EE201 Lecture 32 P 1 Phasor Impedance Admittance Definitions Impedance Admittance ZR j R Resistors YR j 1 R ZL j j L Inductors YL j 1 j L ZC j 1 j C Capacitors 1 Admittance Impedance mho 1 ohm I Y j V YC j j C EE201 Lecture 32 P 2 For any 2 terminal device with input voltage and current Vin and I in respectively Zin j V in I in Yin j I in V in Impedance and admittance are complex quantities that can be separated into real and imaginary parts Z j Re Z j Resistance Im Z j Reactance Y j Re Y j Conductance Im Y j Susceptance EE201 Lecture 32 P 3 Equivalent impedance for 2 terminal devices in series I in From KVL Vin V1 V2 I1 V1 Z1 j Vin V2 Z2 j I2 Z in j I2 V2 Z2 j Iin I1 I2 Vin Z1 j Z2 j Iin Zin j Z1 j Z2 j Voltage division Vi j V1 Z1 j I1 Impedances add in series Zi j Vin Z1 j Z2 j Zi j V in Zeq j EE201 Lecture 32 P 4 Equivalent admittance for 2 terminal devices in parallel I in Yin j I1 I2 V in Y1 j V1 Y2 j V2 Vin V1 V2 Iin I1 I2 Yin j Iin Vin I1 I 2 Vin I1 V1 Yin j Y1 j Y2 j For parallel configuration of elements admittances add I2 V2 EE201 Lecture 32 P 5 Current division formula Ii Ii Yi j Yeq j Yi j Yk j Iin Iin k Impedances are manipulated in same way as resistances Admittances are manipulated in same way as conductances EE201 Lecture 32 P 6 Example Compute Zeq for the series circuit shown below 200 rad sec 4mH Z1 j 1 2 2mF Z2 j Z3 j Z1 j j L j 200 0 004 j0 8 Z2 j R 1 2 Z3 j 1 j C j c j2 5 Zeq j Z1 Z2 Z3 Zeq j 1 2 j1 7 EE201 Lecture 32 P 7 Example Find value of L to yield Zin j 0 25 j0 25 when 4rad sec L 0 5 0 5F Zin j Step 1 Find Z11 for parallel elements 1 1 Z11 1 R j C 1 0 5 j2 Z11 2 j2 8 1 2 j2 EE201 Lecture 32 Step 2 Add ZL and solve for L Zin j ZL j Z11 j Zin j j L 0 25 j0 25 Grouping j terms j 4L 0 25 j 0 25 L 0 125 H P 8
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