|
I
I
I
I
I
I
I
j
j
j
j
(
)
.
Recasting these equations in matrix form,
1
100
50
50
25 25
25
300 0
0
+
−
−
+
=
∠
j
j
j
j
.
I
I
1
2
V
O
V
S
= 300
∠
0°
I
x
I
1
I
y
+
–
+
–
+
–
+
–
1
Ω
25
Ω
0.25
Ω
j
25
Ω
j
100
Ω
j
50
I
y
j
50
I
x
I
2
Fig. 7.6-6
The phasor equivalent circuit for the circuit in Fig. 7.6-5
7.24
The Sinusoidal Steady-State Response
Solving for
I
1
and
I
2
, we get
I
1
=
∠ ×
+
+
+
−
=
∠ ×
300 0
25 25
25
1
100 25 25
25
50
50
300 0
( .
)
(
)( .
) (
)(
)
(
j
j
j
j
j
225 25
25
25 25
2550
300 0 35 53 44 71
2550 13 89 43
.
)
( .
)
.
.
.
.
+
+
=
∠ ×
∠
°
∠
j
j
A
°°
=
∠ −
°
=
∠ ×
∠
°
=
∠
°
4 18
44 72
300 0
50
2550 13 89 43
5 88 0 57
.
.
.
.
.
.
A
A
I
2
j
\
V
o
=
25
×
5.88
∠
0.57
°
=
147
∠
0.57
°
V
\
i
1
(t)
=
4.18 cos(100t – 44.72
°
) A
i
2
(t)
=
5.88 cos(100t
+
0.57
°
) A
v
o
(t)
=
147 cos(100t
+
0.57
°
) V
Source power
=
0.5
×
300
×
4.18
×
cos(0
-
(
-
44.72
°
))
=
445.5 W.
Average power delivered to 25
Ω
=
0.5
×
5.88
×
5.88
×
25
=
432.2 W.
[This is an equivalent circuit for a 2:1 two-winding transformer using dependent sources to model
the mutual coupling between the windings. Note that the output voltage is almost equal to half of the
input voltage and almost in phase with it. The presence of the 1
Ω
and 0.25
Ω
resistors make the output
amplitude slightly less than 150 V and phase of output slightly different from zero. These resistors
model the inevitably present winding resistances.]
example: 7.6-4
A pair of AC voltage sources with same frequency connected through an inductance is called a
synchronous link in Electrical Power System Engineering. The sources are generating stations and
the inductance is that of the high voltage transmission line that links up the stations. When two DC
sources are linked together by means of a resistance, the higher voltage source sends power to the
lower voltage source. But when two AC sources at same frequency are linked together, it is not the
magnitude of voltage that decides the magnitude and direction
of power flow.
Show that in the synchronous link in Fig. 7.6-7, the
leading voltage source sends average power to the lagging
voltage source and that for small phase difference between
the two sources the power exchanged is proportional to phase
difference in radians.
Solution
The frequency is not given specifically, however, the reactance value of the inductor is directly
specified as X and the word synchronous implies that both sources are at the same frequency. The source
voltages are specified as rms values and this is a common practice in Electrical Power Engineering.
Fig. 7.6-7
A synchronous link
+
–
+
–
V
1
∠
1
V
rms
δ
δ
jX
V
2
∠
2
V
rms
I
Sinusoidal Steady-State Response from Phasor Equivalent Circuit
7.25
Electronics and Communication Engineers prefer to specify amplitude rather than rms value. In this
book, if rms value is specified, it will be explicitly mentioned after the unit of the quantity.
The current phasor
I
from first source to second source is found in the following:
I
=
∠ − ∠
=
−
+
−
=
V
V
jX
V
V
j V
V
jX
V
1
1
2
2
1
1
2
2
1
1
2
2
1
d
d
d
d
d
d
( cos
cos
)
( sin
sin
)
( ssin
sin
)
( cos
cos
)
( sin
sin
)
d
d
d
d
d
d
1
2
2
1
1
2
2
1
1
2
2
0
−
−
−
=
−
∠ °
V
X
j
V
V
X
V
V
X
−−
−
∠ °
∴
=
−
−
( cos
cos
)
( )
( sin
sin
)
cos
( c
V
V
X
i t
V
V
X
t
V
1
1
2
2
1
1
2
2
1
90
d
d
d
d
w
oos
cos
)
cos(
)
( sin
sin
)
cos
( cos
d
d
w
d
d
w
d
1
2
2
1
1
2
2
1
90
−
+ °
=
−
+
V
X
t
V
V
X
t
V
11
2
2
−
V
X
t
cos
)
sin
d
w
The source voltage time-function for first source is
v t
V
t
1
1
1
2
( )
cos(
)
=
+
w d
We find the average power delivered by the first source by taking the two terms in current one by
one. The first component is a cos
w
t component and the phase angle by which the voltage leads this
component is
d
1
. Therefore, average power delivered through this current, P
1
, is
P
V
V V
X
V
V V
1
1
2
1
1
1 2
2
1
1
2
1
1
1 2
0 5 2
=
×
−
=
−
.
sin
cos
sin
cos
sin
cos
si
d
d
d
d
d
d
nn
cos
d
d
2
1
X
The second current component is a sin
w
t i.e., a cos(
w
t – 90
°
) component. The phase angle by
which the voltage leads this component is 90
°
+
d
1
. Therefore, the average power delivered through
this current, P
2
, is
P
V
V V
X
V
1
1
2
1
1
1 2
2
1
1
2
0 5 2
90
90
=
×
° +
−
° +
=
−
.
cos
cos(
)
cos
cos(
)
cos
d
d
d
d
d
11
1
1 2
2
1
sin
cos
sin
d
d
d
−
V V
X
Adding P
1
and P
2
to get the total average power delivered by the first source,
P
V
V V
X
V
V V
=
−
+
−
+
1
2
1
1
1 2
2
1
1
2
1
1
1 2
2
sin
cos
sin
cos
cos
sin
cos
si
d
d
d
d
d
d
d
nn
d
1
X
7.26
The Sinusoidal Steady-State Response
∴ =
−
[
]
=
−
≈
−
P
V V
X
V V
X
P
V V
X
1 2
1
2
1
2
1 2
1
2
1 2
1
sin
cos
cos
sin
sin(
)
(
d
d
d
d
d d
d dd
d d
2
1
2
1
)
(
)
if
in radians <<
−
Therefore, in the synchronous link shown in Fig. 7.6-7, average power flows from leading voltage
source to lagging voltage source quite independent of the voltage magnitude relationship between
them. Moreover, for small phase difference between the two sources, the power flow is proportional
to the phase difference in radians.
example: 7.6-5
The source current in the circuit in Fig. 7.6-8 is i
S
(t)
=
I
m
cos
w
t A. Find
w
and k such that the current
i
y
is in phase with i
S
(t) and has the same amplitude as that of i
S
(t).
i
s
i
x
k
i
x
R
a
R
b
R
c
i
y
R
R
R
C
C
C
Fig. 7.6-8
Circuit for Example 7.6-5
Solution
The phasor equivalent circuit is shown in Fig. 7.6-9.
I
s
I
x
kR
c
I
x
R
a
R
b
R
R
R
–
+
I
1
R
c
1
j
C
I
2
I
3
I
y
1
j
C
1
j
C
ω
ω
ω
Fig. 7.6-9
The phasor equivalent circuit for circuit in Fig. 7.6-8
The dependent source current value is
=
+
kR I
R
R
a s
a
b
. This source is transformed into a dependent
voltage source of value
=
+
=
kR R
R
R
a
c s
a
b
I
I
s
a
in series with R
c
. Source transformation theorem is
applicable under sinusoidal steady-state condition. Three meshes and the mesh current phasors are
as shown in Fig. 7.6-9. The mesh impedance matrix can be written by inspection. Hence, the mesh
equations in matrix form will be as below.
Sinusoidal Steady-State Response from Phasor Equivalent Circuit
7.27
R
R
j C
R
R
R
j C
R
R
R
j C
c
+ +
−
−
+
−
−
+
1
0
2
1
0
2
1
1
2
3
w
w
w
I
I
I
=
−
a
I
s
0
0
The determinant of the mesh impedance matrix after some simplification is
∆ =
+
−
+
−
+
−
Z
R
R R
R
R
C
j
R R
R
C
C
3
2
2
3
3
5
6
4
1
c
c
c
(
)
(
)
(
)
(
)
w
w
w
We need to solve for
Do'stlaringiz bilan baham: |
