332:221 Principles of Electrical Engineering I Fall Hourly Exam 2 November 6, 2006

Transcription

1 2:221 Principles of Electricl Engineering I Fll 2006 Nme of the student nd ID numer: Hourly Exm 2 Novemer 6, 2006 This is closed-ook closed-notes exm. Do ll your work on these sheets. If more spce is required, do your work on the ck side of the sheets nd indicte ccordingly so tht the grder does not miss it. Prolem Pge Mximum Points # Points erned Description V oc y Nodl Anlysis i sh y Nodl Anlysis 1 V oc y Mesh Anlysis i sh y Mesh Anlysis R Th y Test Method Op Amp Anlysis Op Amp Anlysis Totl mximum points = 100 Totl points erned y the student = Unless you mster Node Voltge nd Mesh current Methods, you will not shine in Electricl Engineering.

2 2 Exm Info: The sme circuit is used in ll the first five prolems. The gol here is to see how well you know Node Voltge Method nd Mesh current method, nd determining the Thevenin resistnce y test voltge method. You need to know super node s well s super mesh concepts. Alger is simple. In fct you cn cross-check nswers etween different wys of doing the sme prolem. (Prolem 1) Determintion of v Th y node voltge method: Consider the circuit given (not given here ut will of course e given in exm). Our im in this prolem is to determine the open circuit voltge v oc t the terminls nd y utilizing the node voltge method. (Prolem 2) Determintion of i sh y node voltge method: Consider the circuit given in Prolem 1. Our im in this prolem is to determine the short circuit current i sh through the terminls nd y utilizing the node voltge method. (Prolem ) Determintion of v Th y mesh current method: Consider the given in Prolem 1. Our im in this prolem is to determine the open circuit voltge v oc t the terminls nd y utilizing the mesh current method. (Prolem 4) Determintion of i sh y mesh current method: Consider the circuit given in Prolem 1. Our im in this prolem is to determine the short circuit current i sh through the terminls nd y utilizing the mesh current method. (Prolem 5) Determintion of R Th y test voltge nd test current method: We first construct n independent source-less circuit of the circuit given in Prolem 1. Apply test voltge v t t the terminls nd, nd determine the test current supplied y it to the circuit. Then we recognize tht the Thevenin resistnce R Th is the rtio vt. Utilize ny method to determine the rtio vt. (Prolem 6) This is prolem on Op-Amp circuits. (Prolem 7) This is prolem on Op-Amp circuits.

3 1 (Prolem 1, 1 points) Determintion of v Th y node voltge method: Consider the circuit shown elow. Our im in this prolem is to determine the open circuit voltge v oc t the terminls nd y utilizing the node voltge method. v d d v e e 2A 2Ω v 1 v f Ω f 4V v oc Consider s the reference node, nd mrk the voltges of nodes d, e, nd f s v d, v e, nd v f s shown. We note tht v oc = v e. Step 1: Write down reltionship tht might exist etween the voltges v d nd v 1. We note tht v d = v 1. Step 2: Do we know the vlue of v f? If so, wht is its vlue? We note tht v f = 4 V. Step : Write down reltionship tht might exist etween the node voltges v d nd v e. We note tht v e = v d = v d. Step 4: Write super node eqution involving the nodes d nd e. 2 v d 2 v e (4) = 0 2 v d 2 v d 4 = 0. Step 5: Solve the ove equtions to determine v e which equls v oc. The ove implies tht v d = 20 V nd hence v oc = v e = v d = 20 V. The following ws not sked; however it might help to know tht v Th = v oc.

4 2 (Prolem 2, 14 points) Determintion of i sh y node voltge method: Consider the circuit shown elow. Our im in this prolem is to determine the short circuit current i sh through the terminls nd y utilizing the node voltge method. Note tht some of the steps re sme s those in Prolem 1, ut repeted here for completeness of this prolem. v d d v e e 2A 2Ω v 1 v f Ω f 4V i sh Consider s the reference node, nd mrk the voltges of nodes d, e, nd f s v d, v e, nd v f s shown. Step 1: Write down reltionship tht might exist etween the voltges v d nd v 1. We note tht v d = v 1. Step 2: Do we know the vlue of v f? If so, wht is its vlue? We note tht v f = 4 V. Step : Write down reltionship tht might exist etween the node voltges v d nd v e. We note tht v e = v d = v d. Step 4: Write super node eqution involving the nodes d nd e. 2 v d 2 v e (4) v e 10 = 0 2 v d 2 v d 4 v d 10 = 0. Step 5: Solve the ove equtions to determine v e. The ove implies tht v d = 50 V nd hence v e = v d = 50 V. Step 6: Once v e is known, determine i sh. Thus, i sh = v e 10 = = 5 A. The following ws not sked; however it might help to know the nswer for Prolem 5. R Th = v oc i sh = 20 5 = 4 Ω.

5 (Prolem, 1 points) Determintion of v Th y mesh current method: Consider the circuit shown elow. Our im in this prolem is to determine the open circuit voltge v oc t the terminls nd y utilizing the mesh current method. d e 2A Ω i 1 2Ω v 1 i 2 f 4V v oc Mrk the mesh currents i 1 nd i 2 s shown. Step 1: Do we know the vlue of i 1? If so, wht is its vlue? We note tht i 1 = 2 A. Step 2: Determine v 1 in terms of i 1 nd i 2. We note tht v 1 = (i 1 i 2 )2. Step : In view of steps 1 nd 2, we need one mesh eqution to solve for oth the mesh currents. Write n pproprite mesh eqution. We write the mesh eqution for i 2 s v 1 i 2 4 = 0 (i 1 i 2 )2 2(i 1 i 2 )2 i 2 4 = 0. Step 4: Solve the ove equtions to determine i 2. The ove implies tht i 2 = 8 A. Step 5: Determine v oc = v Th y knowing i 2. We note tht v oc = v Th = 4 i 2 = 4 24 = 20V. This grees with wht hs een computed in Prolem 1.

6 4 (Prolem 4, 15 points) Determintion of i sh y mesh current method: Consider the circuit shown elow. Our im in this prolem is to determine the short circuit current i sh through the terminls nd y utilizing the mesh current method. Note tht some of the steps given elow re sme s those in Prolem, ut repeted here for completeness of this prolem. d e 2A Ω i 1 2Ω v 1 i 2 f i sh 4V i sh Mrk the mesh currents i 1, i 2, nd i sh s shown. Step 1: Do we know the vlue of i 1? If so, wht is its vlue? We note tht i 1 = 2 A. Step 2: Determine v 1 in terms of i 1 nd i 2. We note tht v 1 = (i 1 i 2 )2. Step : In view of steps 1 nd 2, we need two mesh equtions to solve for ll the mesh currents. Write pproprite mesh equtions. We write the mesh eqution for i 2 s v 1 (i 2 i sh ) 4 = (i 1 i 2 )2 2(i 1 i 2 )2 (i 2 i sh ) 4 = 0 i 2 = i sh 8 A. We write the mesh eqution for i sh s 4 (i sh i 2 ) i sh 10 = 0 1i sh i 2 4 = 0. Step 4: Solve the ove equtions to determine i sh. Solving the ove equtions, we get i sh = 5 A nd i 2 = i sh 8 A = 2 A. The vlue of i sh grees with wht hs een computed in Prolem 2.

7 5 (Prolem 5, 20 points) Determintion of R Th y test voltge nd test current method: We first construct n independent source-less circuit. Apply test voltge v t t the terminls nd, nd determine the test current supplied y it to the circuit. Then we recognize tht the Thevenin resistnce R Th is the rtio vt. Utilize ny method to determine the rtio vt. The independent source-less circuit long with the test source is s shown elow. 2Ω v 1 i 1 Ω i g v t Independent source-less circuit We cn solve the ove prolem y utilizing formlly the node voltge method or mesh current method. Do y these methods s home-work (not collected). We will solve the prolem here y kind of mixed method (some of the node voltges re mentlly simplified). It is esy to see tht i g = v t 10. Thus, i 1 = i g = v ( t 10 nd hence v 1 = i 1 2 = 2 v ) t 10. We cn now write the KVL eqution round the externl loop in order to relte v t nd, ( v 1 10 v t = 2 v ) t v t = 0. We cn rewrite the ove eqution s [ ( ) [ 10 ] = v t 1 2 ]. By multiplying throughout with, the ove eqution cn e simplified s We cn now determine R Th s 4 = v t. R Th = v t = 4 Ω.

8 6 v c v R f (Prolem 6, 12 points) Consider the ctive summing or dder circuit shown on the left. Design the vlues for resistnces R,,, nd R f so tht the output voltge v out is the verge of v, v, nd v c except for the sign inversion. Tht is, we need v R V CC N i n v i o n P v p i p V CC R L v out v out = 1 (v v v c ). Assume tht the Op-Amp is idel. To get the design eqution, write node eqution t the negtive input node N of Op-Amp. Note: In Op-Amp circuits, one should choose resistors lrge enough not to lod the outputs significntly ut smll enough tht stry cpcitnces do not cuse prolems. A rule of thum is to choose resistor vlues in the rnge, 500 Ω to 50 KΩ. Tke s the reference node. Then, to strt with, we oserve tht the node voltges v n is zero s the Op-Amp is ssumed to e idel. Following the given suggestion, in order to get the design eqution, we write node eqution t the negtive input node N of Op-Amp s This cn e simplified s 0 v R 0 v 0 v c 0 v out R f = 0. [ Rf v out = v R f v R ] f v c. R To get the verge of ll the three signls, we need R f R = R f = R f = 1. Tht is R = = = R f.

9 7 (Prolem 7, 1 points) Consider the following Interfce Circuit. Determine the volt-meter reding v o in terms of the sensor voltge v T nd in terms of ll the resistnce vlues nd the is voltge. Assume tht the Op-Amps re idel. Interfce Circuit R Sensor v T R d v o1 v is v o Meter Tke s the reference node. Then, to strt with, we oserve the following s the Op-Amps re ssumed to e idel: The voltge t the negtive input terminl of the first Op-Amp is v T. The voltge t the negtive input terminl of the second Op-Amp is v is. We then write the following node equtions one t the negtive input terminl of the first Op-Amp nd the other t the negtive input terminl of the second Op-Amp, v is v o1 By simplifying the ove equtions, we get v T R d v T v o1 = 0 v is v o R = 0. v o = (1 R )v is R (1 R d )v T.