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Electrical Power and Machines- Lec03_Single Phase Circuit
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Unit 2
Single-Phase Circuits
Lecture O
utlines
Phasor Relationships for Circuit Elements
Im
pedance of Circuit Elements
Components of Im
pedance
Admittance
Admittance
Im
pedance C
ombinations
Three common circuit elements will be studied.
Their characteristics when subjected to
sinusoidal input is to be analysed.
Circuit Elements
Resistors
Inductors
Capacitors
Tim
e D
omain
Phasor Domain θ∠
=mI
)co
s(θ
ω+
=t
I m
Phasor Relationship for Resistors
)co
s(φ
ω+t
Vm
φ∠
mV
The current that
passes through a
resistor is in
phasewith the
voltage across it.
Resistor Phase Diagram voltage across it.
Tim
e
V
I
dt
di
Lv=
IV
Ljω
=
)4
56
0co
s(1
2)
(°
+=
tt
vPhasor Relationship for Inductors
)4
56
0co
s(1
2)
(°
+=
tt
v
H 10.
L=
°−
∠=
°∠
°∠
=
⋅
°∠
==
45
290
6
45
12
1.0
60
45
12 j
LjωV
I
)45
60
cos(
2)
(°
−=
tti
dt
di
Lv=
IV
Ljω
=
)4
56
0co
s(1
2)
(°
+=
tt
vPhasor Relationship for Inductors
)4
56
0co
s(2
)(
°−
=t
ti Note that the current lags the voltage by 90o
Tim
e
VI
dt
iC
v
∫=
CjωI
V=
)3
01
00
cos(
6)
(°
−=
tt
vPhasor Relationship for Capacitors
µF
5
0=
C
°∠
=
°∠
°−
∠=
×⋅
°−
∠=
=−
60
03
.0
)90
005
.0
)(30
6(
)10
50
100
)(30
6(6
j
CjωV
I
)60
100
cos(
03
.0
)(
°+
=t
ti
dt
iC
v
∫=
CjωI
V=
)3
01
00
cos(
6)
(°
−=
tt
vPhasor Relationship for Capacitors
Tim
e
)60
100
cos(
03
.0
)(
°+
=t
ti Note that the current leads the voltage by 90o
VI
Inductor
Capacitor
Phase Diagrams for Inductors &
Capacitors
The current lagsthe
voltage by 90o
The current leads
the voltage by 90o
For the sake of consistency, always m
ake the voltage as the reference.
IV
R=
IV
Ljω
=C
jωI
V=
V“o
ppositio
n s
how
ed b
y
Impedance of Circuit Elements
Impedance
IVZ=
Resistor
R=
ZC
jω1=
ZL
jω=
Z
Inductor
Capacitor
“oppositio
n s
how
ed b
y
the c
ircuit t
o the f
low
of
a s
inusoid
al curr
ent”
Unit: ohm
s (
Ω)
Impedance
jXR+
=Z
resistance
reactance
Com
ponents of Impedance
resistance
reactance
If X > 0, im
pedance is inductive or lagging
If X < 0, im
pedance is capacitive or leading
If X = 0, im
pedance is resistive
Means the
current lags
the voltage
Ljω
=L
Z
0=
ω
Impedance of an Inductor
0=
ω
∞→
ω
Cjω1
=C
Z
Impedance of a Capacitor
0=
ω
∞→
ω
Admittance
VI
ZY
==
1“r
ecip
rocal of im
pedance”
Unit: sie
mens (
S)
Admittance U
nit: sie
mens (
S)
jBG+
=Y
conductance
susceptance
Find v(t)and i(t)in the circuit shown.
4=
ωV
010
s°
∠=
V
Cjω1
5+
=Z
1
Example of Impedance Calculations
5.25
010
j−
°∠
==ZV
Is
CjωI
IZV
C=
=
)4.
63
4co
s(47
.4
)(
°−
=t
tv
)6.
26
4co
s(79
.1
)(
°+
=t
ti
1.0
4
15
×+
=j
Ω 5.
25
j−
=
°∠
=6.
26
79
.1
°−
∠
°∠
=6.
26
59
.5
010
1.0
4
6.26
79
.1
×
°∠
=j
°−
∠=
°∠
°∠
=4.
63
47
.4
90
4.0
6.26
79
.1
Impedance in Series
N3
21
eqZ
ZZ
ZZ
++
++
=..
....
...
equivalent
impedance
Impedance in Parallel
N3
21
eqZ1
Z1
Z1
Z1
Z
1+
++
+=
....
....
.
N3
21
eqY
YY
YY
++
++
=.........
equivalent
impedance
equivalent
admittance
rad/s
50
=ω
Find the input im
pedanceZ
in
Z3
Z1
Z2andZ
3are parallel 10
8
1
23
11
11
++
=+
=j
-jZ
ZZ
Example 1
Z2
Ω−
=
××
=
=
−
10
10
250
11
3
j
j
Cjω
1Z
Ω−
=
××
×+
=−
)23(
10
10
50
13
3
j
j2Z
Ω+
=
×+
=
+=
)1
08(
2.0
50
8
8
jj
Lj ω
3Z
7.1
71
7.
46
0.3
66
0.
13
14
44
81
1
20
30
16
24
23
10
8
10
82
3
2
23
∠∠=
++=
−+
−
−+
+=
++
=+
=
jj
jj
j
jj
j-j
32Z
ZZ
07
.1
22
.3
3.18
39
.3
0.36
60
.13
7.17
17
.46
j−
=°
−∠
=∠∠
=23
Z
Ω−
=−
+−
=
+=
07
.11
22
.3
07
.1
22
.3
10
jj
j
23
1in
ZZ
Z
Ω∠
=−
=
73
.8-
11
.53
0
7.
11
22
.3
jinZ
Find the voltage across
the inductor vO
°−
∠=
⇒
°−
15
20
)15
4co
s(20
S
t
V
Example 2 (cont)
rad/s
4
=ω
Ω=
×=
⇒
20
54
H 5
jj
OZ
Ω−
=×
×=
⇒−
25
10
10
4
1m
F
01
3j
jCZ
A 0
.16
5-
0.0
47
0.
74
17
.0
0.5
96
2.
11
6
15
20
j=°
−∠
=
°∠
°−
∠=
=TOTAL
SSZ
VI
Z1
Z2
I O
I S
I COO
SO
SC
VZVI
II
I−
=−
=
Example 2 (cont)
Ω=
1
00
j2Z
100
20
1
25
11
11 2
j
j-j
−=
+=
+=
OC
ZZ
Z
OO
IV
Ljω
=
Ω°
∠=
+=
+=
0.
59
62
.1
16
10
06
0j
21
TOTAL
ZZ
Z
Z2
I C
20
165
.0
047
.0
jj
OV
−−
=
CC
OO
OZ
IZ
IV
==
20
165
.0
047
.0
25
jj
-j
OO
COC
VV
ZVI
−−
==
=
V 98
.15
16
.17
7.4
5.16
∠=
+=
jOV
)16
4co
s(2.
17
)(
°+
=t
tv
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