quiz08sp01_sols

quiz08sp01_sols - Solutions Part ii -1.6264 1.2088 4.3077...

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Unformatted text preview: Solutions Part ii -1.6264 1.2088 4.3077 -0.4176 Solving nodes with MATLAB V1: -2.1176 V2: -2.3529 V3: 2.5882 V4: -4.4706 2kΩ V3 3kΩ I1 V2 4mA 5kΩ MESH ANALYSIS Mesh 1 : I1 = 4mA I2 Mesh 2 : 2kI 2 − 5ki0 + 1k ( I1 − I 3 ) + 3k ( I 2 − I1 ) = 0 (5k )iO 1kΩ V4 V1 +− iO +− 20V Mesh 3 : 1k ( I 3 − I 2 ) + 5ki0 + 4kI 3 − 20 + 5k ( I 3 − I1 ) = 0 4kΩ I3 Controlling Variable : i0 = I1 − I 3 8 i0 = − = −0.47059mA 17 iV NODE ANALYSIS Node Constraint : V4 − V1 = 5ki0 (= V2 ) V1 + 20 V4 − V2 V1 − V3 + + =0 4k 1k 2k V2 V2 − V4 V2 − V3 Node V2 : + + =0 5k 1k 3k V − V2 V3 − V1 Node V3 : 3 + − 4mA = 0 3k 2k V Controlling Variable : i0 = 2 5ki0 = V2 5k Super - node : SOME ADDITIONAL METHODS - USE LOOP CURRENTS - SOURCE TRANSFORM THE VOLTAGE SOURCE - SOURCE SUPERPOSITION - THEVENIN TO COMPUTE CURRENT THROUGH VOLTAGE SOURCE iO + iV = 4mA The circuit cannot be broken at the 5k!! USING SOURCE TRANSFORMATION V3 3kΩ V2 4mA 5kΩ 2kΩ (5k )iO 1kΩ V4 V1 +− iO 4kΩ 5mA USING SOURCE SUPERPOSITION 2kΩ V3 I2 3kΩ I1 V2 4mA Mesh 2 : (5k )iO 1kΩ V4 V1 +− ' iO 5kΩ MESH ANALYSIS Mesh 1 : I1 = 4mA 4kΩ I3 2kI 2 − 5ki0 + 1k ( I 2 − I 3 ) + 3k ( I 2 − I1 ) = 0 Mesh 3 : 1k ( I 3 − I 2 ) + 5ki0 + 4kI 3 + 5k ( I 3 − I1 ) = 0 Controlling Variable : 52 ' i0 = I1 − I 3 = 17 V3 3kΩ V2 5kΩ MESH ANALYSIS Mesh 2 : 2kΩ I2 (5k )iO 1kΩ V4 V1 +− " iO I3 +− 20V 4kΩ " 2kI 2 − 5ki0 + 1k ( I 2 − I 3 ) + 3k ( I 2 ) = 0 Mesh 3 : 1k ( I 3 − I 2 ) + 5ki0 + 4kI 3 + 5k ( I 3 ) − 20 = 0 Controlling Variable : " i0 = − I 3 = − 60 17 ' " i0 = i0 + i0 = 52 60 8 − = − = −0.47059 17 17 17 USING THEVENIN TO COMPUTE CURRENT THROUGH THE VOLTAGE SOURCE V3 3kΩ I1 V2 4mA 5kΩ 2kΩ Mesh 1 : I1 = 4mA Mesh 2 : ' 2kI 2 − 5ki0 + 1k ( I 2 ) + 3k ( I 2 − I1 ) = 0 I2 (5k )iO 1kΩ V4 V1 +− ' i0 MESH ANALYSIS Controlling Variable : ' i0 = I1 4kΩ − vOC + V3 4mA 2kΩ ' ' vOC = −5ki0 + 1kI 2 + 5ki0 = 1kI 2 vOC = 16 (V ) 3 THIS IS THE SAME CIRCUIT ANALYZED IN SOURCE SUPERPOSITION! 52 i SC + = 4mA 17 3kΩ ' 1kΩ V4(5k )i0 V1 i = 16 mA V2 +− SC ' 17 16 i0 VTH = vOC = (V ) 5kΩ 4kΩ 3 i SC v 17 RTH = OC = k i SC 3 The circuit cannot be broken at the 5k!! RTH iV VTH + - +- 20V iV + iO = 4mA iV = 20 + VTH 76 / 3 76 = = 17 mA RTH 17 / 3 ...
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This note was uploaded on 12/01/2011 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.

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