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**Unformatted text preview: **REVIEW – Phasors Page 1 THINGS YOU SHOULD KNOW ABOUT PHASORS (Lectures 31-36) 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 Node-Voltage and Mesh-Current Methods 4. How to compute the Thevenin equivalent of a complex circuit. PAGE TRANSFORMATIONS AND KCL CALCULATIONS 2-3 PHASE-SHIFT CALCULATIONS 4-5 KCL & KVL 6-10 NODE VOLTAGE & MESH CURRENT 11-12 THEVENIN & NORTON EQUIVALENT 13-15 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 o-j 10 20 200-j125 + - +- 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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