1 ECE 201 – Spring 2010 Exam #2 Tuesday, March 9, 2010 Division 0101: Prof. Capano (9:30am) Division 0201: Prof. Tan (10:30 am) Division 0301: Prof. Jung (7:30 am) Division 0401: Prof. Capano (11:30am) Instructions 1. DO NOT START UNTIL TOLD TO DO SO. 2. Write your Name, division, professor, and student ID# (PUID) on your scantron sheet. 3. This is a CLOSED BOOKS and CLOSED NOTES exam. 4. There is only one correct answer to each question. 5. Calculators are allowed (but not necessary). 6. If extra paper is needed, use back of test pages. 7. Cheating will not be tolerated. Cheating in this exam will result in an F in the course. 8. If you cannot solve a question, be sure to look at the other ones and come back to it if time permits. 9. As described in the course syllabus, we must certify that every student who receives a passing grade in this course has satisfied each of the course outcomes. On this exam, you have the opportunity to satisfy outcomes i, iii, iv and viii. (See the course syllabus for a complete description of each outcome.) On the chart below, we list the criteria we use for determining whether you have satisfied these course outcomes. If you fail to satisfy any of the course outcomes, don’t panic. There will be more opportunities for you to do so. Course Outcome Exam Questions Total Points Possible Minimum Points required to satisfy course outcome i 7-8 14 7 iii 1-5, 14 42 21 iv 9-14 42 21 viii 6 7 7 Potentially useful formula: ( )ott /ox(t) x( ) x(t ) x( ) e−− τ+= ∞+ − ∞ τ = L/R τ = RC2 1. Using source transformations, find the voltage “v” (in V): (1) 12 (2) 24 (3) 36 (4) 48 (5) 56 (6) 60 (7) 72 2. The Norton equivalent network for a more complicated circuit has a load resistor (RL) attached to it as shown. A plot if iL – v relationship is also given. Find VOC and RTH for the corresponding Thevenin equivalent network. (1) 12V; 6Ω (2) 24V; 4Ω (3) 36V; 6Ω (4) 48V; 4Ω (5) 24V; 8Ω (6) 48V; 2Ω (7) 60V; 5Ω3 3. Find the correct Thevenin equivalent network for the circuit shown below. (1) (2) (3) (4) (5) (6) (7)4 4. For the circuit below, the short-circuit current ISC in the Norton equivalent circuit is (in A): (1) 0 (2) 0.5 (3) 1 (4) 1.5 (5) −0.5 (6) −1 (7) −1.5 5. The Thevenin equivalent resistance RTH for the circuit shown below is (in Ω): (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 75 6. Find the maximum power transferred to the load resistor for the circuit below (in W): (1) 150 (2) 300 (3) 450 (4) 600 (5) 750 (6) 900 (7) 1050 7. In the circuit below, the switch has been closed (at the left position as shown) for a very long time. Find the instantaneous stored energy in the capacitor at t = 0 sec (in J): (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 76 8. Compute the equivalent capacitance (Ceq) of the capacitor network shown below. (1) 2 F (2) 4 F (3) 6 F (4) 8 F (5) 10 F (6) 12 F (7) 14 F 9. The switch in the circuit shown below was opened at t = 0 sec. Find iL(t) for t ≥ 0 (in A): (1) 10t10e− (2) 0.2t5e− (3) 0.2t20e− (4) 2t5e− (5) 5t20e− (6) 20t10e− (7) 2.5t5e−7 10. The switch in the circuit shown below was closed for a long time. It opened at t = 0 sec. The initial condition of the inductor current was ( )Li 0 1A−=. Find the inductor current at t = 3 sec, iL(t = 3 sec), in A: (1) 0 (2) e−1 (3) e−2 (4) e−3 (5) e−4 (6) e−5 (7) 1 11. In the circuit shown below, the switch was moved from the left position to the right position at t = 0 sec. Find vc(t) for t ≥ 0 (in V): (1) 2t40 e−+ (2) 0.5t20 20e−+ (3) 0.5t30 20e−+ (4) 2t40 10e−+ (5) 2t30 20e−+ (6) t30 5e−+ (7) t20 10e−+8 12. For the circuit shown below, the input voltage is )(9)(18)( tututvin+−−= V. Compute the input current iin(t) for t ≥ 0 (in A). (1) 1×10−3+1.0×10−3e−t (2) 2×10−3+1.0×10−3e−t (3) 1×10−3+1.5×10−3e−t (4) 2×10−3+1.5×10−3e−t (5) 1×10−3+2.0×10−3e−t (6) 2×10−3+2.0×10−3e−t (7) 1×10−3+2.5×10−3e−t 13. Assuming υC(0) = 0 V, what is the time it takes for υC(t) to rise to 12(1-e-2) V? (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 79 14. For a linear network consisting of resistors and dependent sources, the output voltage υC(t) is computed using the given inputs and initial condition as shown below. Now, the input current is increased from 0.25u(t) to 0.5u(t) as shown below. Compute υC (t). (1) ( ) ( ) 0.06 0.06 0.0610 50 1 30 1 0.5−− −= +− +− −tt tCt e e e utυ (2) ( )( ) 0.06 0.06 0.0620 50 1 30 1 0.5−− −= +− +− −tt tCt e e e utυ (3) ( )( ) 0.06 0.06 0.0610 100 1 30 1 0.5−− − = + − +− − tttCt e e e utυ (4) ( )( ) 0.06 0.06 0.0610 50 1 60 1 0.5−− −= +− +− −tt tCt e e e utυ (5) ( )( ) 0.06 0.06 0.0620 100 1 60 1 0.5−− −= + − +− −tttCt e e e utυ (6) ( ) ( ) 0.06 0.06 0.0620 100 1 30 1 0.5−−− = + − +− − tt tCt e e e utυ (7) ( )( ) 0.06 0.06 0.0610 100 1 60 1 0.5−− −= + − +− −tttCt e e e utυ ( ) ( ) 0.06 0.06 0.0610 50 1 30 1 0.5−− −= +− +− −tt tCt e e e utυ( )Initial condition0 10Cυ−=( )Initial condition0 10Cυ−=(
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