8. ch4.8 t- 4.13 9_29_10 X

8. ch4.8 t- 4.13 9_29_10 X - BASICELECTRICCIRCUITS...

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Click to edit Master subtitle style  10/28/10 BASIC ELECTRIC CIRCUITS George Haines, Instructor Pima Community College Monday 9/25/2010 Chapters 4.8-4.
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 10/28/10 Requires  solving  5  + constraint = 6  equations Example 4.6 Node-Voltage vs the Mesh-Current Method Find the power dissipated in 300 Ω resistor Supernode:  2 + constraint equation = 3 2 + constraint equation = 3
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 10/28/10 With Supernode:  2 +  constraint equation = 3 @ the supernode: v1/100 + (v1 – v2)/250 + v3/200 +(v3-v2)/400 + (v3 – 128 -  v2)/500 +  (v3 + 256)/150 = 0 The supernode constraint equation: v3 = v1 – 50iΔ = v1 – v2  50/300 = 
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 10/28/10 2 + constraint  equation = 3 @va:  va/200 + (va – 256)/150 + (va – vb)/100 + (va –  vc)/300 = 0  @vc:   vc/400 + (vc + 128)/500 + (vc – vb)/250 + (vc – va)/ 300 = 0 Constraint equation: vb = 50iΔ = 50(vc – va)/300 = (vc – 
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 10/28/10 Example 4.7 Node-Voltage vs the Mesh-Current Method
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 10/28/10 Solve for ia and v0: 1.) 193 = 10ia + 10ib + 10ic + 0.8v , Θ 2.) Constraint Eqn: ib – ia = 0.4 vΔ = 0.4(2)ic = 0.8ic; 3.) v  = -7.5ib and 4.)  ic – ib = 0.5 Θ Using a  Supermesh
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 10/28/10 1.) (vo – 193)/10 – 0.4vΔ + (vo – va)/2.5 = 0 2.) (va – vo)/2.5 - 0.5 + [va - (vb + 0.8v )]/10 = 0 Θ 3.) vb/7.5 + 0.5 + (vb + 0.8v  – va)/10 = 0 Θ and the constraint equations 4.) v  = -vb Θ 5.) vΔ = {[va – (vb + 0.8v )]/10} Θ => 5 equations and: vo, vΔ, v , va, vb => 5  Θ Using Node-Voltage Analysis
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 10/28/10 4.9 Source Transformation Even though node-voltage and mesh-current 
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8. ch4.8 t- 4.13 9_29_10 X - BASICELECTRICCIRCUITS...

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