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Unformatted text preview: REVIEW – Phasors Page 1 THINGS YOU SHOULD KNOW ABOUT PHASORS (Lectures 3136) 1. How to transform a circuit to the phasor domain, i.e., convert sources and computer impedances. 2. How to analyze basic circuits using KCL and KVL 3. How to use NodeVoltage and MeshCurrent Methods 4. How to compute the Thevenin equivalent of a complex circuit. PAGE TRANSFORMATIONS AND KCL CALCULATIONS 23 PHASESHIFT CALCULATIONS 45 KCL & KVL 610 NODE VOLTAGE & MESH CURRENT 1112 THEVENIN & NORTON EQUIVALENT 1315 REVIEW – Phasors Page 2 BASIC PHASOR CIRCUIT ANALYSIS v o sin(4000t) 10cost(4000t) 20 200 2uf +  + First, identify the frequency of operation. In this case, w=4000 radians/second. This is an angular frequency. The corresponding linear frequency is given by f = ! 2 " = 4000 2 " = 637 Hz (cycles/second) Second, transform all passive components to phasors. 20 !" 20 ! for the first resistor 200 !" 200 ! for the second resistor 2 μ f ! 1 j " C = 1 j 4000 ( ) 2 # 10 $ 6 ( ) = $ j 125 % for the 2 µ f capacitor Third, transform all sources to phasors. 10cos 4000 t ( ) ! 10 " ° = 10 for the voltage source 1sin 4000 t ( ) = cos 4000 t ! 90 ° ( ) " 1 #! 90 ° = ! j REVIEW – Phasors Page 3 Redraw the circuit as a resistive circuit (with complex resistors and sources). This circuit can now be analyzed in many ways. v oj 10 20 200j125 +  + Consider doing KCL with v 1 being at the top of the current source. The KCL equation is then + ! j ( ) ! ˆ v 1 20 " # $ % & ’ ! ˆ v 1 ! 10 ( ) ! j 125 + 200 " # $ % & ’ = This is a single equation which can be solved to give ˆ v 1 = 0.719 ! j 19.55 We can then calculate the current through the 200 Ω resistor to be ˆ i 200 = ˆ v 1 + 10 200 ! j 125 = 0.719 ! j 19.55 ( ) + 10 200 ! j 125 = 0.082 ! j 0.046 Using Ohm’s Law we get the output voltage as ˆ v o = 0.082 ! j 0.046 ( ) 200 ( ) = 16.5 ! j 9.2 = 18.9 "! 29.3 ° We inverse phasor transform to get the output voltage as a function of time. v o t ( ) = 18.9cos 4000 t ! 29.3 ° ( ) REVIEW – Phasors Page 4 BASIC PHASOR CIRCUIT ANALYSIS (HW11 S07) (a) The frequency of the source voltage in the circuit shown above is adjusted until i g is in phase with v g . What is the value of ! in radians per second? SOLUTION As is usually with problems of this type we convert the problem to phasors. The equivalent impedance seen by the voltage source is Z eq = 500 + 1 j ! C + 1000 ! j ! L = 500 + 1 j ! (10 ! 6 ) + 1000( j ! 0.5) 1000 + j ! 0.5 Z eq =500 10 6 ! j+ 1000 ! 2 +2 " 10 6 j ! 2000 2 + ! 2 For ˆ i g to be in phase with ˆ v g the imaginary components of Z eq need to sum to zero which requires that ! 10 6 ! + 2 " 10 6 ! 2000 2 + ! 2 = . Simple algebra gives ! = 2000 rad / s for this to be true. REVIEW – Phasors Page 5 (b) If v g = 20cos !...
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 Fall '09
 Impedance, Volt, Thévenin's theorem, Norton's theorem, PHASOR CIRCUIT ANALYSIS, BASIC PHASOR CIRCUIT

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