Nilsson riedel electric circuits 9th solutions

INSTRUCTOR'S SOLUTION MANUAL www.elsolucionario.net http://www.elsolucionario.net LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS www.elsolucionario.net Circuit Variables Assessment Problems AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters per second to miles per second: × 108 m 100 cm in ft mile 124,274.24 miles · · · · = 1s 1m 2.54 cm 12 in 5280 feet 1s Now set up a proportion to determine how long it takes this signal to travel 1100 miles: 124,274.24 miles 1100 miles = 1s xs Therefore, x= 1100 = 0.00885 = 8.85 × 10−3 s = 8.85 ms 124,274.24 AP 1.2 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond We begin by expressing $10 billion in scientific notation: $100 billion = $100 × 109 Now we determine the number of milliseconds in one year, again using a product of ratios: year day hour sec year · · · · = 365.25 days 24 hours 60 mins 60 secs 1000 ms 31.5576 × 109 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: $100 × 109 year 100 · = = $3.17/ms year 31.5576 × 10 ms 31.5576 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval 1–1 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 1–2 CHAPTER Circuit Variables AP 1.3 Remember from Eq (1.2), current is the time rate of change of charge, or i = dq In this problem, we are given the current and asked to find the total dt charge To this, we must integrate Eq (1.2) to find an expression for charge in terms of current: t q(t) = i(x) dx We are given the expression for current, i, which can be substituted into the above expression To find the total charge, we let t → ∞ in the integral Thus we have ∞ qtotal = = 20e−5000x dx = 20 −5000x ∞ 20 e = (e−∞ − e0) −5000 −5000 20 20 (0 − 1) = = 0.004 C = 4000 µC −5000 5000 AP 1.4 Recall from Eq (1.2) that current is the time rate of change of charge, or i = dq In this problem we are given an expression for the charge, and asked to dt find the maximum current First we will find an expression for the current using Eq (1.2): i= = dq d t = − + e−αt 2 dt dt α α α d d t −αt d −αt − e − e dt α dt α dt α2 −αt t e − α e−αt − −α e−αt α α α = 0− = − 1 −αt +t+ e α α = te−αt Now that we have an expression for the current, we can find the maximum value of the current by setting the first derivative of the current to zero and solving for t: di d = (te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = dt dt Since e−αt never equals for a finite value of t, the expression equals only when (1 − αt) = Thus, t = 1/α will cause the current to be maximum For this value of t, the current is i= −α/α e = e−1 α α © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems 1–3 Remember in the problem statement, α = 0.03679 Using this value for α, i= e−1 ∼ = 10 A 0.03679 AP 1.5 Start by drawing a picture of the circuit described in the problem statement: Also sketch the four figures from Fig 1.6: [a] Now we have to match the voltage and current shown in the first figure with the polarities shown in Fig 1.6 Remember that 4A of current entering Terminal is the same as 4A of current leaving Terminal We get (a) v = −20 V, (c) v = 20 V, i = −4 A; (b) v = −20 V, i = −4 A; (d) v = 20 V, i = 4A i = 4A [b] Using the reference system in Fig 1.6(a) and the passive sign convention, p = vi = (−20)(−4) = 80 W Since the power is greater than 0, the box is absorbing power [c] From the calculation in part (b), the box is absorbing 80 W AP 1.6 [a] Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig 1.5, p = vi To find the time at which the power is maximum, find the first derivative of the power with respect to time, set the resulting expression equal to zero, and solve for time: p = (80,000te−500t)(15te−500t) = 120 × 104 t2 e−1000t dp = 240 × 104 te−1000t − 120 ì 107 t2e1000t = dt â 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 1–4 CHAPTER Circuit Variables Therefore, 240 × 104 − 120 × 107 t = Solving, t= 240 × 104 = × 10−3 = ms 120 × 107 [b] The maximum power occurs at ms, so find the value of the power at ms: p(0.002) = 120 × 104 (0.002)2 e−2 = 649.6 mW [c] From Eq (1.3), we know that power is the time rate of change of energy, or p = dw/dt If we know the power, we can find the energy by integrating Eq (1.3) To find the total energy, the upper limit of the integral is infinity: ∞ wtotal = 120 × 104 x2e−1000x dx 120 × 104 −1000x = e [(−1000)2 x2 − 2(−1000)x + 2) (−1000)3 =0− ∞ 120 × 104 e (0 − + 2) = 2.4 mJ (−1000)3 AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and thus entering the lower terminal where the polarity marking of the voltage is negative Thus, using the passive sign convention, p = −vi Substituting the values of voltage and current given in the figure, p = −(800 × 103 )(1.8 × 103 ) = −1440 × 106 = −1440 MW Thus, because the power associated with the Oregon end of the line is negative, power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems 1–5 Chapter Problems P 1.1 [a] We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First, recall that mm = 103 µm Let’s also express the rate of growth of bamboo using the units mm/s instead of mm/day Use a product of ratios to perform this conversion: 250 mm day hour 250 10 · · · = = mm/s day 24 hours 60 60 sec (24)(60)(60) 3456 Use a ratio to determine the time it takes for the bamboo to grow 10 µm: 10/3456 × 10−3 m 10 × 10−6 m = 1s xs [b] P 1.2 so x= 10 × 10−6 = 3.456 s 10/3456 × 10−3 cell length 3600 s (24)(7) hr · · = 175,000 cell lengths/week 3.456 s hr week Volume = area × thickness Convert values to millimeters, noting that 10 m2 = 106 mm2 106 = (10 × 106 )(thickness) ⇒ thickness = 106 = 0.10 mm 10 × 106 P 1.3 (260 × 106 )(540) = 104.4 gigawatt-hours 109 P 1.4 [a] 20,000 photos x photos = (11)(15)(1) mm mm3 x= [b] 16 × 230 bytes x bytes = (11)(15)(1) mm3 (0.2)3 mm3 x= P 1.5 (20,000)(1) = 121 photos (11)(15)(1) (16 × 230 )(0.008) = 832,963 bytes (11)(15)(1) (480)(320) pixels bytes 30 frames · · = 9.216 × 106 bytes/sec frame pixel sec (9.216 × 106 bytes/sec)(x secs) = 32 × 230 bytes x= 32 × 230 = 3728 sec = 62 ≈ hour of video 9.216 ì 106 â 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 1–6 CHAPTER Circuit Variables 5280 ft 2526 lb kg · · = 20.5 × 106 kg mi 1000 ft 2.2 lb P 1.6 (4 cond.) · (845 mi) · P 1.7 w = qV = (1.6022 × 10−19 )(6) = 9.61 × 10−19 = 0.961 aJ P 1.8 n= P 1.9 C/m3 = 35 × 10−6 C/s = 2.18 × 1014 elec/s 1.6022 × 10−19 C/elec 1.6022 × 10−19 C 1029 electrons × = 1.6022 × 1010 C/m3 electron 1m Cross-sectional area of wire = (0.4 × 10−2 m)(16 × 10−2 m) = 6.4 × 10−4 m2 C/m = (1.6022 × 1010 C/m3)(6.4 × 10−4 m2 ) = 10.254 × 106 C/m Therefore, i C C m = (10.254 × 106 ) × avg vel sec m s Thus, average velocity = P 1.10 i 1600 = = 156.04 àm/s 10.254 ì 10 10.254 ì 106 First we use Eq (1.2) to relate current and charge: i= dq = 20 cos 5000t dt Therefore, dq = 20 cos 5000t dt To find the charge, we can integrate both sides of the last equation Note that we substitute x for q on the left side of the integral, and y for t on the right side of the integral: q(t) q(0) dx = 20 t cos 5000y dy We solve the integral and make the substitutions for the limits of the integral, remembering that sin = 0: q(t) − q(0) = 20 sin 5000y 5000 t = 20 20 20 sin 5000t − sin 5000(0) = sin 5000t 5000 5000 5000 But q(0) = by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = × 10−3 sin 5000t C = sin 5000t mC © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems P 1.11 1–7 [a] In Car A, the current i is in the direction of the voltage drop across the 12 V battery(the current i flows into the + terminal of the battery of Car A) Therefore using the passive sign convention, p = vi = (30)(12) = 360 W Since the power is positive, the battery in Car A is absorbing power, so Car A must have the ”dead” battery [b] w(t) = t p dx; = 60 s 60 w(60) = 360 dx w = 360(60 − 0) = 360(60) = 21,600 J = 21.6 kJ P 1.12 p = (12)(100 × 10−3 ) = 1.2 W; hr · t w(t) = P 1.13 p = vi; 3600 s = 14,400 s hr 14,400 p dt w= w(14,400) = t 0 1.2 dt = 1.2(14,400) = 17.28 kJ p dx Since the energy is the area under the power vs time plot, let us plot p vs t Note that in constructing the plot above, we used the fact that 40 hr = 144,000 s = 144 ks p(0) = (1.5)(9 × 10−3 ) = 13.5 × 10−3 W p(144 ks) = (1)(9 × 10−3 ) = × 10−3 W w = (9 × 10−3 )(144 × 103 ) + (13.5 × 10−3 − × 10−3 )(144 × 103 ) = 1620 J P 1.14 Assume we are standing at box A looking toward box B Then, using the passive sign convention p = −vi, since the current i is flowing into the − terminal of the voltage v Now we just substitute the values for v and i into the equation for power Remember that if the power is positive, B is absorbing power, so the power must be flowing from A to B If the power is negative, B is generating power so the power must be flowing from B to A © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 1–8 CHAPTER Circuit Variables [a] p = −(125)(10) = −1250 W [b] p = −(−240)(5) = 1200 W [c] p = −(480)(−12) = 5760 W 1250 W from B to A 1200 W from A to B 5760 W from A to B [d] p = −(−660)(−25) = −16,500 W P 1.15 16,500 W from B to A [a] p = vi = (40)(−10) = −400 W Power is being delivered by the box [b] Entering [c] Gaining P 1.16 [a] p = vi = (−60)(−10) = 600 W, so power is being absorbed by the box [b] Entering [c] Losing P 1.17 [a] p = vi = (0.05e−1000t )(75 − 75e−1000t) = (3.75e−1000t − 3.75e−2000t) W dp = −3750e−1000t + 7500e−2000t = dt = e1000t so ln = 1000t so thus 2e−2000t = e−1000t p is maximum at t = 693.15 µs pmax = p(693.15 µs) = 937.5 mW ∞ [b] w = = P 1.18 [3.75e−1000t − 3.75e−2000t] dt = 3.75 −1000t 3.75 −2000t ∞ e − e −1000 −2000 3.75 3.75 − = 1.875 mJ 1000 2000 [a] p = vi = 0.25e−3200t − 0.5e−2000t + 0.25e−800t p(625 µs) = 42.2 mW [b] w(t) = t (0.25e−3200t − 0.5e−2000t + 0.25e−800t ) = 140.625 − 78.125e−3200t + 250e−2000t − 312.5e−800t µJ w(625 às) = 12.14 àJ â 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems 18–29 For V2 = 0: V1 = (50 + j50)I1 − j150I2 = −j150I1 + (400 + j800)I2 50 + j50 ∆= −j150 = 2500(1 + j24) −j150 400 + j800 50 + j50 V1 N2 = I2 = −j150 = j150V1 N2 j150V1 = ∆ 2500(1 + j24) a12 = −V1 I2 = V2 =0 −50 (24 − j1) Ω j150I1 = (400 + j800)I2 a22 = − [b] VTh = I1 I2 = − (2 − j1) V2 =0 Vg 260/0◦ (260/0◦ )6 1560/0◦ = = = a11 + a21Zg (−1 + j1)/3 + j25/150 −2 + j2 + j1 −2 + j3 = 120(−2 − j3) = 432.47/ − 123.69◦ V ZTh = = a12 + a22Zg [−(50/3)(24 − j1)] + [(−8/3)(2 − j1)(25)] = a11 + a21Zg [(−1 + j1)/3] + [(j/150)(25)] −100(24 − j1) − 16(2 − j1)(25) −3200 + j500 = −2 + j2 + j1 −2 + j3 = 607.69 + j661.54 Ω © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 18–30 CHAPTER 18 Two-Port Circuits [c] V2 = 1000 (432.67/ − 123.69◦ ) = 248.88/ − 146.06◦ 1607.69 + j661.54 v2 (t) = 248.88 cos(4000t − 146.06◦ ) V P 18.33 [a] ZTh = g22 − g12 g21 Zg + g11 Zg 1 g12 g21 = − + j 2 1 −j =j 2 + g11Zg = + − j1 = − j1 · ZTh = 1.5 + j2.5 − · ZL = 2.1 − j1.3 Ω j3 = 2.1 + j1.3 Ω − j1 V2 g21 ZL = Vg (1 + g11Zg )(g22 + ZL ) − g12 g21Zg g21 ZL = 1 −j (2.1 − j1.3) = 0.4 − j1.7 2 + g11Zg = + − j1 = − j1 g22 + ZL = 1.5 + j2.5 + 2.1 − j1.3 = 3.6 + j1.2 g12 g21 Zg = j3 V2 0.4 − j1.7 0.4 − j1.7 = = Vg (2 − j1)(3.6 + j1.2) − j3 8.4 − j4.2 V2 = 0.4 − j1.7 (42/0◦ ) = − j6 V(rms) = 7.81/ − 50.19◦ V(rms) 8.4 − j4.2 The rms value of V2 is 7.81 V −V2 −5 + j6 [b] I2 = = = −3 + j1 A(rms) ZL 2.1 − j1.3 P = |I2|2 (2.1) = 21 W © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems [c] 18–31 I2 −g21 = I1 g11ZL + ∆g 1 −j 6 ∆g = = 1 +j − −j 2 2 1 − +j 2 5 +j −j + −j = −j 12 12 12 12 3 g11 ZL = 0.8 3.4 −j (2.1 − j1.3) = −j 6 6 · g11ZL + ∆g = 0.8 3.4 −j + − j = 0.8 − j0.9 6 6 I2 −[(1/2) − j(1/2)] = I1 0.8 − j0.9 · I1 = (0.8 − j0.9)I2 = −0.5 + j0.5 1.6 − j1.8 I2 −1 + j1 = (−1.7 + j0.1)(−3 + j1) = − j2 A (rms) · Pg (developed) = (42)(5) = 210 W % delivered = 21 (100) = 10% 210 P 18.34 V1 = h11I1 + h12 V2 I2 = h21I1 + h22V2 From the first measurement: h11 = V1 = × 103 = 800 Ω I1 h21 = −200 = −40 · V1 = 800I1 + h12 V2 ; I2 = −40I1 + h22V2 From the second measurement: h22V2 = 40I1 h22 = 40(20 ì 106 ) = 20 àS 40 â 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 18–32 CHAPTER 18 Two-Port Circuits 20 × 10−3 = 800(20 × 10−6 ) + 40h12 · h12 × 10−3 = = 10−4 40 Summary: h12 = 10−4 ; h11 = 800 Ω; h21 = −40; h22 = 20 µS From the circuit, Zg = 250 Ω; ZTh = Vg = 5.25 mV h11 + Zg h22 Zg + ∆h ∆h = 800(20 × 10−6 ) + 40 × 10−4 = 20 × 10−3 ZTh = VTh = i= 20 × 800 + 250 = 42 kΩ + 20 × 10−3 10−6 (250) −h21Vg 40(5.25 × 10−3 ) = = 8.4 V h22 Zg + ∆h 25 × 10−3 8.4 = 0.10 mA 84,000 P = (0.10 × 10−3 )2 (42,000) = 420 µW P 18.35 When V2 = V1 = 20 V, I1 = A, I2 = −1 A When I1 = V2 = 80 V, V1 = 400 V, I2 = A © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems h11 = V1 I1 h12 = V1 V2 h21 = I2 I1 h22 = I2 V2 ZTh = = 20 = 20 Ω = 400 =5 80 = −1 = −1 = = 37.5 mS 80 V2 =0 I1 =0 V2 =0 I1 =0 18–33 Zg + h11 = 10 Ω h22 Zg + ∆h Source-transform the current source and parallel resistance to get Vg = 240 V Then, I2 = h21Vg = −1.5 A (1 + h22 ZL )(h11 + Zg ) − h12h21 ZL P = (−1.5)2(10) = 22.5 W P 18.36 [a] g11 = I1 V1 ; I2 =0 V1 = I1 sC = g21 = sL + sC = V2 V1 I2 =0 [sL + (1/sC)](1/sC) sL + (2/sC) sL + (1/sC) s2 LC + (1/C)[s2 + (1/LC)] = = s2LC + sC(s2LC + 2) s[s2 + (2/LC)] Cs[s2 + (2/LC)]) s2 + (1/LC) · g11 = V2 = (1/sC) V1 sL + (1/sC) · g21 = so V2 (1/sC) (1/LC) = = = V1 sL + (1/sC) s LC + s + (1/LC) (1/LC) s2 + (1/LC) © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 18–34 CHAPTER 18 Two-Port Circuits g12 = I1 = I1 I2 ; g22 = V1 =0 −(1/sC) I2 sL + (1/sC) g22 = sL (1/sC) = V2 I2 so V1=0 g12 = −(1/LC) + (1/LC) s2 sL/sC sL (1/C)s = = sL + (1/sC) s LC + s + (1/LC) Summary: g11 = g21 = [b] Cs[s2 + (2/LC)]) ; s2 + (1/LC) s2 (1/LC) ; + (1/LC) g12 = g22 = s2 −(1/LC) + (1/LC) s2 (1/C)s + (1/LC) 109 = = 25 × 106 LC (0.2)(200) g11 = g12 = g21 = g22 = × 10−7 s(s2 + 50 × 106 ) s2 + 25 × 106 −25 × 106 s2 + 25 × 106 25 × 106 s2 + 25 × 106 × 106 s s2 + 25 × 106 V2 g21 ZL = = V1 g22 + ZL 25×106 400 s2 +25×106 5×10 s + 400 (s2 +25×106 ) V2 25 × 106 25 × 106 = = V1 s + 12,500s + 25 × 106 (s + 2500)(s + 10,000) V1 = 30 s V2 = 750 × 106 30 40 10 = − + s(s + 2500)(s + 10,000) s s + 2500 s + 10,000 v2 = [30 − 40e−2500t + 10e−10,000t]u(t) V © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems P 18.37 [a] y11 = I1 V1 ; y21 = V2 =0 s s · I1 s2 + = V1 s(s2 + 2) y11 = I2 = · V2 =0 s(s2 + 1) + s I1 s2 + V1 = s + I1 = I2 V1 18–35 −(1/s) −1 s2 + −1 I1 = · V1 = V1 s + (1/s) s + s(s + 2) s(s2 + 2) y21 = −1 s(s2 + 2) Because the two-port circuit is symmetric, y12 = y21 = [b] −1 s(s2 + 2) and y22 = y11 = s2 + s(s2 + 2) V2 y21 Zg = Vg y12 y21Zg ZL − (1 + y11Zg )(1 + Y22 ZL ) = y21 y12 y21 − (1 + y11)(1 + y22) −1 s(s2 + 2) = s2 + − + s2 (s2 + 2)2 s(s2 + 2) −s(s + 2) = − (s3 + s2 + 2s + 1)2 = = s3 + 2s2 1+ s2 + s(s2 + 2) + 3s + (s + 1)(s2 + s + 2) 50 s(s + 1)(s2 + s + 2) √ =− ±j 2 · V2 = s1,2 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 18–36 CHAPTER 18 Two-Port Circuits V2 = K1 K2 K3 K3∗ √ + √ + + s s+1 s+ −j s+ +j 2 2 K1 = 25; K3 = 9.45/90◦ K2 = −25; · v2(t) = [25 − 25e−t + 18.90e−0.5t cos(1.32t + 90◦ )]u(t) V P 18.38 The a parameters of the first two port are a11 = −5 × 10−3 −∆h = = −125 × 10−6 h21 40 a12 = −h11 −1000 = = −25 Ω h21 40 a21 = −25 −h22 = × 10−6 = −625 × 10−9 S h21 40 a22 = −1 −1 = = −25 × 10−3 h21 40 The a parameters of the second two port are a11 = ; a12 = or a11 = 1.25; 3R ; ; 4R a22 = a21 = mS; 96 a21 = a12 = 54 kΩ; a22 = 1.25 The a parameters of the cascade connection are a11 = −125 × 10−6 (1.25) + (−25)(10−3 /96) = −10−2 24 a12 = −125 × 10−6 (54 × 103 ) + (−25)(1.25) = −38 Ω a21 = −625 × 10−9 (1.25) + (−25 × 10−3 )(10−3 /96) = −10−4 S 96 a22 = −625 × 10−9 (54 × 103 ) + (−25 × 10−3 )(1.25) = −65 × 10−3 Vo ZL = Vg (a11 + a21Zg )ZL + a12 + a22Zg a21Zg = −10−4 −10−2 (800) = 96 12 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems a11 + a21Zg = 18–37 −10−2 −10−2 −10−2 + = 24 12 (a11 + a21Zg )ZL = −10−2 (72,000) = −90 a22Zg = −65 × 10−3 (800) = −52 Vo 72,000 = = −400 Vg −90 − 38 − 52 vo = Vo = −400Vg = −3.6 V P 18.39 [a] From reciprocity and symmetry a = a , ∆a = 1; · 52 − 24a 11 22 21 = 1, a21 = S For network B a11 = V1 V2 I2 =0 V1 = (5 + j15 − j10)I1 = (5 + j5)I1 V2 = (−j10 + j5)I1 = −j5I1 a11 = a21 = + j5 = −1 + j1 −j5 I1 V2 = I2 =0 = j0.2 S −j5 a22 = a11 = −1 + j1 ∆a = = (−1 + j1)(−1 + j1) − j0.2a12 · a12 = −10 + j5 Summary: a11 = a12 = 24 Ω a21 = S a22 = a11 = −1 + j1 a12 = −10 + j5 Ω a21 = j0.2 S a22 = −1 + j1 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 18–38 CHAPTER 18 Two-Port Circuits [b] a11 = a11a11 + a12a21 = −5 + j9.8 a12 = a11a12 + a12a22 = −74 + j49 Ω a21 = a21a11 + a22a21 = −1 + j2 S a22 = a21a12 + a22a22 = −15 + j10 I2 = −Vg = 0.295 + j0.279 A a11ZL + a12 + a21Zg ZL V2 = −10I2 = −2.95 − j2.79 V P 18.40 a11 = z11 35/3 = = 8.75 × 10−3 Ω z21 4000/3 a12 = 25 × 104 /3 ∆z = = 62.5 Ω z21 4000/3 a21 = 1 = = 0.75 × 10−3 Ω z21 4000/3 a22 = z22 10,000/3 = = 2.5 Ω z21 4000/3 a11 −y22 −40 × 10−6 = = 0.05 S = y21 −800 × 10−6 a12 = −1 −1 = = 1250 S y21 −800 × 10−6 a21 = −∆y −4 × 10−8 = = 50 × 10−6 S −6 y21 −800 × 10 a22 = −y11 −200 × 10−6 = = 0.25 S y21 −800 × 10−6 a11 = a11a11 + a12a21 = (8.75 × 10−3 )(0.05) + (62.5)(50 × 10−6 ) = 3.5625 × 10−3 a12 = a11a12 + a12a22 = (8.75 × 10−3 )(1250) + (62.5)(0.25) = 26.5625 a21 = a21a11 + a22a21 = (0.75 × 10−3 )(0.05) + (2.5)(50 × 10−6 ) = 162.5 × 10−6 a22 = a21a12 + a22a22 = (0.75 × 10−3 )(1250) + (2.5)(0.25) = 1.5625 V2 = = ZL Vg (a11 + a21Zg )ZL + a12 + a22Zg [3.5625 × 10−3 (15,000)(0.03) = 3.75 V + (162.5 ì 106 )(10)](15,000) + 26.5625 + (1.5625)(10) â 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems 18–39 P 18.41 [a] At the input port: V1 = h11 I1 + h12V2 ; At the output port: I2 = h21 I1 + h22 V2 [b] V2 + (100 × 10−6 V2 ) + 100I1 = 104 therefore I1 = −2 × 10−6 V2 V2 = 1000I1 + 15 × 10−4 V2 = −5 × 10−4 V2 100I1 + 10−4 V2 + (−2 × 10−6 )V2 = therefore I1 = 205 × 10−10 V2 Vg = 1500I1 + 15 × 10−4 V2 = 3000 × 10−8 V2 V2 105 = = 33,333 Vg P 18.42 [a] V1 = I2 (z12 − z21) + I1 (z11 − z21) + z21(I1 + I2) = I2 z12 − I2 z21 + I1z11 − I1z21 + z21I1 + z21I2 = z11I1 + z12I2 V2 = I2(z22 − z21) + z21(I1 + I2 ) = z21I1 + z22I2 [b] Short circuit Vg and apply a test current source to port as shown Note that IT = I2 We have V V + IT (z12 − z21) − IT + =0 z21 Zg + z11 − z21 Therefore V = Thus z21(Zg + z11 − z12) IT Zg + z11 and VT = V + IT (z22 − z21) VT z12 z21 = ZTh = z22 − IT Zg + z11 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 18–40 CHAPTER 18 Two-Port Circuits For VTh note that Voc = z21 Vg since I2 = Zg + z11 P 18.43 [a] V1 = (z11 − z12)I1 + z12(I1 + I2) = z11I1 + z12I2 V2 = (z21 − z12)I1 + (z22 − z12)I2 + z12(I2 + I1) = z21I1 + z22I2 [b] With port terminated in an impedance ZL , the two mesh equations are V1 = (z11 − z12)I1 + z12(I1 + I2 ) = ZL I2 + (z21 − z12)I1 + (z22 − z12)I2 + z12(I1 + I2) Solving for I1: I1 = V1 (z22 + ZL ) z11(ZL + z22) − z12z21 Therefore V1 z12z21 Zin = = z11 − I1 z22 + ZL P 18.44 [a] I1 = y11V1 + y21V2 + (y12 − y21)V2 ; I1 = y11V1 + y12 V2 ; I2 = y21V1 + y22V2 I2 = y12V1 + y22V2 + (y21 − y12)V1 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems 18–41 [b] Using the second circuit derived in part [a], we have where ya = (y11 + y12) and yb = (y22 + y12 ) At the input port we have I1 = ya V1 − y12(V1 − V2 ) = y11 V1 + y12V2 At the output port we have V2 + (y21 − y12)V1 + yb V2 − y12(V2 − V1 ) = ZL Solving for V1 gives V1 = + y22ZL V2 −y21ZL Substituting Eq (18.2) into (18.1) and at the same time using V2 = −ZL I2, we get y21 I2 = I1 y11 + ∆yZL P 18.45 [a] The g-parameter equations are I1 = g11 V1 + g12 I2 and V2 = g21 V1 + g22 I2 These equations are satisfied by the following circuit: [b] The g parameters for the first two port in Fig P 18.38(a) are g11 = h22 25 × 10−6 = = × 10−3 S ∆h × 10−3 g12 = −h12 −5 × 10−4 = = −0.10 ∆h × 10−3 g21 = −h21 −40 = = 8000 h ì 103 â 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 18–42 CHAPTER 18 Two-Port Circuits g22 = h11 1000 = = 200 kΩ ∆h × 10−3 From Problem 3.65 Ref = 72 kΩ, hence our circuit reduces to V2 = −8000V1 (72) 272 I2 = −V2 8V1 = 72,000 272 vg = mV · V1 − × 10−3 8V1 + V1 (5 × 10−3 ) − 0.1 =0 800 272 V1 − × 10−3 + 4V1 − 80V1 =0 34 · V1 = 3.4 × 10−3 V2 = −8000(72) × 3.4 × 10−3 = −7.2 V 272 From Problem 3.65 Vo = 0.5; · Vo = −3.6 V V2 This result matches the solution to Problem 18.38 P 18.46 [a] To determine b11 and b21 create an open circuit at port Apply a voltage at port and measure the voltage at port and the current at port To determine b12 and b22 create a short circuit at port Apply a voltage at port and measure the currents at ports and [b] The equivalent b-parameters for the black-box amplifier can be calculated as follows: 1 b11 = = −3 = 1000 h12 10 b12 = h11 500 = −3 = 500 kΩ h12 10 b21 = h22 0.05 = −3 = 50 S h12 10 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Problems b22 = 18–43 ∆h 23.5 = −3 = 23,500 h12 10 Create an open circuit a port Apply V at port Then, b11 = V2 V1 b21 = I2 V1 = = 1000 so V1 = mV measured V1 = I2 = 50 S 10−3 I1 =0 I1 =0 so I2 = 50 mA measured Create a short circuit a port Apply V at port Then, b12 = − V2 I1 b22 = − I2 I1 = −1 = 500 kΩ so I1 = −2 µA measured I1 = −I2 = 23,500 so I2 = 47 mA measured −2 × 10−6 V1 =0 V1=0 P 18.47 [a] To determine z11 and z21 create an open circuit at port Apply a current at port and measure the voltages at ports and To determine z12 and z22 create an open circuit at port Apply a current at port and measure the voltages at ports and [b] The equivalent z-parameters for the black-box amplifier can be calculated as follows: ∆h 23.5 z11 = = = 470 Ω h22 0.05 z12 = h12 10−3 = = 0.02 Ω h22 0.05 z21 = − z22 = h21 1500 =− = −30 kΩ h22 0.05 1 = = 20 Ω h22 0.05 Create an open circuit a port Apply mA at port Then, z11 = V1 I1 z21 = V2 I1 = V1 = 470 Ω so V1 = 470 mV measured 0.001 = V2 = −30,000 Ω so V2 = −30 V measured 0.001 I2 =0 I2 =0 Create an open circuit a port Apply A at port Then, z12 = V1 I2 z22 = V2 I2 = V1 = 0.02 Ω so V1 = 0.02 V measured = V2 = 20 Ω so V2 = 20 V measured I1 =0 I1 =0 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net ... Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Simple Resistive Circuits Assessment Problems AP 3.1 Start from the right hand side of the circuit and make series