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Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Practice Examination Questions With Solutions Module 3 – Problem 6 Filename: PEQWS_Mod03_Prob06.doc Note: Units in problem are enclosed in square brackets. Time Allowed: 30 Minutes Problem Statement: Use the mesh-current method to write a complete set of independent equations that could be used to solve this circuit. Do not attempt to solve the equations. Do not attempt to simplify the circuit. iX R1 = 1[W] + R3 = 7[W] vS1= 3[V] R4 = 10[W] iS1= 6[S]vX R2 = 2[W] R6 = 8[W] R5 = 4[W] R7 = 9[W] - iS2= 5iX R8 = 12[W] iS3= 11[A] vX R10= 14[W] vS2= 15[V] R11= 16[W] + + Page 3.6.1 R9 = 13[W] Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Problem Solution: The problem statement was: Use the mesh-current method to write a complete set of independent equations that could be used to solve this circuit. Do not attempt to solve the equations. Do not attempt to simplify the circuit. iX R1 = 1[W] + R3 = 7[W] vS1= 3[V] R4 = 10[W] iS1= 6[S]vX R2 = 2[W] R6 = 8[W] R5 = 4[W] R7 = 9[W] - iS2= 5iX R8 = 12[W] iS3= 11[A] vX R10= 14[W] vS2= 15[V] R9 = 13[W] R11= 16[W] + + - The first step in the solution is to identify the meshes, and define the mesh currents. This is done in the circuit schematic that follows. The mesh currents are marked in red. Page 3.6.2 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. iX R1 = 1[W] iA + R3 = 7[W] iB vS1= 3[V] iC iE iS1= 6[S]vX R2 = 2[W] iD R6 = 8[W] R5 = 4[W] R4 = 10[W] R7 = 9[W] - iF iS2= 5iX vX iS3= 11[A] R10= 14[W] iG R8 = 12[W] vS2= 15[V] iH R9 = 13[W] R11= 16[W] + + - Now we need to write the Mesh-Current Method equations. There will be eight equations plus two more for the dependent source variables iX and vX. We will take them alphabetically. For mesh A, we have a fairly straightforward application of KVL. We write A: iA R1 vS1 iA R2 0. For mesh B, we have a similar case to that we had in mesh A, and we can just write B: vS1 (iB iC ) R3 (iB iF ) R5 0. Mesh C is a supermesh case, with the current source iS1 as a part of the C mesh and the D mesh. The supermesh is drawn using a dashed black line in the circuit diagram that follows. Page 3.6.3 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. iX R1 = 1[W] iA + R3 = 7[W] iB vS1= 3[V] iC iE iS1= 6[S]vX R2 = 2[W] iD R6 = 8[W] R5 = 4[W] R4 = 10[W] R7 = 9[W] - iF iS2= 5iX vX iS3= 11[A] R10= 14[W] iG R8 = 12[W] vS2= 15[V] iH R9 = 13[W] R11= 16[W] + + - We have the supermesh equation, C+D: (iC iB ) R3 (iD iH ) R7 (iC iG ) R6 0, and we have the constraint equation, C+D: iD iC iS1. Now, for mesh E, there is a fairly simple equation, E: iE R4 0. Next, we have the F mesh. We have two current sources in this mesh; one is a part of two meshes and one is a part of only this mesh. The current source that is a part of only this mesh, iS2, determines the mesh current, and we can write F: iF iS 2 . Now, for mesh G, we can write the constraint equation given by the value of the iS3 current source, F+G: iG iF iS 3 . The H mesh is fairly straightforward. We write H: (iH iG ) R8 (iH iD ) R7 iH R9 iH R11 0. Now, we have to write equations for the dependent sources. The current iX is equal to the current in the B mesh, and we can write iX : iB iX . Page 3.6.4 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Finally, we need to write an equation for the voltage, vX. We can pick a closed loop and write a KVL equation. Generally, we want to avoid closed loops that include current sources, since we don’t know the voltage across a current source. We pick a closed loop that is shown in the following circuit with a red dashed line. iX R1 = 1[W] iA + R3 = 7[W] iB vS1= 3[V] iC iE iS1= 6[S]vX R2 = 2[W] iD R6 = 8[W] R5 = 4[W] R4 = 10[W] R7 = 9[W] - iF iS2= 5iX vX iS3= 11[A] R10= 14[W] iG R8 = 12[W] vS2= 15[V] iH R9 = 13[W] R11= 16[W] + + - We can write v X : v X (iG iC ) R6 (iG iH ) R8 vS 2 iF R10 0. This is 10 equations in 10 unknowns, and completes the solution that was requested. Note 1: For clarity in showing this solution, and how it unfolds, we have redrawn the circuit a couple of times. In solving this problem on an examination, we would not redraw each time, but rather make marks on the original circuit. In addition, we would not include all of the text that is present here. With this, it should be possible to complete the problem in the allotted time. Note 2: Some students have difficulty trying to determine whether their solution was a valid one, particularly if they have taken a slightly different approach such as picking different directions for the mesh currents. While it is not requested in this problem, a numerical solution for iX and vX is given here. If you are in doubt about Page 3.6.5 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. the validity of your solution, solve for these quantities, and compare with this solution. If your solution is significantly different, then something must be wrong. Our equations were: A: iA R1 vS 1 iA R2 0 B: vS 1 (iB iC ) R3 (iB iF ) R5 0 C+D: (iC iB ) R3 (iD iH ) R7 (iC iG ) R6 0 C+D: iD iC iS 1 E: iE R4 0 F: iF iS 2 F+G: iG iF iS 3 H: (iH iG ) R8 (iH iD ) R7 iH R9 iH R11 0 iX : iB iX v X : v X (iG iC ) R6 (iG iH ) R8 vS 2 iF R10 0 Now, we are going to substitute in the values that were given in the circuit. We get the following system of equations: A: iA1[W] 3[V] iA 2[W] 0 B: 3[V] (iB iC )7[W] (iB iF )4[W] 0 C+D: (iC iB )7[W] (iD iH )9[W] (iC iG )8[W] 0 C+D: iD iC 6[S]v X E: iE 10[W] 0 F: iF 5iX F+G: iG iF 11[A] H: (iH iG )12[W] (iH iD )9[W] iH 13[W] iH 16[W] 0 iX : iB iX v X : v X (iG iC )8[W] (iG iH )12[W] 15[V] iF 14[W] 0 Now, we will substitute these equations into MathCAD. The solution is given in a MathCAD file called PEQWS_Mod03_Prob06_Soln.mcd. The results are given here: Page 3.6.6 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. iA = -1[A] iB = -1.20387[A] iC = 1.11926[A] iD = 3.44002[A] iE = 0[A] iF = -6.01935[A] iG = 4.98065[A] iH = 1.81456[A] iX = -1.20387[A] vX = 0.38679[V] While the mesh current values depend on how you define these variables, the dependent source variables, iX and vX, should be the same with any approach. You can use these answers to check your work. Problem adapted from ECE 2300, Final Exam, Problem 1, Summer 1998, Department of Electrical and Computer Engineering, Cullen College of Engineering, University of Houston. Page 3.6.7