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Файл:The elements of the electrical circuit theory. Tutorial
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φ is equal to zero and the reactive power is equal to zero. Therefore, the resistor is often called the active resistance.
Then, let us consider the inductance coil (Fig. 18.4). The current lags
behind the voltage across the inductance coil by the angle that is equal to
, that is the angle φ is equal to
2
. Therefore, the active power is equal to
2
zero and the reactive power is positive.
Fig. 18.3 Fig. 18.4
At last, let us consider the capacitor (Fig. 18.5). The voltage across the
capacitor lags behind the current by the angle that is equal to
. Conse-
2
quently, in this case, the angle φ is equal to
. Therefore, the active pow-
2
er is equal to zero and the reactive power is negative.
Fig. 18.5
It should be understood that here the plus or the minus does not mean
the direction or the quantitative of the reactive power, since the power is
scalar in its physical nature. The sign of the reactive power shows only the
position of the voltage phase with respect to the current phase. That is why
it is necessary to differentiate the inductive reactive power from the capacitive reactive power.
101
If the current lags behind the voltage, then it is customary to say that the
E
E
reactive power and the cos are inductive. If the voltage lags behind the
current, then it is customary to say that the reactive power and the cos are
capacitive.
For the alternating current circuits, it is necessary to write two power
balances, i.e. the active power balance and the reactive power balance.
The active power balance is
∑∑ ∑
nk
The EMF sources The current sources
cos cos
IUJ IR
nn n k k k
2
. (18.9)
The reactive power balance is
sin sin
IUJ IX
∑∑ ∑
nn n k k k
nk
The EMF sources The current sources
2
. (18.10)
Nota bene! It is necessary to take into account that, when we write the
reactive power balance, we have to pay attention to the sign of the reactive
power in the right-hand member of the equality (18.10). The inductive
reactive power must be positive but the capacitive reactive power must be
negative.
In addition to the active and reactive power, the total power (sometimes
it is also called the apparent power) is widely used. The total power (hereinafter denoted as S) is determined as
SUI . (18.11)
Thus, the total power is equal to the product of the root-mean-square
value of the voltage to the root-mean-square value of the current in the
branch (the dimensionality of the total power is Volt-Ampere).
The active power, the reactive power and the total power are related to
each other by the following relationship:
222
SPQ. (18.12)
102
The relationship (18.12) can be represented graphically as the right-
Z
angled triangle of the powers (Fig. 18.6).
Fig. 18.6
If we compare this triangle of the powers with the triangle of the currents, the triangle of the voltages and the triangle of the impedance (see
Fig. 17.8), then we can easily find the following relationship:
SUI IZUY (18.13)
22
where
SPQ;
22
22
RX;
Ygb.
22
In addition, the following relationships are obvious (from the triangle of
the powers):
cos
PS
sin
QS
Q
arctan
(18.14)
P
P
cos
S
Nota bene! In the relationships (18.14), the parameter
cos
appears. It
is called the power factor. The power factor is equal to the ratio of the active
power to the total power.
103
The complex formulation of the power balance
j
j
j
Z
It is extremely opportune to calculate the power of the sinusoidal mode
in the complex form. Let, in the certain circuit, the complex voltage across
the branch is
UUe
u
. (18.15)
The complex current in this branch is
IIe
i
. (18.16)
The angle between the voltage and the current
must be
ui
known to calculate the power. If the complex voltage (18.15), is multiplied
by the complex number, which is conjugated to the complex current (18.16),
the resulting complex number will contain the difference (but not the sum)
of the exponents, and this difference will be
. It is this resulting
ui
complex number that will represent the total power on the complex plane:
SUIUe Ie UIe UIe
jj j
ui ui
()
. (18.17)
The sign “~” in the complex number representing the total power on the
complex plane shows that, on the one hand, the total power is not the harmonic function of the time (whose complex numbers are denoted by the
capital letter with the dot on top); and on the other hand, the total power is
not the parameter of the circuit (whose complex numbers are denoted by the
capital letter with the line below).
Like any other complex number, the complex total power can be written
in the exponential form, in the trigonometric form, in the algebraic form:
j
S UIe UI jUI P jQ
. (18.18)
cos sin
Thus, the real part of the complex total power is the active power, the
imaginary part of the complex total power is the reactive power.
Let the complex impedance of the branch is
the branch is I
. Then, according to Ohm’s law, the complex voltage across
, the complex current in
the branch is
UIZ
. (18.19)
104
Let us insert the expression (18.19) into the expression (18.17):
L
L
SUI IIZ
. (18.20)
The product of two conjugate complex numbers is equal to the square of
their modulus. Therefore, the complex total power is equal to the product of
the square of the root-mean-square value of the current (which is the real
number) by the complex impedance:
SUI IIZ IZ
2
. (18.21)
Let us now, in the expression (18.21), write the complex impedance in
the algebraic form:
22 2 2
SIZIRjX IRjIXPjQ
. (18.22)
Now, finally, the power balance in the complex form can be written.
The complex total power of the sources will obviously be determined by the
following expression:
SEIUJ
sources
nn
∑∑
nk
The EMF sources The current sources
kk
. (18.23)
The complex total power of the consumers will be determined as
S IZ IRjX IRjX jX
cons
22 2
∑∑ ∑
. (18.24)
C
The power balance in the complex form is
EI UJ I R jX jX
nn
∑∑∑
nk
The EMF sources The current sources
kk
2
. (18.25)
C
In the complex form, it takes only to write the total power balance (but
not two power balances, for the active power and the reactive power). The
complex total power balance (18.25) includes both the active power and the
reactive power at once. At the same time, the signs of the reactances are automatically taken into account. This means that it is automatically determined whether the reactive power is inductive or capacitive.
105
19. THE SERIES-CONNECTED RLC-ELEMENTS
Let us consider the certain circuit with the series connection of the resistor, the inductance coil, the capacitor and the EMF source (Fig. 19.1) and let
us find the current in the circuit if the EMF source ensures, at its terminals,
the voltage that varies according to the following harmonic law:
() sin
In order to do this, let us write the equation of Kirchhoff’s voltage law
for the instantaneous values of the harmonic functions:
itR L itdt et
The equation (19.2) is the integro-differential equation, and its solution
can be obtained as the sum of the partial solution and the solution of the
homogeneous equation. The homogeneous equation is the equation, which
has the zero second member. The equation will turn into the homogeneous
equation if the electromotive force in the circuit is zero.
If the EMF source ensures, at its terminals, the harmonic voltage, then the
partial solution of the integro-differential equation (19.2) can also be obtained
in the form of the harmonic function with the same angular frequency:
() () ()
et E t. (19.1)
Fig. 19.1
() 1
di t
dt C
m
∫
. (19.2)
() sin
The initial phase of the current in the expression (19.3) is the constant
that is not yet known to us. In this case, it is opportune to denote this con-
it I t. (19.3)
m
106
stant as
x
input current, then the initial phase of the input current will be
the initial phase of the input voltage is zero, i.e.
, because if φ is the angle between the input voltage and the
when
,
u
.
i
In the steady-state mode, the component of the current, which is determined by the solution of the homogeneous equation, tends to zero. Therefore,
we will not consider hereinafter the solution of the homogeneous equation.
Let us write the voltage at the terminals of the EMF source (19.1) in the
form
() sin
and let us transform the sine of the sum of two angles by means of the wellknown formula
Let us insert the expression (19.3) and the expression (19.5) into the integro-differential equation (19.2):
() cos sin sin cos
etEtEt. (19.5)
sin cos cos
IR t I L t t
mm
EtEt
et E t (19.4)
sin sin cos cos sin
mm
cos sin sin cos
mm
m
yxyxy .
,
ui
0
I
m
C
(19.6)
There is no doubt that the equation (19.6) must be true for any arbitrary
point of time. And in particular, if 0
take the following form:
⎛⎞
IL E
mm
⎜⎟
⎝⎠
t , then the equation (19.6) must
1
C
107
. (19.7)
sin
But, on the other hand, if
E
Z
L
X
L
X
L
X
the following form:
t , then the equation (19.6) must take
2
IR E
mm
cos
. (19.8)
First, let us square both the equation (19.7) and the equation (19.8), and
next, let us sum up these equations term by term. And then we will obtain
22
(since there is the following well-known equality
⎡⎤
22 2
IR L E
mm
⎛⎞
⎢⎥
⎜⎟
⎝⎠
⎢⎥
⎣⎦
1
C
sin cos 1 ):
2
. (19.9)
It follows, from the expression (19.9), that
I
m
RL
mm
⎛⎞
2
⎜⎟
⎝⎠
1
C
E
2
. (19.10)
If we divide the equation (19.7) by the equation (19.8), then we will obtain the following expressions:
1
L
C
tan
, (19.11)
X
C
RR
XR
arctan
C
. (19.12)
As a result, the solution of the integro-differential equation (19.2) is
where
L ;
XC
EXX
it t
() sin arctan
1
C
⎛⎞
mLC
⎜⎟
ZR
⎝⎠
;
ZR XX
2
108
2
LC
(19.13)
.
The reactances
L
X
X
L
X
L
X
L
j
L
but the equivalent reactance
and
themselves are always greater than zero,
C
X can be either positive, or negative,
C
or zero. Let us analyze the vector diagrams that correspond to all these cases.
The first case, when
X . Let us analyze drawing the first vector
C
diagram in detail, because the sequence of the vectors drawing is extremely
important in drawing the vector diagram. When the scales of the current and
the voltage are chosen, let us draw, first of all, the vector of the current
I
because there is only one current in the circuit, so all the other vectors will
be oriented relative to the vector of the current. Let the vector of the current
will be directed horizontally from the left to the right (Fig. 19.2 a). It is necessary to leave free space around this vector for all the other vectors.
Next, let us draw the vector of the voltage across the resistor IR
coincides in the phase with the vector of the current I
(Fig. 19.2 b). Let us
, which
draw this vector so that it is slightly shifted (so that it does not merge visually with the vector of the current). Let us do the same in the future, always
when vectors can merge visually.
Then let us draw (from the end of the vector of the voltage across the re-
) the vector of the voltage across the inductance coil LjIX
(Fig. 19.2 c). This vector is directed vertically upward, since the vector of
the current I
IX
coil
lags behind the vector of the voltage across the inductance
by the angle that is equal to
.
2
Then let us draw (from the end of the vector of the voltage across the
inductance coil
jIX
(Fig. 19.2 d). This vector is directed vertically downward, since
C
the vector of the voltage across the capacitor
tor of the current
jIX) the vector of the voltage across the capacitor
jIX
lags behind the vec-
C
.
I
by the angle that is equal to
2
If we now connect the beginning of the first voltage vector to the end of
the last voltage vector, then we will obtain the vector that corresponds to the
sum of all three voltage vectors. This vector is the sum voltage at the terminals of the EMF source. On the other hand, according to the Kirchhoff voltage law (19.2), this sum voltage is equal to the electromotive force of the
,
109
source. Let us draw this vector and denote it as
E
E
E
, but we will always keep
in our mind that this vector is equal to the voltage at the terminals of the
EMF source (Fig. 19.2 e).
The initial phase of the voltage at the terminals of the EMF source is
equal to zero. This means that the vector
must coincide with the real coordinate axis on the complex plane. Let us draw the real coordinate axis and
the imaginary coordinate axis, and also let us denote the initial phase of the
I
current
(Fig. 19.2 f).
a) b)
c) d)
e) f)
Fig. 19.2
In the future (on conditions that we always remember that, in all the
modes, the vector
must coincide with the real coordinate axis on the
110
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