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USC Phasor Method Aug 24, 2011 Outline • Review of analysis of DC (Direct Current) circuits • Analysis of AC (Alternating Current) circuits – Introduction – Challenge of analysis of AC circuits • Phasor method – Idea and concept – Advantage • Conclusions • Next… 2 Review of Analysis of DC circuits • DC circuits di u L Inductor: dt u0 Short du dt i0 Open Capacitor: i C u Resistor: i R L t •Pure Resistive + C - + 0 L + US u i R C US R 3 Review of Analysis of DC circuits • Complete solution for DC circuits Unknown variable: 6 Voltages (b) 12 (2b) 6 Currents (b) G Constraint Equations: As number of braches grows: 6 (b) •TooElements: many variables! KCL: 4-1=3 (n-1) •Too many equations! 6 (b) Network: + E – KVL: 6-3=3 b-(n-1) 12 (2b)=12 (2b) 4 Review of Analysis of DC circuits • Summary of DC circuits analysis methods – Circuit simplification • Equivalent transformation of resistors • Equivalent of sources The purposetransformation of circuit analysis method: – •To General analytical methods reduce the number of variables and • Node-voltage method (suitable for fewer nodes) equations • Mesh-current method (suitable of fewer meshs) – Theorem • Superposition (linear circuits) • Thevenin and Norton equivalent 5 Introduction of AC circuits • AC (Alternating current) u i • Why AC? steady state analysis Sinusoidal – Generation, transmission, distribution and consumption of electric energy are all in steady state sinusoidal. + 0 t – Any signal can be thought of as superposition of sinusoidal signals. f ( x) an sin(n n ) n 0 6 Introduction of AC circuits • Challenge with analysis of AC circuit u(t ) U s sin(t s ) L u L (t ) + u(t ) - + u L (t ) U L sin(t L ) + uC (t ) - C di u L Inductor: dt Capacitor: i C u Resistor: i R du dt uC (t ) operation U C sin(twith C ) +,-,*,/ trigonometric isRnot u R (t ) U function ) easy! R sin(t RThe KVL : u(t ) uL (t ) uC (t ) U S sin(t S ) U L sin(t L ) UC sin(t C ) P uC (t )iC (t ) U S sin(t S ) I S sin(t Si ) 7 Review of Analysis of DC circuits • Summary of DC circuits analysis methods – Circuit simplification • Equivalent transformation of resistors • Equivalent transformation of sources – General analytical methods • Node-voltage method (suitable for fewer nodes) • Mesh-current method (suitable of fewer meshs) – Theorem • Superposition (linear circuits) • Thevenin and Norton equivalent 8 Introduction of AC circuits 9 Phasor Method 10sin(t 300 ) 5 sin(t 600 ) 20sin(t 450 ) Hint: 10 Phasor Method Charles Proteus Steinmetz •In 1893, he introduced the phasor method to calculation of AC circuits GE required him to submit a itemized invoice. They soon received it. It included two items: 1. Marking chalk "X" on side of generator: $1. 2. Knowing where to mark chalk "X": $999. German-American mathematician and engineer (1865 – 1923) 11 Phasor Method Trigonometric function Phasor Domain U U sin(t ) transform 10300 10sin(t 30 ) 0 5 sin(t 600 ) Inverse transform 5600 12 Phasor Method Complex operation: Sum/Subtraction: (a1 jb1 ) (a2 jb2 ) (a1 a2 ) j (b1 b2 ) Multiplication/Division: F11 F22 F1 F21 2 ; F11 F2 2 F1 F2 1 2 13 Phasor Method Time Domain Phasor Domain transform Sinusoidal expression Phasor （Complex） Trigonometric calculation Complex Operation Inverse transform Result (sinusoidal) Result (Phasor) 14 Phasor Method Trigonometric calculation equivalent Complex Operation 10sin(t 300 ) 5 sin(t 600 ) 10sin t cos300 10cost sin 300 5 sin t cos600 5 cost sin 600 0 0 0 0 ( 10 sin 30 5 sin 60 ) cost (10cos30 5 cos60 ) sin t a sin t b cos t 0 0 a 10 cos 30 5 cos 60 R sin(t ) b 10sin 300 5 sin 600 15 Phasor Method Trigonometric calculation equivalent Complex Operation F F cos jF sin 10sin(t 30 ) 5 sin(t 60 ) 10300 5600 10cos300 j10sin 300 5 cos600 j5 sin 600 (10cos300 5 cos600 ) j (10sin 300 5 sin 600 ) a jb 0 0 a 10 cos 30 5 cos 60 R sin(t ) R b 10sin 300 5 sin 600 b 2 2 R a b ; arctan a 0 0 16 Phasor Method Example: 10sin(t 300 ) 5 sin(t 600 ) 20sin(t 450 ) 10300 5600 20 450 (8.66 j5) (2.5 j 4.33) (14.14 j14.14) 25.3 j 4.81 25.75 10.760 25.75sin(t 10.760 ) 17 Conclusions • The trigonometric function involved in the sinusoidal steady-state circuits is not convenient to calculation. • By projecting trigonometric function to phasor domain, the calculation can be dramatically simplified. 18 Quiz 1- problem1 Convert the following instantaneous currents to phasors, using cos(wt) as the reference. Give your answer in polar form. (1). 2). 19 20 Review of Analysis of DC circuits • Summary of DC circuits analysis methods – Circuit simplification • Equivalent transformation of resistors • Equivalent transformation of sources – General analytical methods • Node-voltage method (suitable for fewer nodes) • Mesh-current method (suitable of fewer meshs) – Theorem • Superposition (linear circuits) • Thevenin and Norton equivalent 21 Review of Analysis of DC circuits • Summary of DC circuits analysis methods – Circuit simplification • Equivalent transformation of resistors • Equivalent transformation of sources – General analytical methods • Node-voltage method (suitable for fewer nodes) • Mesh-current method (suitable of fewer meshs) – Theorem • Superposition (linear circuits) • Thevenin and Norton equivalent 22 •For the circuit shown below, compute the voltage across the load terminals. 0.1Ω j0.5Ω I=1250° A + + - 240 0 ° V LOAD LOAD - 0.1 j 0.5 0.5178.7 1250 * 0.5178.7 63.7578.7 2400 63.7578.7 240 12.5 j 62.5 227.5 j 62.5 235.93 15.36 23 USC Power Aug 24, 2011 Review of Phasor Questions: 1. What is the main idea of Phasor method? 2. 10sin(t 300 ) a.1030 0 b.10 60 0 c. 10 600 2 d. 10 300 2 25 Review of Phasor R L u L (t ) + u(t ) - + - - + u R (t ) C + - uC (t ) u(t ) U s sin(t s ) u L (t ) U L sin(t L ) uC (t ) U C sin(t C ) u R (t ) U R sin(t R ) 26 Power Instantaneous Power Average Power Real Power Active Power Reactive Power Complex Power Apparent Power 27 Power 28 Power: Pure Resistive 29 Power: Pure Inductive 30 Power: Pure Capacitive 31 Average Power 32 Example 2.1 33 Complex Power 34 Power Triangle 35 Power Triangle 36