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Файл:The elements of the electrical circuit theory. Tutorial
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Let us, for example, multiply the numerator and denominator of the
R
R
fraction by the conjugate complex number in order to get rid of the imaginary unit in the denominator:
1 ajb ajb a b
In the mathematical manipulations with the complex numbers, it is necessary to take into account that it is most opportune to add and subtract the
complex numbers – in the algebraic form, but to multiply and divide – in the
exponential form. When the addition is made, the real part of one complex
number is added to the real part of another complex number; the imaginary
part of one complex number is added to the imaginary part of another complex number. When the multiplication is made, the moduluses of the complex numbers are multiplied; the exponents of the complex numbers are
added. When the division is made, the modulus of the dividend is divided
by the modulus of the divisor; the exponent of the divisor is subtracted from
the exponent of the dividend.
Nota bene! The important conclusion follows from the properties of
the complex currents and the complex voltages: any complex current and
any complex voltage can be represented as two components (the real component and the imaginary component). The real component corresponds to
the active power; the imaginary component corresponds to the reactive
power:
where
eI
ajb ajbajb
is the active component of the complex current;
22 22 22
ab ab ab
IReI jImI
(17.13)
j
. (17.12)
Im I
is the
reactive component of the complex current.
where
the reactive component of the complex voltage.
If we divide the complex voltage of the branch by the complex current
of the branch, then we will obtain the parameter, which is called the com-
eU
is the active component of the complex voltage;
UReU jImU
91
(17.14)
Im U
is
plex impedance. The complex impedance also has both the active compo-
Z
R
Z
nent and the reactive component:
U
Re Z jIm Z R jX
I
(17.15)
where
the active resistance);
impedance (it is the reactance).
tive and capacitive. In the complex form, the inductive reactance is always
multiplied by
the voltage across the coil by the angle that is equal to
reactance is always multiplied by
voltage across the capacitor lags behind the current in the capacitor by the
angle that is equal to
complex conductance:
where
active component of the complex current and the modulus of the complex
current, then we will obtain the right-angled triangle that is called the triangle of the currents (Fig. 17.8 a). The triangle of the voltages can be obtained
in the same way (Fig. 17.8 b). If we divide the triangle of the voltages by
the triangle of the currents, then we will obtain the right-angled triangle that
consists of the active resistance, the reactance and the impedance
(Fig. 17.8 c).
eZ R is the active component of the complex impedance (it is
Im Z X is the reactive component of the complex
Nota bene! It is necessary to note that the reactance can be both induc-
j , because the current in the inductance coil lags behind
. The capacitive
2
j in the complex form, because the
.
2
The modulus of the complex impedance is
22
RX. (17.16)
The quantity that is the inverse to the complex impedance is called the
11 RjX
Ygjb
ZRjX
g is the active conductance; b is the reactive conductance.
If we put together the active component of the complex current, the re-
22
RX
(17.17)
92
a) b) c)
Fig. 17.8
Nota bene! The complex voltage and the complex current are related to
each other by the complex formulation of Ohm’s law, which is analogous to
Ohm’s law for the direct current circuits. The complex voltage across the
branch, in which there is no the EMF source, is equal to the product of the
complex current in the branch by the complex impedance:
UIZ
. (17.18)
The complex formulation of Kirchhoff’s laws is analogous to Kirchhoff’s laws for the direct current circuits.
The complex formulation of Kirchhoff’s current law as follows:
at the node of the electric circuit, the algebraic sum of the complex currents
is zero.
The complex formulation of Kirchhoff’s voltage law as follows: in the
closed loop of the electric circuit, the algebraic sum of the complex voltages
is equal to the algebraic sum of the complex EMF sources that are in the
closed loop.
The complex formulation of Ohm’s law and Kirchhoff’s laws can be
applied to the analysis of the sinusoidal electric circuits.
Example 17.1 (using the complex equations set of Kirchhoff’s laws)
For the circuit shown in Fig. 17.9, it is necessary to write the complex
equations set of Kirchhoff’s laws and find the unknown currents.
93
Fig. 17.9
The given data are
10 Ohm
R
et t
et t
C
L
141sin 628 V
1
212sin 628 30 V
2
100 F
1
25 mH
2
1
R
R
2
3
12 Ohm
18 Ohm
In the circuit diagram, the arrows denote the conditionally positive
direction of the currents and the EMF sources, which must be adhered during the analysis.
Let us calculate the complex EMF sources (the root-mean-square values). It is most opportune to divide the complex numbers in the exponential
form. Also, for the ease, let us write the exponential form as follows: “the
modulus
the exponent”:
E
1
E
1
141
100 V
2
(Ex. 17.1.1)
150 30 V
212 30
2
Let us calculate the reactances:
X
1
C
XL
22
L
11
C
628 100 10
1
628 25 10 15.7 Ohm
94
16 Ohm
6
3
(Ex. 17.1.2)
Let us change the circuit diagram so that it has the complex EMF
X
L
X
sources, the complex currents and the complex impedances (Fig. 17.10).
Fig. 17.10
Nota bene! The reactance
j (for the ease, let us denote this fact directly in the circuit diagram).
The reactance
is inductive, so it must be multiplied by
2
is capacitive, so it must be multiplied by
1C
j .
Let us choose the independent loops and the ways around the loops
(Fig. 17.11) and write the complex equations set of Kirchhoff’s laws:
⎧
IR jX IR E
11 1 33 1
⎪
⎪
IR jX IR E
⎨
22 2 33 2
C
L
(Ex. 17.1.3)
⎪
⎪
II I
123
⎩
0
Fig. 17.11
95
Let us substitute the numerical data into the complex equations set
(Ex. 17.1.3):
⎧
10 16 18 100
⎪
The complex equations set (Ex. 17.1.4) can be solved by any convenient
method. For example, let us simplify it by eliminating the third equation and
the current
I:
3
⎪
12 15.7 18 150 30
⎨
⎪
II I
⎪
123
⎩
⎧
10 16 18 100
⎪
⎪
12 15.7 18 150 30
⎨
⎪
III
⎪
312
⎩
jI I
13
jI I
jI I
jI I
23
0
13
23
(Ex. 17.1.4)
⎧
10 16 18 100
⎪
⎨
⎪
12 15.7 18 150 30
⎩
⎧
Let us solve the equations set (Ex. 17.1.5) using Cramer’s method.
At the same time, let us take into account that it is most opportune to add
and subtract the complex numbers in the algebraic form, but to multiply and
divide in the exponential form.
⎪
⎨
⎪
18 30 15.7 150 30
⎩
28 16 18
Δ
28 16 30 15.7 18
(32.25 29 45 )(33.86 27 37 ) 18
jI II
jI II
28 16 18 100
jI I
IjI
12
j
18 30 15.7
jj
112
212
12
j
2
96
(Ex. 17.1.5)
2
(1091.2 2 7 ) 18
1090.45 40.37 324
766.45 40.37 767.51 3 Ohm ;
j
j
2
2
100 18
Δ
1
150 30 30 15.7
j
100 30 15.7 18 150 30
661.73 220 697.34 18 23 Volt×Ohm;
Δ
2
4837.5 0 15 1800 3037.5 Volt×Ohm.
The complex currents are
IA
1
IA
2
III
312
j
j
28 16 100
j
18 150 30
32.25 29 45 150 30 1800
Δ
1
Δ
Δ
2
Δ
767.51 3
3037.5
767.51 3
0.906 21 23 3.96 3
0.844 0.331 3.95 0.208
jj
0.906 21 23
3.96 3
(Ex. 17.1.6)
4.8 0.539
j
4.83 6 24
A
97
As a result, let us write expressions for the instantaneous values of the
currents using the solution of the complex equations set (Ex. 17.1.6):
( ) 0.906 2 sin 628 21 23 1.28sin 628 21 23
it t t A
1
( ) 3.96 2 sin 628 3 5.6sin 628 3
it t t A
2
( ) 4.83 2 sin 628 6 24 6.83sin 628 6 24
it t t A
3
(Ex. 17.1.7)
18. THE POWER OF THE SINUSOIDAL MODE
Let us consider the certain passive two-terminal circuit that contains the
resistive, inductive and capacitive elements (Fig. 18.1 a). Let us connect the
sinusoidal EMF source to the input of this two-terminal circuit. At the same
time, let us have the opportunity to measure the amplitude, angular frequency and initial phase of the input voltage and current (Fig. 18.1 b).
a) b)
Fig. 18.1
Thus, we know that the input voltage and current of the two-terminal
circuit are determined by the following expressions:
() sin
where
phase of the input voltage;
()
ut is the input voltage; ()
in
ut U t, (18.1)
in m u
() sin
it I t (18.2)
in m i
is the initial phase of the input current.
i
it is the input current; u is the initial
in
98
It takes only to multiply the expression (18.1) by the expression (18.2),
x
x
in order to find the instantaneous power that comes to the two-terminal
circuit:
Let us transform the product of two harmonic functions in the expression (18.3) by means of the well-known formula
In the expression (18.4), the difference
the input voltage and the input current of the two-terminal circuit. Let us
hereinafter denote this angle as φ.
There is no doubt that, in the expression (18.4), half the product of the
amplitudes of the voltage and the current is equal to the product of the rootmean-square values of the voltage and the current:
Thus, the expression (18.4) takes the following form:
() () () sin sin
pt utit UI t t. (18.3)
in in in m m u i
1
sin sin cos cos
pt t
() cos cos 2
in u i u i
pt UI t
in i
yxyxy ⎡ ⎤
⎣⎦
2
UI
mm
⎡⎤
⎣⎦
2
UI U I
mm m m
2
⎡⎤
() cos cos 2 2
⎣⎦
⎡⎤
cos cos 2
UI t
⎣⎦
is the angle between
ui
22
UI.
.
. (18.4)
(18.5)
i
Let us transform, in the expression (18.5), the cosine of the sum of two
angles by means of the well-known formula
cos cos cos sin sin
( ) cos cos 2 cos sin 2 sin
pt UI t t
in i i
⎡⎤
⎣⎦
⎡⎤⎡⎤
cos 1 cos 2 sin sin 2
UI t UI t
pt p t
⎣⎦⎣⎦
'( ) "( )
yxyxy .
ii
99
(18.6)
The curves of two components of the instantaneous power are shown
p
p
in Fig. 18.2. The component '( )
effective work. The component "( )
t (see Fig. 18.2 a) is the power of the
t (see Fig. 18.2 b) is the power of the
electromagnetic interchange between the energy source and the energy
receiver.
a) b)
Fig. 18.2
In order to estimate the power of the sinusoidal mode, only the integral
values are required (but not instantaneous). As these integral values, the active power and the reactive power are used.
The active power (hereinafter denoted as P) is the average value of the
power of the effective work during the period:
T
1
Pptdt
'( )
∫
T
0
T
cos 2
UI
TT
⎡⎤
1 cos 2 cos
∫
⎢⎥
⎣⎦
0
⎛⎞
tdtUI
⎜⎟
⎝⎠
i
. (18.7)
The reactive power (hereinafter denoted as Q) is the maximum value
that the power of the electromagnetic interchange can reach:
sinQUI (18.8)
Let us consider the resistor (Fig. 18.3). The voltage across the resistor
and the current in the resistor coincide in the phase. Consequently, the angle
100
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