MAE140 - Linear Circuits - Fall 10 Midterm, October 28

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1 M140 - Lnear rcuts - Fall 10 Mdterm, October 28 nstructons () 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 () o not forget to wrte your name, student number, and nstructor B 1 s Fgure 1: rcut for all questons. 1. Node Voltage nalyss [6 ponts] Formulate node-voltage equatons for the crcut n Fgure 1. hoose a ground node so that you do not have to wrte supernode equatons. Use the node labels provded n the fgure and clearly ndcate the fnal equatons, crcut unknowns and how you handle the presence of the voltage source. lso show how the voltage s related to the node voltages. o not modfy the labels or the locaton of the ground. No need to solve any equatons! Soluton: n order to deal wth the voltage source wthout wrtng supernode equatons, you can chose Node or Node as the ground node. We wrte the rest of the soluton n terms of Node here. n ths way, we can say that V V S (1) and not wrte KL at node (ths s method # 2).

2 To fnd the values of the other voltages, we wll use Krchoff urrent Law. enote G 1/. Then, by nspecton: Node : G(V V B ) G(V V ) GV 0 Node B: G(V B V ) G(V B V ) S Node : G(V V ) G(V V ) GV S. The set of equatons n V, V B, V can be rewrtten n matrx form as: 3G G G V 0 G 2G 0 V B S GV S (2) G 0 3G V S GV S quaton (2) needs to be solved n the unknowns V, V B, V whereas V s provded by quaton (1). (3 pts) Once we have solved for the node voltages, we can obtan V O as V O V V. 2. Mesh urrent nalyss [6 ponts] Formulate mesh-current equatons for the crcut n Fgure 1. Use the mesh currents shown n the fgure and clearly ndcate the fnal equatons, crcut unknowns and how you handle the presence of the current source. lso show how the voltage s related to the mesh currents. o not modfy the crcut dagram or the labels. No need to solve any equatons! Soluton: The current source belongs to two meshes and, n ths exam, the crcut cannot be rearranged to change ths. Therefore, we wll need combne Mesh 1 and Mesh 2 nto one supermesh. Ths wll gve us two equatons as follows: KVL appled to the supermesh: 1 ( 1 3 ) 2 ( 2 4 ) 0, Supermesh constrant equaton: 2 1 S. We also need to wrte KVL equatons for meshes 3 and 4: (2 pts) Mesh 3: 3 ( 3 1 ) ( 3 4 ) 0, Mesh 4: ( 4 3 ) ( 4 2 ) V S 0 The above four equatons need to be solved for the four mesh currents 1, 2, 3 and 4. Once we have those, we can compute the voltage usng Ohm s Law as ( 1 3 ). (2 pts) Page 2

3 3. Lnearty nalyss [6 ponts 2 bonus ponts] For the crcut shown n Fgure 1, use lnearty to solve for the unkonw voltage. Soluton: We are more nterested n your reasonng than your calculatons. You wll get all your ponts for explanng how you solve the problem, not for computng the soluton. We wll use superposton to compute as the sum of two voltages V0 1 V0 2 whch we wll compute for the followng two crcuts. n the frst crcut, settng 0 leads to the dagram: B s We can assocate the two resstors between and and and n parallel and rearrange the crcut to avod a supermesh as n the next dagram: Page 3

4 B 3 s 1 2 /2 We wll use method #2 not to wrte KVL at mesh 3 and obtan by nspecton mesh-current equatons: [ ] 3 1 ( ) , 0 3 s 3 fter you solve for 1, 2 and 3 you should be able to compute n the second crcut, settng V 1 0 ( 3 2 ) (1/2 pt) s 0 leads to the dagram: B We can assocate the two resstors on the top n seres to elmnate Node B, choose the ground to be at node leadng to the dagram: Page 4

5 2 We use method #2 not to wrte KL at node and obtan the remanng node-voltage equatons by nspecton: [ ] 2.5G G 0.5G ( ) V V G 3G G 0, V 0 V fter you solve for V and V you should be able to compute V 2 0 V V (1/2 pt) f you care for the gruelng computatons... n the frst crcut, dvde all by and substtute 3 to get [ ] ( ) ( ) s s then 2 ( 1 2 ) V 1 0 ( 3 2 ) s (8/13 1) (5/13) s n the second crcut, dvde by G and substtute V to get [ ] ( ) ( ) Vs 1 3 then Fnally V ( V V V 2 0 V V (6/13 5/13) (1/13) V 1 0 V 2 0 (1/13) (5/13) s ) ( ) 7/13 8/13 s ( ) 5/13 V 6/13 s f you got ths far you deserve two more bonus ponts. One pont f you got one correct partal answer, ether V0 1 or 2. (2 bonus pts) Page 5

6 4. Nodal nalyss Mesh nalyss wth ependent Source [2 bonus ponts] Suppose the voltage source n the crcut 1 s now a dependent voltage source wth value V S k. Show how the mesh-current equatons you found n Problem 2 should be modfed n order to take ths dependency nto account? learly ndcate the fnal equatons and crcut unknowns. Use the results you obtaned n Problem 1. No need to solve any equatons! Soluton: We should start by the soluton of problem 2, whch gves us the equatons: , , V S S, The only equaton that nvolves V S s the thrd so we modfy t to nclude the dependence on : k 0 and use the expresson for ( 1 3 ) also obtaned n problem 2 to wrte (1 bonus pt) k( 1 3 ) 0. The fnal equatons to be solved for 1, 2, 3 and 4 are , , k 1 2 (1 k) , 2 1 S (1 bonus pt) Page 6

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