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Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Practice Examination Questions With Solutions Module 4 – Problem 1 Filename: PEQWS_Mod04_Prob01.doc Note: Units in problem are enclosed in square brackets. Time Allowed: 30 Minutes Problem Statement: For the circuit shown, calculate the Thévenin equivalent resistance with respect to terminals a and b. vS2= iX R3 = 30[] a + R1 = 10[] iX + - R2 = 20[] iS1= 5iX vS1= 100[V] - R4 = 40[] b Page 4.1.1 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Problem Solution: For the circuit shown, calculate the Thévenin equivalent resistance with respect to terminals a and b. vS2= iX R3 = 30[] a + R1 = 10[] iX + - R2 = 20[] iS1= 5iX vS1= 100[V] - R4 = 40[] b The first step in the solution is to decide what items should be solved for. Note that the only thing we have been asked for is the Thévenin resistance. Specifically, we could solve for the open-circuit voltage and short-circuit current, and then take the ratio to get the solution. Another possibility is to find the Thévenin resistance directly. Here, and generally when we are asked for only the equivalent resistance, it is fastest to solve for it directly. This is true even when there are dependent sources present, as here. Therefore, we will set the independent sources equal to zero, and then apply a test source. We will choose a 1[V]-voltage source, since it looks as though that would allow us to find the solutions quickly. We have done so in the circuit that follows. Page 4.1.2 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. + iX vS2= iX R3 = 30[] R2 = 20[] a + R1 = 10[] iT vC iS1= 5iX + R4 = 40[] - vT= 1[V] b - Notice that in addition, we have defined a current iT that we will solve, to find the Thévenin resistance. Also, we have defined a reference node, and a node voltage vC, that we will use to find the solution. In fact, we will start by writing the KCL for node C, and the dependent source variable equation for iX. We have vC vC vC vS 2 vT iS 1 0, and R2 R1 R3 vC iX . R1 Now, we can substitute in values, vC v v 10[]iX 1[V] C C 5iX 0, and 20[] 10[] 30[] vC iX . 10[] We plug the second equation into the first, and get v vC 10[] C 1[V] v vC v 10[] C 5 C 0. 20[] 10[] 30[] 10[] Solving for vC, we get v vC v 1[V] C 5 C 0, or 20[] 10[] 30[] 10[] vC 650[mS] 33.3[mA], which means that vC 51.3[mV]. From this, it is clear that Page 4.1.3 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. vC 51.3[mV] 5.13[mA]. 10[] 10[] Now, we can solve for iT by applying KCL at the a node, and get vT vT vS 2 vC iT 0, which is R4 R3 iX 1[V] 1[V] 10[] 5.13[mA] 51.3[mV] iT . 40[] 30[] Thus we have that 1[V] 1[V] iT 58.3[mA]. 40[] 30[] Finally, we note that the equivalent resistance, which is the Thévenin resistance, is the ratio of vT/iT, and so we have v 1[V] REQ RTH T 17.1[]. iT 58.3[mA] Thus, our answer is RTH 17.1[]. We could have found the Thévenin voltage, which is the open-circuit voltage. Without showing the work here, we note that this would have been 57.1[V]. Then, we could have found the short-circuit current, which would have been 3.33[A]. We leave the work for this to the interested student. Taking the ratio of these, we would get the same answer as above. In our opinion, though, the approach that we took was quicker, and therefore better. Problem adapted from ECE 2300, Exam 2, Problem 1, Spring 2001, Department of Electrical and Computer Engineering, Cullen College of Engineering, University of Houston. Page 4.1.4