CHAPTER 9 Series Parallel Analysis Of AC Circuits Chapter .

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CHAPTER 9Series–Parallel Analysis of AC CircuitsChapter Outline9.1AC Series Circuits9.2AC Parallel Circuits9.3AC Series–Parallel Circuits9.4Analysis of Multiple-Source AC Circuits Using Superposition9.1 AC SERIES CIRCUITSIn DC analysis of series-parallel circuits, what type of numbers was utilized?If the signal source is AC instead of DC and the circuit contains inductors and/or capacitors, what type ofnumbers would you expect to be utilized? Why?Good news: the circuit analysis techniques are the same.What is the phase shift of the voltage with respect to the current for an inductor? Why?What is the phase shift of the voltage with respect to the current for a capacitor? Why?What is the phase shift of the voltage with respect to the current for a resistor? Why?What would you expect the phase shift to be for AC signals in series, parallel, and series–parallel circuits withresistances, inductors, and capacitors? Why?Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 1

How is series–parallel circuit analysis with AC sources approached? Review the circuit laws, concepts, and results for DC circuit analysis by stating in words each relation that follows:1.I1 I2 I3 . IN (9.1)2. VNn 1n Vrise1 Vrise2 Vrise3 Vdrop1 Vdrop2 Vdrop3 0 (9.2)How are the signs for voltage rises and voltage drops assigned (Figure 9.1 below)?Is the key to the sign of the voltage rise (or drop) the current direction or the direction that KVL is applied?N3.RT R1 R2 R3 RN Rn (9.3)4.RVx VT x (9.4)RTn 1Which of these laws and concepts are applicable to AC series circuits? Answer the following questions:Does constant current throughout a series circuit change for any signal?Does KVL change for any signal? Why or why not?Are 3 and 4 above valid? Why or why not?If the same DC circuit analysis techniques are valid for AC, and resistances are used in DC circuit analysis, thenwhat must be used in AC circuit analysis? Why?State in words the circuit laws, concepts, and results for AC series circuits:1.I 1 I 2 I 3 I N (9.5)2. V Nn 1n V rise1 V rise2 V rise3 V drop1 V drop2 V drop3 0 (9.6)Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 2

How is a voltage polarity assigned with AC signals considering the polarity is alternating?N3.Z T Z 1 Z 2 Z 3 Z N Z n (9.7)n 14.Z V x V T xZ (9.8)TExample 9.1.1(Explain each step.)For the circuit shown in Figure 9.2 below, determine (a) Z T , (b) I , (c) V R and V L by the VDR, and (d) V C byOhm’s law. (e) Verify KVL.:vS(t) 10 sin(827t) VR 35 C 20 FL 100 mH:a.b.c.d.e.Z TI V R and V L by VDRV by Ohm’s lawCVerify KVLStrategy:Determine Z R , Z L , and Z CZ T Z R Z C Z LLabel a circuit diagram with voltage polarities and current directions.V I SZ TZ V R V S R , V L V SZ TZ LZ TV C I Z C V S V R V C V L 0Solution:Z R R 35 Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 3

11Z C j j j 60.460 C(827)(20 )Z L j L j (827)(0.1) j82.7 a.Z T Z R Z C Z L 35 j 60.46 j82.7 35.000 j 22.240 35.0 j 22.2 Note: All voltages and currents are expressed as peak values in this example. See Figure 9.3 below.b.V 10 0 I S 0.24115 32.434 0.241 32.4 A ZT 35 j 22.24c.Z 35V R V S R 10 0 8.4402 32.433 8.44 32.4 V 35 j 22.24ZTDetermine V L in a similar manner. Answer: V L 19.943 57.567 19.9 57.6 Vd.V C I Z C (0.24115 32.434 )( j 60.46) 14.579 122.434 14.6 122.4 Ve.(Refer to Figure 9.3) V S V R V C V L 0Perform KVL.Is KVL satisfied?Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 4

9.2 AC PARALLEL CIRCUITSState the important laws, concepts, and results that were found for DC parallel circuits in words:1.VS V1 V2 V3 VN (9.9)2. INn 1n I1 I 2 I 3 I N 0 (9.10)How is the sign of the current assigned to if it is entering the node? Leaving the node?3.RT 1111 R1 R2RNRT R1 R2 in general (9.11)R1 R2for two parallel resistorsR1 R2(9.12)Ix IT RTin general (9.13)RxI1 IT R2I R, I 2 T 1 for two parallel resistors (9.14)R1 R2R1 R25.G I 1 V R6.GT G1 G2 . GN (9.16)4.(9.15)Which of these laws and concepts are applicable to AC series circuits? Answer the following questions:Does constant voltage across a parallel circuit change for any signal?Does KCL change for any signal? Why or why not?Are 3 - 6 above valid? Why or why not?If the same DC circuit analysis techniques are valid for AC, and resistances are used in DC circuit analysis, thenwhat must be used in AC circuit analysis? Why?Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 5

State in words the circuit laws, concepts, and results for AC parallel circuits:1.V S V 1 V 2 V 3 V N2. I I 1 I 2 I 3 I N 0 (9.18)Z T 1(general or special case?)111 Z 1 Z 2Z NNn 13.n(9.17)(9.19)Z Z Z T Z 1 Z 2 1 2 (general or special case?) (9.20)Z 1 Z 24.I Z I x T T (general or special case?) (9.21)Z xI Z I Z I 1 T 2 , I 2 T 1 (general or special case?) (9.22) Z1 Z 2Z1 Z 2Example 9.2.1(Explain each step.)For the circuit shown in Figure 9.4 below, determine (a) Z T , (b) V , (c) I R by the CDR, and (d) I C by Ohm’slaw. (e) Verify KCL.::R 10 XC 18 I T 8 20 Aa.Z TTb.c.d.e.:V I R by the CDRI C by Ohm’s lawVerify KCLDetermine Z R and Z CZ R Z CZ T Z R Z CLabel a circuit diagram with voltage polarities and current directions.V I T Z TI Z I Z I R T T or I R T CZ RZ R Z C(which equation is easier?note the previous step)Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 6

V I C Z CApply KCL to the “top” node.Solution:Z R R 10 Z C jX C j18 a.Z R Z C(10)( j18)Z T 8.7416 29.055 8.74 29.1 10 j18Z R ZCb.See Figure 9.5 below for voltage polarity and current direction assignments.V I T Z T (8 20 )(8.7416 29.055 ) 69.933 49.055 69.9 49.1 Vc.I Z (8 20 )(8.7416 29.055 )I R T T 10Z R 6.9933 49.055 6.99 49.1 Ad.V 69.933 49.055 I C 3.8851 40.945 3.89 40.9 A j18ZCe.KCL at node: I T I R I C 0Perform KCL:Is KCL satisfied?Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 7

What is the inverse of resistance (G 1/R)?What is the inverse of reactance? B Identify:BL 1X(9.23)11BC (9.24)XLXCIf susceptance is the inverse of reactance, what is the inverse of impedance?1Y Z (9.25)Is Y a complex number? Why or why not?How does admittance (Y ) relate to conductance (G) and susceptance (B)? Explain each step:Z R jX(9.26)Why is the plus sign used with inductive reactance?Why is the minus sign used with capacitive reactance?Y G jB (9.27)Conductance is the part of admittance and susceptance is the part of admittance.1 I IY ( I V ) (9.28)Z V VWhy is the sign minus used with inductive susceptance?Why is the plus sign used with capacitive susceptance?What is the unit of admittance? abbreviation?5.I 1Y V Z 6.Y T Y 1 Y 2 Y N Y n(S) (9.29)N(9.30)n 1Example 9.2.2(Explain each step.)For the circuit shown previously in Figure 9.4 (repeated below), determine (a) Y T from G and B, and (b) Y Tfrom Z .TGiven:R 10 XC 18 Z T 8.7416 29.055 7.6415 j 4.24535 (from Example 9.2.1)Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 8

Desired:a.b.Strategy:a.b.Y T from G and BY T from Z TG 1 R , B 1 X , Y T G jBY 1 Z TTSolution:a.G BC 1 1 0.1SR 1011 0.055556 SX C 18Y T G jB 0.1 j 0.05556 0.100 j 0.0556 Sb.11Y T 0.114396 29.055 Z 8.7416 29.055 0.0999993 j0.055556 0.100 j0.0556 SWhy do the results from parts (a) and (b) match?Observation about the impedance–admittance relationship:start with Z T of the circuit in Figure 9.4(repeated above). Explain each step:Z T 7.6415 j 4.24535 R jXConsider:11G 0.13086 0.131SR 7.6415BC 11 0.23555 0.236 SX C 4.24535Compare these results with those of Example 9.2.2:1 1G 0.1 0.100 SR 10BC 11 0.055556 0.0556 SX C 18(9.31)(9.32)(9.33)(9.34)These answers do not match! Why not? Which approach is correct?Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 9

Example 9.2.3(fill in)Determine the (a) total equivalent admittance and (b) parallel equivalent circuit for the series circuit shown inFigure 9.6a (below).:::RS 10 XS 13 (inductive)a.Y Tb.RP and XP1Y T , express in rectangular forma.Z Tb.RP 1 Gp , and X P 1 BPSolution: (fill in)Answers:a. Y T 0.037175 j 0.048327 S 0.0372 j 0.0483 Sb.RP 26.9 , X P 20.7 Example 9.2.4(fill in)Determine the resistance and reactance values of the series equivalent circuit for the parallel circuit shown inFigure 9.7a (below).Given:Desired:Strategy:Solution:Answer: RS 2.000 ; XS 4.000 Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 10

Example 9.2.5(fill in)Determine the component values for the parallel equivalent circuit of the series circuit shown in Figure 9.8a(below) if 20 krads/s.Given:Desired:Strategy:Solution:Answer: RP 12.5 , CP 2.00 FExample 9.2.6(fill in)For the circuit in Figure 9.9 (below), determine (a) the total equivalent admittance, and (b) the values for thecomponents of the series equivalent circuit. The frequency is 1 kHz.Given:Desired:Strategy:a.11Y T Y R Y C Y L GP jBC jBL j C jRP Lb.1Z T RS jX S , X S CY TSolution:Answers: a.b.Y T (8.3333 j3.1001) mS (8.33 j3.10) mSZ T 105.41 j39.215 , RS 105 ; XS 39.215 (capacitive), CS 4.06 FWhy are there only two reactive components in the series equivalent circuit?Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 11

Generalization of Series–Parallel Conversions to ―Handbook‖ Equations(Explain each step.)Case 1: Series-to-Parallel Conversion—see Fig. 9.10:ZT RS jX S , YT GP jBP(9.35)Y T 1( RS jX S )Y T ( R jX S ) RS jX S1 S ( RS jX S ) ( RS jX S ) RS2 X S2(9.37)Y T RSX j 2 S 2RS2 X S2RS X S(9.38)GP RS,R X S22S jBP (9.36)RP jX S,2RS X S2RS2 X S2RSXP (9.39)RS2 X S2XS(9.40)Case 2: Parallel-to-Series Conversion—see Fig. 9.11:Z T RP ( jX P ), Z T RS jX S(9.41)R ( jX P )Z T P( RP jX P )(9.42)R ( jX P ) ( RP jX P ) jX P RP2 jX P ( jX P ) RPZ T P ( RP jX P ) ( RP jX P )RP2 X P2(9.43)R X 2 jRP2 X PRP X P2RP2 X PZ T P P2 jRP X P2RP2 X P2RP2 X P2(9.44)RS RP X P2RP2 X P2 jX S jRP2 X PRP2 X P2(9.45),XS RP2 X PRP2 X P2Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and Lokken(9.46)Class Notes Ch. 9 Page 12

9.3 AC Series–Parallel CircuitsKVL, total equivalent impedance, the VDR, and the fact that the current is the same are generally applied towhat type of circuit?KCL, total equivalent impedance and admittance, the CDR, and the fact that the voltage is the same aregenerally applied to what type of circuit?The following general guidelines for DC series–parallel circuit analysis are extended to AC:1.The general approach to analyzing an AC series–parallel circuit is to identify the groups ofthat are in series and those that are in parallel and to apply the appropriate series and parallel circuit laws tothe impedances in that group.2.There are usually multiple approaches to analyzing an AC series–parallel circuit. think about your strategy formulate an efficient approach3.A single-source series–parallel circuit is fundamentally a series circuit or fundamentally a parallel circuit. The source is the key. If a single impedance is in series with the source on either side (or both sides) of the source, then thecircuit is fundamentally a circuit. If the current path from the source divides on both sides of the source, then the circuit is fundamentallya circuit.4.A group of series impedances, usually a “leg” in a parallel circuit, is called a .5.In the VDR and the CDR, the total voltage or current becomes the total voltage or the current for that groupof impedances, not necessarily the entire circuit.Example 9.3.1(Explain each step.)Determine the total equivalent admittance of the circuit in Figure 9.12 (below).Given:circuit in Figure 9.12Desired:Strategy:1 111Y 1 , Z T j8 7, Y T 2 j4Y1ZTSolution:11Y 1 0.5 j 0.25 S (identify group 1)2 j4Z T 1 j8 7 8.6000 j 7.2000 0.5 j 0.2511Y T 0.08916 39.936 8.6 j 7.2Z T 0.0892 39.9 S 0.0684 j0.0572 SContemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 13

Example 9.3.2(Explain each step.)Determine (a) I x and (b) V o for the circuit shown in Figure 9.13 (below).Given:circuit in Figure 9.13Desired: a. I xV oStrategy:b.Z x [5 ( j3)], Z T j 6 Z x 4 (identify Z x )V I T SZ TV o I T Z xV I x oZ CSolution:(5) ( j3)Z x 2.5725 59.036 5 j3Z T j 6 2.5725 59.036 4 6.5372 35.478 V 10 0 I T S 1.5297 35.478 A ZT 6.5372 35.478 V o I T Z x (1.5297 35.478 )(2.5725 59.036 ) 3.9352 94.514 3.94 94.5 VV 3.9352 94.514 I x o 1.3117 4.514 A 1.31 4.5 A3 90 Z CExample 9.3.3Redraw the circuit in Figure 9.14 (below), clearly showing series and parallel groups of components.Solution: (Try to redraw the circuit first, then check it with the answer in Figure 9.15 on the next page.)Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 14

Example 9.3.4(fill in steps)Determine (a) I x and (b) V ab for the circuit shown in Figure 9.16 (below).Given:circuit in Figure 9.16Desired:a.b.Strategy:I xV abRedraw and label the circuit (next to Fig. 9.16) Solution: (use separate paper)Answers:a.I x 1.41 95.0 Ab.Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenV ab 12.1 152.4 VClass Notes Ch. 9 Page 15

9.4 ANALYSIS OF MULTIPLE-SOURCE AC CIRCUITS USING SUPERPOSITIONConsider AC series–parallel circuits that contain several sources as well as several impedances.What condition on the sources must hold true if superposition with phasors is to be used? Why? (Hint:reactances)In superposition the circuit is analyzed one source at a time. The other sources are .How does one deactivate an AC voltage source? Why does it guarantee 0 V?How is an AC current source deactivated? Why does it guarantee 0 A?Write the general superposition procedureThis last step is actually the superposition step: the phasor voltages and currents are superposed.Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 16

Example 9.4.1Determine the current I a and ia(t) in Figure 9.18 (below).Peak values for sources are shown.Given:circuit in Figure 9.18Desired:I a and ia(t)I a uses the peak value of ia(t).Strategy:superpositionSolution:Deactivate the 15 V source; redraw the circuit (see Figure 9.19 below).Explain each step:substrategy: Z T j3 5 j 6 (4 j 2) 10 30 I T Z T CDR I a Z T 5 j 3 111 j6 4 j2 5 j 3 4.7434 18.435 9.6177 8.973 I T 10 30 1.0398 21.027 A9.6177 8.973 I a (1.0398 21.027 ) (6 90 ) 1.1028 23.973 A( j 6) (4 j 2)Fill in the rest of the solution steps on separate paper:Deactivate the 10 V source; redraw the circuit; determine the substrategy; determine I a and I a .Answers: I a 1.6076 66.991 A , I a 1.9342 32.234 1.93 32.2 A , ia(t) 1.93 sin( t 32.2 ) AContemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 17

AC sourcesThe properties of ideal AC sources are An ideal AC voltage source has a constant and can deliver any An ideal AC current source delivers a constant at anyExplain a practical AC voltage source model shown in Figure 9.21b (below).How was the following equation obtained?V t V S V Zint V S I t Z int(9.47)What does the previous equation predict about the terminal voltage of a practical AC source?Explain a practical AC current source model shown in Figure 9.22b (below).How was the following equation obtained?V I t I S I Zint I S tZ (9.48)intWhat does the previous equation predict about the terminal current of a practical AC source?Which quantity was not calculated in this chapter that was determined in previous chapters (conspicuous by itsabsence)?Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 18

Learning ObjectivesDiscussion: Can you perform each learning objective for this chapter? (Examine each one.)As a result of successfully completing this chapter, you should be able to:1.Utilize phasors in AC series–parallel circuit voltage and current calculations.2.Describe the fundamental properties of series and parallel circuits and subcircuits.3.Calculate all voltages and currents in single-source AC series, parallel, and series–parallel circuits.4.Perform series–parallel circuit conversions.5.Describe why and how superposition is used in multiple-source AC series, parallel, and series–parallelcircuits.6.Calculate all voltages and currents in multiple-source AC series, parallel, and series–parallel circuits.7.Explain what an AC current source is and how to analyze AC series–parallel circuits that contain a currentsource.Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008Strangeway, Petersen, Gassert, and LokkenClass Notes Ch. 9 Page 19

Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008 Class Notes Ch. 9 Page 1 Strangeway, Petersen, Gassert, and Lokken CHAPTER 9 Series–Parallel Analysis of AC Circuits Chapter Outline 9.1 AC Series Circuits 9.2 AC Parallel Circuits 9.3 AC Series–Parallel Circuits 9.4 Analysis of Multiple-Source AC Circuits Using Superposition 9.1 AC SERIES CIRCUITS