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Файл:The elements of the electrical circuit theory. Tutorial
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On the other hand, the current 2I can be found using Ohm’s law:
R
23
E
.
R
I
2
Finally, let us find the real currents (see Fig. 8.1):
IIIJ
111
III
222
IIIII
33332
k
The superposition method cannot be used to calculating the power, since
the power is not proportional to the current, but to the square of the current.
It is necessary to note also, that the superposition method is only applicable
to the linear circuits.
9. THE FORMULAS FOR CONVERTING Y-Δ AND Δ-Y
The connection of three branches in one node is usually called the Yconnection (Fig. 9.1 a). The connection of three branches so that they present the triangle is usually called the Δ-connection (Fig. 9.1 b). In node 1, in
node 2, and in node 3, both the Y-connection and the Δ-connection are connected to the rest of the circuit that is not shown in Fig. 9.1.
a) b)
Fig. 9.1
41
The currents, which enter in the node 1, in the node 2 and in the node 3,
R
R
R
are designated as
the node 3 are denoted as
I , 2I , 3I . The potentials of the node 1, the node 2 and
1
, 2 , 3 .
1
Often, in the analysis of the electric circuits, it is necessary to convert
the Y-connection into the Δ-connection or the Δ-connection into the
Y-connection. If the conversion is performed so that the potentials of the
node 1, the node 2 and the node 3 and the currents, which enter in these
nodes, remain unchanged, then all the external circuit “will not sense” this
replacement. Let us get the formulas for converting Y-Δ and Δ-Y using
Ohm’s law and Kirchhoff’s laws.
Kirchhoff’s current law for the Y-connection is (hereinafter, the letter Y
above the current means that the formula refers to the Y-connection.)
. (9.1)
123
0III
However, on the other hand, Ohm’s law gives the following expressions:
Ig
1101
Ig
2202
Ig
3303
(9.2)
where
1
g
1
,
1
1
g
2
,
2
1
g
3
are the conductances of the branches in
3
the Y-connection (see Fig. 9.1 a).
The potential of the central node
in the Y-connection (see Fig. 9.1 a)
0
can be found by substituting formulas (9.2) into formula (9.1):
11 2 2 3 3 0 1 2 3
ggg
11 2 2 3 3
0
gg g
123
0ggg ggg
(9.3)
Since there is no the central node in the Δ-connection, let us remove the
potential of this node
from formulas (9.2) by substituting formula (9.3)
0
into formulas (9.2):
42
g
g
g
Υ
Ig g
11011 1
⎛⎞
⎜⎟
⎝⎠
ggg
11 2 2 3 3
gg g
123
, (9.4)
Υ
Ig g
22022 2
⎛⎞
⎜⎟
⎝⎠
Υ
Ig g
33033 3
⎛⎞
⎜⎟
⎝⎠
ggg
11 2 2 3 3
gg g
123
ggg
11 2 2 3 3
gg g
123
, (9.5)
. (9.6)
Kirchhoff’s current law for the Δ-connection is (hereinafter, the letter Δ
above the current means that the formula refers to the Δ-connection.)
Δ
III
11231
1 2 12 3 1 13
112 13 212 313
Δ
III
22312
2 3 23 1 2 12
212 23 112 323
Δ
III
33123
3 1 13 2 3 23
gg
ggg
gg
ggg
gg
(9.7)
(9.8)
(9.9)
where
gR
12
1
12
313 23 313 223
gR
,
23
ggg
1
gR
,
23
13
1
are the conductances of the
13
branches in the Δ-connection (see Fig. 9.1 b).
Since the currents, which enter in the Y-connection and in the Δ-
connection from the outside, are equal to each other, let us set equal (9.4)
with (9.7), then (9.5) with (9.8), then (9.6) with (9.9):
43
Δ
g
g
g
R
R
R
R
R
R
R
II
11
Δ
II
22
II
33
Δ
After all the conversions, the expressions can be obtained that relate the
conductances of the Δ-connection to the conductances of the Y-connection:
gg
g
12
g
13
g
23
12
gg
123
gg
13
gg
123
gg
23
gg
123
, (9.10)
, (9.11)
. (9.12)
The formulas for resistances in converting Y-Δ can be obtained from the
expressions (9.10)-(9.12):
R
RRR
12 1 2
12
, (9.13)
3
R
, (9.14)
RRR
13 1 3
RRR
23 2 3
13
2
R
23
. (9.15)
1
The inverse formulas for resistances in converting Δ-Y can be obtained
from the formulas (9.13)-(9.15):
RR
R
1
12 13
RR
12 13 23
44
, (9.16)
RR
R
R
R
2
R
3
12 23
RR
12 13 23
RR
13 23
RR
12 13 23
, (9.17)
. (9.18)
Nota bene! The following method can be recommended in order to re-
member and use the formulas (9.13)-(9.18) correctly.
Converting Y-Δ, put two fingers in the nodes to which the branch of the
Δ-connection will be connected (Fig. 9.2). The resistance of the Δ-branch,
which has to be found, is equal to the sum of the resistances of the Ybranches, which touch with the fingers, plus the product of these resistances
divided by the resistance of the third branch.
Fig. 9.2
Converting Δ-Y, put the finger in the node to which the branch of the
Y-connection will be connected (Fig. 9.3). The resistance of the Y-branch,
which has to be found, is equal to the product of the resistances of the
Δ-branches, which touch with the finger, divided by the sum of the resistances of all three branches of the Δ-connection.
If the resistances of all the branches of the Y-connection or the Δ-con-
nection are equal to each other, then the Y-connection or the Δ-connection
are called symmetric. The branch resistances of the symmetric Y-connection
and the symmetric Δ-connection are related by the expression that can be ob
obtained from the formulas (9.13)-(9.18):
45
Δ
R
R
sym.
Δ
RR
sym. sym.
sym.
3
(9.19)
3
Fig. 9.3
Example 9.1 (converting Y-Δ)
In the circuit shown in Fig. 9.4, it is necessary to find the unknown cur-
I
rent
.
7
Fig. 9.4
46
The given data are
R
R
R
R
R
R
4Ohm
R
R
R
E
4
5
6
1
3Ohm
2Ohm
10 V
ab
,
4
,
,
bc
ac
2.26 Ohm
R
1
R
2
R
3
3Ohm
2.17 Ohm
Let us convert the Y-connection of the resistors
equivalent Δ-connection with the resistors
der to do this, let us use formulas (9.13)-(9.15) and Fig. 9.2:
,
into the
5
6
(Fig. 9.5). In or-
RR
RRR
ab
46
RRR
bc
56
RRR
ac
45
46
R
5
RR
56
R
4
RR
45
R
6
42 8.67Ohm
3 2 6.5 Ohm
43 13Ohm
42
3
32
4
43
2
Fig. 9.5
47
Let us convert the resistances of the parallel branches (see the for-
E
R
mula (3.4)):
RR
1
R
ab
1
R
ac
2
R
bc
3
ab
RR
1
RR
2
RR
2
RR
3
RR
3
ab
ac
ac
bc
bc
2.26 8.67
2.26 8.67
313
2.44 Ohm
313
2.17 6.5
2.17 6.5
1.79 Ohm
1.63 Ohm
Let us further convert all the resistances into the equivalent resistor:
RR R
12 2
R
equ
ab ac bc
RR R
ab ac bc
122
1.79 2.44 1.63
1.79 2.44 1.63
At last, let us find the current
E
I
.
7
R
equ
1.24 Ohm
I using Ohm’s law:
7
10
7
8.06 A
1.24
10. THE FORMULAS FOR CONVERTING
THE PARALLEL BRANCHES INTO
ONE EQUIVALENT BRANCH
Let several parallel branches (with the sources and without the sources)
are located between two nodes (Fig. 10.1 a). The current
without) into the node
a and the same current I exits out the node b. Let us
I enters (from
convert these parallel branches into one equivalent branch, in which there
are the EMF source
and the resistor
equ
48
(Fig. 10.1 b).
equ
a) b)
Fig. 10.1
For this, let us write the equation of Kirchhoff’s current law and the
equations of Ohm’s law for the parallel branches:
IIIIJ , (10.1)
123 k
IUEg , (10.2)
111ab
IU Eg , (10.3)
222ab
IUg . (10.4)
33ab
Let us insert the equations (10.2)-(10.4) into the equation (10.1):
IUEgUEgUgJ
IU g g g Eg Eg J
ab k
⎡⎤
IU ggg
⎢⎥
ab
⎣⎦
11 2 2 3
ab ab ab k
123 1122
Eg E g J
11 2 2
gg g
123
k
123
(10.5)
Now, let us write the equation of Ohm’s law for the equivalent branch
(see Fig. 10.1 b):
IU E g . (10.6)
ab equ equ
49
At last, let us compare the formula (10.5) with the formula (10.6). There
g
E
g
E
are the obvious relationships for the parameters of the equivalent branch:
E
equ
gg g , (10.7)
123equ
gEg J
11 2 2
gg g
123
k
. (10.8)
Let us generalize the formula (10.7) and the formula (10.8) so that these
formulas correspond to the arbitrary number of the branches. There is no
doubt that when the parallel branches are converted into one equivalent
branch, the conductance of the equivalent branch must be equal to the
arithmetic sum of all the branches conductances:
g
equ n
. (10.9)
∑
The equivalent EMF source is the fraction whose denominator is the
sum of all the conductances of the branches (these are both passive and active branches). The numerator of this fraction is the algebraic sum of the
current sources plus the algebraic sum of the products of each EMF source
on the conductance of its branch:
where
p is the number of the branches with the EMF sources; k is the num-
E
equ
∑
g
n
ber of the branches with the current sources;
(10.10)
n is the number of all the
gJ
∑∑
pp k
branches.
In formula (10.10), those EMF sources and current sources are positive,
which coincide with the equivalent EMF source in the direction, and those
sources are negative, which are opposite to the equivalent EMF source.
Example 10.1 (converting the parallel branches into one equivalent
branch)
In the circuit shown in Fig. 10.2, it is necessary to find the unknown
current
I .
3
50
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