Branch 1 i1 va vc0017 ω branch 2 i2 va3000 ω

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Branch 1: i1 = (Va -Vc)/.0017 Branch 2: i2 = Va/3000 Branch 3: i3 = Va/10000 Branch 4: i4 = (Va- Vd)/500 Branch 4: -i4 = (Vd-Va)/500 Branch 5: i5 = (Vd- 0)/2000 KCL Node a : (Va-Vc)/0.0017 + Va/3000 + Va/10000 + (Va- Vd)/500 = 0; KCL Node d : (Vd-Va)/500 + Vd/2000 = 0 Node c : Vc= -12V . These three (3) equations depend only on the node i1 node d i5 500 2000 i4
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Solving a problem using the node voltage method 10000 i2 node b 12 volts (voltage source) i3 node a + - 3000 0.0017 node c Define the voltage at node b to equal 0 volts. Write an equation for the current in each separate branch that is leaving node a. Do the same for node d Branch 1: i1 = (Va -Vc)/.0017 Branch 2: i2 = Va/3000 Branch 3: i3 = Va /10000 Branch 4: i4 = (Va- Vd)/500 Branch 4: -i4 = (Vd-Va)/500 Branch 5: i = 10 x i2 KCL Node a : (Va-Vc)/0.0017 + Va/3000 + Va/10000 + (Va- Vd)/500 = 0; KCL Node d : (Vd-Va)/500 + 10 x i2 = (Vd-Va)/500 + 10 x Va/3000 = 0 i1 node d i = 10 x i2 500 i4
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  • Fall '08
  • PIETRUCHA
  • Mesh Analysis, Kirchhoff's circuit laws, Vab

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