EE40_Fall08_Lecture4 - EE40 Lecture 4 Connie Chang-Hasnain...

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Slide 1 EE40 Fall 08 Connie Chang-Hasnain EE40 Lecture 4 Connie Chang-Hasnain 9/5/08 Reading: Chap. 2 Topics: Voltage Divider Current Divider Node-Voltage Analysis
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Slide 2 EE40 Fall 08 Connie Chang-Hasnain Consider a circuit with multiple resistors connected in series. Find their “equivalent resistance”. • KCL tells us that the same current ( I ) flows through every resistor • KVL tells us the sum of the voltage drops across the resistors equals V SS We conclude that the equivalent resistance of resistors in series is the sum of the individual resistances R 2 R 1 V SS I R 3 R 4 - + Resistors in Series
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Slide 3 EE40 Fall 08 Connie Chang-Hasnain I = V SS / ( R 1 + R 2 + R 3 + R 4 ) Voltage Divider + V 1 + V 3 R 2 R 1 V SS I R 3 R 4 - + SS 4 3 2 1 1 1 V R R R R R V + + + = SS 4 3 2 1 3 3 V R R R R R V + + + = Etc.
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EE40 Fall 08 Connie Chang-Hasnain SS 4 3 2 1 2 2 V R R R R R V + + + = Correct, if nothing else is connected to nodes Why? What is V 2 ? SS 4 3 2 1 2 2 V R R R R R V + + + When can the Voltage Divider Formula be Used? + V
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EE40_Fall08_Lecture4 - EE40 Lecture 4 Connie Chang-Hasnain...

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