EE 201 Selected Homework Answers

Spring 1997
Electric Circuits, Fifth Edition by Nilsson and Riedel, Addison-Wesley, Solutions Manual Copyright 1996, third Printing (ISBN 0-201-84711-6)

1.6 Ans: w=qV=1.92 aJ, 1.9 Ans: [a] p=-400 W, power is delivered by box, [b] Entering, [c] Gain

1.11 Ans: [a] p=-3000 W from B to A, [b] p=1200 W from A to B, [c] p=1080 W from A to B, [d] p=-9600 W from B to A

1.21 Ans: [a] pmax=2000 W, [b] pmax(extracting)=2000 W, [c] pavg=0, [d] pavg=254.65 W

1.26 Ans: SumPabs=Pdel, yes

2.2 Ans: [a] 8, [b] 6 [c] 4 [d] va-R1, vb-R3, [e] 6, [f] (1)va-R1-R4-R2, (2)vb-R2-R5-R3, (3)R4-R6-R5, (4)va-R1-R6-R3-vb (5)va-R1-R4-R5-R3-vb, (6)va-R1-R6-R5-R2, (7) vb-R2-R4-R6-R3

2.4 Ans: [a] Yes, independent v-source can carry whatever current connection requires; independent i-source can support any voltage required by the connection, [b] 30 V-source: abs, 10 V-source: del, 8 A-source: del, [c] P30V=240 W (abs), P10V=-80 W (del), P8A=-160W (del), SumPabs=SumPdel=240W, [d] Yes, 30V is del, 10V is del, 8A is abs, P30V=-240W del, P10V=-80W del, P8A=320W abs

2.13 Ans: [a] ig=9A, [b] p15ohm=240W, [c] vg=190 V, [d] SumPdiss=1710W, SumPdel=1710W, SumPdiss=SumPdel

2.22 Ans: [a] R=7 ohms, [b] pg(supplied)=1000 W

2.27 Ans: [a] vy=3.70 V, [b] SumPgen=109.38 mW, SumPdiss=109.38 mW

3.1 Ans: [a] i1=4 A, i2=8 A, i3=12 A, [b] v1=48 V, v2=72 V, v3=24 V, v4=48 V

3.2 Ans: [a] p4ohm=576 W, p3ohm=192 W, p18ohm=288 W, p6ohm=384 W, [b] p120V(delivered)=1440 W, [c] pdiss=1440 W

3.3 Ans: [a] Rab=15 ohms, [b] Req=24 ohms, Rab=20 ohms

3.4-(circuit b only) Ans: [a] Re=2 ohms, Rab=24 ohms, [b] Pb=864 W

3.8 Ans: [a] vg=252 V, i40ohm=3.6 A, vo=144 V, io=7.56 A, [b] P12ohm=846.72 W, [c] P12A(dev)=3024 W

3.16 Ans: [a] vo=47 V, [b] PR1=330 mW, PR2=470 mW, [c] R2=8836 ohms, R1=6204 ohms

3.26 Ans: [a] RA=20,008 uohms, [b] RA=50.05 mohms, [c] RA=100.2 mohms, [d] RA=2.83 ohms

3.30 Ans: ig=5.96 A, imeas=3.95 A, True: ig=6.0 A, itrue=4 A, %error=-5%

3.33 Ans: [a] RV=49,980 ohms, [b] RV=4980 ohms, [c] RV=230 ohms, [d] RV=5 ohms

3.34 Ans: [a] vmeter=126 V, [b] vmeter=72 V, [c] vmeter=18 V, [d] vmetera=126 V, vmeterb+vmeterc=90 V, no, because loading effect

3.35 Ans: [a] R1=29,950 ohms, R2=120 kohms, R3=150 kohms, [b] imove=0.96 mA, v1=144 V, i1=0.192 mA, i2=1.152 mA, vmeas=316.8 V [c] v1=150 V, i1=0.20 mA, i2=1.20 mA, vmeas=vx=330 V

3.43 Ans: [a] Rx=2000 ohms,[b] is=7.5 mA, [c] P=36 mW, P=94.5 mW

3.51 Ans: To balance the bridge, the ratio R2/R1 should be set to 0.001

3.54 Ans: [a] ig=12 A, i2=4 A, [b] i12.64=8 A, io=-160 mA, i1=7.84 A, [c] v=-19.84 V, [d] Psupplied=6480 W

4.3 Ans: v1=50 V, v2=180 V

4.7 Ans: v1=144 V, v2=64 V, ia=5.6 A, ib=0.6 A, ic=5 A, id=0.2 A, ie=4.8 A

4.9 Ans: [a] v1=101.24 V, v2=10.66 V, v3= -106.57 V, i1=23.76 A, i2=5.33 A, i3=18.43 A, i4=15 A, i5=9.77 A, i6=8.66 A [b] SumPdev=5273.09 W, SumPdiss=5273.09 W

4.11 Ans: v1=20 V, v2=80 V, pdel=180 W

4.15 Ans: vo=3.2 V

4.18 Ans: [a] 25vo-225+5vo=225, vo=15 V, [b] io=1.5 A, pds=11.25 W (abs), [c]p450mA(dev)=6.75 W, p45V(dev)=54 W, Sumpdev=60.75 W

4.19 Ans: p150isigma=-750 W, the dependent v-source delivers 750 W to ckt, 4.20 Ans: [a] io=3.4 A, pdev=8800 W

4.21 Ans: vdelta=15 V

4.27a Ans: [a] ia=9.8 A, ib=-0.2 A, ic=-10 A

4.35 Ans: P7iphi(developed)=2016 W

4.36 Ans: [a] vo=8 V, [b] p4idelta(deliver)=-1 W

4.72 Ans: [a] Ro(ohms) 0 2 6 10 15 20 Po(W): 0 200 337.5 360 345.6 320 Ro(ohams): 30 40 50 70 Po(W): 270 230.4 200 157.5 [b] Plot the above; P vs. R [c] Ro=10 ohms, Po(max)=360 W

4.73 Ans: Ro=Rth=5 kohms, [b] P=45 mW

4.82 Ans: [a] Since 0<=Ro<=infinity maximum power will be delivered to the 6 ohm resistor when Ro=0, [b] P=150 W

4.83 Ans: vo'=-90 V, io'=-2.25 A, vo''=360 V, ia=-4.8 A, io''=4.8 A, vo=270 V, io=2.55 A

4.88 Ans: vo'=30 V,vo''=20 V, vo=50 V

5.1 Ans: [a] va=1.0V, vb=0V, vo=-10V, [b] va=2.0V, vb=0V, vo=-12V (sat), [c] va=2.5V, vb=3V, vo=8V, [d] va=3.0V, vb=2V, vo=-8V, [e] va=1.5V, vb=2.5V, vo=12V (sat) [f] 3.2<=va<=5.6V

5.2 Ans: [a] vo=-4.7 V, [b] io=2.1 mA

5.4 Ans: io=-3.26 mA

5.6 Ans: [a] ia=18 uA, va=-720 mV, [b] vo=-5.76 V, [c] ia=18 uA, [d] ib=42 uA, io=54 uA

5.19 Ans: Ra=20kohms, Rb=12kohms, Rc=15kohms, Rd=30kohms, 5.20 Ans: [a] vo=10.54 V, [b] -4.55<=vg<=4.55 V, [c] Rf=181.76 kohms

5.21 Ans: [a] vo=(1+(R2/R1))vs, [b] vo=vs, [c] Because vp=vs, thus the output voltage follows the signal voltage

5.22 Ans: [a] p12kohms=19.2 uW, [b] v12kohms=72mV, p12kohms=0.432 uW, [c] pa/pb=44.44, [d] Yes power amplification, voltage follower, current amplification

5.27 Ans: Ra=200kohms, Rb=1.0Mohms

5.32 Ans: [a] 91.36vn-vo=90vg, (9E6)vn+37vo=0, vo/vg=-89.966, [b] vg=100mV, vn=36.986uV, [c] Rg=2000.74ohms, [d] vo/vg=90, vn=0V, Rg=2000 ohms

5.34 Ans: [a] 55v1-4vo=vg, vo/vg=13.49, [b] vn=999.45mV, vp=999.83mV, [c] vp-vn=387.78 uV, [d] ig=692.47 pA, [e] vo=13.5vg, vn=vp=1V, ig=0A

5.36 Ans: [a] -20vth+341vn=64, 26vth+249,999vn=0, vth=-3.1943V, isc=-2.3460A, Rth=1.3616 ohms

5.38 Ans: i1=1mA, i2=4mA, i3=2mA, i4=3mA, vo1=12.40 V

6.2 Ans: [a] i=0, t<=0, i=50t A, 0<=t<=10 ms, i(t)=1-50t A, 10 ms<=t<=20 ms, i(t)=0, 20 ms<=t<=infinity, [b] v=0, t<=0, v=2V, 0<t<10ms, v=-2V, 10ms<t<20ms, v=0, 20ms<t<infinity, p=0, t<=0, p=100t W, 0<t<10ms, p=-2+100t W, 10ms<t<20ms, p=0, 20ms<=t<=0, w=50t^2, 0<=t<=10ms, w=0.02-2t+50t^2, 10ms<=t<=20ms, w=0, 20ms<=t<=infinity

6.16 Ans: [a] q=CV=5 uC, [b] v(25 us)=20 V, [c] w=32 uJ

6.20 Ans: Lab=15H,

6.22 Ans: [a] io(0)=i1(0)+i2(0)=2A, [b] io=2e^-2.5t A, t=0 [c] i1=1.6e^-2.5t+4.4 A, [d] i2=0.4e-2.5t-4.4 A, [e] w(0)=282J, [f] wdel=40J, [g] wtrapped=242 J

7.1 Ans: [a] R=v/i=16 ohms, [b] tau=0.05 s=50 ms, [c] l=R/20=0.8 H, [d] w(0)=1/2(0.8)(12.5)^2=62.5 J, [e] to=12.77 ms

7.7 Ans: iL=4e^-200t A, t=0+, vo=-40e^-200t V, t=0+

7.8 Ans: wstored=0.60 J, % diss=33.33

7.17 Ans: [a] iL=-2.4e^-100t A, t=0, [b] vL=60e^-100t V, t=0+, [c] idelta=0.625iL=-1.5e^-100t A, t=0+

7.19 Ans: [a] iab=90 A, t=0+, [b] iab=240/2=120 A, t=infinity, [c] t=(1/2500)ln(3)=400ln(3) us; t=439.44 us

7.20 Ans: [a] R=v/i=20 kohms, [b] c=(1/(10^3)(20000))=0.05 uF, [c] tau=1 ms, [d] w(0)=250 uJ [e] to=~804.72 us

7.23 Ans: [a] io(t)=24e^-5000t mA, t=0+, [b] vo=(-8e^-5000t+80) V, t=0, [c] wtrapped=2880 uJ

7.27 Ans: [a] alpha=160 uA/V, [b] vphi=5.4e^-40t V, t=0+

7.33 Ans: [a] v(0+)=200=va, R=20 ohms, L=40 mH [B] to=2ln(2) ms=1.39 ms

7.47 Ans: [a] vo=-12+60e^-1250t V, t=0, [b] io=6e^-1250t mA, t=0+, [c] vg=-12+45e^-1250t V [d] vg(0+)=-12+45=33 V

7.36 Ans: io=8+2e^-125,000t mA, t=0, vo=-50e^-125,000t V, t=0+

7.50 Ans: vth=45 V, Rth=4 kohms, vo(t)=45-90e^-4000t V, t=0

7.67 Ans: vo(200 us)=13.24e^-0.5=8.03 V

7.80 Ans: R=30 kohms

Chapter 8:

8.1:

a.) s 1 = - 10,000 rad/s ; s 2 = -40,000 rad/s

b.) Overdamped

c.) R = 3125 W

d.) s 1,2 = - a ± j w d rad/s & s 1,2 = - 16,000 ± j 12,000 rad/s

e.) R = 2500 W

8.3:

a.) iL(0) = -45 mA, ; .) iR(0) = 75 mA ; .) iC(0) = -30 mA

b.) v(t) = 10 e –5,000 t + 5 e –20,000 t V, t ³ 0

c.) iL(t) = -40 e –5,000 t -5 e –20,000 t mA, t ³ 0

8.10:

    1. v(t) = 10 e –0.5 t sin(5t) V, t ³ 0

8.12:

a.) R = 2500 W

b.) v(t) = (80,000 t - 15) e –2,000 t V, t ³ 0

8.21:

iL(t) = (4 - 12 e – 40 t + 3 e – 160 t ) mA, t ³ 0

8.24:

iL(t) = 60 – 750,000t e –10,000 t - 105 e –10,000 t mA, t ³ 0

8.32:

iL (t) = 20 + 80 e –800 t cos(600t)- 127.5 e –800 t sin(600t) mA, t ³ 0

8.33:

a.) R = 25,000 W ; L = 5 H ; w o = 2,000 rad/s

  1. b.) vc(0) = 60 V Þ di(0)/dt = vc(0)/L = 12 A/s

    2. c.) i (t) = 4 e –1,000 t - 4 e –4,000 t mA, t ³ 0

  2. d.) 462.10 m s

e.) imax = 1.89 mA

f.) vi(t) = 5 [di(t)/dt] = [- 20 e –1,000 t + 80 e –4,000 t ] V, t ³ 0

8.36:

    1. R = 1250 W
    2. di(0)/dt = 60 A/s

c.) vc(t) = 56,250t e –5,000 t + 15 e –5,000 t V, t ³ 0

8.38:

i (t) = 280 e –1000 t cos(7000t) + 40 e –1000 t sin(7000t) mA, t ³ 0

8.42:

vc(t) = 52 + 78 e –2,000 t - 34 e –6,000 t V, t ³ 0

8.43:

vc(t) = 250 – (5 x 10 6) t e –25,000 t - 200 e –25,000 t V, t ³ 0

Solutions to assigned homework problems:

9.1:

    1. Vm = 80 V
    2. f = 1000p /2p = 500 Hz
    3. w = 1000p = 3141.59 rad/s
    4. q = - p /6 = - 0.5236 radians
    5. q = - 30°
    6. T = 2 ms
    7. 1000p t = p /6 , so t = 1/6 ms

9.4:

    1. w = 400p rad/s
    2. i = 20 cos (400p - 45° ) A

9.6:

a.) Y = 111.60 – j 6.70 = 111.80 Ð -3.43°

y = 111.80 cos (500t - 3.43° )

b.) Y = 78.56 + j 66.61 = 102.99 Ð 40.29°

y = 102.99 cos (377t + 40.29° )

c.) Y = 139.99 - j 80.71 = 161.59 Ð -29.96°

y = 161.59 cos (100t – 29.96° )

d.) Y = 0

y = 0

9.8

Vm = Ö 2Vrms = Ö 2(120) = 169.71 V

9.9

    1. 400 Hz
    2. 0°
    3. I = (200/w L) Ð - 90° \ q i = - 90°
    4. w L = 5W
    5. w = 800p rad/s ; L = 1.99 mH
    6. ZL = j 5 W

9.10:

    1. w = 502,654.82 rad/s
    2. I = (25 x 10 -3)w C Ð 90° , \ q i = 90°
    3. 1/w C = 39.79 W
    4. C = (0.05 x 10 -6) = 0.05 m F
    5. Zc = -j 39.79 W

9.11:

    1. V = 300Ð - 30° V
    2. I = 2.12Ð 15° A
    3. i = 2.12 cos (4000t + 15° ) A

      4. 9.12:

    4. I = 2.12Ð 15° A

9.15:

Z ab = 20 + j 15 W = 25 Ð 36.87° W

9.32:

    1. V g = 140 + j 330 V = Va = 5 Ic = 358.47Ð 67.01° V
    2. ib = 2.5 cos(800t + 90° ) A

ic = 10 cos(800t + 36.87° ) A

vg = 358.47 cos(800t + 67.01° ) V

9.34:

Y = 3 – j1 mmho ; Z = 300 + j 100 W

vo = 474.34 cos(8000t + 18.43° ) V

9.38:

Vo = 188.43Ð -42.88° V

9.40:

Vo = 0 + j 80 V

9.43:

io (t)= 28.28 cos(50,000t - 45° ) A

9.46:

vo(t)= 56.57 cos(10,000 t - 45° ) V

9.51:

VTh = 72 – j 24 V

ZTh = 1.8 + j 5.4 W

9.52:

IN = 6.4 – j 4.8 A

ZN = 50 - j 25 W

9.62:

Vo = Vm /2 – I Rx ; I = Vm/ (Rx – j Xc)

9.64:

vo(t)= 3.16 cos(107 t + 55.30° ) V

10.1:

    1. P = 1044.42 W (absorbed)

      2. Q = 731.31 VAR (absorbed)

    2. P = 85.51 W (absorbed)

      3. Q = - 234.92 VAR (delivered)

    3. P = - 573.58 W (delivered)

      4. Q = - 819.15 VAR (delivered)

    4. P = - 225 W (delivered)

Q = 389.71 VAR (absorbed)

10.4:

Load #1: Z = 13.27 Ð 25.23° W

q = 25.23°

pf = 0.90 Lagging

rf = 0.43

Load #2: Z = 184.08 Ð - 64.24° W

q = - 64.24°

pf = 0.43 Leading

rf = - 0.90

Load #3: Z = 696.90 Ð - 54.97° W

q = - 54.97°

pf = 0.57 Leading

rf = - 0.82

10.5:

I rms = 11.55 A

10.6:

R = 9.6 W

10.9:

    1. P g = 1084.24 W

      2. Q g = 271.06 VAR

    2. Delivering
    3. Absorbing
    4. Q cap = -1355.29 VAR

      5. P 12.5W = 271.06 W

      P 15W = 813.18 W

      |I o| = 7.36 A

      Q ind = 1084.24 VAR

    5. S P developed = 1084.24 W

      6. S P dissipated = 1084.24 W

    6. S Q developed = 1355.29 W

S Q absorbed = 1355.29 W

10.17:

    1. S T = s1 + s2 + s3 = 6000 + j 2000 VA

      2. I = 24 – j 8 A

      Z = 9.88 Ð 18.43° W

    2. pf = 0.9487 Lagging

      3. 10.21:

      Ig1 = 49.6 – j 11.92 A ; Ig2 = 47.68 – j 7.76 A

      Sg1 = -6200 – j 1490 VA ; Sg2 = 5960 – j 970 VA

      10.23:

      a.) S T = 6000 + j 2000 VA ; I = 24 – j 8 A (rms) Þ

      Vg = 253.52 + j 7.36 V = 253.63 Ð 1.66 ° V

      b.) Pl = 25.60 W

      Ql = 204.80 VAR

      c.) Ps = 6000 + 25.60 = 6025.60 W

      Qs = 2000 + 204.8 = 2204.80 VAR

    3. h = efficiency = 99.58 %

10.29:

    1. jw L = j 40 kW

      2. ZTh = 20 - j 20 kW

      ZL = ZTh* = 20 + j 20 kW

    2. VTh = 40 - j 40 V

IL = Ö 2 Ð -45° mA

P = 20 mW

10.30:

a.) Io = 242.29 Ð 12.99° mA

P = 2.94 W

b.) = RL = 200 W

RL

 

10.39:

    1. 1/w C = 100 Þ C = 26.53 m F
    2. Before removal: Vs = 4069.71 Ð 5.64° V (rms)

      3. After removal: Vs = 4463.73 Ð 4.50° V (rms)

      Percent change is: 9.68 %

    3. P LOSS (with) = 2000 W

P LOSS (without) = 4000 W

Percent increase is: 100 %

10.40:

Vg = - j 2 V

Vo = 4 + j 2 V = 4.47 Ð 26.57° V

P = 12.5 x 10 –3 W = 12.5 mW