RLC Series AC Circuits

Sixth Edition, last update July 25, 2007 2 Lessons In Electric Circuits, Volume II – AC By Tony R. Kuphaldt Sixth Edition, last update July 25, 2007 i c 2000-2008, Tony R. Kuphaldt This book is published under the terms and conditions of the Design Science License. These terms and conditions allow for free copying, distribution, and/or modification of this document by the general public. The full Design Science License text is included in the last chapter. As an open and collaboratively developed text, this book is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the Design Science License for more details. Available in its entirety as part of the Open Book Project collection at: www.ibiblio.org/obp/electricCircuits PRINTING HISTORY • First Edition: Printed in June of 2000. Plain-ASCII illustrations for universal computer readability. • Second Edition: Printed in September of 2000. Illustrations reworked in standard graphic (eps and jpeg) format. Source files translated to Texinfo format for easy online and printed publication. • Third Edition: Equations and tables reworked as graphic images rather than plain-ASCII text. • Fourth Edition: Printed in November 2001. Source files translated to SubML format. SubML is a simple markup language designed to easily convert to other markups like L A T E X, HTML, or DocBook using nothing but search-and-replace substitutions. • Fifth Edition: Printed in November 2002. New sections added, and error corrections made, si nce the fourth edition. • Sixth Edition: Printed in June 2006. Added CH 13, sections added, and error corrections made, figure numbering and captions added, since the fifth edition. ii Contents 1 BASIC AC THEORY 1 1.1 What is alternating current (AC)? . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 AC waveforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Measurements of AC magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.4 Simple AC circuit calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.5 AC phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.6 Principles of radio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.7 Contributors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2 COMPLEX NUMBERS 27 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.2 Vectors and AC waveforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.3 Simple vector addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.4 Complex vector addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.5 Polar and rectangular notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.6 Complex number arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 2.7 More on AC ”polarity” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.8 Some examples with AC circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 2.9 Contributors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3 REACTANCE AND IMPEDANCE – INDUCTIVE 57 3.1 AC resistor circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.2 AC inductor circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.3 Series resistor-inductor circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 3.4 Parallel resistor-inductor circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 3.5 Inductor quirks . . . . . . . . . . . . . . . . . . . . . . . . . . . RLC Series AC Circuits RLC Series AC Circuits Bởi: OpenStaxCollege Impedance When alone in an AC circuit, inductors, capacitors, and resistors all impede current How they behave when all three occur together? Interestingly, their individual resistances in ohms not simply add Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other’s effect [link] shows an RLC series circuit with an AC voltage source, the behavior of which is the subject of this section The crux of the analysis of an RLC circuit is the frequency dependence of XL and XC, and the effect they have on the phase of voltage versus current (established in the preceding section) These give rise to the frequency dependence of the circuit, with important “resonance” features that are the basis of many applications, such as radio tuners An RLC series circuit with an AC voltage source The combined effect of resistance R, inductive reactance XL, and capacitive reactance XC is defined to be impedance, an AC analogue to resistance in a DC circuit Current, voltage, and impedance in an RLC circuit are related by an AC version of Ohm’s law: I0 = V0 Z or Irms = Vrms Z Here I0 is the peak current, V0 the peak source voltage, and Z is the impedance of the circuit The units of impedance are ohms, and its effect on the circuit is as you might 1/14 RLC Series AC Circuits expect: the greater the impedance, the smaller the current To get an expression for Z in terms of R, XL, and XC, we will now examine how the voltages across the various components are related to the source voltage Those voltages are labeled VR, VL, and VC in [link] Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in R, L, and C are equal and in phase But we know from the preceding section that the voltage across the inductor VL leads the current by one-fourth of a cycle, the voltage across the capacitor VC follows the current by one-fourth of a cycle, and the voltage across the resistor VR is exactly in phase with the current [link] shows these relationships in one graph, as well as showing the total voltage around the circuit V = VR + VL + VC, where all four voltages are the instantaneous values According to Kirchhoff’s loop rule, the total voltage around the circuit V is also the voltage of the source You can see from [link] that while VR is in phase with the current, VL leads by 90º, and VC follows by 90º Thus VL and VC are 180º out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude Since the peak voltages are not aligned (not in phase), the peak voltage V0 of the source does not equal the sum of the peak voltages across R, L, and C The actual relationship is V0 = √V0R 2+(V0L − V0C)2, where V0R, V0L, and V0C are the peak voltages across R, L, and C, respectively Now, using Ohm’s law and definitions from Reactance, Inductive and Capacitive, we substitute V0 = I0Z into the above, as well as V0R = I0R, V0L = I0XL, and V0C = I0XC, yielding I0Z = √I02R2 + (I0XL − I0XC)2 = I0√R2 + (XL − XC)2 I0 cancels to yield an expression for Z: Z = √R2 + (XL − XC)2, which is the impedance of an RLC series AC circuit For circuits without a resistor, take R = 0; for those without an inductor, take XL = 0; and for those without a capacitor, take XC = 2/14 RLC Series AC Circuits This graph shows the relationships of the voltages in an RLC circuit to the current The voltages across the circuit elements add to equal the voltage of the source, which is seen to be out of phase with the current Calculating Impedance and Current An RLC series circuit has a 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF capacitor (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for L and C are the same as in [link] and [link] (b) If the voltage source has Vrms = 120 V, what is Irms at each frequency? Strategy For each frequency, we use Z = √R2 + (XL − XC)2 to find the impedance and then Ohm’s law to find current We can take advantage of the results of the previous two examples rather than calculate the reactances again Solution for (a) At 60.0 Hz, the values of the reactances were found in [link] to be XL = 1.13 Ω and in [link] to be XC = 531 Ω Entering these and the given 40.0 Ω for resistance into Z = √R2 + (XL − XC)2 yields Z = √R2 + (XL − XC)2 = √(40.0 = 531 Ω at 60.0 Hz Ω )2 + (1.13 Ω − 531 Ω )2 Similarly, at 10.0 kHz, XL = 188 Ω and XC = 3.18 Ω , so that 3/14 RLC Series AC Circuits Z Ω )2 + (188 Ω − 3.18 Ω )2 = √(40.0 = 190 Ω at 10.0 kHz Discussion for (a) In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values It is clear that XL dominates at high frequency and XC dominates at low frequency Solution for (b) The current Irms can be found using the AC version of Ohm’s law in Equation Irms = Vrms / Z: Irms = Vrms Z = 120 V ... Electric Circuits part 1 Using PSpice Dr. Ngo Van Sy University of Dannang ngvnsy@yahoo.com Mb: 0913412123 Chương 3 CÁC MẠCH RLC ĐƠN GIẢN DƯỚI TÁC ĐỘNG DC VÀ AC  MẠCH RLC NỐI TIẾP  MẠCH RLC SONG SONG  CÁC MẠCH DAO ĐỘNG THỰC TẾ  MẠCH DAO ĐỘNG LC  MẠCH DAO ĐỘNG BA ĐIỂM ĐIỆN CẢM  MẠCH DAO ĐỘNG BA ĐIỂM ĐIỆN DUNG  MẠCH GHÉP HỖ CẢM  CÔNG SUẤT TRONG MẠCH DƯỚI TÁC ĐỘNG ĐIỀU HÒA (AC) MẠCH RLC NỐI TIẾP + R e (t) = co s (wt) L + C ).2)(( 1 )( ) 1 )(( 1 ) 1 )(( )( )( )( )()( 1 )()( )cos()()( 1)( )( 222 0 2 2 22 0 2 2 2 0 2 2 0 2 0 CH sss s L sI LC s L R ss s L sC sLRs s sZ sE sI s s sEsI sC ssLIsRI ttedtti Cdt tdi LtRi ωαω ωω ω ω +++ = +++ = +++ == + ==++ ==++ ∫  Dùng công thức Heavisaid tìm về hàm gốc i(t) trong miền thời gian       ∆ −−− ∆ − ∆+ = ∆+ −= ∆+ −≈ +− +− = ∆+ = ∆+ ≈= +++++= =−−=+−= =−== −=∆≈== ∆ − ∆ − )cos()exp()cos( 2 1 )( 4 1 )(4 1 )( )( 4 1 )(4 1 )( )( ))((2).2(2)( 1 2 0 22 )( 22 ' 2 1 3 )( 22 0 ' 2 01 1 2 0 222' 2 * 343 * 10201 00 α ω ωα α ω ω ωα ωα ωαωα ωα ωα ωαω ω ωαωα ωαωα ωω ωωωωωωα α ω α ω arctgttarctgt L ti e jjH jH A e jjH jH A ssssssH sjsjs sjsjs LC L R r jarctg r r jarctg CH rr CHCHCH  Thành phần cưỡng bức  Tần số  Biên độ  Biên độ tại cộng hưởng  Biên độ tại cạnh dải thông  Dải thông  Pha α ω ϕ ω ωα ω ωα ω α ω ω ωα ∆ = =∆ == ∆+ =∆ = ∆+ = ∆ − ∆+ = arctg L R R I L I R I L I arctgt L ti qđ i d m d dm m m cb 2 2 1 2 )0( 2 1 )( 1 )0( 2 1 )cos( 2 1 )( 2 2 22 0 0 22  Thành phần quá độ  Tần số  Biên độ  Thời gian tắt  Pha α ω ϕ αα τ α ωα ω α ω ωα ωα ∆ = === − ∆+ = −= ∆ −− ∆+ −= arctg R L t L I L R LC arctgtt L ti qđ i m r rqđ 6,4 3,210ln )exp( 2 1 4 1 )cos()exp( 2 1 )( 22 2 2 22  Chế độ xác lập điều hòa trong mạch dao động đơn  Trở kháng Z  Điện kháng X  Cảm kháng X L  Dung kháng |X| C . C X LX XX C LX jXR C LjR Cj LjRZ EIZ EI Cj LjR EI Cj ILjIR C L CL ω ω ω ωω ω ω ω ω ωω ωωω ωω ω ω ωω ω ωωω 1 1 )( )() 1 ( 1 )( )()().( )()() 1 ( )()( 1 )(.)(. = = −=−= +=−+=++= = =++ =++  Trở kháng đặc tính  Hệ số phẩm chất  Độ lệch cộng hưởng tổng quát  Độ lệch cộng hưởng tương đối 0 0 ) 1 ( 1 1 )()( ω ω ω ω ν ω ωξ ρ ωωρ CH CH CHCCHL C L RR X C L RR Q C L XX −= −== == === MẠCH RLC SONG SONG  Tính chất đối ngẫu  Dòng điện – điện áp  Vòng – Nút  Hệ pt Iv – Hệ pt Un  Nối tiếp – Song song  Các thông số đối ngẫu:  R-G  L-C  E-Ing  Bảng so sánh các đặc trưng của mạch dao động đơn nối tiếp và song song (xem trang 96-97) [...]...Bảng các thông số của mạch RLC nối tiếp và GLC song song ωCH NT = 1 LC | ωCH SS = 1 LC CÁC MẠCH DAO ĐỘNG THỰC TẾ MẠCH DAO ĐỘNG LC (trang 97) 2 2 RL' RC' 1k 2 1 L C' 1 1 1 C Rtd + L' 2 2 1 +   MẠCH DAO ĐỘNG BA ĐIỂM ĐIỆN CẢM (trang 99)  MẠCH DAO ĐỘNG BA ĐIỂM ĐIỆN DUNG (trang 100) i(t) R L Mạch RL   Phương trình mạch điện Biểu thức dòng điện trong miền biến đổi Laplace e(t)=cos(ω0t)... exp( )+ 2 2 2 R 1 + ω0 R C RC 1 + 1 −t exp( ) R RC τ = RC i (t ) = 1 1 2 2 2 ω0 R C cos(ωot − arctg 1 )] ωo RC MẠCH GHÉP HỖ CẢM  Trang 105 CÔNG SUẤT TRONG Solve problems in single and threephase low voltage circuits Part A Topic 3: Resistance in AC Circuits Resistance in AC Circuits  Resistive Component – –  Any device that consists of a ‘conductor’ with an atomic structure that does not allow electrons to move as easily as a normal conductor A resistive component’s function is to convert electrical energy into heat energy Key Characteristic – – A resistive component will OPPOSE THE FLOW OF CURRENT It achieves this by slowing the flow of electrons by not allowing them to cascade through as quickly Resistance in AC Circuits  Symbol ~ R  Unit of measurement ~ OHMS (Ω) Resistance in AC Circuits  In a purely resistive AC circuit, the circuit can be analysed using the same principles as DC circuits – – – – – Ohm’s Law Resistors in Series Resistors in Parallel Kirchoff’s Voltage Law Kirchoff’s Current Law Resistance in AC Circuits: Ohm’s Law VR R VR IR IR R VS ƒ Simple Resistive Circuit Ohm’s Law – Resistive component Resistance in AC Circuits: Resistance in Series and Parallel R1 R1 R2 R2 VS ƒ IS VS ƒ Resistance in Parallel Resistance in Series RTotal= R1+R2+… = + +… RTotal R1 R2… Kirchoff’s Voltage Law: Series Resistive Circuit VR1 VR2 R1 R2 Kirchoff’s Voltage Law •The ‘sum’ of the voltage drops in the circuit will equal the supply voltage VS Vs = VR1 + VR2+… ƒ [Purely resistive circuit only] Parallel Resistive Circuit IR1 R1 Kirchoff’s Current Law R2 IR2 IS VS ƒ •The ‘sum’ of the currents entering a junction will be equal to the sum of the currents exiting the junction Is = IR1 + IR2+… [Purely resistive circuit only] Power in a Resistive Circuit   Since a resistor converts electrical energy into heat, then POWER is dissipated This can be calculated by: P = VI Or P = I2R Or P = V2/R Resistance In AC Circuits: Exercises Question 1: R1 R2 50Ω 80Ω IR3 IS VS=230V ƒ=50Hz R3 100Ω Determine: •RTotal •IS •VR2 •IR3 •PR2 Resistance In AC Circuits: Exercises  Answers  RTotal = 94.4Ω  IS = 2.44A  VR2 = 108V  IR3 = 1.08A  PR2 = 146W Resistance In AC Circuits: Exercises Question 2: R1 IR2 5Ω R2 ?Ω IR3 IS=4A VS=50V ƒ=50Hz Determine: •IR2 R3 •R2 20Ω •PR1 Resistance In AC Circuits: Exercises  Answers  IR2 = 2.5A  R2 = 12Ω  PR1 = 80W Resistance In AC Circuits: Exercises Question 3: R2 IR2 R4 ?Ω 5Ω IR3=1.5A R1 IS •VS IR4=2A R3 10Ω Determine: 25Ω VS=? V ƒ=50Hz •IS •R2 Resistance In AC Circuits: Exercises  Answers  VS = 25V  IS = 3A  R2 = 30Ω Review  At this stage, you should have a clear understanding of resistance in AC circuits, including: – – – – – Understand the concept of resistance in AC circuits; Understand the application of Ohm’s Law; Understand and be able to apply Kirchoff’s Voltage law to a purely resistive circuit; Understand and be able to apply Kirchoff’s current Law to a purely resistive circuit; How to make calculations involving V, I, R, and P in combined series/parallel circuits [...].. .Resistance In AC Circuits: Exercises  Answers  RTotal = 94.4Ω  IS = 2.44A  VR2 = 108V  IR3 = 1.08A  PR2 = 146W Resistance In AC Circuits: Exercises Question 2: R1 IR2 5Ω R2 ?Ω IR3 IS=4A VS=50V ƒ=50Hz Determine: •IR2 R3 •R2 20Ω •PR1 Resistance In AC Circuits: Exercises  Answers  IR2 = 2.5A  R2 = 12Ω  PR1 = 80W Resistance In AC Circuits: Exercises Question 3: R2 IR2 R4 ?Ω 5Ω IR3=1.5A... Exercises Question 3: R2 IR2 R4 ?Ω 5Ω IR3=1.5A R1 IS •VS IR4=2A R3 10Ω Determine: 25Ω VS=? V ƒ=50Hz •IS •R2 Resistance In AC Circuits: Exercises  Answers  VS = 25V  IS = 3A  R2 = 30 Ω Review  At this stage, you should have a clear understanding of Solve problems in single and threephase low voltage circuits Part A Topic 4: Inductance in AC Circuits Inductance in AC Circuits  Inductive Component – – Any device that consists of a conductor formed into a coil consisting of many turns, normally wound around some form of iron core An inductive component’s function is to set up (and store) energy in an electromagnetic field Inductance in AC Circuits  Key Characteristic – – – An inductive component will ACT TO OPPOSE A CHANGE IN CURRENT It achieves this by setting up a self-induced emf to either oppose an increase in current, or resist a decrease in current How strongly it does this is indicated by the coil’s INDUCTANCE, symbol L, measured in Henries (H) L Inductance in AC Circuits  Effect in an AC circuit – – Since the current in an AC circuit is continually changing (ie alternating between positive flow and negative flow), an inductive component in an AC circuit continually acts to oppose the change in current As a result, the inductive component produces a constant opposition to the flow of alternating current Inductance in AC Circuits  This opposition to alternating current flow is called INDUCTIVE REACTANCE XL = 2π ƒ L  Where: – XL is the inductive reactance in Ohms (Ω) 2π is a constant – ƒ is the frequency in Hertz (Hz) – L is the inductance in Henries (H) – Inductance in AC Circuits  Important! – – INDUCTIVE REACTANCE is an opposition to current flow like RESISTANCE, but it is NOT the same as resistance, even though both are measured in Ohms As a result, INDUCTIVE REACTANCE and RESISTANCE can NOT be simply added to find the total opposition to current flow in a circuit Inductance in AC Circuits  Key Advantages – – Ability to store energy in the form of an electromagnetic field Ability to limit the flow of AC current to a low value WITHOUT consuming a large quantity of power (not heating excessively), unlike a resistor Inductance in AC Circuits  Examples of Inductive Components – – – – – – Transformer windings Motor windings Ballasts for discharge lighting Solenoids / Contactor coils Fault current limiting devices (surge arrestors) Filter and tuning circuits Inductance in AC Circuits: Ohm’s Law VL VL L IL IL XL VS ƒ Simple Inductive Circuit Ohm’s Law – Inductive component Inductive Reactance: Exercises L=? L=0.25H IL=7.32A IL=? VS=230V VS=230V ƒ=50Hz ƒ=50Hz Q1 Determine: Q2 Determine: Inductive reactance XL, and Inductive reactance XL, and Current though inductor IL Inductance of inductor L Inductive Reactance: Exercises  A 230V, 50Hz AC supply is to be applied to a fluorescent light fitting If the current for the lamp needs to be limited to 0.8A max, determine the appropriate value of inductance required by the ballast Inductance in AC Circuits: Inductance in Series and Parallel L1 L2 VS ƒ L1 L2 IS VS ƒ Inductive Reactance in Parallel Inductive Reactance in Series X L Total= X L1+X L2+… X L Total = + +… X L1 X L2… Series Inductive Circuit VL1 VL2 Kirchoff’s Voltage Law L1 L2 •The ‘sum’ of the voltage drops in the circuit will equal the supply voltage Vs = VL1 + VL2+… VS ƒ [Purely Inductive circuit only] Parallel Inductive Circuit IL1 IL2 L1 L2 IS Kirchoff’s Current Law •The ‘sum’ of the currents entering a junction will be equal to the sum of the currents exiting the junction VS Is = IL1 + IL2+… ƒ [Purely Inductive circuit only] Inductance in AC Circuits: Exercises L1=0.25H L2=0.1H IL1 L1=1.6H L3=0.2H IS IL2 L2=2.4H VS=230V ƒ=50Hz Determine: •XL Total •IS •VL3 IS IL3 L3=1.2H VS=230V ƒ=50Hz Determine •XL Total •IS •IL2 Inductance in AC Circuits: Exercises  Q1 Answers: Series – XL1 = 78.54 Ω – XL2 = 31.42 Ω – XL3 = 62.83 Ω – XL Total = 172.79 Ω – IS = 1.33 A – VL3 = 83.6 V Inductance in AC Circuits: Exercises  Q2 Answers: Parallel – XL1 = 502.7 Ω – XL2 = 754.0 Ω – XL3 = 377.0 Ω – XL Total = 167.6 Ω – IS = 1.37 A – IL2 = 0.305 A Review  At this stage, you should have a clear understanding of inductance in AC circuits, including: – – – – – Understand the Solve problems in single and threephase low voltage circuits Part A Topic 5: Capacitance in AC Circuits Capacitance in AC Circuits  Capacitive Component – – A device that consists of a two metal plates separated by a dielectric (insulator) A capacitive component’s function is to store energy in an electrostatic field Capacitance in AC Circuits  Key Characteristic – – – A capacitive component will ACT TO STORE ELECTRICAL CHARGE THAT WILL OPPOSE THE APPLIED VOLTAGE (and therefore current) It achieves this by storing a negative charge on one plate and a positive charge on the other How strongly it does this is indicated by the capacitor’s CAPACITANCE, symbol C, measured in Farads (F) C Capacitance in AC Circuits  Effect in an AC circuit – – Since the voltage in an AC circuit is continually changing (ie alternating between positive voltage and negative voltage), a capacitor in an AC circuit continually charges and discharges which then opposes the change in voltage (and current) As a result, the capacitor produces a constant opposition to the flow of alternating current Capacitance in AC Circuits  This opposition to alternating current flow is called CAPACITIVE REACTANCE XC =  Where: – 2π ƒ C Xc is the capacitive reactance in Ohms (Ω) 2π is a constant – ƒ is the frequency in Hertz (Hz) – C is the capacitance in Farads (F) – Capacitance in AC Circuits  Important! – – CAPACITIVE REACTANCE is an opposition to current flow like RESISTANCE, but it is NOT the same as resistance, even though both are measured in Ohms As a result, CAPACITIVE REACTANCE and RESISTANCE can NOT be simply added to find the total opposition to current flow in a circuit Capacitance in AC Circuits  Key Advantage – – Ability to produce better operation for circuits containing highly inductive loads Ability to limit AC current flow without consuming any power Capacitance in AC Circuits  Examples of Capacitive Components – – – Capacitor Capacitor banks (for power factor correction) Filter or tuning circuits Capacitance in AC Circuits: Ohm’s Law VC IC VC C IC XC VS ƒ Simple Capacitive Circuit Ohm’s Law – capacitive component Capacitive Reactance: Exercises C=47µF C =? IC=0.723A IC=? VS=230V VS=230V ƒ=50Hz ƒ=50Hz Q1 Determine: Q2 Determine: Capacitive reactance XC, and Capacitive reactance XC, and Current though capacitor Ic Capacitance of capacitor C Series Capacitive Circuit VC1 VC2 Kirchoff’s Voltage Law C1 C2 •The ‘sum’ of the voltage drops in the circuit will equal the supply voltage Vs = VC1 + VC2+… VS ƒ [Purely capacitive circuit only] Parallel Capacitive Circuit IC1 IC2 IS C1 Kirchoff’s Current Law C2 •The ‘sum’ of the currents entering a junction will be equal to the sum of the currents exiting the junction VS Is = IC1 + IC2+… ƒ [Purely capacitive circuit only] Capacitance in AC Circuits  Exercises on series / parallel combinations Capacitance in AC Circuits: Capacitance in Series and Parallel C1 = 15μF VS=230V ƒ=50Hz Determine the following: •XC Total •IS •VC2 C1 = 220μF C2=33μF C2 = 680μF IS VS = 32V ƒ=50Hz Determine the following: •XC Total •IS •IC1 Capacitance in AC Circuits  Series Answers – XC1 = 212.2 Ω – XC2 = 96.5 Ω – XC Total = 308.7 Ω – IS = 0.75A – VC2 = 71.9V Capacitance in AC Circuits  Parallel Answers – XC1 = 14.47 Ω – XC2 = 4.68 Ω – XC Total = 3.54 Ω – IS = 9.04 A – IC1 = 2.21 A Review  At this stage, you should have a clear understanding of Capacitance in AC circuits, including: – – – – – Understand the concept of capacitive reactance; Understand the application of Ohm’s Law to capacitive circuits; Understand and be able to apply Kirchoff’s Voltage law to a purely capacitive circuit; Understand and be able to apply Kirchoff’s current Law to a purely capacitive circuit; How to make calculations involving V, I, and XC in series/parallel circuits Phase Relationship between Voltage and Current Resistive Circuit Phase Relationship between Voltage and Current: Resistive Circuit V R ‘V’ Waveform ‘I’ Waveform I I V VS ƒ Simple ... impedance of an RLC series AC circuit For circuits without a resistor, take R = 0; for those without an inductor, take XL = 0; and for those without a capacitor, take XC = 2/14 RLC Series AC Circuits. .. (AC+ DC), Virtual Lab 10/14 ϕ RLC Series AC Circuits Section Summary • The AC analogy to resistance is impedance Z, the combined effect of resistors, inductors, and capacitors, defined by the AC. .. must the capacitance be at this frequency? (a) 1.31 μH (b) 1.66 pF 12/14 RLC Series AC Circuits An RLC series circuit has a 2.50 Ω resistor, a 100 μH inductor, and an 80.0 μF capacitor.(a) Find