ac1 - ELEC 101 AC Circuits 1 2009/10 Spring Sinusoidal (AC)...

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Unformatted text preview: ELEC 101 AC Circuits 1 2009/10 Spring Sinusoidal (AC) Circuits 1 Basics A Sinusoids are important AC has advantages over DC for simple transmission and low transmission loss. 1. Most signals are sinusoids (sine/cosine waves) 2. Sinusoids are building blocks of all periodic signals, e.g. rectangular wave, triangular wave, etc. 3. Responses (I, V) of sinusoids (integral or differential) are also sinusoids 4. Responses (I, V) of sinusoids have the same frequency with some phase difference. response xr = x2sin(t + ) input xi = x1sin(t) x(t) 2 t The response x2sin(t + ) leading the input x1sin(t) by radian. ELEC 101 AC Circuits 2 comparing cos t component cos t cos 40 R cos t cos10 Lcos t sin10 Find R and L. vR(t) i(t) cos(t 10 )A Example v(t) cos(t 40 )V R 2009/10 Spring comparing sin t component sin t sin 40 R sin t sin10 Lsin t cos10 L vL(t) cos 40 R cos10 sin 40 R sin10 sin10 cos10 cos 40 Lsin10 Lcos10 sin 40 R cos10 sin10 L Use normal method Given Not required in exam sin sin() cos(90o ) v(t) cos(t 40 ) (cos t cos 40 sin t sin 40 ) v R (t) i(t)R R cos(t 10 ) R(cos t cos10 sin t sin10 ) di(t) v L (t) L L( )sin(t 10 ) dt L(sin t cos10 cos t sin10 ) vL (t) ωLsin(ωt 10) ωLsin( ωt 10) ωLcos(90 ωt 10) ωLcos(ωt 80) cos10(cos 40 Rcos10) sin10(sin 40 Rsin10) R(cos2 10 sin2 10) cos10 cos 40 sin10 sin 40 cos(10 40) R cos 50 sin10(cos 40 Lsin10 ) cos10(Lcos10 sin 40 ) L(cos 2 10 sin 2 10 ) sin10 cos 40 cos10 sin 40 sin(10 40 ) L sin 50 v(t) v R (t) v L (t) cos(t 40 ) R cos(t 10 ) Lcos(t 80 ) Conclusion: May not be a good method! ELEC 101 B AC Circuits 3 2009/10 Spring Complex Method Complex method can greatly simply the AC calculation. 1 40 V 1 10 A 50 Find R and L. vR(t) i(t) cos(t 10 )A R 140V 10A R 10 A jL R jL Same Example v(t) cos(t 40 )V V VR VL cos 50 jsin 50 L vL(t) Compare the real and imaginary part Use complex method I 1 10A V 140V Details will be discussed later VR R L jL VL R cos 50 L sin 50 ELEC 101 AC Circuits 4 Basic RCL Theory 2 a Study the relationship of the Magnitude and Phase between the current and voltage across the R, C, and L. A 2009/10 Spring Resistance R Vm Im R b V I Example for resistance v(t) and i(t) are in phase Find i(t). Consider i(t) Im cos(t I ) R v(t) Vm cos(t V ) From Ohm’s Law v(t) i(t)R Vm cos(t V ) Im cos(t I ) R Equate the magnitude and phase, Hence, Vm Im R V I i( t ) v(t) 10V 5A 2 i(t) v(t) 10cos(10t)V 10V cos(10t) 5Acos(10t) 2 ELEC 101 B AC Circuits 5 Capacitance C 2009/10 Spring Vm Im b V I 90 Consider i(t) Im cos(t I ) 1 C a C v(t) Vm cos(t V ) From Faraday’s Law v(t) 1 i(t)dt C cos(t 90) cos(t )cos(90) sin(t )sin(90) sin(t ) 1 I cos(t I )dt C m 1 sin(t I ) 1 cos(t I 90) Im Im C C for capacitance i(t) leads v(t) by 90º Find i(t) . Example i( t ) 50mF v(t ) 10cos(10t)V Vm cos(t V ) Equate the magnitude and phase, Hence, Vm Im 1 C V I 90 v(t) (10V 10rad / s 0.05F)cos(10t 90) 5Acos(10t 90) 10V 5A 90 i(t) (Vm C)cos(10t 90)A 0 ELEC 101 C AC Circuits 6 Inductance L a Consider i(t) Im cos(t I ) L v(t) Vm cos(t V ) From Faraday’s Law v(t) L di(t) dt Vm cos(t V ) L 2009/10 Spring di(t) dt Vm Im L b V I 90o Find i(t) . Example dI cos(t I ) L m dt LIm sin(t I ) i( t ) Hence, Vm Im L V I 90o 0.2H v(t) 10cos(10t)V Vm )cos(10t 90 )A L 10V )cos(10t 90 ) ( 10rad / s 0.2H 5Acos(10t 90 ) i(t) ( Im Lcos(t I 90) Equate the magnitude and phase, for inductance i(t) lags v(t) by 90o 10V v(t) 5A 0 90 ELEC 101 AC Circuits 7 Example + i(t) = 5cos10tA Find X and v(t) = 10cos10t V sketch i(t) vs v(t). _ Vm 10V 2 Im 5A Example + Find X and sketch i(t) vs v(t). i(t) v(t) and i(t) are in phase X R X 2009/10 Spring v(t) 5cost A 10 sint V _ i(t) = 5cost 90o 0o v(t) = 10sint i(t) leads v(t) by 90º X is a capacitive load X C Im 5A 0.5F Vm 10V 1rad / s X ELEC 101 3 AC Circuits 8 Complex Number Z B There are many ways to represent a complex number 2009/10 Spring Trigonometry form of Z: Z M(cos j sin ) Euler’s Identity: ej cos j sin A Rectangular form of Z: Z = a + jb a is real part of Z b is imaginary part of Z C (i is used for current in Electronics, hence j is used for imaginary number) j j 1 j 1 j Z = M e j M: magnitude of Z 1 j M Imaginary axis a2 b2 1 : phase of Z tan Z jb Exponential form of Z: b a M 0 a Real axis D Polar form of Z: Z M ELEC 101 4 AC Circuits 9 2009/10 Spring Complex Method Magnitude and phase change for L, C and R can be included as impedance (complex resistance) for resistance V Vm V R I I m I magnitude of R Vm I m R phase of R V I magnitude of 1/jC Vm I m phase of 1/jC V I 90 magnitude of jL Vm I m L phase of jL V I 90 for capacitance 1 V Vm V I I m I jC 1 90 C 1 C for inductance V Vm V jL I I m I L90 ELEC 101 5 AC Circuits 10 2009/10 Spring Summary a Transform i(t), v(t) into V, I phasors i (t ) I m cos(t I ) v(t ) Vm cos(t V ) I phasor I m I transform to V phasor Vm V b Transform L, C and R into impedance Z R R transform to C 1 jC transform to L jL transform to c Hence the sinusoidal circuit is transformed into a complex circuit d Apply DC circuit analysis e Transform V, I phasor back to v(t), i(t) ELEC 101 Example AC Circuits 11 b Find i(t). C = 0.2mF Transform L, C and R into impedance Z Hence L = 5mH i(t) v(t)=100cos1000t V 2009/10 Spring Transform into 10 1 1 jC j1krad/ s 0.2mF 0.2mF a Transform i(t), v(t) into V, I phasor Hence 10 10 1 5j j 0.2 1 j j Transform into 5mH v(t ) 100V cos (1000t 0) jL j1krad/ s 5mH j5 V phasor 100V0 V phasor 100 0 V 100(cos 0 jsin 0 )V 100(1 0)V 100V c Hence the sinusoidal circuit is transformed into a complex circuit 5j Since is a constant, can be excluded from the V (or I) phasor. 100V I 5j 10 ELEC 101 d AC Circuits 12 V 100V 1000V Z 2 j 526.6 100 1 26.6A 26.6 5 26.6 Apply DC circuit analysis I I Z 100V 5j 10 5j 10 10j (2 j) 10j 5j 5j j 2 (2 j) (2 j) Z 5j (5j) // 10 5j 20 j 10 j2 5j 4 j2 e I j 1 1 2 1 tan (1 / 2) 2 2 526.6 Transform V,I phasors back to v(t), i(t) 100 A 26.6 5 i(t) 2 20 j 10 4 1 5j 4 j 2 b 2 j a jb a2 b2 tan1 5j 2009/10 Spring a 100 cos (1000 t 26.6 ) A 5 include t in i(t) ELEC 101 6 AC Circuits 13 Phasor in cosine form Example a V1 (t ) 10 cos ( 260t ) V 4 V2 (t ) 20 sin ( 260t ) V 3 Transform V1(t) into V1 phasor V1 phasor 10 VO(t) =V1(t) + V2(t). Find VO(t). b 2009/10 Spring V 1045V 4 Transform V2(t) into V2 phasor The convention is to transform cos t to phasor. Hence transform V2(t) into cosine form first. sin cos(90o ) cos( 90o ) a Transform V1(t) into V1 phasor b Transform V2(t) into V2 phasor c VO phasor = V1 phasor + V2 phasor d Transform VO phasor into VO(t) V2 (t ) 20sin (260t )V 3 20cos (260t 90)V 3 20cos (260t 30)V Transform V2(t) into V2 phasor V2 phasor 20 30V ELEC 101 AC Circuits 14 VO phasor = V1 phasor + V2 phasor c All symbols can be used VO phasor 10 45 V 20 30 V 10(cos 45 jsin 45 )V 20(cos 30 jsin 30 )V 7.07 j7.07 17.32 j10 24.39 j2.93 a jb a2 b2 tan1 o 24.57 6.84 V d Transform VO phasor into VO(t) 2009/10 Spring + _ commonly used b a Voltage and current can be represented in difference forms. Example Time form i(t) 1 2 cos(250t)A L v(t ) + vL (t) 173 2 cos(250t 90)V _ + 173 H 250 R 100 VO phasor 24.57 6.84 V v R (t ) _ vR (t) 100 2 cos(250t)V v(t) 200 2 cos(250t 60)V VO (t ) 24.57 cos ( 260t 6.84) V + _ Complex form (Phasor form) I 10Arms + VL 17390Vrms _ j173 V 20060Vrms = 250 + 100 VR 1000Vrms _ ELEC 101 7 AC Circuits 15 2009/10 Spring Phasor Diagram To display the phase difference of components in a circuit To provide a nice graphical way of thinking of the solution ― phasor addition and subtraction Y phasor = Y1 phasor + Y2 phasor y(t) = y1(t) + y2(t) (t) (t) ELEC 101 AC Circuits 16 Example Find and plot all I and V in a phasor diagram. I 4 VR VL VO 12 2cos(377t 90)V 1.33mF VC Find Z 1 1 j2 jC j 377rad / s 1.33mF Find I I VO 12 290V 12 290V 345A Z 4 j4 4 245 Find all V + I= 345o A VR = I 4 = 1245o V 12 290o V _ _ jL j 377rad / s 15.9mH j6 Z R jXL jXC 1 R jL 4 j6 j2 4 j4 jC b c + 15.9mH a 2009/10 Spring d VL = I j6 = 18135oV VC = I 2j = 645o V Plot the phasor diagram 377rad / s VL 18V VO VR 12V I 45º VC 6V ELEC 101 AC Circuits 17 Impedance Diagram 8 Z jX z Z R Z and Y are called Immittance Z Z : Impedance R : Resistance X : Reactance Z : Impedance angle Same example with =337 rad/s. I 4 ZR=4 z : magnitude of Z 15.9mH 1 : Admittance Z 1 : Conductance G R 1 B : Susceptance X 1.33mF Z ZL= j6 ZC=j2 Y V VmV z( V I ) z Z R jX I I m I X z R2 X2 and Z tan1 R X = XL - XC 2009/10 Spring XL = L : Inductive Reactance XC = 1/C : Capacitive Reactance 377rad / s ZL= j6 Z = ZR + ZL + ZC = 4+j4 45º ZR=4 ZC=j2 ELEC 101 AC Circuits 18 Find the unknown elements in Z. Example 2009/10 Spring V 283150 V I 11.3140 A 2510 Z b Find Z c Z = a + jb , I lags V by 10º Hence Z is an inductive element (Z is R and L in series or in parallel) i(t) v(t) = 283 cos (800t + 150º) V v(t) i(t) = 11.3 cos (800t + 140º) A Z a Transform i(t), v(t) into V, I phasor b V Find Z I 150 283V i(t) v(t) 11.3A d a Z a jb Z is R and C in series or parallel 140 Z is R and L in series or parallel Z a jc c a=R b L c d 1 C Assume Z is R and L in series (can also assume in parallel) Z = 25 (cos 10º + j sin 10º) = 24.6 + j4.35 R + jL Transform v(t) and i(t) into V and I phasors V phasor 283150 V I phasor 11.3140 A e Equate real and imaginary parts R = 24.6 L = 4.35 / = 4.35/800rad/s = 5.44 mH ELEC 101 9 AC Circuits 19 Circuit with Two Frequencies Example VXi 330A Find VX(t). 1H 2j 2 2j 1 2j 2 2j 1 230V VXi (t) 2 cos(2t 30)V v(t) 6 cos (3t) V i(t) 3cos (2t 30) A VX(t) 2009/10 Spring b 2 0.25F i(t) 1 Kill i(t) and find VX(t) due to v(t) alone VXv 3j v(t) 2 1 j1.33 60 V Kill i(t) We can use superposition to separate the sources. Complex method can be used for the circuit with the same frequency. a Use V division 1 1 3j 2 1.33j 1 1 60V 60V 3 1.67j 3.4329.1 1.75 29.1V VXv 60V Kill v(t) and find VX(t) due to i(t) alone VXi 3 30 A 2j 1 Use I division 2 VXv (t ) 1.75 cos(3t 29.1)V -2j c Kill v(t) VX (t) VXi(t) VXv(t) 2cos(2t 30)V 1.75cos(3t 29.1)V ELEC 101 AC Circuits 20 2009/10 Spring VX Use source transform to find VX. Example 20j 5j V 20 0º V X 2jA -20j -5j jA -30V/ 20j = 1.5jA -5j -15V / -5j = - 3jA 20j // -20j = 20j(-20j) / (20j - 20j) = 15j 15j -20j 5j -5j VX 20V 2jA15j -20j = -30V VX jA -j = 5V 200º V = 20 (cos 0º + j sin 0º) = 20V 1.5jA -5j -3jA VX -5j 20j -30V VX 1.5j - 3j = -1.5jA -5j -20j 5-20 = -15V VX = -1.5jA -5j = -7.5 V ELEC 101 Example AC Circuits 21 100 0 V 1000 V 5 j 590 20 90 A I SC Use Norton’s Theorem to find V(t). 5mH 100 cos 1000t V 2009/10 Spring V(t) 0.2mF 10 c Find ZN Kill 100V j5 a Find Norton equivalent for the network -5j Z N 5 j / / 5 j 5 j ( 5 j) 5j 5j V(t) network ISC ZN d Replace network with Norton equivalent 10 V load 20-90ºA b Find ISC 100 0º V 10 jL j 1krad / s 5mH j5 V = 20-90º A 10 = 200 -90ºV j5 -5j ISC replace the 1 1 5 j load by a jC j 1krad / s 0.2mF short e Convert phasor V into v(t) V(t) = 200 cos (1000t - 90º) V ELEC 101 Example AC Circuits 22 Find mesh currents I1 and I2. 3 10 cos 1000t V re-arrange (3 4 j ) I1 j4 I 2 10V 0.5mF I1 4 mH I2 ( 2 j4 ) I1 j2 I 2 0 2I1 c a Use complex method 1 1 jC j 1krad / s 0.5mF j2 100o V I1 -j2 I2 I1 14 j8 1.2429.7A 13 20 j30 2.7756.3A 13 2I1 j4 b Solve for I1 and I2 phasors I 2 jL j 1krad / s 4mH j4 3 2009/10 Spring Apply KVL to meshes 10V 3 I1 (I1 I 2 ) j4 0 j4 (I 2 I1 ) ( j2 )I 2 2I1 0 d Convert I1 and I2 phasors into I1(t) and I2(t) I1(t) 1.24 cos (1000t 29.7) A I2 (t) 2.77 cos (1000t 56.3) A ELEC 101 Example AC Circuits 23 Find Thevenin & Norton equivalent. j4 VX I 2 a Ohm’s law for 2: 2009/10 Spring -j2 -4V I c Find ISC VX VX I 2 -j2 ISC VX ' VOC j4 -VX' -4V 2 Apply KVL VX ' I j4 ( 4V) VX ' 0 (I ISC ) 2 ( 4V) ISC j2 0 VX ' 2 (I ISC ) I SC 1A j2A Apply KVL to find I VX I j4 ( 4V) VX 0 I 2 I j4 4V I 2 0 I 4V 1V 1 A 4 j4 1 j 2 45 d Find ZTH Z VOC 3 j 1 j TH ISC 1 j2 Thevenin equivalent ZTH = 1 - j b Find VOC VOC 2 I 4V 2 45 4V 3 jV 3.1618.4V VOC = - 3V - j V Norton equivalent ZTH = 1 - j ISC = - 1A - j2 A ...
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