Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Operational amplifier wikipedia , lookup
Integrated circuit wikipedia , lookup
Rectiverter wikipedia , lookup
Regenerative circuit wikipedia , lookup
Surface-mount technology wikipedia , lookup
Charlieplexing wikipedia , lookup
Opto-isolator wikipedia , lookup
Current mirror wikipedia , lookup
RLC circuit wikipedia , lookup
Dave Shattuck University of Houston Practice Examination Questions With Solutions Module 1 – Problem 8 Filename: PEQWS_Mod01_Prob08.doc Note: Units in problem are enclosed in square brackets. Time Allowed: 30 minutes*. Problem Statement: In the circuit shown below, find the voltage vX. R1 = 1.0[kW] R3 = 3.3[kW] R4 = 2.7[kW] R5 = 2.2[kW] iS= 5[mA] + R7 = 4.7[kW] vX R6 = 5.9[kW] - R8 = 8.2[kW] R2 = 1.2[kW] * This “time allowed” figure is based on the student’s circuit analysis skills at this point. Once the student has learned the material in the second module, this problem should be completed in significantly less time than what is shown here. Page 1.8.1 Dave Shattuck University of Houston Problem Statement: In the circuit shown below, find the voltage vX. R1 = 1.0[kW] R3 = 3.3[kW] R4 = 2.7[kW] R5 = 2.2[kW] iS= 5[mA] + R7 = 4.7[kW] vX R6 = 5.9[kW] - R8 = 8.2[kW] R2 = 1.2[kW] Solution: The first step in the solution of such problems is to define the variables that we are going to need. However, beginning students often have the problem of trying to decide what to name. It is not always clear which voltages and which currents will be of interest in a problem. In addition, we can waste time if we always label everything, everywhere, in a problem. One approach is to just start somewhere, and name (define) variables as needed. This is the approach that we will use here. I will begin by defining the voltages across resistors R5 and R6, and the currents through these same two resistors. I have chosen a particular polarity for each of these voltages, but this choice is arbitrary. How did I know that I needed to define these two voltages? The key is that I looked at this circuit, and decided that I was going to need to write Page 1.8.2 Dave Shattuck University of Houston KVL around the loop that is marked with a red dashed line in the figure that follows. R1 = 1.0[kW] R3 = 3.3[kW] R4 = 2.7[kW] R5 = 2.2[kW] iS= 5[mA] + v5 + - i5 R7 = 4.7[kW] vX + R6 = 5.9[kW] v6 i6 R8 = 8.2[kW] R2 = 1.2[kW] When we look at this circuit, we may notice that these two resistors, R5 and R6, are connected to the rest of the circuit at only one end. Stated another way, the right hand side of R5 is not connected to anything. If we were to write the KCL for the right hand “node” of R5, we would get i5 0, which is equivalent to i5 0. We have put “node” in quotation marks because it is not really a connection point between two parts of the circuit. However, it does not matter whether this is a “node” or not; we can write KCL for it anyway. We can write KCL for any closed surface. Using Ohm’s law, then, we can show that v5 is also zero. A similar set of arguments holds for R6, and we can show that v6 is zero as well. Page 1.8.3 Dave Shattuck University of Houston For our next step, we want to work towards finding v7 and v8, which are defined in the circuit diagram that follows. If we can find v7 and v8, we will be able to find vX. We will find them by first finding i7 and i8. Note that i7 is also the current through the R3 resistor, because i5 is zero. Similarly, we recognize that i8 is the current through the R4 resistor because i6 is zero. Note also that the current through the R1 resistor is iS, since this is the only path available for the current. Thus, we can write KCL for the closed surface shown as a red dashed line below, and write iS i7 i8 0, or i7 i8 5[mA]. (Eq. 1) R1 = 1.0[kW] R3 = 3.3[kW] R4 = 2.7[kW] R5 = 2.2[kW] iS= 5[mA] + v5 + - vX + v6 i5 + v7 + i6 i8 i7 R7 = 4.7[kW] R2 = 1.2[kW] R6 = 5.9[kW] v8 R8 = 8.2[kW] - - This is one equation with two unknowns. If we can write another equation using just these two unknowns, we could solve. There is a tendency to figure that since KCL worked so well, we should just do it again. However, when we try this, by writing KCL for the bottom node, a closed surface shown as a red dashed line in the circuit below, we get the following equation, Page 1.8.4 Dave Shattuck University of Houston iS i7 i8 0, or i7 i8 5[mA]. This equation is simply the same equation we had before, after multiplying both sides by minus one. This is not an independent equation, and it will not help us. Later in this course we will show ways to systematically write independent equations. Until then, we simply continue to write equations until we have the same number of independent equations and unknowns. R1 = 1.0[kW] R3 = 3.3[kW] R4 = 2.7[kW] R5 = 2.2[kW] iS= 5[mA] + v5 + - vX + v6 i5 + v7 + i6 i8 i7 R7 = 4.7[kW] R2 = 1.2[kW] R6 = 5.9[kW] v8 R8 = 8.2[kW] - - It makes sense that if we use KVL, we might get an independent equation. Therefore, we write KVL for the closed loop drawn in the circuit diagram that follows. The closed loop is shown as a red dashed line. Page 1.8.5 Dave Shattuck University of Houston R1 = 1.0[kW] + + i7 i8 v4 R3 = 3.3[kW] R4 = 2.7[kW] v3 iS= 5[mA] R5 = 2.2[kW] - + v5 + - vX R6 = 5.9[kW] + - v6 i5 i6 + + v7 i8 i7 v8 R7 = 4.7[kW] R2 = 1.2[kW] R8 = 8.2[kW] - - Writing the equation, we have v7 v3 v4 v8 0. This doesn’t seem to help, since we have just introduced four more unknowns. However, we can use Ohm’s law to express each of these voltages in terms of a current, and get i7 R7 i7 R3 i8 R4 i8 R8 0, or with values, i7 4.7[kW] i7 3.3[kW] i8 2.7[kW] i8 8.2[kW] 0. (Eq. 2) Taken together, the two equations Eq. 1 and Eq. 2 are two equations in two unknowns. I can solve these, and get i7 2.88[mA], and i8 2.12[mA]. Using these two values, I can find the voltages v7 and v8, as Page 1.8.6 Dave Shattuck University of Houston v7 2.88[mA]R7 13.5[V], and v8 2.12[mA]R8 17.4[V]. With this information, I am ready to solve for vX. Using KVL around the loop shown with a red dashed line in the figure that follows, I can write v7 v5 vX v6 v8 0, which upon substituting gives 13.5[V] 0 vX 0 17.4[V ] 0. R1 = 1.0[kW] + + i7 i8 v4 R3 = 3.3[kW] R4 = 2.7[kW] v3 R5 = 2.2[kW] - iS= 5[mA] + v5 + - vX R6 = 5.9[kW] + - v6 i5 i6 + + v7 i8 i7 v8 R7 = 4.7[kW] R2 = 1.2[kW] R8 = 8.2[kW] - - We can solve for vX and get vX 3.9[V]. Some notes are worth making about this solution. You may have noticed that several of the resistor values were not used in the solution. Specifically, resistors R1, R2, R5 and R6 do not affect the solution. If you put in different values for these resistors, the solution does not change. This is common. It might be asked why these resistors are included if they do not affect the solution. There are Page 1.8.7 Dave Shattuck University of Houston many possible answers to this question. One answer is that they are used for other applications of the circuit. The resistors R5 and R6 in this circuit may be used to limit the current when something is placed across the terminals of the voltage vX. Another answer is that they do affect other values that were not solved for in this problem. The resistors R1 and R2 do affect the amount of power delivered by the current source. In any case, our major concern is to be able to solve the circuits as they are given, which generally includes the ability to recognize what parameters we will use, and what parameters we will not use. You will become better at this as your skill in solving circuits improves. Problem adapted from Final Exam, Problem 1, Summer 2000, University of Houston, Department of Electrical and Computer Engineering, Cullen College of Engineering. Page 1.8.8