CIRCUIT IMPEDANCE. By: Enzo Paterno Date: 05/2007

Transcription

1 IMPEDANE IUIT IMPEDANE By: Enzo Paterno Date: 05/007 5/007 Enzo Paterno

2 Inductors Source Acknowledgement: 3//03

3 IMPEDANE - IMPEDANE OF EEMENTS 5/007 Enzo Paterno 3

4 IMPEDANE - esistance: [ Ω ] ESISTIVE EEMENT Since v and i are in phase, has an angle of 0 V θ [ Ω] I θ 0 j0 e j0 5/007 Enzo Paterno 4

5 IMPEDANE - Inductance: Since v leads i by 90, has an angle of 90 [ H ] INDUTIVE EATANE V θ 90 ω [ Ω] I θ 5/007 Enzo Paterno 5 j 90 e π j

6 IMPEDANE - apacitance: Since I leads v by 90, has an angle of - 90 [ F ] APAITIVE EATANE V θ [ Ω] ω I θ π j 5/007 Enzo Paterno 6 j e j

7 IMPEDANE DIAGAM The impedance diagram is a complex plane that is used to plot the impedance of the various components (i.e.,, ) of A circuits. This diagram is used to compute the total impedance of a particular circuit. j Time domain frequency Domain V P.707 V p V p / sqrt() j 90 0 c jc 90 5/007 Enzo Paterno 7 -j

8 EAMPE Using complex algebra, find the current i for the circuit below. Sketch the waveforms of v and i. 5/007 Enzo Paterno 8

9 EAMPE Find the voltage v for the circuit below: i.5sin(377t 60 ) 0.06 H v - 5/007 Enzo Paterno 9

10 EAMPE i.5sin(377t 0.06 H 60 ) Note : ω 377 (0.06) 6.03 Ω I (.707)(.5) 60º.06 60º º V I (.06) (6.03) 90 º 60 º V º v (.44) (6.39) sin( 377 t 50º ) v 9.04 sin( 377t 50 ) 5/007 Enzo Paterno 0

11 EAMPE Using complex algebra, find the voltage v for the circuit below. Sketch the waveforms of v and i. 5/007 Enzo Paterno

12 EAMPE Find the impedance in Ohms of the series circuit elements that must be enclosed in the box for the indicated voltage and current to exist at the input terminals: I 0 ma 40º E 80 V 30º - Give answer in polar and rectangular forms! 5/007 Enzo Paterno

13 EAMPE I 0 ma 40º E 80 V 30º - E I 80v 30 4kΩ 80 0mA 40 4kΩ 80 -Y -80º 4k cos 80 j4k 5/007 Enzo Paterno 3 sin kω J kω

14 EAMPE Using complex algebra, find the voltage v for the circuit below. Sketch the waveforms of v and i. 5/007 Enzo Paterno 4

15 IUIT IMPEDANE IMPEDANE OF SEIES IUITS 5/007 Enzo Paterno 5

16 IMPEDANE OF SEIES IUITS I. N T I The total impedance of a series configuration circuit is the sum of individual impedances: T 3 N 5/007 Enzo Paterno 6

17 IMPEDANE OF SEIES IUITS Example: circuit A B T AB AB AB AB 0 j 90 tan tan tanθ tan( θ ) tanθ tan( θ ) -θ θ Odd function 5/007 Enzo Paterno 7 a - a

18 IMPEDANE OF SEIES IUITS Example: circuit A B T AB AB AB 0 j 90 tan 5/007 Enzo Paterno 8

19 IMPEDANE OF SEIES IUITS Example: circuit A B T AB AB AB AB 0 j j( ( 3 90 j ) ) 5/007 Enzo Paterno 9 tan 90

20 IMPEDANE OF SEIES IUITS j T j is always positive T tan T j -j T tan 5/007 Enzo Paterno 0

21 SEIES IUITS T 3 N I V T. N I The current is the same through each element just as for dc circuits: I V T 5/007 Enzo Paterno

22 SEIES IUITS T 3 N V I T. N I The voltage across each element can be found using Ohm s law just as for dc circuits: V i V I i,,3, i I, V I, V 3 I 3 5/007 Enzo Paterno

23 SEIES IUITS T 3 N V I T. N I The voltage across each element can be found using the VD just as for dc circuits: V i V i i T,,3, 5/007 Enzo Paterno 3

24 SEIES IUITS T 3 N V I T. N I Kirchhoff s voltage law can be applied in the same manner just as for dc circuits: V i N V i i 5/007 Enzo Paterno 4

25 IMPEDANE - T 3 N I V T. N I Example: Average Power to the circuit can be obtained using: IV p T cos θ θ T is the angle difference between E and I 5/007 Enzo Paterno 5 T V 00 0 I P 00(0) cos 53.3 P W

26 EAMPE Draw the impedance diagram for the circuit below and find the total impedance. 5/007 Enzo Paterno 6

27 EAMPE Determine the input impedance to the series network given below. Draw the impedance diagram. 5/007 Enzo Paterno 7

28 EAMPE Using the voltage divider rule, find the voltage across each element of the circuit below. 5/007 Enzo Paterno 8

29 EAMPE 5/007 Enzo Paterno 9

30 KT FEQUENY ESPONSE FEQUENY ESPONSE OF THE IUIT 5/007 Enzo Paterno 30

31 5/007 Enzo Paterno 3 Vin ] [ ) ( Ω f f π ω Angle Magnitude j j T T T T T tan tan ω KT T FEQUENY ESPONSE T Vout

32 KT G FEQUENY ESPONSE Vout G G Vin ( jc) Jc Vout Vin c c c c c Vin ( c 90 ) c tan c 90 c c tan c c c c c 5/007 Enzo Paterno 3

33 5/007 Enzo Paterno 33 ( ) 4 f f G c G π π ω KT G FEQUENY ESPONSE Frequency cutoff G f G f G f π

34 PF - G FEQUENY ESPONSE 0 0(.707) 4.4 v 6 Voltage Magnitude ,000 0,000 00,000,000,000 0,000,000 Frequency (Hz) f Vc 5/007 Enzo Paterno 34

35 KT FEQUENY ESPONSE FEQUENY ESPONSE OF THE IUIT 5/007 Enzo Paterno 35

36 5/007 Enzo Paterno 36 Vin ] [ ) ( Ω f f π ω Angle Magnitude j j T T T T T tan tan ω KT T FEQUENY ESPONSE T Vout

37 5/007 Enzo Paterno 37 G Vin Vout G J j Vin Vout tan 90 tan 90 ) ( KT G FEQUENY ESPONSE

38 5/007 Enzo Paterno 38 f G π ω KT G FEQUENY frequency cutoff 0 G f G f G f π

39 HPF - G FEQUENY ESPONSE 0 0(.707) 4.4 v 6 Voltage Magnitude ,000 0,000 00,000,000,000 0,000,000 Frequency (Hz) f Vr V 5/007 Enzo Paterno 39

40 IUIT IMPEDANE IMPEDANE OF PAAE IUITS 5/007 Enzo Paterno 40

41 ADMITTANE In ac circuits, we define admittance (Y) as being equal to /. The unit of measure for admittance as defined by the SI system is Siemens (S) and it s a measure of how well an ac circuit allows (admit), current to flow in the circuit. Example: For a resistor, we get: Example: // 5/007 Enzo Paterno 4

42 SUSEPTANE Susceptance is the reciprocal of reactance (/) is called and is a measure of how susceptible an element is to the passage of current through it. Susceptance is also measured in siemens and is represented by the capital letter B. Susceptance is analogous to the conductance of a resistor: Inductor: apacitor: 5/007 Enzo Paterno 4

43 PAAE IUITS For the network below: a. Find the admittance of each parallel branch. 5/007 Enzo Paterno 43

44 PAAE IUITS For the network below: b. Determine the input admittance. 5/007 Enzo Paterno 44

45 PAAE IUITS For the network below: c. alculate the input impedance. // // 3 5/007 Enzo Paterno 45

46 PAAE IUITS For the network below: d. Draw the admittance diagram. Y T Y Y Y 5/007 Enzo Paterno 46

47 PAAE ac NETWOKS I j i j T IY Y T j I N j i j 5/007 Enzo Paterno 47