Homework
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Modifications
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Answers:
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Due
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Problem Set #1 Solution
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1.a. i1=-1.5A, i2=-0.5A, Vo=-20V, b. p_2A=-130W, p_10ohm=40W,
p_30ohm=67.5W, p_40ohm=10W, p_50ohm=12.5W. 2. i1=-2A, i2=-1A, Vo=-10V. 3. vx=-500V, i1=30mA, power=-30W.
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T: May 22,2007
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Problem Set #2 Solution
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Take v1 to be
across resistor R1 for #3.
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1. V1=60V. 2. i1=150microA.
3. V1=(iaR3-va)*(R1/(R1+R2+R3)).
4. V1=-Va*R1/(R1+(1+alpha)R2) 5. vo=vs+isR2
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W: May 23, 2007
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Problem Set #3 Solution
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#4 Hint: remember to
replace Vx with an equation containing your mesh current
variables.
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1. Vo=150/67V, i1=50/67A; 2. Vx=(-6/44)V, Vy=-30V, power=36/85,184W;
3. V1=2V, V2=5V, iL=1mA; 4. i1=(21/8k)A, power=-21mW
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W: June 6, 2007
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Problem Set #4 Solution
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#4 Change the dependent
current source into a dependent voltage
source of value aix.
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4. Rth=R1+R2+alpha;
Vth=isR1; 5.
power=-27mW;
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W: June 6, 2007
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Problem Set #5 Solution
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1. R2=4k, R3=2k; 2. i(t)=e^(-9tsec);
3. vo(t)=24+36e^(-t/12);
4. vo(t)=27-9e^(-t/(3/8))
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W: 6/27/2007
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Problem Set #6 Solution
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1.a. wL=5.625nJ, b. i(t)=3.75e-3e^(-t/250n);
2.a. Vc(t)=-Vs+Vs(R2/(R1+R2))*[1-e^(-t/(R1||R2+R3)C];
3.RL=1.8k, power=293.9mW; 4.i=is/(1-alphaR2)
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W: 6/27/2007
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Problem Set #7 Solution
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1. Zab=(3.8+0.6j)ohms; 2. omega=447 rad/sec.; 3. I=4e^(-j39); 4. i_x(t)=1.92cos(4t+14)A
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W: 7/18/2007
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Problem Set #8 Solution
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Note that the
magnitude of a complex number is
taken without a sign. (e.g. take the absolute
value of the vector length).
(i.e. It is the
vector length not the direction of the vector)
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1. a. .352 –j0.936; b. 2e^(-j30); c. 3e^j150 or 3e^-j210; d. 10; e. 2; 2. a. R=30k or L=1.2H (this is
a huge value – not real practical); 3.
R=30k => Io=3sqrt(2)mA; L=1.2H => Io=12sqrt(2)mA. 4. Vth=5e^j0
V, Z_Th=1.25k-j3k
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W: 7/18/2007
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Problem Set #9 Solution
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1. a. vo=-Vs1(R2/R1)+Vs2(1+R2/R1); 2.
a.R2=300kohm, b.Vo=vsum – Vdelta(1+2R2/R1),
c.Rin=R3; 3. a.C=0.1microF;
5.a. Vth=12,
Rth=2.4k, b. RL=2.4k, c.
Pmax=15mW; 6. a.
I1=1/2(1+j)milliamp, b. i(t)=sqrt(2)/2cos(20kt+45)milliamp
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W: 8/1/2007
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