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mesh_analysis_problem5

# mesh_analysis_problem5 - Problem 1 Consider the circuit...

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Unformatted text preview: Problem 1 Consider the circuit below. a) Use mesh analysis to find the power dissipated in the 50 k resistor through which the current i flows. b) Find the same quantity using nodal analysis. c) Which method was easier? Why? 1 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 Writing KVL around each mesh, we obtain the following system of equations. Note that writing (1) around the outer mesh rather than mesh A results in a "cleaner" equation with fewer variables. 25 - 100000(IB ) - 100000(IC ) = 0 50000(IC - IB ) + 50000(IA - IB ) - 100000(IB ) = 0 -50000(IC - IB ) - 100000(IC ) + 75000(IA - IC ) = 0 Collect terms and write (1), (2), and (3) in matrix form. (1) (2) (3) 100000(IB ) + 100000(IC ) = 25 50000(IA ) - 200000(IB ) + 50000(IC ) = 0 75000(IA ) + 50000(IB ) - 225000(IC ) = 0 2 25 0 100000 100000 IA 50000 -200000 50000 IB = 0 0 75000 50000 -225000 IC Solving for IA , IB , and IC yields the following: IA 326.9 A IB = 115.4 A IC 134.6 A Finally, calculate the power dissipated in the 50k resistor. P50k = 50000i2 = 50000(IC - IB )2 = 50000(.0000192)2 = 18.4 W b) To solve using nodal analysis, redraw the circuit labeling nodes and voltage polarities. By observation we have VA = 25 V. Since we know VA , we need only write KCL at nodes B and C. Thus, the problem reduces to a system of two equations and two unknowns. 3 VB - VC VB - VA VB + + =0 75000 50000 50000 VC VC - VB VC - VA + + =0 100000 50000 100000 Substitute VA = 25, collect terms, and write in matrix form. 8VB - VC 3 -2VB + 4VC 8 -1 VB 3 -2 4 VC VB VC = 25 = 25 25 = 25 = 14.4231 V 13.4615 V Again, calculate the power dissipated in the 50k resistor. (VB - VC )2 50000 0.96162 = 50000 = 18.4 W P50k = c) In this case, nodal analysis is easier because the 25 V source eliminates an unknown node voltage. 4 ...
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