1 0 eq2 eq3 constraint1 2 2 2 2 constraint2 0

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Unformatted text preview: 2 2 Constraint# 2 : 0 1 Eq. 4 Eq. 5 Five equations, 5 unknowns => solve for Ia,Ib,Ic,Id, and Ie c) Nodal analysis is easier in this case since there are only two node equations to solve as compared to 5 equations in the mesh analysis technique. d) Superposition Voltage source : 10V Node a: 10 5 10 0 5 Node b: 5 , 10 20 2 20 5 5 10 4 5 0 , : 5 Voltage source : 5V Node a: 5 10 0 5 Node b: 5 , 5 5 2 20 20 5 5 10 4 5 0 Current source : 1A Mesh a: 10 5 20 0 Supermesh b & c: 20 10 0 Supermesh d & e: 5 5 0 Constraint# 1: 2 2 2 2 0 Constraint# 2 : 1 Solve for Ia, Ib, Ic, Id, and Ie. ,1 Therefore, , 10 ,5 ,1 e) Need to make sure that all the sources are either independent current sources or independent voltage sources. Let’s convert all the sources into independent currents sources using source transformation and perform nodal analysis by inspection. Conductance Matrix: 11 1 10 22 1 5 1 5 12 21 1 5 1 5 1 2 1 20 1 10 1 5 12 1 5 2 5 11 20 Source Vector: 1 1 1 0 2 2 1 0.5 3.5 1 0 = 2 3.5...
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This note was uploaded on 01/24/2014 for the course EE 100 taught by Professor Boser during the Summer '07 term at University of California, Berkeley.

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