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ELECTRICAL TECHNOLOGY
EET 103/4
 Define and explain sine wave, frequency,
amplitude, phase angle, complex number
 Define, analyze and calculate impedance,
inductance, phase shifting
 Explain and calculate active power,
reactive power, power factor
 Define, explain, and analyze Ohm’s law,
KCL, KVL, Source Transformation,
Thevenin theorem.
1
SERIES AND PARALLEL
AC CIRCUITS
(CHAPTER 15)
2
15.1 Introduction
In this chapter, phasor algebra will be used to
develop a quick, direct method for solving both
the series and the parallel ac circuits.
3
15.2 Impedance and the Phasor diagram
Resistive Elements
v and i are in phase
Vm
Im 
R
Or;
Vm  I m R
In phasor form;
Where;
v  Vm sin t
V  V0 
Vm
V
2
4
15.2 Impedance and the Phasor diagram
Resistive Elements
Applying Ohm’s law;
V
V0
I

R R R R
V
 0    R
R
Since v and i are in phase;
Hence;
0   R  0
R  0
5
15.2 Impedance and the Phasor diagram
The boldface Roman quantity ZR, having both
magnitude and an associate angle, is referred to as
the impedance of a resistive element.
ZR is not a phasor since it does not vary with time.
Even though the format R0° is very similar to the
phasor notation for sinusoidal current and voltage, R
and its associated angle of 0° are fixed, non-varying
quantities.
6
15.2 Impedance and the Phasor diagram
Example 15.1
Find i in the figure
Solution
v  100 sin t
V  70.70 V
Applying Ohm’s law;
V 70.70

I


1
4
.
14

0
A

ZR
50
7
15.2 Impedance and the Phasor diagram
Example 15.1 – solution (cont’d)
Inverse-transform;
I  14.140 A
i  20 sin t A
I = 14.14 A
V = 70.7 V
8
15.2 Impedance and the Phasor diagram
Example 15.2
Find v in the figure
Solution


i  4 sin t  30 A
I  2.8330 A
9
15.2 Impedance and the Phasor diagram
Example 15.2 – solution (cont’d)
Applying Ohm’s law;
V  IZ R
 2.8330  20
 5.6630 V
10
15.2 Impedance and the Phasor diagram
Example 15.2 – solution (cont’d)
Inverse-transform;
V  5.6630 V



v  8 sin t  30 V
V = 5.66 V
I = 2.83 A
30
11
15.2 Impedance and the Phasor diagram
Inductive Reactance
X L  L  2fL
ohm ()
By phasor transformation;
V  V0 
Where;
Vm
V
2
12
15.2 Impedance and the Phasor diagram
Inductive Reactance
By Ohm’s law;
V
I
X L  L
V0

X L  L
V

0   L
XL
13
15.2 Impedance and the Phasor diagram
Inductive Reactance
Since v leads i by 90;
0   L  90
Hence;
 L  90
Therefore;
Z L  X L 90
14
15.2 Impedance and the Phasor diagram
Inductive Reactance
Example 15.3
Find i in the figure
Solution
v  24 sin t V
V  16.970 V
15
15.2 Impedance and the Phasor diagram
Inductive Reactance
Example 15.3 – solution (cont’d)
Applying Ohm’s law;
V 16.970
I

ZL
390
 5.66  90 A
Inverse-transform;
I  5.66  90 A


i  8 sin t  90 A
16
15.2 Impedance and the Phasor diagram
Inductive Reactance
Example 15.3 – solution (cont’d)
I  5.66  90 A


i  8 sin t  90 A
V = 16.97 V
I = 5.66 A
17
15.2 Impedance and the Phasor diagram
Inductive Reactance
Example 15.4
Find v in the figure
Solution


i  5 sin t  30 A
I  3.5430 A
18
15.2 Impedance and the Phasor diagram
Inductive Reactance
Example 15.4 – solution (cont’d)
Applying Ohm’s law;
V  IZ L  3.5430  490
 14.14120 A
Inverse-transform;
V  14.14120 V


v  20 sin t  120 A
19
15.2 Impedance and the Phasor diagram
Inductive Reactance
Example 15.4 – solution (cont’d)
Inverse-transform;
V  14.14120 V


V  20 sin t  120 V
V = 14.14 V
I = 3.54 A
120
30
20
15.2 Impedance and the Phasor diagram
Capacitive Reactance
1
1
XC 

C 2fC
ohm ()
By phasor transformation;
V  V0 
Where;
Vm
V
2
21
15.2 Impedance and the Phasor diagram
Capacitive Reactance
By Ohm’s law;
V
I
X C  C
V0

X C  C
V

0   C
XC
22
15.2 Impedance and the Phasor diagram
Capacitive Reactance
Since i leads v by 90;
0   C  90
Hence;
 L  90
Therefore;
Z C  X C   90
23
15.2 Impedance and the Phasor diagram
Capacitive Reactance
Example 15.5
Find i in the figure
Solution
v  15 sin t V
V  10.60 V
24
15.2 Impedance and the Phasor diagram
Capacitive Reactance
Example 15.5 – solution
Use phasor transformation;
v  15 sin t V
V  10.60 V
By Ohm’s law;
V
V
I

Z C X C   90
10.60



5
.
3

90
A

2  90
25
15.2 Impedance and the Phasor diagram
Capacitive Reactance
Example 15.5 – Solution (cont’d)
Inverse-transform;
I  5.390 A


i  7.5 sin t  90 A
I = 5.3 A
V = 10.6 V
26
15.2 Impedance and the Phasor diagram
Capacitive Reactance
Example 15.6
Find v in the figure
27
15.2 Impedance and the Phasor diagram
Capacitive Reactance
Example 15.6 – solution
Use phasor transformation;


i  6 sin t  60 A
I  4.24  60 A
By Ohm’s law;
V  IZC
 4.24  60  0.5  90
 2.12  150 V
28
15.2 Impedance and the Phasor diagram
Capacitive Reactance
Example 15.6 – solution
Inverse- transform; V  2.12  150 V


v  3 sin t  150 V
29
15.2 Impedance and the Phasor diagram
Impedance Diagram
For any configuration (series, parallel,
series-parallel, etc.), the angle associated
with the total impedance is the angle by
which the applied voltage leads the
source current. For inductive networks, T
will be positive, whereas for capacitive
networks, T will be negative.
30
15.2 Impedance and the Phasor diagram
31
15.3 Series Configuration
The overall properties of series ac circuits are
the same as those for dc circuits.
For instance, the total impedance of a system is
the sum of the individual impedances:
32
15.3 Series Configuration
In a series ac configuration having two
impedances, the current is the same through each
element (as it was for the series dc circuit) and is
determined by Ohm’s Law:
33
15.3 Series Configuration
Example 15.7
Draw the impedance diagram and find the total impedance
34
15.3 Series Configuration
Example 15.7 – solution
Impedance diagram
ZT  Z1  Z2
 4  j8
ZT  4  j8  8.9463.34 
35
15.3 Series Configuration
Example 15.8
Draw the impedance diagram and find the total impedance
36
15.3 Series Configuration
Example 15.8 – solution
Impedance diagram
ZT  Z1  Z 2  Z3
 6  j10  j12
 6  j2
ZT  6  j 2  6.32  18.43 
37
15.3 Series Configuration
Kirchhoff’s voltage law can be applied in the
same manner as it is employed for a dc circuit
The power to the circuit can be determined by:
Where T is the phase angle between E and I.
38
15.3 Series Configuration
R-L
Phasor Notation
e  141.4 sin t V
E  1000 V
39
15.3 Series Configuration
R-L
 ZT
ZT  Z1  Z2
 3  j4
ZT  3  j 4  553.13 
Impedance diagram:
40
15.3 Series Configuration
R-L
 I
E
1000
I

ZT 553.13
I  20  53.13 A
41
15.3 Series Configuration
R-L
 VR and VL
VR  IZR
 20  53.13  30
VR  60  53.13 V
VL  IZL  20  53.13  490
VL  8036.87 V
42
15.3 Series Configuration
R-L
 Kirchhoff’s voltage law
V  E  V
R
 VL  0
Or;
E  VR  VL
43
15.3 Series Configuration
R-L
 Kirchhoff’s voltage law
From the above calculation;
VR  60  53.13  36  j 48 V
VL  8036.87  64  j 48 V
E  VR  VL
 36  j 48  64  j 48
 100  j 0  1000 V
44
15.3 Series Configuration
R-L
Phasor
diagram
45
15.3 Series Configuration
R-L
Power
P  EI cosT
 100  20  cos 53.13
 1200 W
Or;
P  I 2 R  20 2  3  1200 W
46
15.3 Series Configuration
R-L
Power factor
Fp  cosT  cos 53.13
Fp  0.6 lagging
47
15.3 Series Configuration
R-C
Phasor Notation


i  7.07 sin t  53.13 A
I  553.13 A
48
15.3 Series Configuration
R-C
 ZT
ZT  Z1  Z2
 6  j8
ZT  6  j8  10  53.13 
Impedance diagram:
49
15.3 Series Configuration
R-C
 E
E  IZT
 553.13 10  53.13
E  500 V
50
15.3 Series Configuration
R-C
 VR and VC
VR  IZR
 553.13  60
VR  3053.13 V
VC  IZC  553.13  8  90
VC  40  36.87 V
51
15.3 Series Configuration
R-C
 Phasor diagram
52
15.3 Series Configuration
R-C
 Time domain
E  500 V
VR  3053.13 V
VC  40  36.87 V
e  70.7 sin t V




vR  42.42 sin t  53.13 V
vC  56.56 sin t  36.87  V
53
15.3 Series Configuration
R-C
 Time domain plot
54
15.3 Series Configuration
R-C
 Kirchhoff’s voltage law
V  E  V
R
 VC  0
Or;
E  VR  VC
55
15.3 Series Configuration
R-C
Power
P  EI cosT
 50  5  cos 53.13
 150 W
Or;
P  I 2 R  52  6  150 W
56
15.3 Series Configuration
R-C
Power factor
Fp  cosT  cos 53.13
 cos 53.13
Fp  0.6 leading
57
15.3 Series Configuration
R-L-C
TIME DOMAIN
PHASOR DOMAIN
58
15.3 Series Configuration
R-L-C
ZT  Z1  Z 2  Z 3  3  j 7  j 3  3  j 4  553.13
59
15.3 Series Configuration
R-L-C
Impedance diagram:
60
15.3 Series Configuration
R-L-C
ZT  553.13
E
500

I


10


53
.
13
A

ZT 553.13
61
15.3 Series Configuration
R-L-C
ZT  553.13
VR  IZ R  10  53.13  30  30  53.13 V
62
15.3 Series Configuration
R-L-C
ZT  553.13
VL  IZ L  10  53.13  790  7036.87 V
63
15.3 Series Configuration
R-L-C
ZT  553.13
VC  IZ C  10  53.13  3  90  30  143.13 V
64
15.3 Series Configuration
R-L-C
Phasor
diagram
65
15.3 Series Configuration
R-L-C
Time-domain plot
66
15.3 Series Configuration
R-L-C
Total Power
PT  EI cos T  50 10 cos 53.13  300 W
Or;
PT  I 2 R  102  3  300 W
Power factor
Fp  cosT  cos 53.13  0.6 lagging
67
15.3 Series Configuration
R-L-C
Example 15.11(a)
Calculate I, VR, VL and VC;
68
15.3 Series Configuration
R-L-C
Example 15.11(a) – solution
Combined the R’s, L’s and C’s;
RT  R1  R2
 6  4  10 
RT
LT
CT
10 
0.1 H
100 mF
v
202sin377t
LT  L1  L2
i
 0.05  0.05  0.1 H
1
1
1
 
CT C1 C2
C1C2
200  200
CT 

 100 mF
C1  C2 200  200
69
15.3 Series Configuration
R-L-C
Example 15.11(a) – solution (cont’d)
Transform the circuit into phasor domain;
v  20 2 sin 377t V
i
I
X L  LT
 377 0.1
V  200 V
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I
 37.7 
1
1
XC 

 26.53 
6
CT 377 100 10
70
15.3 Series Configuration
R-L-C
Example 15.11(a) – solution (cont’d)
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I
Find the total impedance;
ZT  RT  jX L  jX C  10  j37.7  j 26.53
ZT  10  j11.17  1548.16 
71
15.3 Series Configuration
R-L-C
Example 15.11(a) – solution (cont’d)
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I
Apply Ohm’s law;
V
20
I

ZT 1548.16
I  1.33  48.16 A
72
15.3 Series Configuration
R-L-C
Example 15.11(a) – solution (cont’d)
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I
Apply Ohm’s law;
VR  IZ R  1.33  48.16 100
VR  13.3  48.16 V
73
15.3 Series Configuration
R-L-C
Example 15.11(a) – solution (cont’d)
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I
Apply Ohm’s law;
VL  IZ L  1.33  48.16  37.790
VL  50.1441.84 V
74
15.3 Series Configuration
R-L-C
Example 15.11(a) – solution (cont’d)
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I
Apply Ohm’s law;
VC  IZ C  1.33  48.16  26.53  90
VC  35.28  138.16 V
75
15.3 Series Configuration
R-L-C
Example 15.11(b)
Calculate the total power factor.
76
15.3 Series Configuration
R-L-C
Example 15.11(b) – solution
The total power factor;
Fp  cosT  cos 48.16
Fp  0.667 lagging
77
15.3 Series Configuration
R-L-C
Example 15.11(c)
Calculate the total average power delivered to the circuit.
78
15.3 Series Configuration
R-L-C
Example 15.11(c) – solution
The total average power delivered to the circuit.
PT  EI cosT  20 1.33 cos 48.16
PT  17.74 W
79
15.3 Series Configuration
R-L-C
Example 15.11(d)
Draw the phasor diagram;
80
15.3 Series Configuration
R-L-C
Example 15.11(d) – solution
The phasor diagram;
I  1.33  48.16 A
VR  13.3  48.16 V
VL  50.1441.84 V
VC  35.28  138.16 V
81
15.3 Series Configuration
R-L-C
Example 15.11(e)
Obtain the phasor sum of VR, VL and VC and show that it
equals the input voltage E.
82
15.3 Series Configuration
R-L-C
Example 15.11(e) – solution
VR  13.3  48.16 V  8.894  j9.933 V
VL  50.1441.84 V  37.355  j33.446 V
VC  35.28  138.16 V  26.284  j 23.534 V
E  VR  VL  VC
 8.894  37.355  26.284
 j 9.933  j 33.446  j 23.534
 19.965  j 0.021  20  j 0  200 V
83
15.3 Series Configuration
R-L-C
Example 15.11(f)
Find VR and VC using voltage divider rule.
84
15.3 Series Configuration
R-L-C
Example 15.11(f) – solution
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I
ZT  10  j11.17  1548.16 

ZR
20

0

VR  E
 200

13
.
3


48
.
16
V

ZT
1548.16
85
15.3 Series Configuration
R-L-C
Example 15.11(f) – solution
RT
XL
XC
10 
37.7 
26.53 
V
200 V
I

ZC
26
.
53


90

VC  E
 200

35
.
37


138
.
16
V

ZT
1548.16
86
15.6 Summaries of Series ac Circuits
 For a series ac circuit with reactive
elements:
The total impedance will be frequency
dependent.
The impedance of any one element can be
greater than the total impedance of the network.
The inductive and capacitive reactances are
always in direct opposition on an impedance
diagram.
Depending on the frequency applied, the same
circuit can be either predominantly inductive or
predominantly capacitive.
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15.6 Summaries of Series ac Circuits
 For a series ac circuit with reactive
elements:
At lower frequencies the capacitive elements
will usually have the most impact on the total
impedance, while at high frequencies the
inductive elements will usually have the most
impact.
The magnitude of the voltage across any one
element can be greater than the applied
voltage.
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15.6 Summaries of Series ac Circuits
 The magnitude of the voltage across an element
as compared to the other elements of the circuit
is directly related to the magnitude of its
impedance; that is, the larger the impedance of
an element , the larger the magnitude of the
voltage across the element.
 The voltages across a coil or conductor are
always in the direct opposition on a phasor
diagram.
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15.6 Summaries of Series AC Circuits
 The current is always in phase with the voltage
across the resistive elements, lags the voltage
across all the inductive elements by 90°, and
leads the voltage across the capacitive elements
by 90°.
 The larger the resistive element of a circuit
compared to the net reactive impedance, the
closer the power factor is to unity.
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