MAE140 - Linear Circuits - Fall 13 Midterm, October 31

Transcription

1 Instructons ME140 - Lnear Crcuts - Fall 13 Mdterm, October 31 () Ths exam s open book. You may use whatever wrtten materals you choose, ncludng your class notes and textbook. You may use a hand calculator wth no communcaton capabltes () You have 70 mnutes () Do not forget to wrte your name, student number, and nstructor R 1 1 S2 2 R 5 R 2 D v S E 3 3 v x R 3 4 S1 C R 4 (a) Queston 1. (b) Questons 2-4. Fgure 1: Crcuts for questons Equvalent crcuts Part I: [2 ponts] Turn off all the sources n the crcut of Fgure 1(a) and fnd the equvalent resstance as seen from termnals and. Part II: [3 ponts] Fnd the Thévenn equvalent as seen from termnals and. Hnt: If you want, you can use the result obtaned n Part I. When you are decdng what source transformaton to use, be careful not to lose track of termnals and! Part III: [1 pont] Show that the power absorbed by a 40 Ω resstor that s connected to termnals and s 0.4W. Soluton: Part I: We start by swtchng off the sources. We substtute the voltage source by a short crcut, and the current source by an open crcut. Then, we get the crcut on the rght (+.5 pont)

2 Next, we combne the two resstances n parallel n the upper left corner to get the crcut (+.5 pont) Next, we combne the two remanng resstances that are stll n parallel to get the crcut (+.5 pont) 5Ω The fnal equvalent resstance s gven by summng the two resstances n seres R fnal = = (+.5 pont) Part II: From Part I, we can say that R T =. The Thévenn voltage v T s equal to the open crcut voltage seen from termnals -. (+.5 pont) To fnd the open-crcut voltage, we frst note that, snce the 5Ω resstor does not see any current, ts voltage drop s zero, and hence t can be elmnated. 3 (+.5 pont) Page 2

3 Combnng the two resstors n parallel, we obtan the equvalent crcut: 3 Now we transform the voltage source n seres wth the resstor to obtan (+.5 pont) 2 3 (+.5 pont) (Note that, f one nstead transforms the current source n parallel wth a resstor nto a voltage source of 30V n seres wth a resstor, then one loses track of the termnals -. Ths would most lkely lead you to thnk that the voltage drop s 30V, whch s ncorrect because the nodes connected to ths voltage source are not, anymore.) Combnng all the elements, we arrve at -1 (+.5 pont) The voltage drop that the 5Ω resstor sees s then 5V, whch s precsely the open-crcut voltage from termnals -,.e., v T = 5V. Part III: We compute the voltage across the 40Ω resstor wth the Thévenn equvalent as: (+.5 pont) 40 v = v T = 40 ( 5) = 4V (+.5 pont) 40 + R T 50 Then, the power absorbed by a 40Ω resstor s p = v2 40 = = 0.4W. (+.5 pont) 2. Node voltage analyss [6 ponts] Formulate node-voltage equatons for the crcut n Fgure 1(b). Use the node labels through E Page 3

4 provded n the fgure and clearly ndcate how you handle the presence of a voltage source. The fnal equatons must depend only on unknown node voltages and the source values v S, S1, and S2. Do not modfy the crcut or the labels. No need to solve any equatons! Soluton: There are fve nodes n ths crcut and the ground node (E) s drectly connected to the voltage source. Therefore, ths helps us takng care of t by settng v D = v S (ths s method 2). (+ 1.5 pont) We need to derve equatons for the other three unknown node voltages v, v, and v C. We do ths usng KCL and wrte equatons by nspecton. The matrx s 3 4 and the ndependent vector has 3 components. We wrte, G 1 + G 2 G 1 0 G 2 G 1 G 1 + G G 3 + G 4 G 3 v v v C v D = S1 S2 S1 (+.5 pont), (+ 3 ponts) where, as we do usually, G = 1/R. Snce we know that v D = v S, we can rewrte the equatons above as G 1 + G 2 G 1 0 v S1 + G 2 v S G 1 G 1 + G 5 0 v = S2 (+ 1 pont) 0 0 G 3 + G 4 v C S1 + G 3 v S 3. Mesh current analyss [6 ponts] Formulate mesh-current equatons for the crcut n Fgure 1(b). Use the mesh currents shown n the fgure and clearly ndcate how you handle the presence of each of the current sources. The fnal equatons should only depend on the unknown mesh currents and the source values v S, S1, and S2. Do not modfy the crcut or the labels. No need to solve any equatons! Soluton: There are four meshes n ths crcut. We can easly take care of the current source S1 by realzng that t only belongs to one mesh, and therefore 3 = S1. (+ 1 pont) The other current source, S2, belongs to two meshes and s not n parallel wth any resstor, so we need to use a supermesh (combnng meshes 1 and 2) to deal wth t. The current source mposes the constrant KVL for the supermesh reads lke 2 1 = S2. (+ 1 pont) (+ 1 pont) R R 5 2 v S + R 2 ( 1 3 ) = 0, (+ 1 pont) The remanng equaton comes from KVL for mesh 4, v S + R R 3 ( 4 3 ) = 0 (+ 1 pont) Page 4

5 Fnally, usng that 3 = S1, the equatons can be rewrtten as: 2 1 = S2 (R 1 + R 2 ) 1 + R 5 2 = v S R 2 S1 (R 3 + R 4 ) 4 = v S R 3 S1 (+ 1 pont) 4. onus queston [1 pont] Express v x n the crcut of Fgure 1(b) n terms of the node voltage varables of Queston 2 and also n terms of the mesh current varables of Queston 3. Soluton: In terms of the node voltage varables, we have v x = v C v (+.5 pont) and n terms of the mesh current varables, we realze that the voltage drop that the current source sees s equal to the sum of the voltage drops of the resstors R 3 and R 2, v x = R 3 ( 4 3 ) + R 2 ( 1 3 ) (+.5 pont) Page 5