Download Linear Circuits Midterm Exam - Fall 09 and more Exams Introduction to Sociology in PDF only on Docsity! MAE140 - Linear Circuits - Fall 09 Midterm, November 10 Instructions (i) This exam is open book. You may use whatever written materials you choose, including your class notes and textbook. You may use a hand calculator with no communication capabilities (ii) You have 70 minutes (iii) Do not forget to write your name, student number, and instructor B + − 10 V 0.5 A 10 40 20 10 Ω Ω Ω Ω A (a) Question 1. A v S i S B DC R 2 R 4 R 3 R 1 i 1 i 2 (b) Question 2. Figure 1: Circuits for questions 1 and 2. 1. Equivalente Circuits Part I [4 points] Use source transformations and association of resistors to find the Thevenin equiva- lent to the circuit in Fig. 1(a) as seen from terminals A and B. Part II [4 points] A classmate was arguing with you that one would “extract” the most power out of this circuit by connecting a very small resistance between terminals A and B. You argued that he/she was wrong and that a 20 Ω resistor would be a better choice. Who is right and why? Hint #1: Don’t complicate: all you need to do is compare the two options! Hint #2: Use the Thevenin equivalent computed in Part I to answer the question. Solution: Part I: A possible series of source transformations is shown here: 10 V + − +− 10 V 40 20 Ω Ω A B 20 Ω (+ 1 point) 20 V + − 40 40 Ω Ω A B Ω 40 Ω A B 0.5 A 40 (+ 1 point) 0.5 A 20 Ω A B (+ 1 point) Ω + − A B 10 V 20 (+ 1 point) Part II: Let’s investigate the two scenarios. In the case of a small resistance, say R = Ω, using the Thevenin equivalent and voltage division we have that vR = 20 + 10 ≈ /2 V, iR = vR/ ≈ 1/2 A, (+ 1 point) The power absorbed by the small resistor is then pR = vRiR ≈ /4 W. which is close to zero when is small. (+ 1 point) When R = 20Ω, using the Thevenin equivalent and voltage division we have that vR = 20 20 + 20 10 = 5 V, iR = vR/20 = 1/4 A. (+ 1 point) Page 2 Part II [4 points] Design a circuit using a single OpAmp that would realize the exact same function using a single 1 V source. Name one advantage/disadvantage of your design versus the design of Figure 2. Solution: Part I: Most of the circuit is standard with the exception of the 1 V source connected to the ‘+’ terminal of the second opamp stage. So if we label the voltages as in the figure, we have, using the non-inverting opamp gain, V1 = R + R R Vin = 2Vin, Vout = R + R R V4 = 2V4, (+ 1 point) Also, from KCL at the ‘−’ terminal of the second opamp, V2 − V3 R = V1 − V2 R (+ 1 point) which can be rearranged as V3 = 2V2 − V1 Because V2 = VN = VP = 1 V we have that V3 = 2− V1 = 2− 2Vin (+ 1 point) Finally, V4 = V3 because there is a zero current going into the ‘+’ terminal of the opamp. Summa- rizing: V1 = 2Vin, V2 = 1V, V4 = V3 = 2− 2Vin (+ 1/2 point) Hence Vout = 2V4 = 4− 4Vin. (+ 1/2 point) Part II: A standard subtractor would do the job, that is the following circuit (+ 2 points) 4 R + − + − R 4 R Vin Vout 1 V R Note that this is very similar to the circuit in the middle of Figure 2. This circuit has gains Vout = 4R 4R + R 4R + R R 1 V − 4R R Vin = 4− 4Vin (+ 1 point) Page 5 One disadvantage is that the input impedance as seen from Vin is not large, potentially loading the input circuit. One advantage is that we have used fewer components. (Either observation gets + 1 point) Page 6
