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# solutions > Solution_7

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11/12/2012

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Page | 1
ELEC2501: Assignment 7
7.1 Determine the phase angles by which
)
(
1t
v
)
(
1t
i
and
)
(
1t
v
)
(
2t
i
, where :
V
t
t
v
)
25
377
sin(
6
)
(
1
o
+
=
A
t
t
i
)
10
377
cos(
04
.
0
)
(
1
o
\u2212
=
A
t
t
i
)
75
377
sin(
2
.
0
)
(
2
o
\u2212
\u2212
=
Solution:
A
A
t
A
t
t
i
)
80
377
sin(
04
.
0
)
90
10
377
sin(
04
.
0
)
10
377
cos(
04
.
0
)
(
1
o
o
o
o
+
=
+
\u2212
=
\u2212
=
It means that
)
(
1t
v
)
(
1t
i
by
o
o
o
55
80
25
\u2212
=
\u2212
.
A
A
A
t
t
i
)
105
377
sin(
2
.
0
)
180
75
377
sin(
2
.
0
)
75
377
sin(
2
.
0
)
(
2
o
o
o
o
+
=
+
\u2212
=
\u2212
\u2212
=
It means that
o
o
o
80
105
25
\u2212
=
\u2212
7.2 Calculate
)
(t
i, the time-domain current in the resistor in the following circuit if the input
voltage is:
1)
V
t
t
v
)
180
377
cos(
12
)
(
1
o
+
=
2)
V
t
t
v
)
45
377
sin(
16
)
(
2
o
+
=
Solution:
This is purely Resistive network:)
(t
iand
)
(t
v
are in phase
1)
A
t
A
t
R
t
v
t
i
)
180
377
cos(
3
4
)
180
377
cos(
12
/
)
(
)
(
1
1
o
o
+
=
+
=
=
2)
A
t
A
t
R
t
v
t
i
)
45
377
sin(
4
)
45
377
sin(
4
16
/
)
(
)
(
2
2
o
o
+
=
+
=
=
7.3C al cu l at e
)
(t
i, the time-domain current in the capacitor in the following circuit if the input
voltage is:
1)
V
t
t
v
)
30
377
cos(
8
)
(
1
o
\u2212
=
2)
V
t
t
v
)
60
377
sin(
4
)
(
2
o
+
=
Solution:
This is purely Capacitive network: )
(t
i
)
(t
v
by 90\u00b0.
Page | 2
dtt
dv
C
t
i
)
(
)
(=
, However, this equation can be rewritten as
CV
j
I
\u03c9
=
(eq.7.29 Irwin)
1)
,
)
30
8
(
1
A
C
j
I
o
\u2212
\u2220
=\u03c9
where
V
t
t
v
)
30
377
cos(
8
)
(
1
o
\u2212
=
A
A
I
o
o
o
60
34
.
4
30
8
*
90
10
*
1440
*
377
1
6
\u2220
=
\u2212
\u2220
\u2220
=
\u2212
A
t
i
)
60
377
cos(
34
.
4
)
(
1
o
+
=
2)
V
V
t
V
t
t
v
)
30
377
cos(
4
)
90
60
377
cos(
4
)
60
377
sin(
4
)
(
2
o
o
o
o
\u2212
=
\u2212
+
=
+
=
A
A
A
C
j
I
o
o
o
o
60
17
.
2
30
4
*
90
10
*
1440
*
377
)
30
4
(
2
6
\u2220
=
\u2212
\u2220
\u2220
=
\u2212
\u2220
=
\u2212
\u03c9
A
t
i
)
60
377
cos(
17
.
2
)
(
2
o
+
=
7.4C al cu l at e
)
(t
i, the time-domain current in the inductor in the following circuit if the input
voltage is:
1)
V
t
t
v
)
12
377
cos(
24
)
(
1
o
+
=
2)
V
t
t
v
)
48
377
sin(
18
)
(
2
o
+
=
Solution:
This purely Inductive network:
)
(t
v
)
(t
i
by 90\u00b0
dtt
di
L
t
v
)
(
)
(=
This can be simplified to
LI
j
V
\u03c9
=
1)
V
t
t
v
)
12
377
cos(
24
)
(
1
o
+
=
o
o
o
o
o
78
6
)
90
12
(
6
90
10
*
61
.
10
*
377
12
24
3
1
\u2212
\u2220
=
\u2212
\u2220
=
\u2220
\u2220
=
=
\u2212
L
jV
I
\u03c9
A
t
i
)
78
377
cos(
6
)
(
o
\u2212
=
2)
V
t
t
v
)
48
377
sin(
18
)
(
2
o
+
=
V
V
)
42
377
cos(
18
)
90
48
377
cos(
18
o
o
o
\u2212
=
\u2212
+
=
;
o
o
o
o
o
132
5
.
4
)
90
42
(
5
.
4
90
10
*
61
.
10
*
377
42
18
3
2
\u2212
\u2220
=
\u2212
\u2212
\u2220
=
\u2220
\u2212
\u2220
=
\u2212
A
I
A
t
i
)
132
377
cos(
5
.
4
)
(
o
\u2212
=
7.5 Find the frequency-domain impedance Z of the following network:
Page | 3
Solution:
This is a series RLC network:
Impedance
L
C
R
Z
Z
Z
Z
+
+
=
o
54
.
80
04
.
3
)
5
.
03
(
tan
3
5
.
0
)
(
3
5
.
0
5
2
5
.
0
1
2
2
\u2220
=
\u2220
+
=
\u03a9
+
=
\u03a9
+
\u03a9
\u2212
\u03a9
=
+
+
=
\u2212
j
j
j
Z
Z
Z
Z
L
C
R
o
54
.
80
04
.
3\u2220
=
Z
7.6 Find the frequency-domain impedance Z of the following network:
Solution:
This is a Parallel RLC network
Impedance
1
)
1
1
1
(
\u2212
+
+
=
L
C
R
Z
Z
Z
Z
25
.
0
1
5
.
0
25
.
0
1
21
4
1
11
1
j
j
j
j
j
Z
\u2212
=
\u2212
+
=
\u03a9
+
\u03a9
\u2212
+
\u03a9
=
\u03a9
\u2220
=
\u03a9
\u2212
\u2220
\u2248
\u03a9
\u2212
\u2220
\u2212
+
=
\u03a9
\u2212
=
\u2212
o
o
04
.
14
97
.
0
04
.
14
03
.
1
1
)
1
/
25
.
0
(
tan
)
25
.
0
(
1
1
25
.
0
1
1
1
2
j
Z
\u03a9
\u2220
=
o
04
.
14
97
.
0
Z
or
o
04
.
14
03
.
1
1
\u2212
\u2220
\u2248
=
\u2212
Y
Z
7.7 The voltages
)
(t
vR
and
)
(t
vL
in the following circuit can be redrawb as phasors in a phasor
diagram. Use a phasor diagram to show that
)
(
)
(
)
(
t
v
t
v
t
v
s
L
R
=
+
where
V
t
t
vs
)
377
cos(
120
)
(=
Solution:
o
o
o
33
.
79
22
.
22
67
.
10
79
.
11
*
90
10
*
5
*
377
3
\u2220
=
\u2212
\u2220
\u2220
=
=
\u2212
LI
j
VL
\u03c9
o
33
.
79
22
.
22\u2220
V
t
t
vL
)
33
.
79
377
cos(
22
.
22
)
(
o
+
=

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