EE40 Lecture 15 Venkat Anantharam 2 29 08 Reading Chap 5 phasors EE40 Spring 2008 Slide 1 Venkat Anantharam Impedance AC steady state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm s law V IZ Z is called impedance EE40 Spring 2008 Slide 2 Venkat Anantharam Some Thoughts on Impedance Impedance depends on the frequency Impedance is often a complex number Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state EE40 Spring 2008 Slide 3 Venkat Anantharam Resistor I V relationship vR iRR VR IRR where R is the resistance VR phasor voltage IR phasor current hence ZR R Capacitor I V relationship iC CdvC dt Phasor current IC phasor voltage VC capacitive impedance ZC IC VC ZC where ZC 1 j C Inductor I V relationship vL LdiL dt Phasor voltage VL phasor current IL inductive impedance ZL VL ILZL where ZL j L EE40 Spring 2008 Slide 4 Venkat Anantharam Example Single Loop Circuit 10V 0 20k 1 F VC f 60 Hz VC How do we find VC First compute impedances for resistor and capacitor ZR R 20k 20k 0 ZC 1 j 2 f x 1 F 2 65k 90 EE40 Spring 2008 Slide 5 Venkat Anantharam Impedance Example 20k 0 10V 0 VC 2 65k 90 Now use the voltage divider to find VC 2 65k 90 VC 10V 0 2 65k 90 20k 0 VC 1 31V 82 4 EE40 Spring 2008 Slide 6 Venkat Anantharam What happens when changes 10V 0 20k 1 F VC 10 Find VC EE40 Spring 2008 Slide 7 Venkat Anantharam Circuit Analysis Using Complex Impedances Suitable for AC steady state KVL v1 t v2 t v3 t 0 V1 cos t 1 V2 cos t 2 V3 cos t 3 0 Re V1e j t 1 V2 e j t 2 V3e j t 3 0 Phasor Form KVL V1e j 1 V2 e j 2 V3e j 3 0 V1 V2 V3 0 I1 I 2 I 3 0 Phasor Form KCL Use complex impedances for inductors and capacitors and follow same analysis as in chap 2 EE40 Spring 2008 Slide 8 Venkat Anantharam Steady State AC Analysis 5mA 0 0 1 F 1k V Find v t for 2 3000 5mA 0 V 1k EE40 Spring 2008 Slide 9 j530k Venkat Anantharam Find the Equivalent Impedance 5mA 0 Zeq V Z eq 1000 j 530 10 3 0 530 90 1000 j 530 1132 27 9 Z eq 468 2 62 1 V IZ eq 5mA 0 468 2 62 1 V 2 34V 62 1 v t 2 34V cos 2 3000t 62 1 EE40 Spring 2008 Slide 10 Venkat Anantharam Change the Frequency 0 1 F 1k 5mA 0 V Find v t for 2 455000 j3 5 5mA 0 V 1k EE40 Spring 2008 Slide 11 Venkat Anantharam Find an Equivalent Impedance 5mA 0 Zeq V Z eq 1000 j 3 5 10 3 0 3 5 90 1000 j 3 5 1000 0 2 Z eq 3 5 89 8 V IZ eq 5mA 0 3 5 89 8 V 17 5mV 89 8 v t 17 5mV cos 2 455000t 89 8 EE40 Spring 2008 Slide 12 Venkat Anantharam Series Impedance Z1 Z2 Zeq Z3 Zeq Z1 Z2 Z3 For example L1 C1 L2 1 1 Z eq j C1 j C2 Zeq j L1 L2 EE40 Spring 2008 C2 Slide 13 Venkat Anantharam Parallel Impedance Z1 Z2 Z3 Zeq 1 Zeq 1 Z1 1 Z2 1 Z3 For example L1 L2 C1 1 Z eq j C1 C2 L1 L2 Z eq j L1 L2 EE40 Spring 2008 C2 Slide 14 Venkat Anantharam
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