Phasors. Phasors. by Prof. Dr. Osman SEVAİOĞLU Electrical and Electronics Engineering Department. ^ V cos (wt + θ) ^ V sin (wt + θ)

Transcription

1 V cos (wt θ) V sin (wt θ) by Prof. Dr. Osman SEVAİOĞLU Electrical and Electronics Engineering Department EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 1

2 Vector Definition A vector is a magnitude directed in a certain direction (angle) A vector is shown as r where, r is the magnitude, known as radius, is the angle The above representation is known as the polar representation of a vector EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 2

3 j Operator Definition j operator is a vector with unity magnitude, directed in vertical direction, i.e. in 90 o angle o j j = 1 90 o EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 3

4 Complex Number Definition A complex number is a number with two components; Graphical Representation Imaginary component Real component (an ordinary number), Imaginary component (a number multiplied by the j operator) b r a a j b The above representation is known as the rectangular representation Real component EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 4

5 Polar Representation of Complex Number Definition A complex number may be expressed in polar representation by employing the following conversion Graphical Representation Radius r = a 2 b 2 = Tan -1 (b / a) b r θ r a Angle Polar representation of a vector EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 5

6 Conversion from Rectangular Representation to Polar Representation Rule Graphical Representation A complex number expressed in rectangular coordinates can be converted into a number expressed in polar coordinates as follows; Let the complex number expressed in rectangular coordinates be Then where, a j b a j b = r r = a 2 b 2 θ = Tan -1 ( b / a ) θ b r θ a EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 6

7 Conversion from Polar Representation to Rectangular Representation Rule Graphical Representation A complex number expressed in polar coordinates can be converted into a number expressed in rectangular coordinates as follows; Let the complex number expressed in polar (phasor) coordinates be Then r r = a j b b r θ a where, a = r cos b = r sin EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 7

8 Polar and Rectangular Representations - Summary Conversion Rules Polar Representation Rectangular Representation r a = r cos, b = r sin r = a 2 b 2, = Tan -1 (b / a) a j b b r θ a EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 8

9 Addition of two Complex Numbers Method Isn t there an easier way of doing that? Suppose that two phasors are to be added r 1 1 r 2 2 = r tot tot First express the phasors in rectangular coordinates, and then perform the addition Polar Representation a = r cos, b =r sin Rectangular Representation r 1 1 a 1 j b 1 r 2 2 a 2 j b 2 a tot j b tot r = a 2 b 2, =Tan -1 ( b/a ) r tot tot EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 9

10 Subtraction of two Complex Numbers Method I m afraid NOT! Suppose that two phasors are to be subtracted r r 2 2 = r tot tot First express the phasors in polar coordinates, and then perform the subtraction Polar Representation a = r cos, b =r sin Rectangular Representation r 1 1 a 1 j b 1 r a 2 j b 2 (a 1- a 2 ) j (b 1 -b 2 ) r = a 2 b 2, =Tan -1 (b/a) r tot tot EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 10

11 Multiplication of two Complex Numbers Method Suppose that two phasors are to be multiplied ( a 1 j b 1 ) x ( a 2 jb 2 ) = a result j b result First express the phasors in polar coordinates, and then perform the multiplication Rectangular Representation r = a 2 b 2, =Tan -1 (b/a) Polar Representation a 1 j b 1 r 1 1 a 2 j b 2 X r 2 2 r 1 r a = r cos, b = r sin a result j b result EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 11 11

12 Division of two Complex Numbers Method Suppose that two phasors are to be divided ( a 1 j b 1 ) / ( a 2 jb 2 ) = a result j b result First express the phasors in polar coordinates, and then perform the division Rectangular Representation r = a 2 b 2, =Tan -1 (b/a) Polar Representation a 1 j b 1 a 2 j b 2 r 1 1 r 2 2 r 1 / r a = r cos, b = r sin a result j b result EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 12 12

13 Properties of j Operator Definition Multiplication of a complex number by j operator shifts (rotates) the angle of vector by 90 o, while the magnitude is unchanged 2 x 1 90 o 60 o = o 150 o 2 60 o j = 1 90 o 60 o j = 1 90 o Example Multiply complex number 2 60 o by j 2 60 o x j = 2 60 o x 1 90 o = 2 x 1 60 o 90 o = o EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 13 13

14 Properties of j Operator Powers of j j = 1 90 o j 2 = 1 90 o x 1 90 o = o = - 1 j 3 = o = 1-90 o = - j j 4 = 1 4 x 90 o = o = 1 1/j = 1 / 1 90 o = 1-90 o = - j j 4 = 1 j 2 = o j 360 o 180 o - 90 o j 3 = 1 / j = -j EE EE Fundamentals Fundamentals of of Electrical Electrical and and Electronics Electronics Engineering, Engineering, Prof. Prof. Dr. Dr. O. O. SEVAİOĞLU, SEVAİOĞLU, Page Page 14 14

15 Euler s Identity Graphical Representation e jθ = cos θ j sin θ e jθ = cos θ j sin θ = cosθ 2 sin θ 2 = 1 sin θ r b θ a a 2 b 2 = r 2 r = a 2 b 2 cos θ EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 15

16 Leonhard EULER ( ) Swiss Mathematician Leonhard Euler was born in Basel, Switzerland, but the family moved to Riehen when he was one year old and it was in Riehen, not far from Basel, that Leonard was brought up. Paul Euler had, as we have mentioned, some mathematical training and he was able to teach his son elementary mathematics along with other subjects. Euler made substantial contributions to differential geometry, investigating the theory of surfaces and curvature of surfaces. Many unpublished results by Euler in this area were rediscovered by Gauss. Other geometric investigations led him to fundamental ideas in topology such as the Euler characteristic of a polyhedron. In 1736 Euler published Mechanica which provided a major advance in mechanics. EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 16

17 Euler s Identity V e jθ = V (cos θ j sin θ) Graphical Representation V e jθ = V cos θ j sin θ = V cos 2 θ sin 2 θ = V j V sin θ θ V cos θ EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 17

18 Definition of Basic Terms Graphical Representation Now, Let θ be a linear function of time t, i.e. rotate it clockwise θ = wt w = 2 f = 2 x x 50 = 314 Radians/sec ( f = 50 Hz ) 1 Radian = 360 o / ( 2 ) = o EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 18

19 e jθ Euler s Identity = cos θ j sin θ V e jwt = V ( cos wt j sin wt ) Graphical Representation V = Magnitude V (t) = V sin θ = V sin wt θ V(t) V (t) = V sin θ V (t) = V cos θ θ V (t) = V cos θ θ = wt V (t) = V cos wt wt (radians) V EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 19

20 Symbolic Representation Mathematical Notation Graphical Representation Now let a phasor be located at an angular position initially, i.e. V (t) = V cos (wt ) t = 0 In other words the phasor is at an angular position at t = 0 The phasor on the RHS is then represented mathematically by the following notations; θ = wt t = 0 = V sin θ = V sin V = Magnitude V cos Initial phase angle = EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 20

21 Angular Displacement Total Angular Displacement Graphical Representation Total angular displacement at time t = t 1 may then be expresssed as; θ( t 1 ) = w t 1 V sin θ = V sin (wt 1 ) Angle swept during t = t 1 Then the horizontal component becomes; V (t) = V cos (wt ) t = t 1 = V cos (wt 1 ) Initial phase angle = In other words, the phasor will be at an angular position wt 1 at t = t 1 wt 1 V cos θ = V cos (wt 1 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 21

22 Mathematical Notation Notation The phasor on the RHS may be represented mathematically as; V e jθ( t 1 ) = V e j (w t 1 ) Graphical Representation V sin θ = V sin (wt 1 ) Angle swept during t = t 1 Initial phase angle = wt 1 V e j( w t 1 ) V Amplitude Phase angle V cos (wt 1 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 22

23 Waveform Representation of Resistive Circuits Waveform Representation of Resistive Circuits Consider the resistive circuit shown on the RHS V(t) = V cos wt I(t) = I cos wt V(t) = V cos wt _ R where, I(t) = V s (t) / R = V cos wt / R = I cos wt I = V / R Ohm s Law V I 2,0 1,5 1,0 0,5 0-0,5-1,0-1,5 Angle (Radians) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 23

24 Phasor Representation of Resistive Circuits Phasor Representation of Resistive Circuits V I Consider the resistive circuit shown on the RHS Waveform Representation V(t) = V cos ( wt 0 o ) I(t) = I cos ( wt 0 o ) 2,0 1,5 1,0 0,5 0-0,5 Angle (Radians) Phasor Representation V = V I = I 0 o 0 o Ohm s Law V 0 o Z = = Z 0 o = R I 0 o Z = R 2 = R Z = Tan -1 ( 0 / R ) = 0-1,0-1,5 I V EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 24

25 Waveform Representation of Inductive Circuits Waveform Representation of Inductive Circuits Consider the inductive circuit shown on the RHS I(t) = I sin wt V(t) = V cos wt L Let now, I(t) = I sin wt 2,0 _ V(t) = L di(t) / dt = L w I where, V = L w I V cos wt = V cos wt 0 1 V I 1,5 1,0 0,5 0 Angle (Radians) cos ( wt - 90 o ) = coswt cos 90 o sin wt sin 90 o I(t) = I sin wt = I cos (wt -90 o ) -0,5-1,0-1,5 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 25

26 Phasor Representation of Inductive Circuits Phasor Representation of Inductive Circuits V Waveform Representation V(t) = V cos wt I(t) = I cos (wt - 90 o ) 2,0 1,5 Phasor Representation V = V I = I 0 o -90 o Ohm s Law V 0 o Z = = Z 90 o = Lw 90 o I -90 o I 1,0 0,5 0 Angle (Radians) I -90 o Z = Lw 90 o = j wl Z = R 2 (wl) 2 = wl -0,5-1,0-1,5 V Z V = L w I = Tan -1 ( wl / 0 ) = 90 o EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 26

27 I(t) = C dv s (t) / dt = - C w V sin wt Waveform Representation of Capacitive Circuits Waveform Representation of Capacitive Circuits Consider the capacitive circuit shown on the RHS V(t) = V cos wt I 0 1 V 1,5 1,0 I(t) = I cos( wt90 o ) V(t) = V cos wt _ C cos (wt 90 o ) = coswt cos 90 o - sin wt sin 90 o 0,5 where, = I cos ( wt 90 o ) I 0-0,5-1, Angle (Radians) I = C w V -1,5 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 27

28 Phasor Representation of Capacitive Circuits Phasor Representation of Capacitive Circuits V I Waveform Representation V(t) = V cos wt I(t) = I cos ( wt 90 o ) 1,5 1,0 0,5 0-0,5-1,0-1, Angle (Radians) Phasor Representation I V = V I = I 0 o 90 o 90 o V Ohm s Law Z = R 2 (1/Cw) 2 Z V 0 o Z = = Z -90 o = 1 / Cw -90 o I 90 o Z I = C w V = 1 / jwc = 1-90 o / Cw = - j / Cw = 1 / Cw = Tan -1 ( -Cw / 0 ) = -90 o EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 28

29 Phasor Representation of AC Circuits Phasor Representation of R-L Circuits Consider the following R-L circuit V = V Z = R j wl Z = R 2 (wl) 2 0 o Z θ R R 2 (wl) 2 = Z 2 Z = R 2 (wl) 2 jwl V(t), I(t) Z = Tan -1 ( wl / R) = I = V 0 o / ( R 2 (wl) 2 ) = I - R 2 (wl) 2 I = V / I(t) = I cos (wt - ) V(t) = V cos wt _ L wt (Radians) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 29

30 Phasor Representation of AC Circuits Phasor Representation of R-C Circuits Consider the R-C circuit shown below V = V 0 o Z = R (1 / jwc) = R j (1 / wc) Z = R 2 (1 / wc) 2 Z = Tan -1 (1 / (wcr)) = < 0 Z θ R 2 (1/wC) 2 = Z 2 R -j(1/wc) Z = R 2 (1/wC) 2 V(t), I(t) I = V 0 o / ( R 2 (1 / wc) 2 - ) = I I = V / R 2 (1 / wc) 2 I(t) = I cos (wt - ) V(t) = V cos wt _ C wt (Radians) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 30

31 Phasor Representation of AC Circuit Elements Resistance R-L Element R-C Element R R j wl R -j /(wc) Z = R 2 = R Z = R 2 (wl) 2 Z = R 2 (1/wC) 2 Z = Tan -1 (0 / R) = 0 Z = Tan -1 ( wl / R ) Z = Tan -1 ( (1/wC) / R ) Z = R 0 o Z = R 2 (wl) 2 Tan -1 ( wl / R ) Z = R 2 (1/wC) 2 -Tan -1 ((1/wC)/R ) X = 0 X > 0 X < 0 V V I o V I 0 o I o EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 31

32 Phasor Representation of R-L-C Circuits Phasor Representation of R-L-C Circuits Consider the R-L-C circuit shown on the RHS 1 / j = - j V = V 0 o Z = R jwl - j / wc = R j (wl- 1/wC) I(t) = I cos (wt - ) R L Z = R 2 (wl- 1/wC) 2 Z = Tan -1 ((wl - 1/(wC))/R) = V(t) = V cos wt C I = V 0 o / ( R 2 (wl - 1/wC) 2 ) = I - R 2 (wl - 1/wC) 2 I = V / Ohm s Law: I = V / Z _ EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 32

33 Phasor Representation of R-L-C Circuits R-L-C Circuits I(t) = I cos (wt - ) R = 2 L = 0.01 H Solve the R-L-C circuit shown on the RHS for current phasor V = V 0 o = o V(t) = cos (wt 0 o ) C = 2 mf Z = R jwl - j / wc = 2 j( ) = 2 j 1.55 = 0 o _ Z = R 2 (wl- 1/wC) 2 = Z Z = = Tan -1 ((wl - 1/(wC))/R) = Tan -1 (1.55 / 2) = = o = o w = 2 f =2 x 3.14 x 50 = 314 radians /sec X L = j wl = j 0.01 x 314 = j 3.14 X C = 1 / (jwc) = - j / wc = - j / (314 x 2 x 10-3 ) = - j 1.59 > 0 (Inductive) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 33

34 Phasor Representation of R-L-C Circuits Example I(t) = I cos (wt - ) R = 2 L = 0.01 H Solve the R-L-C circuit shown on the RHS for current phasor I = V 0 o / ( R 2 (wl - 1/wC) 2 ) V(t) = cos (wt 0 o ) C = 2 mf = o / o = I - = 0 o _ = o Amp EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 34

35 METU Phasor Representation of R-L-C Circuits Example (Continued) V = Volts Draw the voltage and current phasors and waveforms V = o Volts (peak) I = o Amp o I = o Amp o R = 2 L = 0.01 H I(t) = cos (wt o ) V(t) = cos (wt 0 o ) C = 2 mf V(t) 0 Angle (Radians) = 0 o _ I(t) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 35

36 Problem Example R = 3 I(t) Solve the circuit on the RHS for current waveform by using the phasor method Waveform Representation V(t) = V cos wt = 220 x 2 cos wt Solution Phasor Representation V = V 0 o = 220 x 2 Z = R 2 X 2 = = 5 0 o Volts V(t) = V cos wt = 220 x 2 cos wt Z = o _ j wl = jx = 4 Z = Tan -1 ( 4 / 3 ) = o I = 220 x 2 0 o o Amp = o Amp EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 36

37 Problem Example Draw the waveforms and phasors for the previous problem I = o Amp V s = V 0 o = 220 x 2 0 o Waveform Representation Phasor Representation V(t) = V cos wt = 220 x 2 cos wt V = V 0 o = 220 x 2 0 o Volts o o I(t) = cos ( wt o ) I = o Amp 2,0 1,5 1,0 I 0,5 Angle (Radians) V(t) I(t) -0,5-1,0 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 37-1,5

38 Example Calculate the equivalent impedance seen between the terminals A and B of the AC circuit given on the RHS (w = 314 rad / sec) A First, let us calculate impedances Z C = 1 / ( jwc ) = 1 / ( j314 x 1x 10-3 ) = - j R 1 = 1 C=1 mf R 2 = 2 Z L = jwl = j 314 x 0.01 = j 3.14 Z C // R 1 = 1 / ( 1/Z c 1/R 1 ) = 1 / (1/-j /1) = 1 /( 1 j0.314 ) = 1 / ( o ) = o = j L=0.01 H ( Z C // R 1 ) Z L = j j 3.14 = j B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 38

39 Example (Continued) (Continued) A Now, let us calculate: ( R eq j Z Leq ) // R 2 A ( R eq j Z Leq ) // R 2 = ( j ) // 2 = o // 2 0 o R eq = R 2 = x o = ( ) j o = o X Leq = B B = a = r cos, b = r sin j EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 39

40 Maximum Power Transfer Condition in AC Circuits Question Calculate the value of the impedance Z L of the load in the AC circuit shown on the RHS, in order to transfer maximum power from source to load Given Circuit Load Impedance: Z L = R L j X L R 1 A V s X 3 X 2 R 4 B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 40

41 Maximum Power Transfer Condition in AC Circuits Solution First simplify the AC circuit to its Thevenin Equivalent Form as shown on the RHS Given Circuit Thevenin Equivalent Circuit R 1 A Z eq = R eq jx eq A Load Impedance Z L = R L jx L Load Impedance Z L = R L jx L V s X3 X 2 V eq R 4 B B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 41

42 Maximum Power Transfer Condition in AC Circuits Solution (Continued) Load Impedance: Z L = R L jx L Then the problem reduces to the determination of the load impedance in the simplified circuit shown on the RHS Z eq = R eq jx eq A P = R L I 2 I 2 = ( V eq. / Z total ) 2 = ( V eq. / ( Z eq. Z L )) 2 Hence, P = R L V eq.2 / ( Z eq. Z L ) 2 = V eq.2 R L / (( R eq. R L ) 2 ( X eq. X L ) 2 ) V eq B Load Impedance Z L = R L jx L EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 42

43 Maximum Power Transfer Condition in AC Circuits Solution (Continued) Load Impedance: Z L = R L jx L Let us now first maximize P wrt X L by differentiating P with respect to X L Z eq = R eq jx eq A dp / dx L = 0 d/dx L V eq.2 R L / (( R eq. R L ) 2 ( X eq. X L ) 2 ) = 0 or V eq. 2 R L ( -2 (X eq. X L ) / [( R eq. R L ) 2 ( X eq. X L ) 2 ] 2 = 0 or V eq.2 R L ( -2 (X eq. X L ) = 0 The above expression becomes zero when; X eq. = - X L V eq B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 43

44 Maximum Power Transfer Condition in AC Circuits Solution (Continued) Load Impedance: Z L = R L jx L Hence, power becomes the same as that for the DC case; Z eq = R eq jx eq A P = V eq.2 R L / (( R eq. R L ) 2 ( X eq. X L ) 2 ) = 0 V eq P = R L V eq.2 / ( R eq. R L ) 2 B EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 44

45 Maximum Power Transfer Condition in AC Circuits Solution (Continued) Thevenin Equivalent Circuit Now, we must maximize P wrt R L by differentiating P with respect to R L dp / dr L = 0 d/dr L V eq.2 R L / ( R eq. R L ) 2 = 0 V eq. 2 [( R eq. R L ) 2 2 ( R eq. R L ) V eq.2 R L ] / d 2 = 0 where, d = ( R eq. R L ) 2 or V eq. 2 [( R eq. R L ) 2 2 ( R eq. R L ) R L ] = 0 ( R eq. R L ) 2 2 ( R eq. R L ) R L = 0 ( R eq. R L ) 2 R L = 0 R eq. = R L V eq Z eq = R eq jx eq Conclusions: For maximum power transfer; (a) X L = - X eq (b) Load resistance must be equal to the Thevenin Equivalent Resistance of the simplified circuit; R eq. = R L or * (c) Z L = Z eq A B Load Impedance Z L = Z eq * EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 45

46 The Principle of Superposition in AC Circuits Question I 1 Solve the AC circuit shown on the RHS by using The Principle of Superposition I 5 I S I 2 Solution V s a) Kill all sources, except one, b) Solve the resulting circuit, c) Restore back the killed source and kill all sources, except another one, d) Repeat the solution procedure (a) - (c) for all sources, e) Then, sum up algebraically all the solutions found I 6 I 4 I 3 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 46

47 The Principle of Superposition in AC Circuits Procedure SC I 5 I 1 I S SC I 2 SC I 5 V I s 1 OC I 2 V s I 6 I 4 I 3 I 2 I 4 I 3 I 2 I 6 I 7 I 5 I 1 OC SC I 2 V s I 4 I 6 I 7 I 3 I 2 EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 47

48 Example 1 - The Principle of Superposition in AC Circuits Question Find the current I Z2 flowing in impedance Z 2 in the following circuit by using the Principle of Superposition Solution Kill all sources except one and solve the resulting circuit V s Z 1 I Z2 V s I s Z 2 Z 1 OC I Z2-1 Z 2 This method is particularly useful when there are sources with different frequencies I Z2-1 = V s / ( Z 1 Z 2 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 48

49 Example 1 - The Principle of Superposition in AC Circuits Solution Sum up the resulting currents Kill all sources except one, sequentially and solve the resulting circuits I Z2 = I Z2-1 I Z2-2 Z 1 I Z2-2 SC I s Z 2 I Z2-2 = ( I S / Z 2 ) / [ ( 1 / Z 1 ) ( 1 / Z 2 ) ] = I S g 2 / (g 1 g 2 ) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 49

50 Example 2 - Sources with Mixed Frequencies Now, find the steady-state current waveform flowing in impedance Z 2 in the following circuit by using the Principle of Superposition Z 1 = 1 j1 V s = 310 coswt Question I s = 100 A (DC) I Z2 Z 2 = 2 j5 This method is particularly useful when there are sources with different frequencies Kill all sources except one, sequentially and solve the resulting circuits Z 1 = 1 j1 OC V s = 310 coswt I Z2-1 = V s 0 o / ( Z 1 Z 2 ) = V s 0 o / (1 j1 2 j5) I Z2-1 Z 2 = 2 j5 = V s 0 o / ( 3 j4 ) = V s 0 o / o = 310 / o = o Amp EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 50

51 Example 2 - Sources with Mixed Frequencies Solution I 2 = 0 Kill all the sources except one, sequentially and solve the resulting circuits Z 1 = 1 j1 OC I Z2-2 Z 2 = 2 I s = 100 A (DC) SC I s = 100 A (DC) I Z2-2 Z 2 = 2 j5 I 2 = 100 Amp (DC) Sum up the resulting currents Please note that the capacitor and inductor in the above circuit respond to DC current source as OC and SC, respectively I Z2 = I Z2-1 I Z2-2 I Z2 = o 100 Amp (DC) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 51

52 Example 2 - Sources with Mixed Frequencies Waveforms Let us now draw the resulting current and voltage waveforms Z 1 = 1 j1 I Z2 = o I = 62 Amp 100 Amp (DC) V s1 (t) = 310 cos wt I DC = 100 Amp V s = 310 coswt I s = 100 A (DC) I Z2 Z 2 = 2 j5 0 /2 3 /2 2 Angle (Radians) EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 52

53 Any questions please... EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 53