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Lect03-Ch2 Pt1 - Resistive Circuits Part I Hambley’s Text...

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Unformatted text preview: Resistive Circuits Part I Hambley’s Text Chapter 2 1.Review & Analyze Series & Parallel Circuits 2. Voltage & Current Divider 3. Node Voltage Analysis 4. Mesh Current Analysis Resistances in Series ADD • common current: • KVL: • Equivalent: v = v 1 + v 2 + v 3 = iR 1 + iR 2 + iR 3 = i ( R 1 + R 2 + R 3 ) R eq = v i = ( R 1 + R 2 + R 3 ) i = i 1 = i 2 = i 3 ME-340 Circuits & Mechatronics Jan. 10, 2010 2 Resistances in Parallel ADD INVERSELY • common voltage: • KCL: • Equivalent R: i = i 1 + i 2 + i 3 = v 1 R 1 + v 2 R 2 + v 3 R 3 = v ( 1 R 1 + 1 R 2 + 1 R 3 ) v = v 1 = v 2 = v 3 R eq = v i = 1 1 R 1 + 1 R 2 + 1 R 3 ME-340 Circuits & Mechatronics Jan. 10, 2010 3 R eq = R 1 || R 2 || R 3 Two Resistances in Parallel 2 1 2 1 2 1 e 1 1 1 R R R R R R R q + = Ο Π Ξ Μ Ν Λ + Ο Π Ξ Μ Ν Λ = Series: add Parallel: add inversely ME-340 Circuits & Mechatronics Jan. 10, 2010 5 Ex 2.1 Find equivalent R: Ex 2.2- Find all currents & voltages: ME-340 Circuits & Mechatronics Jan. 10, 2010 6 i = v s R eq = 90 V 30 Ω = 3 A Equivalent Circuit i ME-340 Circuits & Mechatronics...
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Lect03-Ch2 Pt1 - Resistive Circuits Part I Hambley’s Text...

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