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Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Practice Examination Questions With Solutions Module 4 – Problem 6 Filename: PEQWS_Mod04_Prob06.doc Note: Units in problem are enclosed in square brackets. Time Allowed: 20 Minutes Problem Statement: Find the Thévenin equivalent of the circuit shown, as seen at terminals A and B. Draw your equivalent circuit, showing terminals A and B. iS1= 3iX iX A R1= 20[W] R2= 15[W] + - vS1= 50[V] R3= 10[W] R4= 45[W] B Page 4.6.1 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Problem Solution: Find the Thévenin equivalent of the circuit shown, as seen at terminals A and B. Draw your equivalent circuit, showing terminals A and B. iS1= 3iX iX A R1= 20[W] R2= 15[W] + - vS1= 50[V] R3= 10[W] R4= 45[W] B The first step in the solution is to find one of the three quantities, opencircuit voltage, short-circuit current, and equivalent resistance. In this case, it appears that finding the open-circuit voltage would be fairly simple. One reason for this is that the resistor R2 will have no current through it, and thus will have no effect, and can be removed. Perhaps more significant is that this will also set the current iX equal to zero, which will remove the dependent source. Let’s define the open-circuit voltage, and then solve for it. The voltage is defined in the circuit that follows. Page 4.6.2 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. iS1= 3iX iX A R1= 20[W] R2= 15[W] + - vS1= 50[V] R3= 10[W] R4= 45[W] + vOC B Now, as we noted, this open circuit makes the current iX equal to zero, which means that the dependent source has a value of zero. In addition, the resistor R2 has no current through it, and so we can remove it as well. With these simplifications we have the circuit that follows. A + R1= 20[W] + - vS1= 50[V] R3= 10[W] R4= 45[W] vOC B Page 4.6.3 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. In this circuit, there is only one connection between the right half of the circuit and the left half, and there is no source in the right half. Thus, the voltages and currents in the right half will be zero. We could go ahead as apply KVL and KCL, but we will find that vOC 0. This gives us the Thevenin voltage, which is zero. Now, we need to find either the equivalent resistance or the short-circuit current. The short-circuit current will have to be zero, though, since the opencircuit voltage is zero. We can see this from the relationship vOC iSC REQ . This is actually based on the assumption that the equivalent resistance is not zero, which we will show is true. Thus, if we try to find the REQ from the ratio of the open-circuit voltage to the short-circuit current, we will have an undefined expression. So we will choose to find the equivalent resistance. We set the independent sources equal to zero, and apply a test source. We will choose a 1[A]-current source, since that will make finding iX easy. We have the circuit that follows. iS1= 3iX iX A + R1= 20[W] R2= 15[W] R3= 10[W] R4= 45[W] iT= 1[A] vT B Let’s find iX first. We can write Page 4.6.4 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. iX 1[A]. Thus, iS1 3iX 3[A]. To make the solution a little easier, we will define a new variable, iR4, as shown in the figure that follows. iS1= 3iX iX A + R1= 20[W] R2= 15[W] iR4 R3= 10[W] R4= 45[W] iT= 1[A] vT B Now, we write KVL around the right-most mesh, and we get iR 4 R4 iX R2 vT 0. From KCL, we get iR 4 iS1 iX 0. From the second equation, and the previous values, we have iR 4 iS 1 iX 4[A]. Plugging these values into the first equation, we get vT iR 4 R4 iX R2 4[A] 45[W] 1[A] 15[W] vT 195[V]. Finally, to get the equivalent resistance, we can write REQ vT 195[V] 195[W]. iT 1[A] Page 4.6.5 Dave Shattuck University of Houston © Brooks/Cole Publishing Co. Thus, the Thevenin equivalent is just a resistance. We can redraw it as shown in the figure that follows. A RTH= 195[W] B Note: In this solution, as in most of the solutions in this project, we have taken a very slow and careful approach, redrawing the circuit several times. While it is a good idea to redraw whenever it helps simplify the circuit, it is generally not necessary to redraw the circuit just to define a new variable. We do it here to be very clear about the order of steps. When you work a problem, just mark new variables on your existing diagrams. Only redraw the circuit when you can simplify it to where it makes it easier for you to look at it. Be careful, but don’t waste time. Problem adapted from ECE 2300, Quiz 3, Spring 1992, Department of Electrical and Computer Engineering, Cullen College of Engineering, University of Houston. Page 4.6.6