{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz08sp01_sols

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}