Kvl mesh 1 0017 ω x i1 3000 ω x i1 i3 12v 0 kvl

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KVL Mesh 1: .0017 x i1 + 3000 x (i1 - i3) + 12V= 0 KVL Mesh 2: 10000 x i3 + 3000 x (i3 - i1) = 0 We have two (2) equations that depend only upon
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Mesh 1: .0017 x i1 + 3000 x (i1 - i3) + 12V= 0 Mesh 2: 10000 x i3 + 3000 x (i3 - i1) = 0 3000.0017 x i1 -3000 x i3 = -12V -3000 x i1 +13000 x i3 = 0 13x 3000.0017 x i1 - 13x 3000 x i3 = -12V x 13 3x (-3000 Ω29 x i1 + 3x 13000 x i3 = 0 x3 30000.0221 x i1 +0 =-156V i1= -156V/30000.0221 = - .005199996A; i3 = - . 001199999A Vab + .0017 x i1 +12V = 0 .0017 x i1 =-8.83999 microVolts Vab = -11.99999116 Volts Simple follow-on calculations, based on having obtained the solution for the mesh currents i1-i3= -3.999997mA Vab =3000 x -3.999997mA =- 11.999991V
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Example: Solving a problem using the mesh current method i1 node b 12 volts (voltage source) i3 node a + - 3000 0.0017 There are two (2) meshes. Label and name the mesh currents. Apply Kirchhoff’s voltage law in each mesh . KVL Mesh 1: .0017 x i1 + 3000 x (i1 - i3) + 12V= 0 i3 = 3A We have two (2) equations that depend only upon 3 amperes
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Example: Solving a problem using the mesh current method i1 node b 12 volts (voltage source) i3 node a + - 3000 0.0017 There are two (2) meshes. Label and name the mesh currents. Apply Kirchhoff’s voltage law in each mesh . KVL Mesh 1: .0017 x i1 + Vab + 12V= 0 KVL Mesh 2: 3000 x i3 + Vba = 0 i1-i3 = 3A = the branch current flowing through the current source We have three (3) equations that depend upon the two (2) mesh currents: i1 and i3, and upon Vab 3 amperes (current source)
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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.
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  • Fall '08
  • PIETRUCHA
  • Mesh Analysis, Kirchhoff's circuit laws, Vab

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