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ELECTRICAL TECHNOLOGY
ET 201
 Define series impedances and analyze
series AC circuits using circuit
techniques.
1
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current (review)
FIG. 15.46 Reviewing the frequency response of the basic elements.
2
SERIES
AC CIRCUITS
(CHAPTER 15)
3
15.3 Series Impedances
• 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:
[Ω]
4
15.3 Series Impedances
Example 15.7
Draw the impedance diagram
and find the total impedance.
Solution
ZT  Z1  Z 2
 R0   X L 90
 R  jX L
 4  j8
ZT  8.9463.34 
5
15.3 Series Impedances
Example 15.8
Draw the impedance diagram
and find the total impedance.
Solution
ZT  Z1  Z 2  Z 3
 R0   X L 90   X C   90
 R  jX L  jX C
 6  j10  j12
 6  j2
ZT  6.32  18.43 
6
15.3 Series AC Circuit
• In a series AC configuration having two
impedances, the current I is the same through
each element (as it was for the series DC circuit)
• The current is determined by Ohm’s Law:
ZT  Z1  Z2
V1  ??, V2  ??
7
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
E, I : effective values (Erms, Irms)
θT : phase angle between E and I
8
14.5 Power Factor
P  Erms I rms cos T
Power factor  Fp  cos T
• For a purely resistive load;
T  0
Hence;
FP  cosT  1
P  Erms I rms cos T  Erms I rms
• For purely inductive or purely capacitive load;
T  90
Hence;
FP  cosT  0
P  Erms I rms cosT  0
9
14.5 Power Factor
• Power factor can be lagging or leading.
– Defined by the current through the load.
• Lagging power factor:
– Current lags voltage
– Inductive circuit
• Leading power factor:
– Current leads voltage
– Capacitive circuit
10
15.3 Series Configuration
R-L
1. Phasor Notation
Series R-L circuit
Apply phasor notation
e  141.4 sin t
E  100 V0
11
15.3 Series Configuration
R-L
2. ZT
ZT  Z1  Z 2
 (30)  (490)
 3  j4
ZT  5 53.13
Impedance diagram:
12
15.3 Series Configuration
R-L
3. I
E
100 V0
I

ZT 5 53.13
I  20 A   53.13
13
15.3 Series Configuration
R-L
4. VR and VL
Ohm’s Law:
VR  IZR
 (20 A   53.13 )(3 0)
VR  60  53.13 V
VL  IZL
 (20 A   53.13 )( 4 90 )
VL  8036.87 V
14
15.3 Series Configuration
R-L
Kirchhoff’s voltage law:
V  E  V
R
 VL  0
E  VR  VL
Or;
In rectangular form,
VR  60 V  53.13  36  j 48 V;
VL  80 V36.87  64  j 48 V
E  VR  VL
 (36  j 48)  (64  j 48)  100  j 0
 100 V0
15
15.3 Series Configuration
R-L
Phasor diagram:
E  100 V0
I  20 A   53.13
VR  60  53.13 V
VL  8036.87 V
I is in phase with the VR and lags the VL by 90o.
I lags E by 53.13o.
16
15.3 Series Configuration
R-L
Power:
The total power delivered to the circuit is
PT  EI cos T
 (100)( 20) cos 53.13
 1200 W
Where
E, I : effective values;
θT : phase angle between E and I
Or;
PT  I 2 R  202  3  1200 W
17
15.3 Series Configuration
R-L
Power factor:
Fp  cos T
 cos 53.13
Fp  0.6
lagging
P  EI cos 
P I 2 R IR
R
R
cos  




EI
EI
E E I ZT
R
FP  cos T 
ZT
18
15.3 Series Configuration
R-C
1. Phasor Notation
Apply phasor notation
Series R-C circuit


i  7.07 sin t  53.13 A
I  553.13 A
19
15.3 Series Configuration
R-C
2. ZT
Impedance diagram:
ZT  Z1  Z 2
 (60)  (8  90)
 6  j8
ZT  10  53.13 
20
15.3 Series Configuration
R-C
3. E
E  IZT
 (553.13 )(10  53.13 )
E  500 V
21
15.3 Series Configuration
R-C
4. VR and VC
Ohm’s Law:
VR  IZ R
 (553.13 )(60 )
VR  3053.13 V
VC  IZC
 (553.13 )(8  90 )
VC  40  36.87 V
22
15.3 Series Configuration
R-C
Kirchhoff’s voltage law:
V  E  V
R
Or;
 VC  0
E  VR  VC
23
15.3 Series Configuration
R-C
Phasor diagram:
I  553.13 A
E  500 V
VR  3053.13 V
VC  40  36.87 V
I is in phase with the VR and leads the VC by 90o.
I leads E by 53.13o.
24
15.3 Series Configuration
R-C
e  70.7 sin t V
Time domain: E  500 V


 56.56 sin t  36.87  V
VR  3053.13 V
vR  42.42 sin t  53.13 V
VC  40  36.87 V
vC


25
15.3 Series Configuration
R-C
Power:
The total power delivered to the circuit is
P  EI cos T
 (50)(5) cos 53.13
 150 W
Or;
P  I 2 R  52  6  150 W
26
15.3 Series Configuration
R-C
Power factor:
Fp  cos T
 cos 53.13
Fp  0.6
Or;
leading
R
FP  cos T 
ZT
6
FP 
 0.6 leading
10
27
15.3 Series Configuration
R-L-C
1. Phasor Notation
TIME DOMAIN
PHASOR DOMAIN
28
15.3 Series Configuration
R-L-C
Impedance diagram:
2. ZT
ZT  Z1  Z 2  Z 3
 R0   X L 90   X C   90
 3  j 7  j3
 3  j4
ZT  553.13
29
15.3 Series Configuration
R-L-C
3. I
E
500
I

ZT 553.13
I  10  53.13 A
30
15.3 Series Configuration
R-L-C
4. VR , VL and VC
Ohm’s Law:
VR  IZ R  (10  53.13 )(30 )
VR  30  53.13 V
VL  IZ L  (10  53.13 )(790 )
VL  7036.87 V
VC  IZ C  (10  53.13 )(3  90 )


VC  30  143.13 V
31
15.3 Series Configuration
R-L-C
Kirchhoff’s voltage law:
V  E  V
R
Or;
 VL  VC  0
E  VR  VL  VC
32
15.3 Series Configuration
R-L-C
Phasor diagram:
E  500 V
I  10  53.13 A
VR  30  53.13 V
VL  7036.87 V
VC  30  143.13 V
I is in phase with the VR , lags the VL by 90o, leads the VC by 90o
I lags E by 53.13o.
33
15.3 Series Configuration
R-L-C
Time domain:
34
15.3 Series Configuration
R-L-C
Power:
The total power delivered to the circuit is
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
Fp  0.6
lagging
35
15.4 Voltage Divider Rule
• The basic format for the VDR in AC circuits is
exactly the same as that for the DC circuits.
Zx
Vx 
E
ZT
Where
Vx : voltage across one or more elements in a series that
have total impedance Zx
E : total voltage appearing across the series circuit.
ZT : total impedance of the series circuit.
36
15.3 Series Configuration
Example 15.11(a)
Calculate I, VR, VL and VC in phasor form.
37
15.3 Series Configuration
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
ve
202sin377t
i
LT  L1  L2
 0.05  0.05  0.1 H
1
1
1
 
CT C1 C2
C1C2
200  200
CT 

 100 mF
C1  C2 200  200
38
15.3 Series Configuration
Example 15.11(a) – Solution (cont’d)
Find the reactances.
X L  LT  377(0.1)
 37.7 
RT
XL
XC
10 
37.7 
26.53 
VE
200 V
I
1
1
XC 

 26.53 
6
CT 377(100 10 )
1. Transform the circuit into phasor domain.
e  20 2 sin 377t V
i
E  200 V
I
39
15.3 Series Configuration
Example 15.11(a) – Solution (cont’d)
2. Determine the total impedance.
ZT  RT  jX L  jX C
 10  j37.7  j 26.53
 10  j11.17
ZT  1548.16 
RT
XL
XC
10 
37.7 
26.53 
V
E
200 V
I
3. Calculate I.
E
200
I

ZT 1548.16
I  1.33  48.16 A
40
15.3 Series Configuration
Example 15.11(a) – Solution (cont’d)
4. Calculate VR, VL and VC
VR  IZ R
RT
XL
XC
10 
37.7 
26.53 
VE
200 V
I
 (1.33  48.16 )(100 )
VR  13.3  48.16 V
VL  IZ L
 (1.33  48.16 )(37.790 )
VC  IZ C
 (1.33  48.16 )( 26.53  90 )
VL  50.1441.84 V
VC  35.28  138.16 V
41
15.3 Series Configuration
Example 15.11(b)
Calculate the total power factor.
Solution
I  1.33  48.16 A
E  200 V
Angle between E and I is 48.16
Fp  cosT  cos 48.16
Fp  0.667
lagging
42
15.3 Series Configuration
Example 15.11(c)
Calculate the average power delivered to the circuit.
Solution
E  200 V
I  1.33  48.16 A
PT  EI cos T  (20)(1.33) cos 48.16
PT  17.74 W
43
15.3 Series Configuration
Example 15.11(d)
Draw the phasor diagram.
Solution
E  200 V
I  1.33  48.16 A
VR  13.3  48.16 V
VL  50.1441.84 V
VC  35.28  138.16 V
44
15.3 Series Configuration
Example 15.11(e)
Obtain the phasor sum of VR, VL and VC and show
that it equals the input voltage 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  j33.446  j 23.534
E  19.965  j 0.021  20  j 0  200 V
45
15.3 Series Configuration
Example 15.11(f)
Find VR and VC using voltage divider rule.
Solution
ZT  1548.16 

RT
XL
XC
10 
37.7 
26.53 
VE
200 V
I
ZR
100

VR 
E
(
20

0
)

ZT
1548.16
VR  13.3  48.16 V
ZC
26.53  90

VC 
E
(
20

0
)

ZT
1548.16
VC  35.37  138.16 V
46
15.6 Summaries of Series AC Circuits
For a series AC circuits 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.
47
15.6 Summaries of Series AC Circuits
(continued…)
• At lower frequencies, the capacitive elements
will usually have the most impact on the total
impedance.
• At high frequencies, the inductive elements will
usually have the most impact on the total
impedance.
• The magnitude of the voltage across any one
element can be greater than the applied voltage.
48
15.6 Summaries of Series AC Circuits
(continued…)
• 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 an inductor or capacitor are
always in direct opposition on a phasor diagram.
49
15.6 Summaries of Series AC Circuits
(continued…)
• 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.
50