Instructor | TA | |
Dr. Angela Rasmussen Office Hours: R 9:30-11am |
Shomit Das Email: das.shomit@gmail.com Office: MEB 4504 Office Hours: R 11-12 |
On the following practice examples for the final, use R_T for the op-amp problems as a symbolic value and solve for an equation for Vo. Ignore other parts to that particular problem.
Unit 1 | Unit 2
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Unit 3
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Unit 4 |
The following is NOT free from errors! Please email me if you find
discrepancies in your work af
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1. i1=7.5A, i2=2.5A, i3=2.5A; 4. Ix=2.5A; 5. I1=8A, I2=2A; 6. Vc=50V; 7. i1=(iaR2+Va)/(R2+R3); 8. ia=-Va/((1-alpha)R1); 9. i1=250mA, i2=150mA, i3=100mA, Vo=1.5V, p_24ohm=1.5W, p6V=-3W; 10. i1=6A, i2=3.6A, i3=2.4A, Vo=31.2V | R, May 28 | |
+is+I1=0 instead of -is+I1=0 |
1. i1=2.5mA, 2. V1=6V, 3. Req=10kW, 4. i1=(vs+isR2)/(R1+R2+R3); 6. V2=(iaR1R2)/(R1+R2-alpha); 8. i1=(iaR2+Va)/(R2+R3) ; 9. vo=-isR3(1+R2/R1); 10. vo=-is(R1+R2) | R, May 28 | |
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3. i_8k=-3.125m, i_10k=-2.5m, i_5k=-0.8m, i_6k=-2.5m, i_2k=3m, i_7k=-0.43m; 4. (a) 1A, (b) -1V, (c) -1A; 6. i1=4A; 7. Ix=100mA; 8. v1=0, v2=28, P=-18W; 9. v1=2, v2=20; 10. vb=9.3, ve=8.6, iL=20mA | T, June 9 | |
#6 should read V2 instead of R1 at top right node. |
1. i1=2mA, i2=1mA, i3=2A, 2. Vx=108.6, power=72.3W, 3. Vth=-2V, Rth=3ohm; 4. Rth=36/17, Vth=8V; 5. power=2.5mW; 8. Vth=isR2, Rth=((1+alpha)R2+R1)/(1+alpha), 10. power=-60mW | T, June 16 | |
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1.a) 0A, (b) 420e-6A, (c) 5me^(-t/2e-6); 2.(a) 0, (b) 90pV, (c) -480pi*sin(2pi*5*t)microV; 3. t=6.93e-6; 4.i_L(t)=1/L*(6t-100m(e^(-t/50m)-1); 5. v_c(t)=3e^(-t/9ms)V; 6. i_L(t)=3me^(-6t/5e-6); 7. vc(t)=15e^(-t/6m); 8.(a) v_c(t)=9-3e^(-t/10m), (b) wc=31.2e-6J; 9.(a) i_L(t)=900e-6+[6-900e-6]e^(-t/0.1n)A | T, June 30 | |
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1. i1(t)=-30micro(e^-t/150nsec)A, 2. V1(t)=40/3(1-e^-6t/16msec), 3. i1(t)=1me^-t/1microsec, 4. i1(t)=0, 5. V=isR2+(alphaVsR2)/(alphaR3+R3+R4), 6. wL=30pF, v(t)=0.24+0.96e^-t/10microsec, 7. Vc(t)=is(R2R3)/(R1+R2+R3)+[isR3-is(R2R3)/(R1+R2+R3)]e^-t/(R3||(R1+R2)C, 8. RL=15k 9. power=60mW, 10. i=(alpha(vs+isR4))/(R1+R2)(R3+R4) | R, July 2 | |
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2a. 0.3520-j0.936, b. 2e^(-j30), c.3e^(j150), d. 10, e. 2 4. 3.8+j0.6, 6.V=120e^(-30), 7. I=4e^(-39), 8. V=120, 9. I=8e^(-j0.473), 10. i(t)=0.936cos(10kt+40) | T, July 21 | |
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1. R=30k or L=1.2H, 2. (R=30k): Io=3sqrt(2)mA, (L=1.2H): Io=12sqrt(2)mA, 3. L=2.3microH, 4. Io=5.2A, 6. Vth=5, Zth=1.25k-j3k, 8. Vth=1kV, Zth=50sqrt(2)e^(-135), 9. Vth=4sqrt(2), Zth=sqrt(2)ke^(-45), 10 L=27.32H, Io=10mA | R, July 23 | |
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1. Vo=-is1R2+is2*(R3||R4)*(1+R2/R1); 2. R2=-100k; 3. Rin=R3||R4; 4. Vth=2.2, Rth=200; 5. RL=200, Pmax=6.05mW | T, Aug 4 | |
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1. a=C=26.5nF, b=R=3k; 4. Vab=3e^-j60degree; 5. Vab(t)=3cos(500kt-60degrees)V | T, Aug 4 |
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