1 electrical technology et 201 define series impedances and analyze series ac circuits using...

1

ELECTRICAL TECHNOLOGY ET 201

Define series impedances and analyze series AC circuits using circuit techniques.

2

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.

3

(CHAPTER 15)

SERIESSERIESAC CIRCUITSAC CIRCUITS

4

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]

5

Example 15.7

Draw the impedance diagram and find the total impedance.

15.3 Series Impedances

84

90021

j

jXR

XR

L

L

T

ZZZ

34.6394.8 TZ

Solution

6

26

12106

90900321

j

jj

jXjXR

XXR

CL

CL

T

ZZZZ

Example 15.8

Draw the impedance diagram and find the total impedance.

15.3 Series Impedances

43.1832.6 TZ

Solution

7

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:

21 ZZZ T

????, 21 VV

8

• 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

15.3 Series Configuration

9

14.5 Power Factor

• For a purely resistive load;

Hence;

• For purely inductive or purely capacitive load;

Hence;

TpF cos factor Power

0T 1cos TPF

TrmsrmsIEP cos

rmsrmsTrmsrms IEIEP cos

90T

0cos TrmsrmsIEP

0cos TPF

10

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

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R-L1. Phasor Notation

te sin4.141 0V 100 E


Series R-L circuit Apply phasor notation

12

R-L2. ZT

Impedance diagram:

43

)904()03(21

j

T

ZZZ

13.53 5 TZ


13

R-L

3. I

13.53 5

0V 100

TZ

EI

13.53A 20 I


14

R-L

4. VR and VL

Ohm’s Law:

)0 3)(13.53A 20(

RR ZIV

V 13.5360 RV

)90 4)(13.53A 20(

LL ZIV

V 87.3680 LV


15

R-LKirchhoff’s voltage law:

Or;

In rectangular form,

0 LR VVEV

LR VVE


V; 483613.53V 60 jR V

V 486487.36V 80 jL V

LR VVE

0V 100

0100)4864()4836(

jjj

16

R-LPhasor diagram:

I is in phase with the VR and lags the VL by 90o.

I lags E by 53.13o.


13.53A 20 I

V 13.5360 RV

V 87.3680 LV

0V 100 E

17

R-L

Power: The total power delivered to the circuit is

Where

E, I : effective values;

θT : phase angle between E and I

Or;

W1200

13.53cos)20)(100(

cos

TT EIP

W120032022 RIPT


18

R-L

Power factor:

13.53cos

cos

TpF

lagging 6.0pF


TZ

R

IE

R

E

IR

EI

RI

EI

P

EIP

2

cos

cos

TTP Z

RF cos

19

R-C1. Phasor Notation

A 13.53sin07.7 ti A 13.535 I


Series R-C circuit Apply phasor notation

20

R-C2. ZT

Impedance diagram:

86

)908()06(21

j

T

ZZZ

13.5310 TZ


21

R-C

3. E

)13.5310)(13.535(

TIZE

V 050 E


22

R-C

4. VR and VC

Ohm’s Law:

)06)(13.535(

RR ZIV

V 13.5330 RV

)908)(13.535(

CC ZIV

V 87.3640 CV


23

R-CKirchhoff’s voltage law:

Or;


0 CR VVEV

CR VVE

24

R-CPhasor diagram:

I is in phase with the VR and leads the VC by 90o.

I leads E by 53.13o.


A 13.535 I

V 050 E

V 13.5330 RV

V 87.3640 CV

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R-CTime domain: V 050 E

V 13.5330 RV

V 87.3640 CV


V sin7.70 te

V 13.53sin42.42 tvR

V 87.36sin56.56 tvC

26

R-C


Or;

W150

13.53cos)5)(50(

cos

TEIP

W1506522 RIP


27

R-C

Power factor:

Or;

13.53cos

cos

TpF

leading 6.0pF


TTP Z

RF cos

leading 6.010

6PF

28

R-L-C1. Phasor Notation

TIME DOMAIN

PHASOR DOMAIN


29

R-L-C Impedance diagram:

2. ZT

43

373

90900321

j

jj

XXR CL

T

ZZZZ


13.535TZ

30

R-L-C

3. I

13.535

050

TZ

EI


A 13.5310 I

31

R-L-C

4. VR , VL and VC

Ohm’s Law:

V 13.5330 RV


V 13.14330 CV

)907)(13.5310( LL IZV

)03)(13.5310( RR IZV

)903)(13.5310( CC IZV

V 87.3670 LV

32

R-L-CKirchhoff’s voltage law:

Or;


0 CLR VVVEV

CLR VVVE

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R-L-CPhasor diagram:

I is in phase with the VR , lags the VL by 90o, leads the VC by 90o

I lags E by 53.13o.


A 13.5310 I

V 13.5330 RV

V 13.14330 CV

V 87.3670 LV

V 050 E

34

R-L-CTime domain:


35

R-L-C


Or;

Power factor:

W30013.53cos)10)(50(cos TT EIP

W30031022 RIPT

13.53coscos TpF


lagging 6.0pF

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• The basic format for the VDR in AC circuits is exactly the same as that for the DC circuits.

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.

15.4 Voltage Divider Rule

EZ

ZV

T

xx

37

Example 15.11(a)

Calculate I, VR, VL and VC in phasor form.


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Example 15.11(a) - Solution

Combined the R’s, L’s and C’s.R T L T C T

10 0.1 H 100 mF

202sin377tv

i

H 1.005.005.021

LLLT

104621 RRRT

21

111

CCCT F 100

200200

200200

21

21 m

CC

CCCT


e

39

Example 15.11(a) – Solution (cont’d)

Find the reactances.

1. Transform the circuit into phasor domain.

7.37

)1.0(377TL LX

53.26)10100(377

116

TC CX

V 377sin220 te V 020 E

i I


R T X L X C

10 37.7 26.53

200 VV

IE

40


2. Determine the total impedance.

3. Calculate I.

17.1110

53.267.3710

j

jj

jXjXR CLTT

Z

16.4815 TZ


R T X L X C

10 37.7 26.53

200 VV

I

16.4815

020

TZ

EI A 16.481.33 I

E

41


4. Calculate VR, VL and VC

V 16.483.13 RV

)010)(16.4833.1(

RR IZV


R T X L X C

10 37.7 26.53

200 VV

I

V 84.4114.50 LV)907.37)(16.4833.1(

LL IZV

V 16.13828.35 CV)9053.26)(16.4833.1(

CC IZV

E

42

15.3 Series ConfigurationExample 15.11(b)

Calculate the total power factor.

Solution

Angle between E and I is

16.48coscos TpF

lagging 667.0pF

A 16.481.33 IV 020 E16.48

43

Example 15.11(c) Calculate the average power delivered to the circuit.

Solution


16.48cos)33.1)(20(cos TT EIP

W74.17TP

A 16.481.33 IV 020 E

44

Example 15.11(d) Draw the phasor diagram.

Solution


A 16.481.33 I

V 16.483.13 RV

V 84.4114.50 LV

V 16.13828.35 CV

V 020 E

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Example 15.11(e) Obtain the phasor sum of VR, VL and VC and show that it equals the input voltage E.

Solution


V 933.9894.8V 16.483.13 jR V

V 446.33355.37V 84.4114.50 jL V

V 534.23284.26V 16.13828.35 jC V

534.23446.33933.9284.26355.37894.8 jjjCLR

VVVE

V 020020021.0965.19 jjE

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Example 15.11(f) Find VR and VC using voltage divider rule.

Solution


R T X L X C

10 37.7 26.53

200 VV

IE

16.4815 TZ

)020(16.4815

010

E

Z

ZV

T

RR V 16.483.13 RV

)020(16.4815

9053.26

E

Z

ZV

T

CC V 16.13837.35 CV

47

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.

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(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.

49


(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.

50


(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.

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