ECE 201 – Spring 2011 Exam #2 Thursday, March 10, 2011 Division 0101: Prof. Tan (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. Calculators are allowed (but not necessary). 5. If extra paper is needed, use back of test pages. 6. Cheating will not be tolerated. Cheating in this exam will result in an F in the course. 7. If you cannot solve a question, be sure to look at the other ones and come back to it if time permits. 8. 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 5-6 15 7.5 iii 1-3, 11 30 15 iv 7-10, 12-13 45 22.5 viii 4 7.5 7.5 ()ot t /ox(t) x( ) x(t ) x( ) e− − τ+ = ∞ + − ∞ τ = L/R τ = RC LC12=ω1. For the circuit shown below, find the Thevenin equivalent resistance Req (in kΩ) as seen by the variable load RL. (1) 2/3 (2) 2/5 (3) 3/5 (4) 2/7 (5) 5/7 (6) 10/7 (7) 2 (8) 52. For the circuit shown below, find ISC (in mA), the short-circuit current for the Norton equivalent circuit. (1) −30 (2) 30 (3) −50 (4) 50 (5) −100 (6) 100 (7) −150 (8) 150 3. Find the Norton equivalent circuit seen by the load resistor. (1) sc thI 4A; R 4= = Ω (2) sc thI 4A; R 5= = Ω (3) sc thI 2A; R 4= = Ω (4) sc thI 2A; R 5= = Ω (5) sc thI 4A; R 4= − = Ω (6) sc thI 4A; R 5= − = Ω (7) sc thI 2A; R 4= − = Ω (8) sc thI 2A; R 5= = − Ω4. For the circuit shown below, find the load resistance RL for maximum power transfer and the corresponding PLmax. (1) 1kΩ; 10W (2) 2kΩ; 10W (3) 3kΩ; 10W (4) 4kΩ; 10W (5) 1kΩ; 20W (6) 2kΩ; 20W (7) 3kΩ; 20W (8) 4kΩ; 20W 5. The current through a 1F capacitor for t ≥ 0 is as shown below. If vc(∞) = 0V, find vc(0+) (in V), the initial condition for the capacitor at t = 0. (1) 1 (2) 2 (3) 3 (4) 4 (5) −1 (6) −2 (7) −3 (8) −46. For the circuit shown below, find (i) Leq (in H), the equivalent inductance; and (ii) v3(t) (in V), the voltage drop across inductor L3. (1) 2.1; 0.4sin(2πt) (2) 1.2; 0.4cos(2πt) (3) 1.2; 1.6sin(2πt) (4) 1.2; 1.6cos(2πt) (5) 3.32; 0.4sin(2πt) (6) 3.32; 0.4cos(2πt) (7) 3.32; 1.6sin(2πt) (8) 3.32; 1.6cos(2πt)7. For the circuit shown below, find vo(t) for t ≥ 0. (1) 36 7( )3 3tov t e−= + (2) 36 7( )3 3tov t e−= − (3) 36 7( )5 5tov t e−= + (4) 36 7( )5 5tov t e−= − (5) 66 7( )3 3tov t e−= + (6) 66 7( )3 3tov t e−= − (7) 66 7( )5 5tov t e−= + (8) 66 7( )5 5tov t e−= − ( ) 10 ( ) 4 ( )inv t u t u t= − − +( )Ov t+−8. In the circuit below, the switch opens at t = 0 sec. Find the voltage across the resistor, vr(t) for t≥ 0 (in V): (1) 10t6e− (2) 10t12e−− (3) 8.33t12e−− (4) 10t6e−− (5) 10t12e− (6) 8.33t6e−− (7) 8.33t12e− (8) 8.33t6e− 9. In the circuit below, ()cv 0 0V−=. Find vc(t) for t > 0 (in V): (1) 3t2e−− (2) t2e 4−− + (3) 1.5t2e 2−− + (4) 3t2e 1−− + (5) 1.5t2e−− (6) t2e−− (7) t2e 2−− + (8) 12e 1−− +10. The switch in the circuit below has been open for a long time before it is closed at time t = 0. Find ix(0+) (in A), ix(∞) (in A), and the time constant τ (in s). (1) ix(0+) = 0 A; ix(∞) = 1.5 A; τ = 4.5 s (2) ix(0+) = −1 A; ix(∞) = 1.5 A; τ = 4.5 s (3) ix(0+) = 0 A; ix(∞) = 1.5 A; τ = 2.5 s (4) ix(0+) = −1 A; ix(∞) = 1.5 A; τ = 2.5 s (5) ix(0+) = 0 A; ix(∞) = 1.0 A; τ = 4.5 s (6) ix(0+) = −1 A; ix(∞) = 1.0 A; τ = 4.5 s (7) ix(0+) = 0 A; ix(∞) = 1.0 A; τ = 2.5 s (8) ix(0+) = −1 A; ix(∞) = 1.0 A; τ = 2.5 s11. Consider the 1st-order circuit below. The response vc(t) for t ≥ 0 under two conditions are shown in the table. Find vc(t) for t ≥ 0 under condition 3 (in V). Condition vc(0-) vs(t) vc(t) 1 5 V 0 2t5e V− 2 0 10u(t) V ()2t4 1 e V−− 3 10 V 15u(t) V ? (1) 4 + e-2t (2) 4 − e-2t (3) 5 + e-2t (4) 5 − e-2t (5) 6 + 4e-2t (6) 6 − 4e-2t (7) 9 + 4e-2t (8) 9 − 4e-2t12. For the circuit shown below, what is the time it takes for vC(t) to rise from 0V to 5V (in second). Assume vC(0+) = 0V. (1) 1⋅ln(1.5) (2) 1⋅ln(2.0) (3) 1⋅ln(2.5) (4) 1⋅ln(3.0) (5) 2⋅ln(1.5) (6) 2⋅ln(2.0) (7) 2⋅ln(2.5) (8) 2⋅ln(3.0) 10V2R k= Ω1C mF=13. For the circuit shown below, the switch is closed for a long time before it opens at time t=0. Find vC(0+), iL(0+), and ω for t ≥ 0. (1) vC(0+) = 0V; iL(0+)=0.5A; ω=0.25 rad/s (2) vC(0+) = 0V; iL(0+)=0.5A; ω=0.50 rad/s (3) vC(0+) = 0V; iL(0+)=0.5A; ω=0.75 rad/s (4) vC(0+) = 0V; iL(0+)=0.5A; ω=1.00 rad/s (5) vC(0+) = 5V; iL(0+)=0.0A; ω=0.25 rad/s (6) vC(0+) = 5V; iL(0+)=0.0A; ω=0.50 rad/s (7) vC(0+) = 5V; iL(0+)=0.0A; ω=0.75 rad/s (8) vC(0+) = 5V; iL(0+)=0.0A; ω=1.00 rad/s 2L H=2C
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