, the power of the sources is
PEIEIEI
sources
22 4 3 5 1
() ()
10 0.224 8 1.51 10 0.634 20.66 .
W
The power of the consumers is
22
PIRRIR
cons 1 1 5 2 2
()
222
()
IRR IRIR
33 4 6677
( 0.634) (5 4) (0.224) 10
( 1.5 1) ( 2 1) ( 0.8 58) 5
(1.734) 2 20.65 W.
Taking into account the computational error,
22
22
2
PP .
sources cons
In general, the equations set of the mesh-current method is
31
IR I R I R I R IR E
R
R
R
E
⎧
11 11 22 12 1 1 1 11
⎪
IR I R I R I R IR E
⎪
11 21 22 22 2 2 2 22
⎪
⎪
⎪
IR I R I R I R IR E
11 1 22 2
⎪
⎨
⎪
⎪
IR I R I R I R IR E
11 1 22 2
⎪
⎪
⎪
⎪
IR
11 1
⎩
k k kk kk mm km nn kn kk
m m kk mk mm mm nn mn mm
IR IR I R IR E
n
kk k mm m nn n
kk k mm m nn n
22 2nkknkmmnmnnnnnn
(7.7)
where
of all the loop resistances and is always positive;
sistance of the adjacent branch, 0
adjacent branch,
cent branch (and the equality
is called the intrinsic resistance of the loop, it is equal to the sum
kk
is called the re-
km
R if the currents are opposite in the
km
0
R
if the currents have the same direction in the adja-
km
R is always true);
km mk
kk
is called the
loop EMF source, it is equal to the algebraic sum of the individual EMF
sources of the loop.
The principal determinant of the equations set (7.7) is
RR R R R
11 12 1 1 1
RR R R R
21 22 2 2 2
kmn
kmn
RR R R R
12
k k kk km kn
Δ
(7.8)
RR R R R
12
m m mk mm mn
RR R R R
12
n n nk nm nn
The principal determinant (7.8) is always symmetric about the principal
diagonal.
32
In order to solve the system (7.7) using Cramer’s method, it is necessary
to find the algebraic complements of the determinant (7.8).
Δ
The algebraic complement
of the determinant (7.8) can be found
km
by deleting from the determinant (7.8) of the k column and the m row and
km
(1)
multiplying the resulting determinant by
If we solve the system (7.7) in general, we get the expression for any
.
k
mesh-current:
ΔΔ Δ
12
IE E E
kk kk
kk kk
11 22
ΔΔ Δ
ΔΔ
EE . (7.9)
km kn
mm nn
ΔΔ
The expression (7.9) has important theoretical significance and will be
used in the future when we consider the methods of the analysis of the electric circuits.
Using the mesh-current method if there are the current sources
in the circuit
If there are the current sources in the circuit, it is impossible to write the
equation of Kirchhoff’s voltage law for the loop that contains the current
source. However, the independent loops may be choosing so that each current source is included in only one independent loop. In this case, the real
existing current of the source is equal to the mesh-current, i.e. this meshcurrent is already known and the equation of Kirchhoff’s voltage law is not
required for it. However, this current of the source will be included in the
equations of other mesh-currents, and when we write the equations set, we
have to transpose it to the right-hand member of the equation as the known
quantity.
Example 7.2 (using the mesh-current method if there are the current
sources)
In the circuit shown in Fig. 7.3, it is necessary to calculate the unknown
currents by means of the mesh-current method.
33
Fig. 7.3
The given data are
10 V
E
R
R
R
R
R
1
2
3
4
5
1Ohm
3Ohm
2Ohm
3Ohm
4Ohm
1
15 V
E
2
J
1
k
J
2
k
2A
1A
There are four independent loops in the circuit. Let us choose the loops
so that each current source is included in only one loop, then the current of
the source will be equal to the mesh-current, i.e.
IJ
11 1
IJ
22 2kk
Let us write the equations set for the mesh-current
current
I
:
44
34
I and the mesh-
33
()
I R R R IR IR IR E
⎧
⎪
⎨
IR I R R R IR IR E
⎪
⎩
33 1 3 4 44 3 11 1 22 4 1
()
33 3 44 2 3 5 11 2 22 5 2
. (Ex. 7.2.1)
In the equations set (Ex. 7.2.1), let us transpose terms with known values to the right-hand member:
()
IRRR IR EIRIR
⎧
⎪
⎨
IR I R R R E IR IR
⎪
⎩
33 1 3 4 44 3 1 11 1 22 4
33 3 44 2 3 5 2 11 2 22 5
()
. (Ex. 7.2.2)
Let us substitute the numerical data into the equations set (Ex. 7.2.2):
629
II
⎧
33 44
⎪
⎨
2913
II
⎪
33 44
⎩
(Ex. 7.2.3)
Let us solve the equations set (Ex. 7.2.3) using Cramer’s method:
62
Δ 69 22 50Ohm
;
2
29
92
Δ 99 213 55Volt×Ohm
;
3
13 9
69
Δ 613 92 60Volt×Ohm
;
4
213
Δ
55
33
3
Δ 50
Δ
4
Δ 50
60
1.1 A
1.2 A
.
I ;
I
44
Let us calculate the real existing currents (see Fig. 7.3):
13311
24411
33344
1.1 2 0.9 AII I ;
1.2 2 3.2 AIII
1.1 1.2 2.3 AII I ;
35
;
43322
1.1 1 2.1 AII I
;
54422
633
744
1.2 1 0.2 AII I ;
1.1 AII ;
1.2 AII
.
Let us now check the analysis using the power balance. When calculating the power, we must direct the currents and the voltages so that the supposed power is positive, i.e. the currents in the branches with the EMF
sources must coincide with the EMF sources in direction, and the voltages
at the terminals of the current sources must be opposite to the currents of the
sources (Fig. 7.4).
Fig. 7.4
The voltages at the terminals of the current sources are
UIRIR
12211
Jk
UIRIR
25544
Jk
3.2 3 0.9 1 10.5 V
0.2 4 2.1 3 5.5 V
36
The power of the sources is
E
E
E
PJUJUEIEI
sources 11 2 2 16 27
kJk k Jk
2 10.5 1 5.5 10 1.1 15 1.2 55.5 W.
()
The power of the consumers is
PIRIRIRIRIR
As a result,
cons 11 22 33 44 55
22222
22
( 0.9) 1 ( 3.2) 3
222
(2.3) 2 (2.1) 3 (0.2) 4 55.5 W.
PP .
sources cons
8. THE SUPERPOSITION METHOD
The rational choose of the loops always ensures that the branch with the
current to be found is included in only one independent loop. In this case,
the real existing current will be equal to the mesh-current, and the expression (7.9) will be true for this real existing current. At the same time, in the
expression (7.9), each of the loop EMF sources can be expressed through
,
,
the individual EMF sources in the branches i.e.
pression (7.9) will take the form:
IaEaEaE III
. (8.1)
11 2 2 33k k k k kkk
1
... Then the ex-
2
3
There is no doubt that in the expression (8.1) each term represents the
part of the total current, and this part is due to only one EMF source. This
leads to the theoretically important conclusion that the current in any branch
is equal to the algebraic sum of the partial currents generated by each of the
sources individually.
Nota bene! The method of analyzing the linear circuits, which is called
the superposition method, is based on this principle. The following algorithm can be proposed for this method. The currents that are generated by
each individual source are calculated one after the other. In this case, the
other sources are removed from the circuit, but the internal resistances of
these sources must remain.
37
Example 8.1 (using the superposition method)
In the circuit shown in Fig. 8.1, it is necessary to find the unknown currents by means of the superposition method.
Fig. 8.1
There are two sources in the circuit. Let us separate this circuit into two
circuits: the circuit with the current source and the circuit with the EMF
source.
Let us find the partial currents that are generated by the current source.
Let us remove the EMF source from the circuit for this. Let us take into account the fact that the internal resistance of the EMF source is zero. This
means that it is necessary to put the jumper between the point c and the
point d after removing the EMF source (Fig. 8.2).
There is no doubt that the current
source:
Fig. 8.2
I
IJ
.
1 k
38
is equal to the current of the
1
Nota bene! In order to find the current
R
R
R
R
R
R
R
R
I
and the current
2
I
, let us use
3
the following method that is known in the circuit theory. Let us consider the
circuit shown in Fig. 8.3, where the known current I enters in the node a
and it is necessary to find the current
I
and the current
1
I
.
2
Fig. 8.3
Let us write the equivalent resistance of two parallel branches (see formula (3.4)):
R
12
In order to find the current
R
equ
12
I
1
. (Ex. 8.1.1)
R
, which flows in the resistor
1
(see Fig. 8.3), it is enough to put the current I (that enters in the node a)
in place of the resistor
in the formula of the equivalent resistance
1
(Ex. 8.1.1):
IR
2
Similarly, the current
I
1
12
I can be found. In this case, the current I (that
2
enters in the node a) must be put in place of the resistor
. (Ex. 8.1.2)
R
in the formula
2
(Ex. 8.1.1):
I
1
I
2
12
39
. (Ex. 8.1.3)
R
Let us use formulas (Ex. 8.1.1)-(Ex. 8.1.3) in order to find the current
R
R
R
R
and the current
I in the circuit shown in Fig. 8.2. For this circuit, the
3
I
2
equivalent resistance of two parallel branches is
R
23
R
equ
23
.
R
The current, which enters in the node a, is undoubtedly the current
IJ
. Therefore
1 k
R
k
23
R
k
23
3
,
R
2
.
R
IJ
2
IJ
3
Thus, the partial currents that are generated by the current source are
found.
Let us find, now, the partial currents that are generated by the EMF
source. For this, let us remove the current source from the circuit shown in
Fig. 8.1. Let us take into account the fact that the internal resistance of the
current source is infinity. This means that it is necessary to leave the breakage between the point a and the point b after removing the current source
(Fig. 8.4).
Fig. 8.4
There is only one loop in the circuit shown in Fig. 8.4. In this circuit,
0I and
1
II
.
23
40
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