MAE140 - Linear Circuits - Winter 16 Midterm, February 5

Transcription

1 Instructons ME140 - Lnear Crcuts - Wnter 16 Mdterm, February 5 () 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 50 mnutes () Do not forget to wrte your name and student number Good luck! R 3 R 1 1 R 2 v x 2 v S S C 20 V 4 v o 3 R 4 x 4 10Ω D R 5 (a) Queston 1 (b) Queston 2 Fgure 1: Crcuts for all questons. 1. Equvalent crcuts Part I: [3 ponts] Turn off all the sources n the crcut of Fgure 1(a) and fnd the equvalent resstance as seen from termnals and. 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 (+ 1 pont) 10Ω 10 Ω Next, we combne the two resstances n parallel n the upper left corner to get the crcut (+.5 pont) 10Ω

2 Next, we combne the two resstances n seres on the left to get the crcut (+.5 pont) 10Ω Next, we combne the two resstances n parallel on the left to get the crcut 15Ω (+.5 pont) 10Ω Fnally, we combne the two resstances n seres to get the equvalent resstance as seen from termnals and. 25Ω (+.5 pont) Part II: [5 ponts] Fnd the voltage v 0 usng only superposton, assocaton of resstors, voltage dvson, and current dvson. Soluton: Part II: To use superposton, we frst turn off the ndependent current source We substtute the current source by an open crcut. Then, we get the crcut on the rght (+.5 pont) 20 V v o,1 10Ω Combnng the two resstors n parallel at the top left corner, we obtan the crcut on the rght (+.5 pont) 20 V 10 Ω 10Ω The 10Ohms resstor on the bottom rght does not have any current gong through t, therefore v 0,1 s smply the voltage drop that the 30Ohms resstor sees. Page 2 v o,1

3 Ths can be easly computed usng voltage dvson as (+.5 pont) v 0,1 = = 10V (+.5 pont) Next, we turn off the ndependent voltage source. We substtute the voltage source by a closed crcut. Then, we get the crcut on the rght (+.5 pont) 4 v o,2 10Ω Combnng the two resstors n parallel n the top left corner, and then the resultng resstor n seres wth the 20Ohms resstor on the bottom left, we get the crcut on the rght 30 Ω 4 v o,2 (+ 1 pont) The 10Ohms resstor on the bottom rght does not have any current gong through t, therefore v 0,2 s smply the voltage drop that any of the 30Ohms resstor sees. (+.5 pont) Ths can be easly computed usng current dvson as y superposton, we conclude that 10Ω 1/30 v 0,2 = 30 ( 4) = 60V (+.5 pont) 1/30 + 1/30 v 0 = v 0,1 + v 0,2 = = 50V (+.5 pont) Part III: [1 pont] What s the Thévenn equvalent of the crcut as seen from termnals and? Soluton: Part III: We have computed the equvalent resstance from termnals and wth all sources turned off n Part I, and the open-crcut voltage n Part II. Therefore, the Thévenn equvalent of the crcut s smply Page 3

4 25 Ω -50 V Part IV: [1 pont] Fnd the power absorbed by a 100 Ω resstor that s connected to termnals and. (+ 1 pont) Soluton: Part IV: We use the Thévenn equvalent of the crcut to obtan the answer n an easy way. Connectng the 100Ohms resstor gves rse to the crcut 25 Ω -50 V 100Ω (+.5 pont) y voltage dvson, the voltage drop across the load s Therefore, the power absorbed s 2. Node voltage and mesh current analyss v = 100 ( 50) = 40V P = v 2 G = ( 40) 2 1 = 16W (+.5 pont) 100 Part I: [4 ponts] Formulate node-voltage equatons for the crcut n Fgure 1(b). Use the node labels through D 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 resstor values R 1 through R 5. Do not modfy the crcut or the labels. No need to solve any equatons! Soluton: Part I: There are four nodes n ths crcut and the ground node D (hence v D = 0), whch has already been chosen for us, s drectly connected to the voltage source. Therefore, we take care of the voltage source usng method 2 and set v C = v S. (+ 1 pont) We need to derve equatons for the other two unknown node voltages v and v. We do ths usng KCL and wrte equatons by nspecton. The matrx s 2 3 and the ndependent vector has 2 components. (+.5 pont) Page 4

5 We wrte, ( G1 + G 2 + G 3 G 3 G 2 G 3 G 3 + G 5 0 ) v v v C = ( 0 S ) (+ 2 ponts) where, as we do usually, G = 1/R. Snce we know that v C = v S, we can rewrte the equatons above as ( G1 + G 2 + G 3 G 3 G 3 G 3 + G 5 ) ( v v ) = ( ) G2 v S S (+.5 pont) Part II: [4 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 the current source. The fnal equatons should only depend on the unknown mesh currents and the resstor values R 1 through R 5. Do not modfy the crcut or the labels. No need to solve any equatons! Soluton: Part II: There are four meshes n ths crcut. The current source, S, belongs to two meshes and s not n parallel wth any resstor, so we need to use a supermesh (combnng meshes 2 and 4) to deal wth t. (+ 1 pont) The current source mposes the constrant 4 2 = S. (+.5 pont) KVL for the supermesh reads lke R R R 4 ( 4 3 ) + R 2 ( 2 1 ) = 0 (+.5 pont) The remanng equatons come from KVL for mesh 1 and KVL for mesh 3 R R 2 ( 1 2 ) v S = 0 R 4 ( 3 4 ) + v S = 0 (+ 1 pont) (+ 1 pont) Part III: [2 ponts] Provde two expressons for the voltage v x and the current x, one n terms of node voltages and the other one n terms of mesh currents. Soluton: Part III: In terms of the node voltages, v x and x can be expressed as v x = v v C (+.5 pont) x = 1 R 4 v C (+.5 pont) In terms of the mesh currents, v x and x can be expressed as v x = R 2 ( 1 2 ) x = 3 4 (+.5 pont) (+.5 pont) Part IV: [2 bonus ponts] Is the ndependent voltage source n parallel wth the resstor R 4? Removng the resstor R 4 and substtutng t by an open crcut would elmnate mesh 3. Would ths removal have any effect on the values of the node voltages or the other mesh currents? Would t have any effect on the current that flows through the ndependent voltage source? Justfy your answers. Page 5

6 Soluton: Part IV: The ndependent voltage source s n parallel wth the resstor (both elements form a loop, mesh 3, that contans no other element). (+.5 extra pont) From our dscusson n class of equvalent sources, we know that a voltage source connected n parallel wth a resstor can be transformed nto just a voltage source, wth the rest of the crcut beng oblvous to ths transformaton. (+.5 extra pont) Therefore, removng the resstor R 4 does not have any effect on the values of the node voltages or the other mesh currents (ths n fact can be verfed by checkng the equatons n Parts I and II above). (+.5 extra pont) Fnally, the removal of R 4 would have an effect on the current that flows through the ndependent voltage source. Wth the resstor, the current s 3 1, and wthout the resstor t would be 4 1. (+.5 extra pont) Page 6