Electrical Circuits EE 223
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This 31 page Class Notes was uploaded by Maudie Quitzon on Saturday September 12, 2015. The Class Notes belongs to EE 223 at West Virginia University taught by Staff in Fall. Since its upload, it has received 83 views. For similar materials see /class/202858/ee-223-west-virginia-university in Electrical Engineering at West Virginia University.
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Date Created: 09/12/15
EE 223 AC Circuit Power Analysis it Instantaneous and average power 21 RMS value It Apparent power and power factor the Complex power Product of timedomain voltage and timedomain current 96 W W Determine maximum value Transients Steadystate As the transient dies out the circuit returns to steadystate operation Since the only source remaining in the circuit is dc the inductor eventually acts as a short circuit absorbing zero power VOMU a Power steadystate Power supphed by source vt Vm COSoatG and it Im COSoat p 39 P Vm m COSoat6 COS03t p trigonometric 12 Vm lm 0089 p 12 Vm Im COS2ot e p Identlty pt 12 Vm m 0056 p 12 Vm m oos2ot 6 p J k 39 J V constant periodic two parts independent oft period is 12T average the quotaveragequot average is zero periodic wanted quotunwantedquot 12 Vn m cos2mt H p Ll I I 8 12 pa 12Vm 39m COSHp Hg PviWV NS 7 1 V V 4 vg voltage V 440 V impedance Z 2460 C a 1 6 rads 2160 A pt24cos7tl360 W r delivered to resistor Avera e owe Example 1 I III V 2445 V R 29 time s I I39ll Ill VI T A I I I III 8m A9 n gt Mu v Q 9 o 2 5 I 0U 4 S S O X m C 2 mlm Im m V v v2 2 V O X P Example 2 Avera e ower delivered to purely reactive elements time s if i 1739 Il quot gt W 13 7 v i39 wiiiui it iiquotiiii if Example 3 Voltage across impedance V 1004250 V and Z 504550 Q Determine active power absorbed V I Z 2 42555 2 430 17 j A lt30 z 287 j419 o P 12 22 287 12 17212 287 574 W Example 4 Chapter 11 Problem 8 In the circuit shown in Fig 1127 find the average power being a dissipated in the 39 resistor b generated by the source 1 Figure 112 ZR3 31j34j3 2 I 01 103 if Ignore30 onVN1R25247241142 I 1 slim t II 7 j til 7 u i Mi a 2 39 x 3 10875 W r r b 1 PSgen 12 Im2 REG 1 0 PW Egtlt13463gtlt500s 5194 2075 W A simple loop circuit used to illustrate the derivation of the maXImum power transfer theorem as it applies to circuits operating in the sinusoidal steady state Zth Rththh ZL RLjXL l zL z i RL Rquot and XLXt A 7 r y mg 1 w J J H M r J Example Chapter 11 Problem 12 For the circuit of Fig 1130 a what value of ZL will absorb a maximum average power b What is the value of this maximum power Vm 2120101 2107334 11657 V 105 m 1010158 4Q 105 Figure 1130 a Z 8 j14 Q um M 139 L rum u 107334 116570 b IL 16 ma a396 T Hu u 2 P l jx8 180W L 2 16 Measure for sinusoidal voltages and currents Power outlets 60 Hz quotvoltage of 115Vquot Not the mean of T T2 Not the Amplitude M 115V Measure of effectiveness of a source in delivering power to a resistive load Effective value of periodic current is equal to the DC value that delivers the same average power to resistor it 2 R gt pt gt PR and compare to IDC2 R Mathematical expression P 131sz zli2dz 12 R 1 T 0 DC 0 I Eliza e TO Square root of the mean of the square current rms value Defined for all periodic signals 2 e R l l r l 21quot 7 U r it Im COSoat p with a period of T 27503 1 T 2 2 a 27ra 1 I I mt dt 1 cos 20t2 dt e 6quot m COS 6quot 2 2 real quantity independent of phase angle equal to 0707 the amplitude example g30 A delivers the same as lDC 1A u 1 3 quot 3 i 4ng 39 i i 1 39 n a r i w l v 39 w r 39 w l la 39 ll l l l r a l y l u l H ll w l 7 l lw Hm In general P 12 Vm m cos6 p Veff eff cos6 p For resistors P Veff Ie Ve 2 R Ie 2 R Note We can use amplitude or rms value Use V and V to designate voltages ITHS Example Chapter11 Problem 30 The series combination of a 1kQ resistor and a 2H inductor must not dissipate more than 250 mW of power at any instant Assuming a sinusoidal current with 0 500 rads what is the largest rms current that can be tolerated The peak instantaneous power is 250 mW The combination of elements yields Z 1000 j1000Q 1414445 Q Arbitrarily designate V Vm 40 so that I Vm40 z sz 45 1414 A and vm 1414Im We may write pt 12 Vm m cos 4 12 Vm m cos 2at 4 where 4 the angle of the current 45 This function has a maximum value of 12 lem cos 4 12 lem Thus 0250 12 lem 1 cos 4 12 1414 Im2 1707 and Im 1439 mA In terms of rms current the largest rms current permitted is l1439m N5 1018 mA rms We had P Vm Im COS6 39 p Veff Ieff COS6 39 p In case of direct current we would use voltage times current 8 Veff Ieff This is not the average power Is the quotapparentquot power 8 or AP Measured in voltampere or VA rather than Wto avoid confusion Magnitude of S is always greater or equal to magnitude of P S 2 P Defined as the ratio of average real power to apparent power PF PS Pve le l In the sinusoidal case the power factor is PF cos6 p 6 p is the angle the voltage leads the current PF angle note Purely resistive load has PF 1 Purely reactive load has PF O PF 05 means a phase angle of i60 Resolve ambiguity PF leading or lagging for capacitive or inductive load or Example Chapter 11 Problem 32 a Find the power factor at which the source in the circuit of Fig 1142 is operating b Find the average power being supplied by the source 0 What size capacitor should be placed in parallel with the source to cause its power factor to be unity d Verify your answers with PSpice a IS 92142 2625 Arms 4 12Jl6 Figure 1142 PFS cos 2625 08969 lag 4 2 b PS 2120 X 9214 X 08969 9917W 1311 v mm H H giro v1quot 33 I21 j48 3j4 ZL4 4192j144 1168 j576 c z 1168 39576QY L J L 116825762 39 1576 1207rC C9009 F J 116825762 Example see also chapter 10 Chapter 11 Problem 30 d Examine simulation output file AC sweep at 60Hz ACMAG120Vrms 42 44 H ACPHASE0 39 m PHASEyes VMN00030 VPN00030 6000E01 1200E02 0000E00 FREQ MV PRINT i PVPRNT1 6000E01 9215E00 2625E01 a and b are correct Next add a 90 09uF capacitor in parallel with the source c is correct 948x10395 degrees is essentially zero for unity PF FREQ MVPRNT1 6000E01 8264E00 PVPRNT1 9774E05 Simplifies power calculations We had pt Veff leff cos6 p 12 Vm lm COS20t 6 p Where the average real power P Veff leff cos6 p Using complex nomenclature P Vet39f Ieff Reeje 39p Vet39f Reeje Ieff Ree H4 H4 phasor voltage complex conjugate of phasor current Hence P ReVeff e S Veff leff can be written as S Veff leff ekeP P jQ Hquot vquotvquotr magnitude equals PF angle reactive apparent power average power Isl 8 real Reactive power Q POWer Imaginary part of complex power Dimensions are those of P S APS S Avoiding confusion by using voltamperereactive or VAr Q Veff leff sin6 p Physical interpretation Time rate of energy flow backampforth between source and reactive loads Reactive components charge and discharge at 20 gt current flows Veff eff COS2mt H p I0t Veff lerr COSW 39 P PavaiWVaA 9 Is 3 Q KI I I I I I I Iquot I S V1 2 3V4yfsV 12 1 voltage V 440 V impedance Z 260 Q 260 A a 11 6 rads pt 2 4cosn360 W voltage v 2830 vrms I Wig60 Arms SPjQV2W j346VAr 4460 VA Commonly used graphical representation of SPjQVI VVZ VVZ V2Z ZZ Useful relationships P S cos Q S sin G power angle tan1QP Q P tan 6 m 3 Quadrant means 1st power factor is lagging inductive load S 4th power factor is leading capacitive load Q Need only two quantities to find third Re 39 gt gt P 4 460 VA 2w j 346VAr Another interpretation of active and reactive power components Current components In phase with voltage leff cos6 p 90 out of phase quadrature component leff sin6 p Multiplied by M results in P and Q Veff lm le cs6 ltpfr5iquot9 39 PI Example Chapter 11 Problem 42 3 5 Both sources are operating at the same frequency Find the complex power generated by each source Hm V 1 39 and the complex power absorbed by each passive quot 3 mm 1quot 4m 1 circuit element T Vx 100 Vx Vx leO 6j4 j10 5 0 Vx 1 j0102 j20 6j4 6j4 Vx 5335 44266 V I 100 5335 z4266 39 1 6 j4 S186quot 2 ix 100 X 9806464440 2115 j4423 VA 2 98064 64440A Example Chapter 11 Problem 42 6 H q u Both sources are operating at the same 1 frequency Find the complex power generated by V U 39 each source and the complex power absorbed by min J 1quot i i I H i 3 1quot each passive circuit element SW ix 6 X 98062 2885 jO VA 514m 21498062 0j1923 VA j100 53357442660 5 SW ix 5 gtlt14992 5615 jOVA 2an j10014994 121570 6384 13923 VA 2 s low l j 110 0 11423 VA 1423go VA 2 10 I2 21499412160 39 i r i n f f i gi Hal Wattmeterhours measures active load Varmeterhours measures reactive load Average PF is used to adjust consumer39s bill industry has to pay for unwanted losses Complex power delivered to individual loads equals their sum no matter how loads are connected S V l V I1 2 V l1 2 Vl1 Vl2 Large industrial consumers pay penalty when PF lt 085 Caused by inductive loads motors Why Causes increased Device ratings Transmission and distribution losses Example 022kVAr above 62 of average real power demand S P jQ P j062P P 1j062 P 1774318 Targets cos318 085 pay penalty when PF lt 085 Use compensation capacitors in parallel with load Value of capacitance C Ptan 601d tan 9 new 2 a V rms corrects the PF angle from old to new at the specified frequency and voltage Derived from AQ VzrIns X0 VzrIns 03C Qold Qnew Vzrms 2 DC Ptan 901d tan Gnew VZmls 03C Example Source of 115 Vrms supplies two loads Loads are connected in parallel 7kW 3 kVAr and 4kVA at 085 pf lagging Find the pf of the equivalent load as seen from the input terminal 81 7000 j 3000 s2 4k 0030 j sin0 4k 085 j sinCOS391085 3400 j 2107 S S1 S2 10400 j 5107 11586 42615 0 tan1Q P tan15107 10400 2615 0036 cos 2615 Z V2 S 1152 11586 114ZQ remember S V2Z z11429 42615 102 j 05 9 Instantaneous power Average power P W sinusoidal steadystate P12 Vm lm cosch pl reactances do Active power not contribute to P Effective or rm s value Vrms or lrms V or A leff Im V2 sinusoidal 1 eff 12 dt Apparent power S VA S Veff leff Power factor PF or cos p None ratio P S unity for purely resistive loads and zero for purely reactive load Reactive power Q VAR Means of measuring energy flow rate to and from reactive loads Complex power 8 VA Contains both average power P and reactive power Q SVPjQ
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