Lecturer: James Grimbleby URL: j.b.grimbleby reading.ac.uk
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- Bernadette Randall
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1 AC Circuit Analysis Module: SEEA5 Systems and Circuits Lecturer: UL: .b.grimbleby reading.ac.uk Number of Lectures: ecommended text book: David Irwin and Mark Nelms Basic Engineering Circuit Analysis (8th edition John Wiley and Sons (5 ISBN: School of Systems Engineering - Electronic Engineering Slide
2 AC Circuit Analysis ecommended text book: David Irwin and Mark Nelms Basic Engineering Circuit Analysis (8th edition Paperback 86 pages John Wiley and Sons (5 ISBN: Price: 36 School of Systems Engineering - Electronic Engineering Slide
3 AC Circuit Analysis Syllabus This course of lectures will extend dc circuit analysis to deal with ac circuits The topics that will be covered include: AC voltages and currents Complex representation of sinusoids Phasors Complex impedances of inductors and capacitors Driving-point impedance Frequency response of circuits Bode plots Power in ac circuits Energy storage in capacitors and inductors Three-phase power School of Systems Engineering - Electronic Engineering Slide 3
4 AC Circuit Analysis Prerequities You should be familiar with the following topics: SEEA5: Electronic Circuits Ohm s Law Series and parallel resistances Voltage and current sources Circuit analysis using Kirchhoff s Laws Thévenin and Norton's theorems The Superposition Theorem SEEC5: Engineering Mathematics Complex numbers School of Systems Engineering - Electronic Engineering Slide 4
5 AC Circuit Analysis Lecture AC Voltages and Currents eactive Components School of Systems Engineering - Electronic Engineering Slide 5
6 AC Waveforms Sine waveform (sinusoid Square waveform Sawtooth waveform Audio waveform School of Systems Engineering - Electronic Engineering Slide 6
7 Frequency The number of cycles per second of an ac waveform is known as the frequency f, and is expressed in Hertz (Hz Voltage or Current 6 cycles f 6 Hz s Time School of Systems Engineering - Electronic Engineering Slide 7
8 Frequency Examples: Electrocardiogram: Hz Mains power: 5 Hz Aircraft power: 4 Hz Audio frequencies: Hz to khz AM radio broadcasting:.5 MHz.5 MHz FM radio broadcasting: 8 MHz MHz Television broadcasting: 5 MHz 8 MHz Mobile telephones:.8 GHz School of Systems Engineering - Electronic Engineering Slide 8
9 Period The period T of an ac waveform is the time taken for a complete cycle: period frequency Voltage or Current T.67 s s Time School of Systems Engineering - Electronic Engineering Slide 9
10 Why Linear? We shall consider the steady-state state response of linear ac circuits to sinusoidal inputs Linear circuits contain linear components such as resistors, capacitors and inductors A linear component has the property that doubling the voltage across it doubles the current through it Most circuits for processing signals are linear Analysis of non-linear circuits is difficult and normally requires the use of a computer. School of Systems Engineering - Electronic Engineering Slide
11 Why Steady-State? State? Steady-state means that the input waveform has been present long enough for any transients to die away Output V V in V Time t V in for t School of Systems Engineering - Electronic Engineering Slide
12 Why Sinusoidal? A linear circuit will not change the waveform or frequency of a sinusoidal input (the amplitude and phase may be altered Power is generated as a sinusoid by rotating electrical machinery Sinusoidal id carrier waves are modulated d to transmit information (radio broadcasts Any periodic waveform can be considered to be the sum of a fundamental pure sinusoid plus harmonics (Fourier Analysis School of Systems Engineering - Electronic Engineering Slide
13 Fourier Analysis A square waveform can be considered to consist of a fundamental sinusoid idtogether th with odd ddharmonic sinusoids id Square wave Sum Fundamental 3dh 3rd harmonic 5th harmonic School of Systems Engineering - Electronic Engineering Slide 3
14 epresentation of Sinusoids A sinusoidal voltage waveform v(t ( of amplitude v, and of frequency f : v t v sinπft v sint or: v ( ( t v cosπft v cost where πf is known as the angular frequency v T / f v t School of Systems Engineering - Electronic Engineering Slide 4
15 epresentation of Sinusoids The sinusoid can have a phase term φ: v( t v sin t ( φ A phase shift φ is equivalent to atimeshift -φ/ v τ φ v sin ( t φ t ( v sin t The phase is positive so the red trace leads the green trace School of Systems Engineering - Electronic Engineering Slide 5
16 esistors v i Ceramic tube coated with Conductive film Metal end cap Film: carbon metal metal oxide esistance v i (Ohm s Law School of Systems Engineering - Electronic Engineering Slide 6
17 esistors School of Systems Engineering - Electronic Engineering Slide 7
18 esistors Suppose that: v ( t v sin ( t v i i, v Ohm s Law: i v v i Then: v( t i( t v sin t ( Current in phase with voltage t School of Systems Engineering - Electronic Engineering Slide 8
19 Capacitors i v Insulating dielectric Conducting electrodes Dielectrics: i air Capacitance C polymer dv ceramic q Cv i C dt Al 3 (electrolytic School of Systems Engineering - Electronic Engineering Slide
20 Capacitors School of Systems Engineering - Electronic Engineering Slide
21 Capacitors Suppose that: ( t v C v ( t v sin Then: d i( t C v sin( t dt Cv cos t ( sin π Cv t i dv i C dt i, v i(t v(t t Current leads voltage by π/ (9 School of Systems Engineering - Electronic Engineering Slide
22 Capacitors Does a capacitor have a resistance? v, i v sin( t i t Cv cos( t v( t ( t t t v( t v v( t i ( t i( t i ( Thus resistance varies between ± : not a useful concept School of Systems Engineering - Electronic Engineering Slide 3
23 Capacitors The reactance X C of a capacitor is defined: X C C v i where v is the amplitude of the voltage across the capacitor and i is the amplitude of the current flowing through it Thus: X v C Cv C π fc The reactance of a capacitor is inversely proportional to its value and to frequency School of Systems Engineering - Electronic Engineering Slide 4
24 Inductors v i Magnetisable core Copper wire Core: air ferrite iron silicon steel Inductance L di v L dt School of Systems Engineering - Electronic Engineering Slide 6
25 Inductors School of Systems Engineering - Electronic Engineering Slide 7
26 Inductors Suppose that: v( t v sin ( t Then: i( t v sin t L v cos t L v sin t L ( ( v L π i di v L dt i, v v (t i(t t Current lags voltage by π/ (9 School of Systems Engineering - Electronic Engineering Slide 8
27 Inductors The reactance X C of an inductor is defined: C X C where v is the amplitude of the voltage across the inductor and i is the amplitude of the current flowing through it v i Thus: X c v L v / LL πfl The reactance of an inductor is directly proportional to its value and to frequency School of Systems Engineering - Electronic Engineering Slide 9
28 esistance and eactance X v f i f esistance Capacitance C open short C circuit circuit Inductance L L short circuit it open circuit it School of Systems Engineering - Electronic Engineering Slide 3
29 AC Circuit Analysis Lecture AC Analysis using Differential Equations Complex Numbers Complex Exponential Voltages and Currents School of Systems Engineering - Electronic Engineering Slide 3
30 AC Circuit Analysis The ac response of a circuit is determined by a differential equation: i(t v in (t C vc(t v v ( t i( t v ( t in dv t i t C c ( ( dt dv ( t ( t C c vc ( t dt c in dvc ( t vc ( t v in ( t dt C C School of Systems Engineering - Electronic Engineering Slide 33
31 AC Circuit Analysis Now suppose that the input voltage v in is a sinusoid of angular frequency The output t voltage v c will be a sinusoid id of the same freqeuncy, but with different amplitude and phase: v in ( t v v c ( t cos v cos t Expanding the expression for v c : c ( t ( t φ v c ( t v cost cosφ vsint sinφ Acost dv c ( t A sint B cost dt B sint School of Systems Engineering - Electronic Engineering Slide 34
32 AC Circuit Analysis The differential equation becomes: A B v A sin t B cos t cos t sin t C C C cos t Comparing the coefficients of sint and cost on both sides of the equation: A C B B C A Solving these simultaneous linear equations in A and B: v A v Cv B C C School of Systems Engineering - Electronic Engineering Slide 35
33 AC Circuit Analysis A Thus: v v cos φ v Cv B v sinφ C C v tanφ C C At an angular frequency /C: v v c v π φ 4 v ( π t cos t 4 The output voltage lags the input voltage by π/4 (45 School of Systems Engineering - Electronic Engineering Slide 36
34 AC Circuit Analysis. Voltage gain v /v.77../c./c /C /C /C Angular frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 37
35 Complex Numbers: ectangular Form Complex numbers can be represented in rectangular, polar or exponential form ectangular form: z x y where x is the real part, y is the imaginary i part (x and y are both real numbers, and Complex numbers are often the solutions of real problems, for example quadratic equations School of Systems Engineering - Electronic Engineering Slide 38
36 Complex Numbers: Argand Diagram Imaginary part z x y Imaginary axis y O x eal axis eal part School of Systems Engineering - Electronic Engineering Slide 39
37 Complex Numbers: Polar Form Polar form: z r θ where r is the magnitude, and θ is the angle measured from the real axis: z Imaginary axis r O θ eal axis School of Systems Engineering - Electronic Engineering Slide 4
38 Complex Numbers: Exponential Form Exponential form: z re θ Euler s identity: θ e cos θ sin θ sin θ O θ cos θ The polar and exponential forms are therefore equivalent School of Systems Engineering - Electronic Engineering Slide 4
39 Complex Numbers: Conversion z O θ r x y ectangular to polar: r z x y y θ z z tanθ θ x Polar to ectangular: x r cos θ y r sinθ School of Systems Engineering - Electronic Engineering Slide 4
40 Complex Numbers: Inversion If the complex number is in rectangular form: z x y x y ( x y ( x y x y x y If the complex number is in polar or exponential form: z Ae φ e φ A School of Systems Engineering - Electronic Engineering Slide 43
41 Complex Numbers: Conversion When using the inverse tangent to obtain θ from x and y it is necessary to resolve the ambiguity of π: y z x y x x tanθ y x y x θ y x x y y w x y School of Systems Engineering - Electronic Engineering Slide 44
42 Complex Numbers: Conversion When using the inverse tangent to obtain θ from x and y it is necessary to resolve the ambiguity of π. Calculate θ using inverse tangent: θ tan y x This should give a value in the range: -π/ θ π/ (-9º θ 9º. If the real part x is negative then add π (8º : θ π tan y x School of Systems Engineering - Electronic Engineering Slide 45
43 Complex Numbers: Conversion Convert z π to rectangular form 3 eal part: cos π x 3 π 3 Imaginary part: y sin 3 3 Thus: z 3 School of Systems Engineering - Electronic Engineering Slide 46
44 Complex Numbers: Conversion.73 z 3 Imaginary axis r O θ π 3 (6 o eal axis School of Systems Engineering - Electronic Engineering Slide 47
45 Complex Numbers: Conversion C t t l ti l f Complex Numbers: Conversion Convert to polar or exponential form: z Magnitude: z r Angle: t t ( π π θ z 4 4 tan tan π π π Thus: or 4 4 π π e z z School of Systems Engineering - Electronic Engineering Slide 48 4
46 Complex Numbers: Conversion.5 eal axis Imaginary axis O θ π 4 (45 o.5 z School of Systems Engineering - Electronic Engineering Slide 49
47 Complex Exponential Voltages We shall be using complex exponential voltages and currents to analyse ac circuits: v( t Ve This is a mathematical ti trick for obtaining i the ac response without explicitly solving the differential equations It works because differentiating a complex exponential leaves it unchanged, apart from a multiplying factor: t d Ve dt t Ve t School of Systems Engineering - Electronic Engineering Slide 5
48 Complex Exponential Voltages Suppose that t a complex exponential voltage is applied across a resistor: v ( t Ve t i(t i ( t v ( t V e t The current through the resistor is also a complex exponential School of Systems Engineering - Electronic Engineering Slide 5
49 Complex Exponential Voltages Suppose that t a complex exponential voltage is applied across a capacitor: v ( t Ve C t i(t i ( t dv( t C dt d C Ve dt CVe t t The current through the capacitor is also a complex exponential School of Systems Engineering - Electronic Engineering Slide 5
50 Complex Exponential Voltages Suppose that t a complex exponential voltage is applied across an inductor: v ( t Ve L t i(t i ( t v( t dt L t Ve dt L t Ve L The current through the inductor is also a complex exponential School of Systems Engineering - Electronic Engineering Slide 53
51 Complex Exponential Voltages A complex exponential input to a linear ac circuits results in all voltages and currents being complex exponentials Of course real voltages are not complex The real voltages and currents in the circuit are simply the real parts of the complex exponentials Complex exponential: v ( t e ( cos t sin t v c t eal voltage: v r ( t cos t School of Systems Engineering - Electronic Engineering Slide 54
52 AC Circuit Analysis Lecture 3 Phasors Impedances Gain and Phase Shift Frequency esponse School of Systems Engineering - Electronic Engineering Slide 55
53 Phasors If the input voltage to a circuit is a complex exponential: Phasors then all other voltages and currents are also complex t cin e v t v ( then all other voltages and currents are also complex exponentials: t t t t t t c I i i t i V e e e v e v t v φ φ φ φ ( ( ( ( t t t c e I e e i e i t i φ φ ( ( where V and I are time-independent voltage and current phasors: φ e V v φ e i I e v V School of Systems Engineering - Electronic Engineering Slide 56
54 Phasors The complex exponential voltages and currents can now be expressed: t v ( t V e i c c t ( I e Phasors are independent d of time, but in general are functions of and should be written: t ( I ( V However, when there is no risk of ambiguity the dependency will be not be shown explicitly Note that upper-case letters are used for phasor symbols School of Systems Engineering - Electronic Engineering Slide 57
55 Impedance The impedance Z of a circuit or component is defined to be the ratio of the voltage and current phasors: For a resistor: v c ( t Ve i t c ( t Z Ie V I t v c ( t i c ( t Ve t Ie V I t So that: V Z I School of Systems Engineering - Electronic Engineering Slide 58
56 Impedance For a capacitor: v c ( t t Ve i ( t C c Ie t i c Ie Ie ( t t t dv t c ( C dt d C Ve dt CVe t t I CV So that: Z C V I C School of Systems Engineering - Electronic Engineering Slide 59
57 Impedance For an inductor: v c t ( t Ve i ( t L c Ie t v c Ve Ve ( t t t di t c( L dt d L Ie dt LIe t t V LI So that: Z L V I L School of Systems Engineering - Electronic Engineering Slide 6
58 Impedance Z V I f f esistance Capacitance C C Z Z Inductance L L Z Z School of Systems Engineering - Electronic Engineering Slide 6
59 Impedance All the normal circuit theory rules apply ppy to circuits containing impedances For example impedances in series: Z Z and impedances in parallel: Z Z 3 Z 4 Z Z Z Z 3 Z4 Other relevant circuit it theory rules are: Kirchhoff s laws, Thévenin and Norton's theorems, Superposition School of Systems Engineering - Electronic Engineering Slide 6
60 Impedance Potential divider: V in Z Z I V out V V V I out out in Z IZ V Z Z V in Z V in Z Z in Z Z School of Systems Engineering - Electronic Engineering Slide 63
61 AC Circuit Analysis Suppose that a circuit has an input x(t ( and an output y(t, where x and y can be voltages or currents The corresponding phasors are X( and Y( The real input voltage x(t is a sinusoid of amplitude x : t x ( t x cos( t re ( x e re ( Xe t and the real output voltage y(t is the real part of the complex exponential output: φ t y( t y cos( t φ re( ye e re( Ye t School of Systems Engineering - Electronic Engineering Slide 64
62 AC Circuit Analysis Thus: y e φ x Y X The voltage gain g is the ratio of the output amplitude to the input amplitude: y Y g x X and the phase shift is: φ Y X School of Systems Engineering - Electronic Engineering Slide 65
63 AC Circuit Analysis Using the potential divider formula: Vin C Vc V V c in Z C Z C Z / C / C C School of Systems Engineering - Electronic Engineering Slide 66
64 AC Circuit Analysis V in C V c V c V C in Voltage gain: g V V c in C Phase shift: φ Vc tan V in tan C tan C School of Systems Engineering - Electronic Engineering Slide 67
65 Frequency esponse (C. Volta age Ga ain Angular Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 68
66 Frequency esponse (C Angular Frequency (rad/s. shift Phase π 4 ( 45 π ( 9 School of Systems Engineering - Electronic Engineering Slide 69
67 Frequency esponse g C φ tan C g φ (( g φ π ( 45 C 4 g φ π ( 9 This is a low-pass response School of Systems Engineering - Electronic Engineering Slide 7
68 Frequency esponse VC V in V C School of Systems Engineering - Electronic Engineering Slide 7
69 AC Circuit Analysis Vin C V V V in V in Z Z Z / C C V V in C C V V Vin i C C School of Systems Engineering - Electronic Engineering Slide 7
70 AC Circuit Analysis C V in V V V in C C Voltage gain: g V C Vin C Phase shift : φ V V in tan tan C π tan C School of Systems Engineering - Electronic Engineering Slide 73
71 Frequency esponse (C. Volta age ga ain Angular Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 74
72 Frequency esponse (C π (9 ift Pha ase sh π 4 (45.. Angular Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 75
73 Frequency esponse g C C φ π tan C g φ π (9 g φ π (45 C 4 g φ ( This is a high-pass response School of Systems Engineering - Electronic Engineering Slide 76
74 Frequency esponse V V V in School of Systems Engineering - Electronic Engineering Slide 77
75 AC Circuit Analysis Lecture 4 Driving-Point Impedance School of Systems Engineering - Electronic Engineering Slide 78
76 Impedance The impedance Z of a circuit or component is defined to be the ratio of the voltage and current phasors: I ( V ( AC Circuit Z ( V( I( Impedance Z is analogous to resistance in dc circuits and its units are ohms When Z applies to a -terminal circuit (rather than simple component it is known as the driving-point impedance School of Systems Engineering - Electronic Engineering Slide 79
77 Impedance Z can be written in rectangular form: Z ( ( X( where is the resistance and X is the reactance Thus: and: Z Z X tan X X Z cos Z Z sin Z School of Systems Engineering - Electronic Engineering Slide 8
78 Symbolic and Numeric Forms Substitute component values Symbolic Form Z C Substitute frequency value Numeric Form Z Value at a given frequency Z 4 4 School of Systems Engineering - Electronic Engineering Slide 8
79 Example Determine the driving-point impedance of the circuit at a frequency of 4 khz: Z Z 5 Z C C π Ω 3 9 C nf 5Ω School of Systems Engineering - Electronic Engineering Slide 8
80 Example Z Ω Z C nf 3.93 Ω 5Ω ZZ 9.89 tan 5.67 ( 38.5 School of Systems Engineering - Electronic Engineering Slide 83
81 Example What will be the voltage across the circuit when a current of 5 A, 4 khz flows through it? V IZ 5 ( V In polar form: V IZ ( ( V.67 ( 38.5 School of Systems Engineering - Electronic Engineering Slide 84
82 Example Determine the driving-point impedance of the circuit at a frequency of Hz: Z Z Z C Z C 8Ω C μfμ / C C School of Systems Engineering - Electronic Engineering Slide 85
83 Example Z C 8 6 π ( Ω 8Ω C μf School of Systems Engineering - Electronic Engineering Slide 86
84 Example Z Ω Z Ω 8Ω C μf ZZ 4. tan ( 45. o School of Systems Engineering - Electronic Engineering Slide 87
85 Example What current will flow if an ac voltage of 4 V, Hz is applied to the circuit? I V Z ( A.454A.788 (45. V I Z A.788 ( A School of Systems Engineering - Electronic Engineering Slide 88
86 Phasor Diagrams Where voltages or currents are summed the result can be represented by a phasor diagram: V V V V 3 Imaginary part V O V V eal part V 3 School of Systems Engineering - Electronic Engineering Slide 89
87 Example I 4.3 A 8 6 I C π 4 C.36 A.3A Imagin nary part.a A.A I C I I I C O I.A.A.3A eal part School of Systems Engineering - Electronic Engineering Slide 9
88 Example 3 Determine the driving-point impedance of the circuit at a frequency of 5 Hz: Z Z ZL ZC 4 Ω L C L 3 4 π mh 6 π Ω C μf 4 5. Ω School of Systems Engineering - Electronic Engineering Slide 9
89 Example 3 Z 4 5. Ω Z Z Z Ω 4 Ω L 5. tan 4 36 mh.565 ( 3.4 C μf 8.4Ω.565 ( 3.4 School of Systems Engineering - Electronic Engineering Slide 9
90 Example 3 What voltage will be generated across the circuit if an ac current of A, 5 Hz flows though it? In polar form: V V ZI A (4 5. Ω 4 5. V ZI ( ( V.565 ( 3.4 School of Systems Engineering - Electronic Engineering Slide 93
91 Example 3 V V V V V L C Imaginary part V L V V 4V eal part V C V -V School of Systems Engineering - Electronic Engineering Slide 94
92 Example 4 Determine the driving-point impedance of the circuit at a frequency of 4 Hz: Z Z Z Z L L Z C C /( L C L C( L L C LC Ω C μf L mh School of Systems Engineering - Electronic Engineering Slide 95
93 Example 4 L Z Example LC C Z π 4 ( π π π (.53 ( School of Systems Engineering - Electronic Engineering Slide 96
94 Example 4 Z Z Z tan ( 53. C μff Ω L mh Z 3.9Ω.98 ( 53. School of Systems Engineering - Electronic Engineering Slide 97
95 Example 4 What current will flow if an ac voltage of V, 4 Hz is applied to the circuit? V V I I Z Z ( A A.98 (53. School of Systems Engineering - Electronic Engineering Slide 98
96 Example 4 I I L L I I L I C 5A Z L Imaginary part I C I Z C O C 6.3 IL 5A eal part School of Systems Engineering - Electronic Engineering Slide 99
97 Admittance The admittance Y of a circuit or component is defined to be the ratio of the current and voltage phasors: I ( I( V ( AC Circuit Y ( V( Z( Admittance Y is analogous to conductance in dc circuits and its unit is Siemens Y ( G ( B ( where G is the conductance and B is the susceptance School of Systems Engineering - Electronic Engineering Slide
98 Admittance Y I V f f esistance Capacitance C C Y Y Inductance L L Y Y School of Systems Engineering - Electronic Engineering Slide
99 Admittance All the normal circuit it theory rules apply to circuits its containing admittances For example admittances in series: Y Y Y and admittances in parallel: Y 3 Y Y Y Other relevant circuit it theory rules are: Kirchhoff s laws, Thévenin and Norton's theorems, Superposition 4 Y Y3 Y4 School of Systems Engineering - Electronic Engineering Slide
100 Example 5 Determine the driving-point ing admittance of the circuit it at a frequency of 4 Hz: Y Y C / Y Y / Ω L C C μf L L mh School of Systems Engineering - Electronic Engineering Slide 3
101 Example 5 Determine the driving-point admittance of the circuit at a frequency of 4 Hz: 6 π 4 Y π 4 3 π Ω C μf.3 L mh S 3 School of Systems Engineering - Electronic Engineering Slide 4
102 AC Circuit Analysis Lecture 5 esonant Circuits School of Systems Engineering - Electronic Engineering Slide 5
103 esonant Circuits Passive resonant circuits must contain a resistor, capacitor and an inductor The behaviour of resonant circuits changes rapidly around a particular frequency (the resonance frequency esonant circuits i can be characterised by two parameters: the resonance frequency and the Q-factor There are two basic resonant circuit configurations: series and parallel School of Systems Engineering - Electronic Engineering Slide 6
104 esonant Circuits θ dθ dt d dt gθ L θ, θ t School of Systems Engineering - Electronic Engineering Slide 7
105 esonant Circuits i L C di L L v dv i C C L v C dt L dt C i, v v C i L t School of Systems Engineering - Electronic Engineering Slide 8
106 Parallel esonant Circuit Parallel esonant Circuit L L C L Z Z Z Z Z L C L LC L Z L C LC L L / L LC L LC L L / School of Systems Engineering - Electronic Engineering Slide
107 Parallel esonant Circuit 5 kω C μf L H Impedance is a maximum (resonant frequency when: Z L L / LC LC School of Systems Engineering - Electronic Engineering Slide 3
108 Parallel esonant Circuit Z Z School of Systems Engineering - Electronic Engineering Slide 4
109 Parallel esonant Circuit Parallel esonant Circuit L L Z L C LC L / L Z L Z Z t f / L L L Z LC esonant frequency: C LC L Z School of Systems Engineering - Electronic Engineering Slide 5
110 Parallel esonant Circuit 5kΩ Z Z π (9 Z. Z π Angular frequency (rad/s ( 9 School of Systems Engineering - Electronic Engineering Slide 6
111 Quality Factor The standard form for the denominator of a second-order system is: / Q / Compare this with the impedance Z: So that: Z L L / LC Q LC LL where Q is the quality-factor and is the resonant frequency School of Systems Engineering - Electronic Engineering Slide 7
112 Quality Factor 5 kω C μf L H 3 LC 6 Q L School of Systems Engineering - Electronic Engineering Slide 8
113 Quality Factor Z max Z Δ Q max Z. Angular frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 9
114 Quality Factor kω Z Z Z QQ π (9 Z Z. π Angular frequency (rad/s ( 9 School of Systems Engineering - Electronic Engineering Slide
115 Quality Factor kω Z Z QQ π (9 Z. Z π Angular frequency (rad/s ( 9 School of Systems Engineering - Electronic Engineering Slide
116 Parallel esonant Circuit I I C I L V 5kΩ C μf L H esonance occurs in parallel resonant circuits because the currents in the capacitor and inductor cancel out 4 I 5 I C C C / I L 6 L L School of Systems Engineering - Electronic Engineering Slide
117 Parallel esonant Circuit At resonance: 3 : ma 4 I A I C Imaginary I 6 part C eal 3 O A I I part ma I L 3 A -ma IL School of Systems Engineering - Electronic Engineering Slide 3
118 Parallel esonant Circuit I Below resonance:.5 I C 4 3 : A A I C Imaginary O part -ma I I ma eal part I L 3 A I L -ma School of Systems Engineering - Electronic Engineering Slide 4
119 Parallel esonant Circuit Above resonance: ma. 3 : IC I I C I L 4 A 6 ma Imaginary 3. A part O.5 3 A I L I I ma eal part School of Systems Engineering - Electronic Engineering Slide 5
120 Series esonant Circuit Z Z Z C Z L L C C C LC C LC C C L School of Systems Engineering - Electronic Engineering Slide 6
121 Series esonant Circuit Series esonant Circuit C LC C Z C C C Z L C C Z LC L LC Z C LC C L C Z School of Systems Engineering - Electronic Engineering Slide 7
122 Series esonant Circuit Z C LC C Ω L H C μf School of Systems Engineering - Electronic Engineering Slide 8
123 Series esonant Circuit kω π (9 Z Z Z Ω Ω Z π Angular frequency (rad/s ( 9 School of Systems Engineering - Electronic Engineering Slide 9
124 Series esonant Circuit The standard form for the denominator of a second-order order system is: / Q / Compare this with the admittance Y ( /Z: So that: Y C C LC Q LC C where Q is the quality-factor and is the resonant frequency School of Systems Engineering - Electronic Engineering Slide 3
125 Series esonant Circuit LC 6 Ω 3 rad/s L H Q C 3 6 C μf 5 School of Systems Engineering - Electronic Engineering Slide 3
126 Series esonant Circuit esonance occurs in series resonant circuits because the voltages across the capacitor and inductor cancel out A V Ω V V L L H V C C C 6 V C C μf V L L School of Systems Engineering - Electronic Engineering Slide 3
127 Series esonant Circuit At resonance: 3 : V V kv V L V C 6 Imaginary part eal O part 3 V kv V V C V L -kv 3 V School of Systems Engineering - Electronic Engineering Slide 33
128 Crystal esonator School of Systems Engineering - Electronic Engineering Slide 34
129 Crystal esonator School of Systems Engineering - Electronic Engineering Slide 35
130 Crystal esonator Equivalent circuit: f 8. MHz 34Ω 3.4 Ω L.86 mh C 46pF 4.6 C 4 pf L C Q LC C C School of Systems Engineering - Electronic Engineering Slide 36
131 AC Circuit Analysis Lecture 6 Frequency-esponse Function First-Order Circuits School of Systems Engineering - Electronic Engineering Slide 37
132 Frequency-esponse Function Input X Y Output Frequency-response function: H ( Y( X( Voltage gain g: Y( g H( X ( Phase shift φ: Y ( φ H( X( School of Systems Engineering - Electronic Engineering Slide 38
133 Frequency-esponse Function The order of a frequency-response function is the highest power of in the denominator: First order: H( / Second order: Third order: H( H ( ( / / / 3 ( / ( / The order is normally equal to (and cannot exceed the number of reactive components School of Systems Engineering - Electronic Engineering Slide 39
134 Example Using the potential divider formula: V V c in Z C Z C Z / C / C Vin C Vc H ( C where : / C School of Systems Engineering - Electronic Engineering Slide 4
135 Example H( / / / Vin C Vc Gain: g H( / Phase shift: φ H ( tan φ / School of Systems Engineering - Electronic Engineering Slide 4
136 Example g g School of Systems Engineering - Electronic Engineering Slide 4
137 Decibel The decibel is a measure of the ratio of two powers P, P : p db log P P It can also be used to measure the ratio of two voltages V, V : db V / log log V / V V db log V V School of Systems Engineering - Electronic Engineering Slide 43
138 Decibel Power ratio 4 / /4.. Decibels 6 db db db 6 db 3 db db -3 db -6 db db -6 db / -3 db - db School of Systems Engineering - Electronic Engineering Slide 44
139 Decibel Voltage ratio /.77 /.5.. Decibels 6 db db db 6 db 3 db db -3 db -6 db db -6 db / 77-3 db - db School of Systems Engineering - Electronic Engineering Slide 45
140 Example Circuit is a first-order low-pass filter: g / φ tan / << g (db φ ( g ( 3dB φ π ( 45 4 >> π g ( 6dB / oct φ ( 9 School of Systems Engineering - Electronic Engineering Slide 46
141 Example C kω CμF Gain: g H( 6 / Phase shift: 3 φ H ( tan φ / School of Systems Engineering - Electronic Engineering Slide 47
142 Bode Plot Gain(dB Phase(rad db -3 db - db - db φ g π 4-3 db -6dB/octave -4 db π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 48
143 Example Using the potential divider formula: Example Using the potential divider formula: L Z Z Z V V L in V V L L in in V V L H L / ( L where : / L where : / School of Systems Engineering - Electronic Engineering Slide 49
144 Example g g School of Systems Engineering - Electronic Engineering Slide 5
145 Example 3 Using the potential divider formula: V V c in Z Z Z C / C Vin C Vc H ( C C / where : / C School of Systems Engineering - Electronic Engineering Slide 5
146 Example 3 H( / / / / / Vin Gain: g H ( / C V Phase shift: φ H tanφ / ( School of Systems Engineering - Electronic Engineering Slide 53
147 Example 3 g g School of Systems Engineering - Electronic Engineering Slide 54
148 Example 4 Using the potential divider formula: Example 4 Using the potential divider formula: Z Z Z V V L L in L L L Z Z V L in L in V L V L H L / ( L H : where / / ( L : where / School of Systems Engineering - Electronic Engineering Slide 55
149 Example 4 g g School of Systems Engineering - Electronic Engineering Slide 56
150 Example 4 Circuit is a first-order high-pass filter: g / φ tan / << g (6dB / oct φ π (9 3dB π g ( φ (45 4 >> g (db φ ( School of Systems Engineering - Electronic Engineering Slide 57
151 Bode Plot Gain(dB db -3 db Phase(rad π - db - db -3 db g 6 db / octave φ π 4-4 db Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 58
152 Example 5 Using the potential divider formula: Example 5 g p / C V / / C C V V in c i V t V C C C in V out V C ( ( C C H where : / ( C ( / C C School of Systems Engineering - Electronic Engineering Slide 59
153 Example 5 Example 5 / / ( H in V out V / ( H g C / ( H g School of Systems Engineering - Electronic Engineering Slide 6
154 Example 5 g g School of Systems Engineering - Electronic Engineering Slide 6
155 Example 5 Assuming that << : g / / << g (db << << g ( 6dB / oct >> g School of Systems Engineering - Electronic Engineering Slide 6
156 Example 5 H( / / C( 6 3 (9 rad/s 9Ω Ω C μf C 6 4 rad/s School of Systems Engineering - Electronic Engineering Slide 63
157 Bode Plot Gain(dB Phase(rad db - db g - db φ -3 db -4 db π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 64
158 AC Circuit Analysis Lecture 7 Second-Order Circuits School of Systems Engineering - Electronic Engineering Slide 65
159 Example V in C C V c This circuit must be simplified before the frequency response function can be determined d AThé Thévenin equivalent circuit it is created of the components to the left of the red line School of Systems Engineering - Electronic Engineering Slide 66
160 Example Thévenin equivalent circuit: Z V in C V V / C C Vin C / C Z V in Z C C School of Systems Engineering - Electronic Engineering Slide 67
161 Example C V in C C VC V C Vin C / C / C C V in C C C C School of Systems Engineering - Electronic Engineering Slide 68
162 Example Frequency-response function: V C Vin C C C C V in ( C ( C C H( 3 C C kω, C μf: μ H( School of Systems Engineering - Electronic Engineering Slide 69
163 Example g g School of Systems Engineering - Electronic Engineering Slide 7
164 Bode Plot Gain(dB Phase(rad db - db g - db φ -3 db -4 db π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 7
165 Example C V in C V This circuit cu must be simplified before e the frequency response se function can be determined A Thévenin equivalent circuit is created of the components to the left of the red line School of Systems Engineering - Electronic Engineering Slide 7
166 Example C C V in V in C V V Vin C / C C V in C C C C C School of Systems Engineering - Electronic Engineering Slide 73
167 Example Frequency-response function: V Vin C C C C C V in C ( C ( C C C H( 3 C C kω, C μf: H( School of Systems Engineering - Electronic Engineering Slide 74
168 Example g g School of Systems Engineering - Electronic Engineering Slide 75
169 Bode Plot Gain(dB db Phase(rad π - db g - db φ -3 db -4 db π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 76
170 Example 3 Using the potential divider formula: H( V C / C V / C L in C LC Vin L C Vc where : /( Q / and : Q LC C L C School of Systems Engineering - Electronic Engineering Slide 77
171 Example 3 g g School of Systems Engineering - Electronic Engineering Slide 78
172 Example 3 Circuit is a second-order low-pass filter: H( ( /( Q / g H( << H ( g (db H ( Q g Q >> H( g ( db / oct School of Systems Engineering - Electronic Engineering Slide 79
173 Example 3 L 4mH Ω Vin C.5μF VC 3 LC Q 3 L C 6.5 School of Systems Engineering - Electronic Engineering Slide 8
174 Bode Plot Gain(dB db Q Q db Q - db - db / octave -4 db Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 8
175 Bode Plot Phase(rad Q π Q Q π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 8
176 Example 4 Using the potential divider formula: H( V L V in / L C L LC C LC Vin C L V L where : / /( Q / and : Q LC C L C School of Systems Engineering - Electronic Engineering Slide 83
177 Example 4 g g School of Systems Engineering - Electronic Engineering Slide 84
178 Example 4 Circuit is a second-order high-pass filter: H ( /( << H( / Q / g g H( ( (db / oct H ( Q g Q >> H ( g (db School of Systems Engineering - Electronic Engineering Slide 85
179 Bode Plot Gain(dB db Q Q db - db Q db / octave -4 db Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 86
180 Bode Plot Phase(rad π Q Q Q π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 87
181 Example 5 Using the potential divider formula: Example 5 V C L L C V V in / in V V LC C C H ( Q Q / /( /( L Q Q and : where : / /( C C Q LC and : where : School of Systems Engineering - Electronic Engineering Slide 88
182 Example 5 g g School of Systems Engineering - Electronic Engineering Slide 89
183 Example 5 Circuit is a second-order band-pass filter: H ( /( /( Q Q Q / g H( ( << H( g (6dB / oct Q Q H ( H g (db >> H ( g ( 6dB / oct Q Q School of Systems Engineering - Electronic Engineering Slide 9
184 Bode Plot Gain(dB db - db Q Q Q -4 db Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 9
185 Bode Plot Phase(rad π Q Q Q π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 9
186 Example 6 Using the potential divider formula: H( V C / C L / C L V in LC C LC Vin C L Vc where : LC /( Q / and : Q LC C L C School of Systems Engineering - Electronic Engineering Slide 93
187 Example 6 g g School of Systems Engineering - Electronic Engineering Slide 94
188 Example 6 Circuit is a second-order band-stop filter: H ( /( / Q / g H( ( << H ( g (db H( g ( db >> H ( g (db School of Systems Engineering - Electronic Engineering Slide 95
189 Bode Plot Gain(dB db Q Q Q - db -4 db Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 96
190 Bode Plot Phase(rad π Q Q Q π Frequency (rad/s School of Systems Engineering - Electronic Engineering Slide 97
191 AC Circuit Analysis Lecture 8 Power in AC Circuits School of Systems Engineering - Electronic Engineering Slide 98
192 Power in AC Circuits To calculate the power in a circuit we shall need to make use of some trigonometric identities: cos( A B cos Acos B sin Asin B cos( A B cos AcosB sin AsinB Adding: so that: cos( A B cos( A cos A cos B B cos AcosB { cos( A B cos( A B } cos A { cosa cos} cos A School of Systems Engineering - Electronic Engineering Slide 99
193 rms Voltages and Currents The average power in a resistor is given by: P T v ( t i ( t dt T T ( t T v dt v(t T T T v ( t dt V T rms where : Vrms T v ( t dt i(t School of Systems Engineering - Electronic Engineering Slide
194 rms Voltages and Currents The root-mean-square voltage V rms determines the power dissipated in a circuit: P rms V There eeis sas similar expression pesso for the epower dsspated dissipated when a current I rms flows through a circuit: P I rms These expressions apply to any waveform School of Systems Engineering - Electronic Engineering Slide
195 rms Voltages and Currents The rms value of a sinusoid of amplitude (peak value v : V rms T T T v ( t dt v cos ( t dt T T v cos(t dt T Averages to zero over v v a complete cycle: T π/ School of Systems Engineering - Electronic Engineering Slide
196 rms Voltages and Currents The UK mains power was until recently supplied at 4 V rms and that in Europe V rms On January 995 the nominal voltage across Europe was harmonised at 3 V rms. This corresponds to an amplitude of: v Vrms 3 35 V School of Systems Engineering - Electronic Engineering Slide 3
197 rms Voltages and Currents A mains power (3 V rms electric fire has a resistance of 5 Ω: P rms V 3.7 kw 5 An audio amplifier which drives a 4 Ω loudspeaker at up to 5 W must supply a sinusoidal output voltage: V rms P V rms 4.5 V This corresponds to a sinusoid of peak value 34.6 V School of Systems Engineering - Electronic Engineering Slide 4
198 rms Voltages and Currents Square wave of amplitude ±v : rms Voltages and Currents v -v T T/ / ( ( dt v dt v dt t v V T T T T / ( ( dt v T dt v T dt t v T V T T rms v v dt T v T School of Systems Engineering - Electronic Engineering Slide 5
199 Crest Factor The ratio between the peak voltage and the rms voltage is known as the crest factor: cf V peak V rms For a sinusoid the crest factor is ; for a square wave the crest factor is For audio signals the crest factor depends on the source but is commonly or higher 5 W of audio into 4 Ω loudspeakers would therefore require peak voltages of 5 V or greater School of Systems Engineering - Electronic Engineering Slide 6
200 Power in a eactive Load Capacitors and inductors store energy, but do not dissipate power V rms 5 Hz I 5Ω I C C μf I I C P 4 A 5 π 5 6 Z C 5 4 W A 6.8 A School of Systems Engineering - Electronic Engineering Slide 7
201 Instantaneous Power For sinusoidal voltages and currents: v t i( t v ( cos i cos t ( ( t ( t φ v(t i(t Z Instantaneous power: p( t v( t i( t ( t i cos ( t φ ( t cos( t φ v cos v i cos v i t { cos( φ cosφ} School of Systems Engineering - Electronic Engineering Slide 8
202 Average Power Average power: P T p( t dt T T v cos( t i cos( t φ dt T T v i cos( t φ dt v i cos φ dt T T T If T >> /: P T vi cosφ dt T v i cosφ School of Systems Engineering - Electronic Engineering Slide 9
203 Average Power P cosφ v i φ i(t v Z cosφ i Z cosφ v(t Z School of Systems Engineering - Electronic Engineering Slide
204 Average Power Average power: For a resistor: For a capacitor: For an inductor: φ P v i cosφ v φ P v i i φ π P π P School of Systems Engineering - Electronic Engineering Slide
205 rms Voltages and Currents Power expressed in terms of rms voltages and currents: P v i V V I rms rms cosφφ rms I rms cosφ cos φ (W P P rms V Z I rms cosφ Z cosφ School of Systems Engineering - Electronic Engineering Slide
206 Example Determine the average power dissipated in the circuit: 3 V rms 5 Hz 8Ω C μf Z C 8 π Ω (633 ( o Ω School of Systems Engineering - Electronic Engineering Slide 3
207 Example 3 V rms 5 Hz 8Ω C μf P rms V cosφ Z 3 cos W School of Systems Engineering - Electronic Engineering Slide 4
208 Example Determine the average power dissipated in the circuit: 8 V rms 4 Hz C μf Ω L mh The driving-point impedance of this circuit at 4 Hz (calculated previously is: Z 3.9Ω.98 School of Systems Engineering - Electronic Engineering Slide 5
209 Example Z 3.9 Ω.98 P rms V cosφ Z 8 V rms 4 Hz C μf Ω L mh 8 cos W.983 School of Systems Engineering - Electronic Engineering Slide 6
210 Example Determine the average power dissipated in the circuit Since no power is dissipated in the capacitor we only need to calculate the power in the inductor-resistor resistor leg Z L 8 V rms L 4 Hz Ω 3 π 4 C.53 μf L mh School of Systems Engineering - Electronic Engineering Slide 7
211 Example Z L P V rms cos φ Z 8 cos W 8 V rms 4 Hz C μff Ω L mh School of Systems Engineering - Electronic Engineering Slide 8
212 AC Circuit Analysis Lecture 9 Power Factor Three-Phase Electric Power School of Systems Engineering - Electronic Engineering Slide 9
213 True and Apparent Power The apparent power P a in a circuit is: P V a rms I rms Apparent power is measured in VA The true power P dissipated in a circuit is: P V rms I rms cosφ True power is measured in W School of Systems Engineering - Electronic Engineering Slide
214 Power Factor The power factor is the ratio of the true power to the apparent power: pf P P a V rms V I rms rms I cosφ cosφ rms where ø is the phase difference between voltage and current. It does not matter whether ø is phase of the current with respect to the voltage, or voltage with respect to the current, since: cos φ cos φ School of Systems Engineering - Electronic Engineering Slide
215 Example Determine the power factor, apparent power and true power power dissipated i d in the circuit: it Z 4 5.8Ω 5 5.6Ω.3 (75. pf P a P cos V V rms rms Irms pf Pa Z 5. W o 8Vrms, 4.3 VA 4Hz 4Ω L 6mH School of Systems Engineering - Electronic Engineering Slide
216 Power Factor Correction Most industrial loads have a poor (pf << power factor Examples are induction motors and inductor-ballast lighting Power factor can be corrected by connecting a reactance in parallel with the load This reduces the apparent power and the rms current without affecting the load This is obviously desirable because it reduces the current rating of the power wiring and supply School of Systems Engineering - Electronic Engineering Slide 3
217 Power Factor Correction Power factor is normally corrected by connecting a reactive element Z C in parallel with the load Z L : IS IC Supply current: I S Load current: I L Correction current: I C V S I L ZL ZC A unity yoverall ea power factor aco will beoba obtained edpo provided that V S and I S are in phase: I V S S G G (real School of Systems Engineering - Electronic Engineering Slide 4
218 Power Factor Correction I V S S Z L I V L S I V C S G Z C Z L Z imag G C imag V S IS I L IC Z L Z C If I L leads V S then an inductor is used for correction If I L lags V S then a capacitor is used for correction School of Systems Engineering - Electronic Engineering Slide 5
219 Power Factor Correction Correction of a lagging power factor load with a capacitor: I I S L I C Current (imaginary part I C IS Current (real part I L Note that the magnitude of the supply current I S is less than that of the load I L School of Systems Engineering - Electronic Engineering Slide 6
220 Example Choose a suitable power factor correction component for the circuit: 8Vrms, 4Hz 4Ω L 6mH Z L 4 5.8Ω Z L Thus:. 695 Z C School of Systems Engineering - Electronic Engineering Slide 7
221 Example 4Ω 8 Vrms, 4Hz C.465 μf L 6mH Z C.695 C.695 C π 4.465μF School of Systems Engineering - Electronic Engineering Slide 8
222 Example I I I L C S 8 Z L 8( Z c I L I C I S I L 8Vrms 4Ω C 4Hz L 6mH.465 μff I C School of Systems Engineering - Electronic Engineering Slide 9
223 Example Imaginary part 5A I C I I S L I C IS 5A eal part I L -5A School of Systems Engineering - Electronic Engineering Slide 3
224 Example 3 An electric motor operating from the 5 Hz mains supply ppy has a lagging current with a power factor of.8 The rated motor current is 6 A at 3 V so that the magnitude of /Z L is: I 6 L.69 Z V 3 and the phase of /Z L is: L ZL S cos.8 ±.6435 Since the current lags the voltage the negative phase is used School of Systems Engineering - Electronic Engineering Slide 3
225 Example 3 Z L Z C CC C π μF Before correction: After correction: P P P a a 4 P pf P.8 38 P 4 a IS VS 3 School of Systems Engineering - Electronic Engineering Slide 3
226 Example 3 Imaginary part 5A I C I I S L I C IS 5A eal part I L -5A School of Systems Engineering - Electronic Engineering Slide 33
227 Three-Phase Electric Power Most ac power transmission systems use a three-phase system Three-phase is also used to power large motors and other heavy industrial loads Three-phase consists of three sinusoids with phases π/3 (º apart This allows more power to be transmitted down a given number of conductors than single phase A three-phase transmission system consists of conductors for the three phases and sometimes a conductor for neutral School of Systems Engineering - Electronic Engineering Slide 34
228 Three-Phase Electric Power Three-phase generator Three-phase load School of Systems Engineering - Electronic Engineering Slide 35
229 Three-Phase Electric Power Phase-to-neutral tral voltage v Phase-to-phase voltage v p π v p v sin 3 v 3 v 3 v p v π (6 3 v v π ( o 3 o School of Systems Engineering - Electronic Engineering Slide 37
230 Three-Phase Electric Power UK domestic supply uses three -phase with a phase-toneutral voltage v of 3 V rms (35 V peak This corresponds to a phase-to-phase voltage v p of 4 V rms (563 V peak Each property is supplied with one phase and neutral If the phases are correctly balanced (similar load to neutral on each then the overall neutral current is zero The UK electricity distribution tion network operates at 75 kv rms and 4 kv rms School of Systems Engineering - Electronic Engineering Slide 38
231 AC Circuit Analysis Lecture Energy Storage School of Systems Engineering - Electronic Engineering Slide 39
232 Energy Storage eactive components (capacitors and inductors do not dissipate power when an ac voltage or current is applied Power is dissipated only in resistors Instead reactive components store energy During an ac cycle reactive components alternately store energy and then release it Over a complete ac cycle there is no net change in energy stored, and therefore no power dissipation School of Systems Engineering - Electronic Engineering Slide 4
233 Energy Storage The voltage across a capacitor is increased from zero to V producing a stored energy E: V v(t T t E T v( t i( t dt E T dv v ( t C dt dt V C v dv CV v C dv i C dt i School of Systems Engineering - Electronic Engineering Slide 4
234 Energy Storage Example: calculate the energy storage in an electronic flash capacitor of μf charged to 4 V E CV J School of Systems Engineering - Electronic Engineering Slide 4
235 Energy Storage The current in an inductor is increase from zero to I producing a stored energy E: I i(t T t E T v( t i( t dt E T v di L i ( t dt L dt di L I i di v L dt LI i School of Systems Engineering - Electronic Engineering Slide 47
236 Energy Storage Example: calculate the energy storage in a mh inductor carrying a current of A E Li.J 3 School of Systems Engineering - Electronic Engineering Slide 48
237 AC Circuit Analysis J. B. Grimbleby 8 February 9 School of Systems Engineering - Electronic Engineering Slide 5
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