mesh_analysis_problem6

mesh_analysis_problem6 - Problem 2 Consider the circuit...

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Unformatted text preview: Problem 2 Consider the circuit below. a) Use mesh analysis to find i. b) Find the same quantity using nodal analysis. c) Which method was easier? Why? 5 Solution: a) Redraw the circuit making sure to label all mesh currents and voltage polarities. By observation we have the following: i = IC - IB i1 = IA - IB i2 = IA - IC IA = 0.2 mA Writing KVL around each mesh, we obtain the following system of equations. Note that knowledge of IA reduces the problem to a system of two equations and two unknowns. 50000(IC - IB ) + 50000(IA - IB ) - 100000(IB ) = 0 -50000(IC - IB ) - 100000(IC ) + 75000(IA - IC ) = 0 Substitute IA = 0.2 mA, collect terms, and write in matrix form. 6 -200000IB + 50000IC = -50000(.0002) 50000IB - 225000IC = -75000(.0002) 4 -1 IB .0002 = 2 9 IC .0006 IB 70.59 A = IC 82.35 A From our above observations we have: i = IC - IB = 82.35 A - 70.59 A = 11.77 A b) To solve using nodal analysis, redraw the circuit labeling nodes and voltage polarities. Writing KVL around each mesh, we obtain the following system of equations. VA - VC VA - VB + 50000 100000 VB VB - VC VB - VA + + 75000 50000 50000 VC VC - VB VC - VA + + 100000 50000 100000 7 = .0002 = 0 = 0 Collect terms, and write in matrix form. 3VA - 2VB - VC -3VA + 8VB - 3VC = 20 = 0 -VA - 2VB + 4VC = 0 3 -2 -1 VA 20 -3 8 -3 VB = 0 -1 -2 4 VC 0 15.2941 V VA VB = 8.8235 V 8.2353 V VC We can now calculate i. VB - VC 50000 = 11.77 A i = c) For this circuit mesh analysis was easier because the current source eliminated an unknown mesh current. 8 ...
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