03-Basic%20Laws%20and%20Nodal%20Analysis

# 03-Basic%20Laws%20and%20Nodal%20Analysis - Lecture#3...

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Unformatted text preview: Lecture #3 OUTLINE • Resistors in series – equivalent resistance – voltage-divider circuit • Resistors in parallel – equivalent resistance – current-divider circuit Examples • Nodal Analysis Reading Sections 2.1-2.6, 3.1-3.3 KVL Example 2.6 0 0 12 4 2 4 0 i v v ! + + ! ! = v = ! 6 i ! 16 + 10 i ! 12 i = " i = ! 8 A v = KVL Example 2.6 0 0 12 4 2 4 0 i v v ! + + ! ! = v = ! 6 i ! 16 + 10 i ! 12 i = " i = ! 8 A v = 48 V Resistor Networks – Resistors In Series Equivalent resistance (series) is sum of resistances N n eq R R R R R + + + = = ! ... 2 1 2 1 R R R eq + = v R R R v 2 1 1 1 + = v R R R v 2 1 2 2 + = Voltage divided over resistors (voltage divider) v R R v eq n n = Voltage Divider + – V 1 + – V 2 R 2 R 1 V I − + = + 1 1 1 2 R V V R R = + 2 2 1 2 R V V R R The voltage across two series resistors divides in proportion to their resistances. Voltage divider application: potentiometer • Used, e.g. in volume control SS 4 3 2 1 2 2 V R R R R R V ⋅ + + + = Correct, if nothing else is connected to nodes because R 5 removes condition of resistors in series SS 4 3 2 1 2 2 V R R R R R V ⋅ + + + ≠ When can the Voltage Divider Formula be Used? + – V 2 R 2 R 1 V SS I R 3 R 4 − + R 2 R 1 V SS I R 3 R 4 − + R 5 + – V 2 Resistor Networks – Resistors In Parallel R eq (parallel) is inverted sum of resistances N n eq R R R R R 1 ... 1 1 1 1 2 1 + + + = = ! 2 1 2 1 2 1 1 1 1 R R R R R R R eq + = + = Current divided over resistors (current divider) n eq n n R R i R v i = = 1 1 R v i = 2 2 R v i = 2 1 i i i + = Current Divider R 2 R 1 I I 2 I 1 = + 2 1 1 2 R I I R R = + 1 2 1 2 R I I R R The current to two parallel resistors divides in inverse proportion to their resistances. Series and Parallel Resistors – Light Bulbs Which is the annoying configuration of some holiday lights? Nodal Analysis Circuits can be complex, where do we start? Examine a powerful technique for systematic analysis using KCL +- +- 2 V 9 A 2 +- v y )v y (2 1 1 (3 i x i x 2 +- a b V ab V1 V3 V4 V2 Objective: find all current flows or all voltage drops 1. Choose a reference node (“ground”) Look for the one with the most connections!Look for the one with the most connections!...
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## This note was uploaded on 05/15/2008 for the course EECS 215 taught by Professor Phillips during the Spring '08 term at University of Michigan.

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03-Basic%20Laws%20and%20Nodal%20Analysis - Lecture#3...

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