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INSTRUCTOR'S SOLUTION MANUAL Circuit Variables 1 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: 2 3 × 108 m 100 cm 1 in 1 ft 1 mile 124,274.24 miles · · · · = 3 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: 1 year 1 day 1 hour 1 min 1 sec 1 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 1 year 100 · = = $3.17/ms 9 1 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. 1–2 CHAPTER 1. 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 do this, we must integrate Eq. (1.2) to find an expression for charge in terms of current: q(t) = t Z i(x) dx 0 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 = = Z ∞ 0 20e−5000x dx = 20 −5000x ∞ 20 e = (e−∞ − e0) −5000 −5000 0 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 1 t 1 = − + e−αt 2 2 dt dt α α α d 1 d t −αt d 1 −αt − e − e = 2 dt α dt α dt α2 1 −αt t 1 e − α e−αt − −α 2 e−αt = 0− α α α = − 1 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 = 0 dt dt Since e−αt never equals 0 for a finite value of t, the expression equals 0 only when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For this value of t, the current is i= 1 −α/α 1 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. Problems 1–3 Remember in the problem statement, α = 0.03679. Using this value for α, i= 1 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 2 is the same as 4A of current leaving Terminal 1. 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 × 107 t2e−1000t = 0 dt © 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. 1–4 CHAPTER 1. Circuit Variables Therefore, 240 × 104 − 120 × 107 t = 0 Solving, t= 240 × 104 = 2 × 10−3 = 2 ms 120 × 107 [b] The maximum power occurs at 2 ms, so find the value of the power at 2 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 = Z ∞ 0 120 × 104 x2e−1000x dx 120 × 104 −1000x = e [(−1000)2 x2 − 2(−1000)x + 2) (−1000)3 =0− ∞ 0 120 × 104 0 e (0 − 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. 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 1 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 1 day 1 hour 1 min 250 10 · · · = = mm/s 1 day 24 hours 60 min 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 1 cell length 3600 s (24)(7) hr · · = 175,000 cell lengths/week 3.456 s 1 hr 1 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 = 3 (11)(15)(1) mm 1 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 2 bytes 30 frames · · = 9.216 × 106 bytes/sec 1 frame 1 pixel 1 sec (9.216 × 106 bytes/sec)(x secs) = 32 × 230 bytes x= 32 × 230 = 3728 sec = 62 min ≈ 1 hour of video 9.216 × 106 © 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. 1–6 CHAPTER 1. Circuit Variables 5280 ft 2526 lb 1 kg · · = 20.5 × 106 kg 1 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 3 1 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 C C m Therefore, i = (10.254 × 106 ) × avg vel sec m s Thus, average velocity = P 1.10 i 1600 = = 156.04 µm/s 6 10.254 × 10 10.254 × 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: Z q(t) q(0) dx = 20 Z t 0 cos 5000y dy We solve the integral and make the substitutions for the limits of the integral, remembering that sin 0 = 0: q(t) − q(0) = 20 sin 5000y 5000 t = 0 20 20 20 sin 5000t − sin 5000(0) = sin 5000t 5000 5000 5000 But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = 4 × 10−3 sin 5000t C = 4 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. 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) = Z t 0 w(60) = p dx; Z 1 min = 60 s 60 360 dx 0 w = 360(60 − 0) = 360(60) = 21,600 J = 21.6 kJ P 1.12 p = (12)(100 × 10−3 ) = 1.2 W; w(t) = P 1.13 p = vi; Z 0 t p dt w= w(14,400) = Z 0 t 4 hr · Z 3600 s = 14,400 s 1 hr 14,400 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 ) = 9 × 10−3 W 1 w = (9 × 10−3 )(144 × 103 ) + (13.5 × 10−3 − 9 × 10−3 )(144 × 103 ) = 1620 J 2 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. 1–8 CHAPTER 1. Circuit Variables [a] p = −(125)(10) = −1250 W [b] p = −(−240)(5) = 1200 W 1250 W from B to A 1200 W from A to B [c] p = −(480)(−12) = 5760 W 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 = 0 dt 2 = e1000t so ln 2 = 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 = Z = P 1.18 ∞ 0 [3.75e −1000t − 3.75e −2000t 3.75 −1000t 3.75 −2000t ∞ ] dt = e − e −1000 −2000 0 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) = Z 0 t (0.25e−3200t − 0.5e−2000t + 0.25e−800t ) = 140.625 − 78.125e−3200t + 250e−2000t − 312.5e−800t µJ w(625 µs) = 12.14 µJ © 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. Problems 1–9 [c] wtotal = 140.625 µJ P 1.19 [a] 0 s ≤ t < 1 s: v = 5 V; i = 20t A; p = 100t W i = 20 A; p=0W 1 s < t ≤ 3 s: v = 0 V; 3 s ≤ t < 5 s: v = −5 V; i = 80 − 20t A; p = 100t − 400 W 5 s < t ≤ 7 s: v = 5 V; i = 20t − 120 A; p = 100t − 600 W t > 7 s: v = 0 V; i = 20 A; p=0W [b] Calculate the area under the curve from zero up to the desired time: P 1.20 w(1) = 1 (1)(100) 2 = 50 J w(6) = 1 (1)(100) 2 − 12 (1)(100) + 12 (1)(100) − 12 (1)(100) = 0 J w(10) = w(6) + 21 (1)(100) = 50 J [a] v(10 ms) = 400e−1 sin 2 = 133.8 V i(10 ms) = 5e−1 sin 2 = 1.67 A p(10 ms) = vi = 223.80 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. 1–10 CHAPTER 1. Circuit Variables [b] = vi = 2000e−200t sin2 200t 1 −200t 1 = 2000e − cos 400t 2 2 = 1000e−200t − 1000e−200t cos 400t p w Z = ∞ 1000e 0 −200t e−200t = 1000 −200 ( ∞ ∞ 0 1000e−200t cos 400t dt 0 e−200t −1000 [−200 cos 400t + 400 sin 400t] 2 + (400)2 (200) 200 = 5 − 1000 = 5−1 4 × 104 + 16 × 104 = 4 J w P 1.21 dt − Z ) ∞ 0 [a] p = vi = [16,000t + 20)e−800t][(128t + 0.16)e−800t ] = 2048 × 103 t2 e−1600t + 5120te−1600t + 3.2e−1600t = 3.2e−1600t[640,000t2 + 1600t + 1] dp dt = 3.2{e−1600t[1280 × 103 t + 1600] − 1600e−1600t [640,000t2 + 1600t + 1]} = −3.2e−1600t[128 × 104 (800t2 + t)] = −409.6 × 104 e−1600tt(800t + 1) dp Therefore, = 0 when t = 0 dt so pmax occurs at t = 0. [b] pmax = 3.2e−0 [0 + 0 + 1] = 3.2 W [c] w w 3.2 = = Z t Z0 t 0 pdx 640,000x2 e−1600x dx + Z 0 t 1600xe−1600x dx + t Z 0 t e−1600x dx 640,000e−1600x = [256 × 104 x2 + 3200x + 2] + −4096 × 106 0 t t 1600e−1600x e−1600x (−1600x − 1) + 256 × 104 −1600 0 0 When t → ∞ all the upper limits evaluate to zero, hence w (640,000)(2) 1600 1 = + + 3.2 4096 × 106 256 × 104 1600 w = 10−3 + 2 × 10−3 + 2 × 10−3 = 5 mJ. © 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. Problems P 1.22 [a] p = dp dt 1–11 vi = 400 × 103 t2 e−800t + 700te−800t + 0.25e−800t = e−800t[400,000t2 + 700t + 0.25] = {e−800t[800 × 103 t + 700] − 800e−800t [400,000t2 + 700t + 0.25]} = [−3,200,000t2 + 2400t + 5]100e−800t dp Therefore, = 0 when 3,200,000t2 − 2400t − 5 = 0 dt so pmax occurs at t = 1.68 ms. [b] pmax = [400,000(.00168)2 + 700(.00168) + 0.25]e−800(.00168) = 666 mW [c] w = w = Z t Z0 t 0 pdx 400,000x2 e−800x dx + Z 0 t 700xe−800x dx + t Z 0 t 0.25e−800x dx 400,000e−800x [64 × 104 x2 + 1600x + 2] + −512 × 106 0 t t 700e−800x e−800x (−800x − 1) + 0.25 64 × 104 −800 0 0 When t = ∞ all the upper limits evaluate to zero, hence (400,000)(2) 700 0.25 w= + + = 2.97 mJ. 512 × 106 64 × 104 800 = P 1.23 [a] p = vi = 2000 cos(800πt) sin(800πt) = 1000 sin(1600πt) W Therefore, pmax = 1000 W [b] pmax (extracting) = 1000 W [c] pavg 2.5×10−3 1 1000 sin(1600πt) dt 2.5 × 10−3 0 2.5×10−3 250 5 − cos 1600πt 4 × 10 = [1 − cos 4π] = 0 1600π π 0 Z = = [d] pavg = = P 1.24 [a] q Z 15.625×10−3 1 1000 sin(1600πt) dt 15.625 ×10−3 0 15.625×10−3 40 3 − cos 1600πt 64 × 10 = [1 − cos 25π] = 25.46 W 1600π π 0 = area under i vs. t plot = h = 1 (5)(4) 2 i + (10)(4) + 12 (8)(4) + (8)(6) + 12 (3)(6) × 103 [10 + 40 + 16 + 48 + 9]103 = 123,000 C © 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. 1–12 CHAPTER 1. Circuit Variables [b] w Z = p dt = Z vi dt v = 0.2 × 10−3 t + 9 0 ≤ t ≤ 15 ks 0 ≤ t ≤ 4000s i = 15 − 1.25 × 10−3 t = 135 − 8.25 × 10−3 t − 0.25 × 10−6 t2 p w1 = 4000 Z 0 (135 − 8.25 × 10−3 t − 0.25 × 10−6 t2) dt = (540 − 66 − 5.3333)103 = 468.667 kJ 4000 ≤ t ≤ 12,000 i = 12 − 0.5 × 10−3 t p = 108 − 2.1 × 10−3 t − 0.1 × 10−6 t2 w2 = 12,000 Z 4000 (108 − 2.1 × 10−3 t − 0.1 × 10−6 t2) dt = (864 − 134.4 − 55.467)103 = 674.133 kJ 12,000 ≤ t ≤ 15,000 i = p = w3 = wT P 1.25 30 − 2 × 10−3 t 270 − 12 × 10−3 t − 0.4 × 10−6 t2 Z 15,000 12,000 (270 − 12 × 10−3 t − 0.4 × 10−6 t2) dt = (810 − 486 − 219.6)103 = 104.4 kJ = w1 + w2 + w3 = 468.667 + 674.133 + 104.4 = 1247.2 kJ [a] We can find the time at which the power is a maximum by writing an expression for p(t) = v(t)i(t), taking the first derivative of p(t) and setting it to zero, then solving for t. The calculations are shown below: p p dp dt dp dt t1 = 0 t < 0, p = 0 t > 40 s = vi = t(1 − 0.025t)(4 − 0.2t) = 4t − 0.3t2 + 0.005t3 W 0 ≤ t ≤ 40 s = 4 − 0.6t + 0.015t2 = 0.015(t2 − 40t + 266.67) = 0 when t2 − 40t + 266.67 = 0 = 8.453 s; t2 = 31.547 s (using the polynomial solver on your calculator) p(t1 ) = 4(8.453) − 0.3(8.453)2 + 0.005(8.453)3 = 15.396 W p(t2 ) = 4(31.547) − 0.3(31.547)2 + 0.005(31.547)3 = −15.396 W Therefore, maximum power is being delivered at t = 8.453 s. [b] The maximum power was calculated in part (a) to determine the time at which the power is maximum: pmax = 15.396 W (delivered) © 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. Problems 1–13 [c] As we saw in part (a), the other “maximum” power is actually a minimum, or the maximum negative power. As we calculated in part (a), maximum power is being extracted at t = 31.547 s. [d] This maximum extracted power was calculated in part (a) to determine the time at which power is maximum: pmax = 15.396 W (extracted) [e] w = Z 0 t pdx = Z 0 t (4x − 0.3x2 + 0.005x3 )dx = 2t2 − 0.1t3 + 0.00125t4 w(0) = 0J w(30) = 112.5 J w(10) = 112.5 J w(40) = 0J w(20) = 200 J To give you a feel for the quantities of voltage, current, power, and energy and their relationships among one another, they are plotted below: © 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. 1–14 P 1.26 CHAPTER 1. Circuit Variables We use the passive sign convention to determine whether the power equation is p = vi or p = −vi and substitute into the power equation the values for v and i, as shown below: pa = vaia = (150 × 103 )(0.6 × 10−3 ) = 90 W pb = vbib = (150 × 103 )(−1.4 × 10−3 ) = −210 W pc = −vcic = −(100 × 103 )(−0.8 × 10−3 ) = 80 W pd = vdid = (250 × 103 )(−0.8 × 10−3 ) = −200 W pe = −veie = −(300 × 103 )(−2 × 10−3 ) = 600 W pf = vf if = (−300 × 103 )(1.2 × 10−3 ) = −360 W Remember that if the power is positive, the circuit element is absorbing power, whereas is the power is negative, the circuit element is developing power. We can add the positive powers together and the negative powers together — if the power balances, these power sums should be equal: X Pdev = 210 + 200 + 360 = 770 W; X Pabs = 90 + 80 + 600 = 770 W Thus, the power balances and the total power developed in the circuit is 770 W. P 1.27 pa = −vaia = −(990)(−0.0225) = 22.275 W pb = −vbib = −(600)(−0.03) = 18 W pc = vcic = (300)(0.06) = 18 W pd = vdid = (105)(0.0525) = 5.5125 W pe = −veie = −(−120)(0.03) = 3.6 W pf = vf if = (165)(0.0825) = 13.6125 W pg = −vgig = −(585)(0.0525) = −30.7125 W ph = vhih = (−585)(0.0825) = −48.2625 W Therefore, X Pabs = 22.275 + 18 + 18 + 5.5125 + 3.6 + 13.6125 = 81 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. Problems X Pdel = 30.7125 + 48.2625 = 78.975 W X Pabs 6= X 1–15 Pdel Thus, the interconnection does not satisfy the power check. P 1.28 [a] From the diagram and the table we have pa = −vaia = −(46.16)(−6) = −276.96 W pb = vbib = (14.16)(4.72) = 66.8352 W pc = vcic = (−32)(−6.4) = 204.8 W pd = −vdid = −(22)(1.28) = −28.16 W pe = −veie = −(33.6)(1.68) = −56.448 W pf = vf if = (66)(−0.4) = −26.4 W pg = vg ig = (2.56)(1.28) = 3.2768 W ph = −vhih = −(−0.4)(0.4) = 0.16 W X X Pdel = 276.96 + 28.16 + 56.448 + 26.4 = 387.968 W Pabs = 66.8352 + 204.8 + 3.2768 + 0.16 = 275.072 W Therefore, X Pdel 6= X Pabs and the subordinate engineer is correct. [b] The difference between the power delivered to the circuit and the power absorbed by the circuit is −387.986 + 275.072 = −112.896 W One-half of this difference is −56.448 W, so it is likely that pe is in error. Either the voltage or the current probably has the wrong sign. (In Chapter 2, we will discover that using KCL at the node connecting components b, c, and e, the current ie should be −1.68 A, not 1.68 A!) If the sign of pe is changed from negative to positive, we can recalculate the power delivered and the power absorbed as follows: X Pdel = 276.96 + 28.16 + 26.4 = 331.52 W X Pabs = 66.8352 + 204.8 + 56.448 + 3.2768 + 0.16 = 331.52 W Now the power delivered equals the power absorbed and the power balances for the circuit. P 1.29 [a] From an examination of reference polarities, elements a, e, f, and h use a + sign in the power equation, so would be expected to absorb power. Elements b, c, d, and g use a − sign in the power equation, so would be expected to supply power. © 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. 1–16 CHAPTER 1. Circuit Variables [b] pa = va ia = (5)(2 × 10−3 ) = 10 mW pb = −vbib = −(1)(3 × 10−3 ) = −3 mW pc = −vcic = −(7)(−2 × 10−3 ) = 14 mW pd = −vdid = −(−9)(1 × 10−3 ) = 9 mW pe = ve ie = (−20)(5 × 10−3 ) = −100 mW pf = vf if = (20)(2 × 10−3 ) = 40 mW pg = −vg ig = −(−3)(−2 × 10−3 ) = −6 mW ph = vhih = (−12)(−3 × 10−3 ) = 36 mW X Pabs = 10 + 14 + 9 + 40 + 36 = 109 mW Pdel = 3 + 100 + 6 = 109 mW Thus, 109 mW of power is delivered and 109 mW of power is absorbed, and the power balances. X [c] Looking at the calculated power values, elements a, c, d, f, and h have positive power, so are absorbing, while elements b, e, and g have negative power so are supplying. These answers are different from those in part (a) because the voltages and currents used in the power equation are not all positive numbers. P 1.30 pa = −vaia = −(1.6)(0.080) = −128 mW pb = −vbib = −(2.6)(0.060) = −156 mW pc = vcic = (−4.2)(−0.050) = 210 mW pd = −vdid = −(1.2)(0.020) = −24 mW pe = veie = (1.8)(0.030) = 54 mW pf = −vf if = −(−1.8)(−0.040) = −72 mW pg = vgig = (−3.6)(−0.030) = 108 mW ph = vhih = (3.2)(−0.020) = −64 mW pj = −vjij = −(−2.4)(0.030) = 72 mW X Pdel = 128 + 156 + 24 + 72 + 64 = 444 mW Pabs = 210 + 54 + 108 + 72 = 444 mW X X Therefore, Pdel = Pabs = 444 mW X Thus, the interconnection satisfies the power check. © 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. Problems P 1.31 pa = vaia = (120)(−10) = −1200 W pb = −vbib = −(120)(9) = −1080 W pc = vcic = (10)(10) = 100 W pd = −vdid = −(10)(−1) = 10 W pe = veie = (−10)(−9) = 90 W pf = −vf if = −(−100)(5) = 500 W pg = vgig = (120)(4) = 480 W ph = vhih = (−220)(−5) = 1100 W 1–17 X Pdel = 1200 + 1080 = 2280 W Pabs = 100 + 10 + 90 + 500 + 480 + 1100 = 2280 W X X Therefore, Pdel = Pabs = 2280 W X Thus, the interconnection now satisfies the power check. P 1.32 [a] The revised circuit model is shown below: [b] The expression for the total power in this circuit is va ia − vb ib − vf if + vg ig + vh ih = (120)(−10) − (120)(10) − (−120)(3) + 120ig + (−240)(−7) = 0 Therefore, 120ig = 1200 + 1200 − 360 − 1680 = 360 so 360 =3A 120 Thus, if the power in the modified circuit is balanced the current in component g is 3 A. ig = © 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. Circuit Elements 2 Assessment a Problems AP 2.1 [a] Note that the current ib is in the same circuit branch as the 8 A current source; however, ib is defined in the opposite direction of the current source. Therefore, ib = −8 A Next, note that the dependent voltage source and the independent voltage source are in parallel with the same polarity. Therefore, their voltages are equal, and ib −8 vg = = = −2 V 4 4 [b] To find the power associated with the 8 A source, we need to find the voltage drop across the source, vi . Note that the two independent sources are in parallel, and that the voltages vg and v1 have the same polarities, so these voltages are equal: vi = vg = −2 V Using the passive sign convention, ps = (8 A)(vi ) = (8 A)(−2 V) = −16 W Thus the current source generated 16 W of power. © 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 2–1 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. 2–2 CHAPTER 2. Circuit Elements AP 2.2 [a] Note from the circuit that vx = −25 V. To find α note that the two current sources are in the same branch of the circuit but their currents flow in opposite directions. Therefore αvx = −15 A Solve the above equation for α and substitute for vx, α= −15 A −15 A = = 0.6 A/V vx −25 V [b] To find the power associated with the voltage source we need to know the current, iv . Note that this current is in the same branch of the circuit as the dependent current source and these two currents flow in the same direction. Therefore, the current iv is the same as the current of the dependent source: iv = αvx = (0.6)(−25) = −15 A Using the passive sign convention, ps = −(iv )(25 V) = −(−15 A)(25 V) = 375 W. Thus the voltage source dissipates 375 W. AP 2.3 [a] The resistor and the voltage source are in parallel and the resistor voltage and the voltage source have the same polarities. Therefore these two voltages are the same: vR = vg = 1 kV © 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. Problems 2–3 Note from the circuit that the current through the resistor is ig = 5 mA. Use Ohm’s law to calculate the value of the resistor: vR 1 kV R= = = 200 kΩ ig 5 mA Using the passive sign convention to calculate the power in the resistor, pR = (vR )(ig ) = (1 kV)(5 mA) = 5 W The resistor is dissipating 5 W of power. [b] Note from part (a) the vR = vg and iR = ig . The power delivered by the source is thus psource −3 W psource = −vg ig so vg = − =− = 40 V ig 75 mA Since we now have the value of both the voltage and the current for the resistor, we can use Ohm’s law to calculate the resistor value: R= 40 V vg = = 533.33 Ω ig 75 mA The power absorbed by the resistor must equal the power generated by the source. Thus, pR = −psource = −(−3 W) = 3 W [c] Again, note the iR = ig . The power dissipated by the resistor can be determined from the resistor’s current: pR = R(iR )2 = R(ig )2 Solving for ig , pr 480 mW = = 0.0016 R 300 Ω Then, since vR = vg i2g = so ig = vR = RiR = Rig = (300 Ω)(40 mA) = 12 V √ 0.0016 = 0.04 A = 40 mA so vg = 12 V AP 2.4 [a] Note from the circuit that the current through the conductance G is ig , flowing from top to bottom, because the current source and the conductance are in the same branch of the circuit so must have the same © 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. 2–4 CHAPTER 2. Circuit Elements current. The voltage drop across the current source is vg , positive at the top, because the current source and the conductance are also in parallel so must have the same voltage. From a version of Ohm’s law, ig 0.5 A = = 10 V G 50 mS Now that we know the voltage drop across the current source, we can find the power delivered by this source: vg = psource = −vg ig = −(10)(0.5) = −5 W Thus the current source delivers 5 W to the circuit. [b] We can find the value of the conductance using the power, and the value of the current using Ohm’s law and the conductance value: pg = Gvg2 so G= pg 9 = 2 = 0.04 S = 40 mS 2 vg 15 ig = Gvg = (40 mS)(15 V) = 0.6 A [c] We can find the voltage from the power and the conductance, and then use the voltage value in Ohm’s law to find the current: pg = Gvg2 Thus so vg = q vg2 = pg 8W = = 40,000 G 200 µS 40,000 = 200 V ig = Gvg = (200 µS)(200 V) = 0.04 A = 40 mA AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every circuit element. Write a KVL equation clockwise around the circuit, starting below the voltage source: −24 V + v2 + v5 − v1 = 0 Next, use Ohm’s law to calculate the three unknown voltages from the three currents: v2 = 3i2 ; v5 = 7i5; v1 = 2i1 © 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. Problems 2–5 A KCL equation at the upper right node gives i2 = i5 ; a KCL equation at the bottom right node gives i5 = −i1; a KCL equation at the upper left node gives is = −i2. Now replace the currents i1 and i2 in the Ohm’s law equations with i5: v2 = 3i2 = 3i5; v5 = 7i5 ; v1 = 2i1 = −2i5 Now substitute these expressions for the three voltages into the first equation: 24 = v2 + v5 − v1 = 3i5 + 7i5 − (−2i5) = 12i5 Therefore i5 = 24/12 = 2 A [b] v1 = −2i5 = −2(2) = −4 V [c] v2 = 3i5 = 3(2) = 6 V [d] v5 = 7i5 = 7(2) = 14 V [e] A KCL equation at the lower left node gives is = i1. Since i1 = −i5, is = −2 A. We can now compute the power associated with the voltage source: p24 = (24)is = (24)(−2) = −48 W Therefore 24 V source is delivering 48 W. AP 2.6 Redraw the circuit labeling all voltages and currents: We can find the value of the unknown resistor if we can find the value of its voltage and its current. To start, write a KVL equation clockwise around the right loop, starting below the 24 Ω resistor: −120 V + v3 = 0 Use Ohm’s law to calculate the voltage across the 8 Ω resistor in terms of its current: v3 = 8i3 Substitute the expression for v3 into the first equation: −120 V + 8i3 = 0 so i3 = 120 = 15 A 8 © 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. 2–6 CHAPTER 2. Circuit Elements Also use Ohm’s law to calculate the value of the current through the 24 Ω resistor: i2 = 120 V = 5A 24 Ω Now write a KCL equation at the top middle node, summing the currents leaving: −i1 + i2 + i3 = 0 so i1 = i2 + i3 = 5 + 15 = 20 A Write a KVL equation clockwise around the left loop, starting below the voltage source: −200 V + v1 + 120 V = 0 so v1 = 200 − 120 = 80 V Now that we know the values of both the voltage and the current for the unknown resistor, we can use Ohm’s law to calculate the resistance: R = v1 80 = = 4Ω i1 20 AP 2.7 [a] Plotting a graph of vt versus it gives Note that when it = 0, vt = 25 V; therefore the voltage source must be 25 V. Since the plot is a straight line, its slope can be used to calculate the value of resistance: 25 − 0 25 ∆v R= = = = 100 Ω ∆i 0.25 − 0 0.25 A circuit model having the same v − i characteristic is a 25 V source in series with a 100Ω resistor, as shown below: © 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. Problems 2–7 [b] Draw the circuit model from part (a) and attach a 25 Ω resistor: To find the power delivered to the 25 Ω resistor we must calculate the current through the 25 Ω resistor. Do this by first using KCL to recognize that the current in each of the components is it , flowing in a clockwise direction. Write a KVL equation in the clockwise direction, starting below the voltage source, and using Ohm’s law to express the voltage drop across the resistors in the direction of the current it flowing through the resistors: 25 −25 V + 100it + 25it = 0 so 125it = 25 so it = = 0.2 A 125 Thus, the power delivered to the 25 Ω resistor is p25 = (25)i2t = (25)(0.2)2 = 1 W. AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0, it = 0.25 A. Therefore the current source must be 0.25 A. Since the plot is a straight line, its slope can be used to calculate the value of resistance: ∆v 25 − 0 25 R= = = = 100 Ω ∆i 0.25 − 0 0.25 A circuit model having the same v − i characteristic is a 0.25 A current source in parallel with a 100Ω resistor, as shown below: [b] Draw the circuit model from part (a) and attach a 25 Ω resistor: Note that by writing a KVL equation around the right loop we see that the voltage drop across both resistors is vt. Write a KCL equation at the top center node, summing the currents leaving the node. Use Ohm’s law to specify the currents through the resistors in terms of the voltage drop across the resistors and the value of the resistors. vt vt −0.25 + + = 0, so 5vt = 25, thus vt = 5 V 100 25 © 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. 2–8 CHAPTER 2. Circuit Elements p25 = vt2 = 1 W. 25 AP 2.9 First note that we know the current through all elements in the circuit except the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ resistor is i1 ; the current in the three elements to the right of the 6 kΩ resistor is 30i1 ). To find the current in the 6 kΩ resistor, write a KCL equation at the top node: i1 + 30i1 = i6k = 31i1 We can then use Ohm’s law to find the voltages across each resistor in terms of i1 . The results are shown in the figure below: [a] To find i1, write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 5V source: −5 V + 54,000i1 − 1 V + 186,000i1 = 0 Solving for i1 54,000i1 + 186,000i1 = 6 V so 240,000i1 = 6 V Thus, i1 = 6 = 25 µA 240,000 [b] Now that we have the value of i1 , we can calculate the voltage for each component except the dependent source. Then we can write a KVL equation for the right-hand loop to find the voltage v of the dependent source. Sum the voltages in the clockwise direction, starting to the left of the dependent source: +v − 54,000i1 + 8 V − 186,000i1 = 0 Thus, v = 240,000i1 − 8 V = 240,000(25 × 10−6 ) − 8 V = 6 V − 8 V = −2 V We now know the values of voltage and current for every circuit element. © 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. Problems 2–9 Let’s construct a power table: Element Current Voltage Power Power (µA) (V) Equation (µW) 5V 25 5 p = −vi −125 54 kΩ 25 1.35 p = Ri2 33.75 1V 25 1 p = −vi −25 6 kΩ 775 4.65 p = Ri2 3603.75 Dep. source 750 −2 p = −vi 1500 1.8 kΩ 750 1.35 p = Ri2 1012.5 8V 750 8 p = −vi −6000 [c] The total power generated in the circuit is the sum of the negative power values in the power table: −125 µW + −25 µW + −6000 µW = −6150 µW Thus, the total power generated in the circuit is 6150 µW. [d] The total power absorbed in the circuit is the sum of the positive power values in the power table: 33.75 µW + 3603.75 µW + 1500 µW + 1012.5 µW = 6150 µW Thus, the total power absorbed in the circuit is 6150 µW. AP 2.10 Given that iφ = 2 A, we know the current in the dependent source is 2iφ = 4 A. We can write a KCL equation at the left node to find the current in the 10 Ω resistor. Summing the currents leaving the node, −5 A + 2 A + 4 A + i10Ω = 0 so i10Ω = 5 A − 2 A − 4 A = −1 A Thus, the current in the 10 Ω resistor is 1 A, flowing right to left, as seen in the circuit below. © 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. 2–10 CHAPTER 2. Circuit Elements [a] To find vs , write a KVL equation, summing the voltages counter-clockwise around the lower right loop. Start below the voltage source. −vs + (1 A)(10 Ω) + (2 A)(30 Ω) = 0 so vs = 10 V + 60 V = 70 V [b] The current in the voltage source can be found by writing a KCL equation at the right-hand node. Sum the currents leaving the node −4 A + 1 A + iv = 0 so iv = 4 A − 1 A = 3 A The current in the voltage source is 3 A, flowing top to bottom. The power associated with this source is p = vi = (70 V)(3 A) = 210 W Thus, 210 W are absorbed by the voltage source. [c] The voltage drop across the independent current source can be found by writing a KVL equation around the left loop in a clockwise direction: −v5A + (2 A)(30 Ω) = 0 so v5A = 60 V The power associated with this source is p = −v5A i = −(60 V)(5 A) = −300 W This source thus delivers 300 W of power to the circuit. [d] The voltage across the controlled current source can be found by writing a KVL equation around the upper right loop in a clockwise direction: +v4A + (10 Ω)(1 A) = 0 so v4A = −10 V The power associated with this source is p = v4A i = (−10 V)(4 A) = −40 W This source thus delivers 40 W of power to the circuit. [e] The total power dissipated by the resistors is given by (i30Ω )2(30 Ω) + (i10Ω)2 (10 Ω) = (2)2 (30 Ω) + (1)2 (10 Ω) = 120 + 10 = 130 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. Problems 2–11 Problems P 2.1 The interconnect is valid since the voltage sources can all carry 5 A of current supplied by the current source, and the current source can carry the voltage drop required by the interconnection. Note that the branch containing the 10 V, 40 V, and 5 A sources must have the same voltage drop as the branch containing the 50 V source, so the 5 A current source must have a voltage drop of 20 V, positive at the right. The voltages and currents are summarize in the circuit below: P50V = (50)(5) = 250 W (abs) P10V = (10)(5) = 50 W P40V = −(40)(5) = −200 W (dev) P5A = −(20)(5) = −100 W X (abs) (dev) Pdev = 300 W P 2.2 The interconnection is not valid. Note that the 10 V and 20 V sources are both connected between the same two nodes in the circuit. If the interconnection was valid, these two voltage sources would supply the same voltage drop between these two nodes, which they do not. P 2.3 [a] Yes, independent voltage sources can carry the 5 A current required by the connection; independent current source can support any voltage required by the connection, in this case 5 V, positive at the bottom. [b] 20 V source: absorbing 15 V source: developing (delivering) 5 A source: developing (delivering) [c] P20V = (20)(5) = 100 W (abs) P15V = −(15)(5) = −75 W (dev/del) P5A = X Pabs = −(5)(5) = −25 W X (dev/del) Pdel = 100 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. 2–12 CHAPTER 2. Circuit Elements [d] The interconnection is valid, but in this circuit the voltage drop across the 5 A current source is 35 V, positive at the top; 20 V source is developing (delivering), the 15 V source is developing (delivering), and the 5 A source is absorbing: P20V = −(20)(5) = −100 W (dev/del) P15V = −(15)(5) = −75 W (dev/del) P5A P 2.4 X = (35)(5) = 175 W Pabs = X (abs) Pdel = 175 W The interconnection is valid. The 10 A current source has a voltage drop of 100 V, positive at the top, because the 100 V source supplies its voltage drop across a pair of terminals shared by the 10 A current source. The right hand branch of the circuit must also have a voltage drop of 100 V from the left terminal of the 40 V source to the bottom terminal of the 5 A current source, because this branch shares the same terminals as the 100 V source. This means that the voltage drop across the 5 A current source is 140 V, positive at the top. Also, the two voltage sources can carry the current required of the interconnection. This is summarized in the figure below: From the values of voltage and current in the figure, the power supplied by the current sources is calculated as follows: P10A = −(100)(10) = −1000 W (1000 W supplied) P5A P 2.5 X = −(140)(5) = −700 W (700 W supplied) Pdev = 1700 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. Problems 2–13 The interconnection is invalid. The voltage drop between the top terminal and the bottom terminal on the left hand side is due to the 6 V and 8 V sources, giving a total voltage drop between these terminals of 14 V. But the voltage drop between the top terminal and the bottom terminal on the right hand side is due to the 4 V and 12 V sources, giving a total voltage drop between these two terminals of 16 V. The voltage drop between any two terminals in a valid circuit must be the same, so the interconnection is invalid. P 2.6 The interconnection is valid, since the voltage sources can carry the 20 mA current supplied by the current source, and the current sources can carry whatever voltage drop is required by the interconnection. In particular, note the the voltage drop across the three sources in the right hand branch must be the same as the voltage drop across the 15 mA current source in the middle branch, since the middle and right hand branches are connected between the same two terminals. In particular, this means that v1(the voltage drop across the middle branch) = −20V + 60V − v2 Hence any combination of v1 and v2 such that v1 + v2 = 40 V is a valid solution. P 2.7 The interconnection is invalid. In the middle branch, the value of the current i∆ must be −25 A, since the 25 A current source supplies current in this branch in the direction opposite the direction of the current i∆ . Therefore, the voltage supplied by the dependent voltage source in the left hand branch is 6(−25) = −150 V. This gives a voltage drop from the top terminal to the bottom terminal in the left hand branch of 50 − (−150) = 200 V. But the voltage drop between these same terminals in the right hand branch is 250 V, due to the voltage source in that branch. Therefore, the interconnection is invalid. P 2.8 First, 10va = 5 V, so va = 0.5 V. Then recognize that each of the three branches is connected between the same two nodes, so each of these branches must have the same voltage drop. The voltage drop across the middle branch is 5 V, and since va = 0.5 V, vg = 0.5 − 5 = −4.5 V. Also, the voltage drop © 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. 2–14 CHAPTER 2. Circuit Elements across the left branch is 5 V, so 20 + v9A = 5 V, and v9A = −15 V, where v9A is positive at the top. Note that the current through the 20 V source must be 9 A, flowing from top to bottom, and the current through the vg is 6 A flowing from top to bottom. Let’s find the power associated with the left and middle branches: p9A = (9)(−15) = −135 W p20V = (9)(20) = 180 W pvg = −(6)(−4.5) = 27 W p6A = (6)(0.5) = 3 W Since there is only one component left, we can find the total power: ptotal = −135 + 180 + 27 + 3 + pds = 75 + pds = 0 so pds must equal −75 W. Therefore, P 2.9 X Pdev = X Pabs = 210 W [a] Yes, each of the voltage sources can carry the current required by the interconnection, and each of the current sources can carry the voltage drop required by the interconnection. (Note that i∆ = −8 A.) [b] No, because the voltage drop between the top terminal and the bottom terminal cannot be determined. For example, define v1, v2 , and v3 as shown: The voltage drop across the left branch, the center branch, and the right branch must be the same, since these branches are connected at the same two terminals. This requires that 20 + v1 = v2 + 100 = v3 But this equation has three unknown voltages, so the individual voltages cannot be determined, and thus the power of the sources cannot be determined. P 2.10 [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. Problems [b] P 2.11 Vbb = no-load voltage of battery Rbb = internal resistance of battery Rx = resistance of wire between battery and switch Ry = resistance of wire between switch and lamp A Ra = resistance of lamp A Rb = resistance of lamp B Rw = resistance of wire between lamp A and lamp B Rg1 = resistance of frame between battery and lamp A Rg2 = resistance of frame between lamp A and lamp B S = switch 2–15 Since we know the device is a resistor, we can use Ohm’s law to calculate the resistance. From Fig. P2.11(a), v = Ri so R= v i Using the values in the table of Fig. P2.11(b), R= −108 −54 54 108 162 = = = = = 27 kΩ −0.004 −0.002 0.002 0.004 0.006 Note that this value is found in Appendix H. P 2.12 The resistor value is the ratio of the power to the square of the current: p R = 2 . Using the values for power and current in Fig. P2.12(b), i 5.5 × 10−3 22 × 10−3 49.5 × 10−3 88 × 10−3 = = = (50 × 10−6 )2 (100 × 10−6 )2 (150 × 10−6 )2 (200 × 10−6 )2 = 137.5 × 10−3 198 × 10−3 = = 2.2 MΩ (250 × 10−6 )2 (300 × 10−6 )2 Note that this is a value from Appendix H. P 2.13 Since we know the device is a resistor, we can use the power equation. From Fig. P2.13(a), p = vi = v2 R so R= v2 p © 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. 2–16 CHAPTER 2. Circuit Elements Using the values in the table of Fig. P2.13(b) R= (−10)2 (−5)2 (5)2 (10)2 = = = 17.86 × 10−3 4.46 × 10−3 4.46 × 10−3 17.86 × 10−3 = (15)2 (20)2 = ≈ 5.6 kΩ 40.18 × 10−3 71.43 × 10−3 Note that this value is found in Appendix H. P 2.14 [a] Plot the v—i characteristic: From the plot: R= ∆v (180 − 100) = = 5Ω ∆i (16 − 0) When it = 0, vt = 100 V; therefore the ideal current source must have a current of 100/5 = 20 A [b] We attach a 20 Ω resistor to the device model developed in part (a): Write a KCL equation at the top node: 20 + it = i1 Write a KVL equation for the right loop, in the direction of the two currents, using Ohm’s law: 5i1 + 20it = 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. Problems 2–17 Combining the two equations and solving, 5(20 + it ) + 20it = 0 so 25it = −100; thus it = −4 A Now calculate the power dissipated by the resistor: p20 Ω = 20i2t = 20(−4)2 = 320 W P 2.15 [a] Plot the v − i characteristic From the plot: R= ∆v (130 − 50) = = 8Ω ∆i (10 − 0) When it = 0, vt = 50 V; therefore the ideal voltage source has a voltage of 50 V. [b] When vt = 0, it = −50 = −6.25A 8 Note that this result can also be obtained by extrapolating the v − i characteristic to vt = 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. 2–18 P 2.16 CHAPTER 2. Circuit Elements [a] [b] ∆v = 25V; ∆i = 2.5 mA; [c] 10,000i1 = 2500is , R= ∆v = 10 kΩ ∆i i1 = 0.25is 0.02 = i1 + is = 1.25is , is = 16 mA [d] vs (open circuit) = (20 × 10−3 )(10 × 103 ) = 200 V [e] The open circuit voltage can be found in the table of values (or from the plot) as the value of the voltage vs when the current is = 0. Thus, vs (open circuit) = 140 V (from the table) [f] Linear model cannot predict the nonlinear behavior of the practical current source. P 2.17 [a] Begin by constructing a plot of voltage versus current: © 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. Problems 2–19 [b] Since the plot is linear for 0 ≤ is ≤ 24 mA amd since R = ∆v/∆i, we can calculate R from the plotted values as follows: R= ∆v 24 − 18 6 = = = 250 Ω ∆i 0.024 − 0 0.024 We can determine the value of the ideal voltage source by considering the value of vs when is = 0. When there is no current, there is no voltage drop across the resistor, so all of the voltage drop at the output is due to the voltage source. Thus the value of the voltage source must be 24 V. The model, valid for 0 ≤ is ≤ 24 mA, is shown below: [c] The circuit is shown below: Write a KVL equation in the clockwise direction, starting below the voltage source. Use Ohm’s law to express the voltage drop across the resistors in terms of the current i: −24 V + 250i + 1000i = 0 Thus, i= so 1250i = 24 V 24 V = 19.2 mA 1250 Ω [d] The circuit is shown below: Write a KVL equation in the clockwise direction, starting below the voltage source. Use Ohm’s law to express the voltage drop across the resistors in terms of the current i: −24 V + 250i = 0 Thus, i= so 250i = 24 V 24 V = 96 mA 250 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. 2–20 CHAPTER 2. Circuit Elements [e] The short circuit current can be found in the table of values (or from the plot) as the value of the current is when the voltage vs = 0. Thus, isc = 48 mA (from table) [f] The plot of voltage versus current constructed in part (a) is not linear (it is piecewise linear, but not linear for all values of is ). Since the proposed circuit model is a linear model, it cannot be used to predict the nonlinear behavior exhibited by the plotted data. P 2.18 [a] Write a KCL equation at the top node: −1.5 + i1 + i2 = 0 so i1 + i2 = 1.5 Write a KVL equation around the right loop: −v1 + v2 + v3 = 0 From Ohm’s law, v1 = 100i1 , v2 = 150i2 , v3 = 250i2 Substituting, −100i1 + 150i2 + 250i2 = 0 so − 100i1 + 400i2 = 0 Solving the two equations for i1 and i2 simultaneously, i1 = 1.2 A and i2 = 0.3 A [b] Write a KVL equation clockwise around the left loop: −vo + v1 = 0 So but v1 = 100i1 = 100(1.2) = 120 V vo = v1 = 120 V [c] Calculate power using p = vi for the source and p = Ri2 for the resistors: psource = −vo(1.5) = −(120)(1.5) = −180 W p100Ω = 1.22 (100) = 144 W p150Ω = 0.32 (150) = 13.5 W p250Ω = 0.32 (250) = 22.5 W X Pdev = 180 W X Pabs = 144 + 13.5 + 22.5 = 180 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. Problems P 2.19 2–21 [a] 20ia = 80ib ia = 4ib ig = ia + ib = 5ib 50 = 4ig + 80ib = 20ib + 80ib = 100ib ib 0.5 A, therefore, ia = 2 A = and ig = 2.5 A [b] ib = 0.5 A [c] vo = 80ib = 40 V [d] p4Ω = i2g (4) = 6.25(4) = 25 W p20Ω = i2a (20) = (4)(20) = 80 W p80Ω = i2b (80) = 0.25(80) = 20 W [e] p50V (delivered) = 50ig = 125 W Check: X P 2.20 X Pdis = 25 + 80 + 20 = 125 W Pdel = 125 W [a] Use KVL for the right loop to calculate the voltage drop across the right-hand branch vo . This is also the voltage drop across the middle branch, so once vo is known, use Ohm’s law to calculate io : vo = 1000ia + 4000ia + 3000ia = 8000ia = 8000(0.002) = 16 V 16 = 2000io 16 io = = 8 mA 2000 [b] KCL at the top node: ig = ia + io = 0.002 + 0.008 = 0.010 A = 10 mA. [c] The voltage drop across the source is v0, seen by writing a KVL equation for the left loop. Thus, pg = −vo ig = −(16)(0.01) = −0.160 W = −160 mW. Thus the source delivers 160 mW. © 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. 2–22 P 2.21 CHAPTER 2. Circuit Elements [a] v2 = 150 − 50(1) = 100V i2 = v2 = 4A 25 i3 + 1 = i2 , i3 = 4 − 1 = 3A v1 = 10i3 + 25i2 = 10(3) + 25(4) = 130V i1 = v1 130 = = 2A 65 65 Note also that i4 = i1 + i3 = 2 + 3 = 5 A ig = i4 + io = 5 + 1 = 6 A [b] [c] p4Ω = 52 (4) = 100 W p50Ω = 12 (50) = 50 W p65Ω = 22 (65) = 260 W p10Ω = 32 (10) = 90 W p25Ω = 42 (25) = 400 W X Pdis = 100 + 50 + 260 + 90 + 400 = 900 W Pdev = 150ig = 150(6) = 900 W P 2.22 [a] icd = 80/16 = 5 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. Problems vac = 125 − 80 = 45 iac + ibc = icd so so 2–23 iac = 45/15 = 3 A ibc = 5 − 3 = 2 A vab = 15iac − 5ibc = 15(3) − 5(2) = 35 V so iab = 35/7 = 5 A ibd = iab − ibc = 5 − 2 = 3 A Calculate the power dissipated by the resistors using the equation pR = Ri2R : p7Ω = (7)(5)2 = 175 W p15Ω = (15)(3)2 = 135 W p30Ω = (30)(3)2 = 270 W p16Ω = (16)(5)2 = 400 W p5Ω = (5)(2)2 = 20 W [b] Calculate the current through the voltage source: iad = −iab − iac = −5 − 3 = −8 A Now that we have both the voltage and the current for the source, we can calculate the power supplied by the source: pg = 125(−8) = −1000 W [c] P 2.23 [a] thus pg (supplied) = 1000 W X Pdis = 175 + 270 + 135 + 400 + 20 = 1000 W Therefore, X Psupp = X Pdis va = (5 + 10)(4) = 60 V −240 + va + vb = 0 so vb = 240 − va = 240 − 60 = 180 V ie = vb /(14 + 6) = 180/20 = 9 A id = ie − 4 = 9 − 4 = 5 A vc = 4id + vb = 4(5) + 180 = 200 V ic = vc /10 = 200/10 = 20 A vd = 240 − vc = 240 − 200 = 40 V ia = id + ic = 5 + 20 = 25 A R = vd /ia = 40/25 = 1.6 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. 2–24 CHAPTER 2. Circuit Elements [b] ig = ia + 4 = 25 + 4 = 29 A pg (supplied) = (240)(29) = 6960 W P 2.24 id = 60/12 = 5 A; therefore, vcd = 60 + 18(5) = 150 V −240 + vac + vcd = 0; therefore, vac = 240 − 150 = 90 V ib = vac/45 = 90/45 = 2 A; therefore, ic = id − ib = 5 − 2 = 3 A vbd = 10ic + vcd = 10(3) + 150 = 180 V; therefore, ia = vbd/180 = 180/180 = 1 A ie = ia + ic = 1 + 3 = 4 A −240 + vab + vbd = 0 therefore, vab = 240 − 180 = 60 V R = vab/ie = 60/4 = 15 Ω CHECK: ig = ib + ie = 2 + 4 = 6 A pdev = (240)(6) = 1440 W X Pdis = 12 (180) + 42 (15) + 32 (10) + 52 (12) + 52 (18) + 22 (45) P 2.25 = 1440 W (CHECKS) [a] Start by calculating the voltage drops due to the currents i1 and i2. Then use KVL to calculate the voltage drop across and 35 Ω resistor, and Ohm’s law to find the current in the 35 Ω resistor. Finally, KCL at each of the middle three nodes yields the currents in the two sources and the current in the middle 2 Ω resistor. These calculations are summarized in the figure below: p147(top) = −(147)(28) = −4116 W p147(bottom) = −(147)(21) = −3087 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. Problems 2–25 [b] X X Pdis = 784 + 98 + 441 + 2205 + 1960 + 1715 = 7203 W Psup Therefore, P 2.26 [a] = (28)2 (1) + (7)2 (2) + (21)2 (1) + (21)2 (5) + (14)2 (10) + (7)2 (35) = 4116 + 3087 = 7203 W X Pdis = X Psup = 7203 W v2 = 100 + 4(15) = 160 V; i1 = v1 100 = = 5 A; 4 + 16 20 v1 = 160 − (9 + 11 + 10)(2) = 100 V i3 = i1 − 2 = 5 − 2 = 3 A vg = v1 + 30i3 = 100 + 30(3) = 190 V i4 = 2 + 4 = 6 A ig = −i4 − i3 = −6 − 3 = −9 A [b] Calculate power using the formula p = Ri2 : p9 Ω = (9)(2)2 = 36 W; p10 Ω = (10)(2)2 = 40 W; p11 Ω = (11)(2)2 = 44 W p5 Ω = (5)(6)2 = 180 W p30 Ω = (30)(3)2 = 270 W; p4 Ω = (4)(5)2 = 100 W p16 Ω = (16)(5)2 = 400 W; p15 Ω = (15)(4)2 = 240 W [c] vg = 190 V [d] Sum the power dissipated by the resistors: X pdiss = 36 + 44 + 40 + 180 + 270 + 100 + 400 + 240 = 1310 W The power associated with the sources is pvolt−source = (100)(4) = 400 W pcurr−source = vg ig = (190)(−9) = −1710 W Thus the total power dissipated is 1310 + 400 = 1710 W and the total power developed is 1710 W, so the power balances. © 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. 2–26 P 2.27 CHAPTER 2. Circuit Elements [a] io = 0 because no current can exist in a single conductor connecting two parts of a circuit. [b] 18 = (12 + 6)ig ig = 1 A v∆ = 6ig = 6V v∆ /2 = 3 A 10i1 = 5i2 , so i1 + 2i1 = −3 A; therefore, i1 = −1 A [c] i2 = 2i1 = −2 A. P 2.28 First note that we know the current through all elements in the circuit except the 200 Ω resistor (the current in the three elements to the left of the 200 Ω resistor is iβ ; the current in the three elements to the right of the 200 Ω resistor is 29iβ ). To find the current in the 200 Ω resistor, write a KCL equation at the top node: iβ + 29iβ = i200 Ω = 30iβ We can then use Ohm’s law to find the voltages across each resistor in terms of iβ . The results are shown in the figure below: [a] To find iβ , write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 15.2V source: −15.2 V + 10,000i1 − 0.8 V + 6000iβ = 0 Solving for iβ 10,000iβ + 6000iβ = 16 V so 16,000iβ = 16 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. Problems 2–27 Thus, iβ = 16 = 1 mA 16,000 Now that we have the value of iβ , we can calculate the voltage for each component except the dependent source. Then we can write a KVL equation for the right-hand loop to find the voltage vy of the dependent source. Sum the voltages in the clockwise direction, starting to the left of the dependent source: −vy − 14,500iβ + 25 V − 6000iβ = 0 Thus, vy = 25 V − 20,500iβ = 25 V − 20,500(10−3 ) = 25 V − 20.5 V = 4.5 V [b] We now know the values of voltage and current for every circuit element. Let’s construct a power table: Element Current Voltage (mA) (V) Power Power Equation (mW) 15.2 V 1 15.2 p = −vi −15.2 10 kΩ 1 10 p = Ri2 10 0.8 V 1 0.8 p = −vi −0.8 200 Ω 30 6 p = Ri2 180 Dep. source 29 4.5 p = vi 130.5 500 Ω 29 14.5 p = Ri2 420.5 25 V 29 25 p = −vi −725 The total power generated in the circuit is the sum of the negative power values in the power table: −15.2 mW + −0.8 mW + −725 mW = −741 mW Thus, the total power generated in the circuit is 741 mW. The total power absorbed in the circuit is the sum of the positive power values in the power table: 10 mW + 180 mW + 130.5 mW + 420.5 mW = 741 mW Thus, the total power absorbed in the circuit is 741 mW and the power in the circuit balances. P 2.29 40i2 + 5 5 + = 0; 40 10 i2 = −15.625 mA v1 = 80i2 = −1.25 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. 2–28 CHAPTER 2. Circuit Elements 25i1 + (−1.25) + (−0.015625) = 0; 20 i1 = 3.125 mA vg = 60i1 + 260i1 = 320i1 Therefore, vg = 1 V. P 2.30 [a] −50 − 20iσ + 18i∆ = 0 −18i∆ + 5iσ + 40iσ = 0 Therefore, so 18i∆ = 45iσ − 50 − 20iσ + 45iσ = 0, so iσ = 2 A 18i∆ = 45iσ = 90; so i∆ = 5 A vo = 40iσ = 80 V [b] ig = current out of the positive terminal of the 50 V source vd = voltage drop across the 8i∆ source ig = i∆ + iσ + 8i∆ = 9i∆ + iσ = 47 A vd = 80 − 20 = 60 V X X Pgen = 50ig + 20iσ ig = 50(47) + 20(2)(47) = 4230 W Pdiss = 18i2∆ + 5iσ (ig − i∆ ) + 40i2σ + 8i∆ vd + 8i∆ (20) = (18)(25) + 10(47 − 5) + 4(40) + 40(60) + 40(20) = 4230 W; Therefore, P 2.31 X Pgen = iE − iB − iC = 0 iC = βiB X Pdiss = 4230 W therefore iE = (1 + β)iB i2 = −iB + i1 Vo + iE RE − (i1 − iB )R2 = 0 −i1R1 + VCC − (i1 − iB )R2 = 0 Vo + iE RE + iB R2 − or i1 = VCC + iB R2 R1 + R2 VCC + iB R2 R2 = 0 R1 + R2 Now replace iE by (1 + β)iB and solve for iB . Thus iB = [VCC R2 /(R1 + R2 )] − Vo (1 + β)RE + R1 R2 /(R1 + R2 ) © 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. Problems P 2.32 2–29 Here is Equation 2.25: iB = (VCC R2 )/(R1 + R2 ) − V0 (R1 R2 )/(R1 + R2 ) + (1 + β)RE VCC R2 (10)(60,000) = = 6V R1 + R2 100,000 (40,000)(60,000) R1 R2 = = 24 kΩ R1 + R2 100,000 iB = 6 − 0.6 5.4 = = 0.18 mA 24,000 + 50(120) 30,000 iC = βiB = (49)(0.18) = 8.82 mA iE = iC + iB = 8.82 + 0.18 = 9 mA v3d = (0.009)(120) = 1.08V vbd = Vo + v3d = 1.68V i2 = vbd 1.68 = = 28 µA R2 60,000 i1 = i2 + iB = 28 + 180 = 208 µA vab = 40,000(208 × 10−6 ) = 8.32 V iCC = iC + i1 = 8.82 + 0.208 = 9.028 mA v13 + (8.82 × 10−3 )(750) + 1.08 = 10 V v13 = 2.305 V P 2.33 [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. 2–30 CHAPTER 2. Circuit Elements [b] P 2.34 [a] From the simplified circuit model, using Ohm’s law and KVL: 400i + 50i + 200i − 250 = 0 so i = 250/650 = 385 mA This current is nearly enough to stop the heart, according to Table 2.1, so a warning sign should be posted at the 250 V source. [b] The closest value from Appendix H to 400 Ω is 390 Ω; the closest value from Appendix H to 50 Ω is 47 Ω. There are two possibilites for replacing the 200 Ω resistor with a value from Appendix H – 180 Ω and 220 Ω. We calculate the resulting current for each of these possibilities, and determine which current is closest to 385 mA: 390i + 47i + 180i − 250 = 0 so i = 250/617 = 405.2 mA 390i + 47i + 220i − 250 = 0 so i = 250/657 = 380.5 mA Therefore, choose the 220 Ω resistor to replace the 200 Ω resistor in the model. P 2.35 P 2.36 [a] p = i2R parm = pleg = ptrunk = 250 650 250 650 2 2 250 650 (400) = 59.17 W (200) = 29.59 W 2 (50) = 7.40 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. Problems [b] dT dt ! = arm ! tleg = dT dt 2.39 × 10−4 parm = 35.36 × 10−4 ◦ C/s 4 5 × 104 = 1414.23 s or 23.57 min 35.36 tarm = dT dt 2–31 = leg 2.39 × 10−4 Pleg = 7.07 × 10−4 ◦ C/s 10 5 × 104 = 7,071.13 s or 117.85 min 7.07 ! = trunk ttrunk = 2.39 × 10−4 (7.4) = 0.707 × 10−4 ◦ C/s 25 5 × 104 = 70,711.30 s or 1,178.52 min 0.707 [c] They are all much greater than a few minutes. P 2.37 [a] Rarms = 400 + 400 = 800 Ω iletgo = 50 mA (minimum) vmin = (800)(50) × 10−3 = 40 V [b] No, 12/800 = 15 mA. Note this current is sufficient to give a perceptible shock. P 2.38 Rspace = 1 MΩ ispace = 3 mA v = ispaceRspace = 3000 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. 3 Simple Resistive Circuits Assessment Problems AP 3.1 Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 Ω resistor and the 10 Ω resistor in series: 6 Ω + 10 Ω = 16 Ω Now combine this 16 Ω resistor in parallel with the 64 Ω resistor: 16 Ωk64 Ω = (16)(64) 1024 = = 12.8 Ω 16 + 64 80 This equivalent 12.8 Ω resistor is in series with the 7.2 Ω resistor: 12.8 Ω + 7.2 Ω = 20 Ω Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor: 20 Ωk30 Ω = (20)(30) 600 = = 12 Ω 20 + 30 50 Thus, the simplified circuit is as shown: © 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 3–1 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. 3–2 CHAPTER 3. Simple Resistive Circuits [a] With the simplified circuit we can use Ohm’s law to find the voltage across both the current source and the 12 Ω equivalent resistor: v = (12 Ω)(5 A) = 60 V [b] Now that we know the value of the voltage drop across the current source, we can use the formula p = −vi to find the power associated with the source: p = −(60 V)(5 A) = −300 W Thus, the source delivers 300 W of power to the circuit. [c] We now can return to the original circuit, shown in the first figure. In this circuit, v = 60 V, as calculated in part (a). This is also the voltage drop across the 30 Ω resistor, so we can use Ohm’s law to calculate the current through this resistor: 60 V iA = =2A 30 Ω Now write a KCL equation at the upper left node to find the current iB : −5 A + iA + iB = 0 so iB = 5 A − iA = 5 A − 2 A = 3 A Next, write a KVL equation around the outer loop of the circuit, using Ohm’s law to express the voltage drop across the resistors in terms of the current through the resistors: −v + 7.2iB + 6iC + 10iC = 0 So 16iC = v − 7.2iB = 60 V − (7.2)(3) = 38.4 V 38.4 = 2.4 A 16 Now that we have the current through the 10 Ω resistor we can use the formula p = Ri2 to find the power: Thus iC = p10 Ω = (10)(2.4)2 = 57.6 W AP 3.2 [a] We can use voltage division to calculate the voltage vo across the 75 kΩ resistor: 75,000 vo (no load) = (200 V) = 150 V 75,000 + 25,000 © 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. Problems 3–3 [b] When we have a load resistance of 150 kΩ then the voltage vo is across the parallel combination of the 75 kΩ resistor and the 150 kΩ resistor. First, calculate the equivalent resistance of the parallel combination: (75,000)(150,000) = 50,000 Ω = 50 kΩ 75,000 + 150,000 Now use voltage division to find vo across this equivalent resistance: 50,000 vo = (200 V) = 133.3 V 50,000 + 25,000 75 kΩk150 kΩ = [c] If the load terminals are short-circuited, the 75 kΩ resistor is effectively removed from the circuit, leaving only the voltage source and the 25 kΩ resistor. We can calculate the current in the resistor using Ohm’s law: 200 V i= = 8 mA 25 kΩ Now we can use the formula p = Ri2 to find the power dissipated in the 25 kΩ resistor: p25k = (25,000)(0.008)2 = 1.6 W [d] The power dissipated in the 75 kΩ resistor will be maximum at no load since vo is maximum. In part (a) we determined that the no-load voltage is 150 V, so be can use the formula p = v 2/R to calculate the power: p75k (max) = (150)2 = 0.3 W 75,000 AP 3.3 [a] We will write a current division equation for the current throught the 80Ω resistor and use this equation to solve for R: R i80Ω = (20 A) = 4 A so 20R = 4(R + 120) R + 40 Ω + 80 Ω 480 Thus 16R = 480 and R= = 30 Ω 16 [b] With R = 30 Ω we can calculate the current through R using current division, and then use this current to find the power dissipated by R, using the formula p = Ri2 : 40 + 80 iR = (20 A) = 16 A so pR = (30)(16)2 = 7680 W 40 + 80 + 30 © 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. 3–4 CHAPTER 3. Simple Resistive Circuits [c] Write a KVL equation around the outer loop to solve for the voltage v, and then use the formula p = −vi to calculate the power delivered by the current source: −v + (60 Ω)(20 A) + (30 Ω)(16 A) = 0 Thus, so v = 1200 + 480 = 1680 V psource = −(1680 V)(20 A) = −33,600 W Thus, the current source generates 33,600 W of power. AP 3.4 [a] First we need to determine the equivalent resistance to the right of the 40 Ω and 70 Ω resistors: 1 1 1 1 1 Req = 20 Ωk30 Ωk(50 Ω + 10 Ω) so = + + = Req 20 Ω 30 Ω 60 Ω 10 Ω Thus, Req = 10 Ω Now we can use voltage division to find the voltage vo : vo = 40 (60 V) = 20 V 40 + 10 + 70 [b] The current through the 40 Ω resistor can be found using Ohm’s law: vo 20 V = = 0.5 A 40 40 Ω This current flows from left to right through the 40 Ω resistor. To use current division, we need to find the equivalent resistance of the two parallel branches containing the 20 Ω resistor and the 50 Ω and 10 Ω resistors: (20)(60) 20 Ωk(50 Ω + 10 Ω) = = 15 Ω 20 + 60 Now we use current division to find the current in the 30 Ω branch: 15 i30Ω = (0.5 A) = 0.16667 A = 166.67 mA 15 + 30 i40Ω = [c] We can find the power dissipated by the 50 Ω resistor if we can find the current in this resistor. We can use current division to find this current © 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. Problems 3–5 from the current in the 40 Ω resistor, but first we need to calculate the equivalent resistance of the 20 Ω branch and the 30 Ω branch: (20)(30) = 12 Ω 20 + 30 Current division gives: 12 (0.5 A) = 0.08333 A i50Ω = 12 + 50 + 10 20 Ωk30 Ω = Thus, p50Ω = (50)(0.08333)2 = 0.34722 W = 347.22 mW AP 3.5 [a] We can find the current i using Ohm’s law: 1V i= = 0.01 A = 10 mA 100 Ω [b] Rm = 50 Ωk5.555 Ω = 5 Ω We can use the meter resistance to find the current using Ohm’s law: imeas = 1V = 0.009524 = 9.524 mA 100 Ω + 5 Ω AP 3.6 [a] Use voltage division to find the voltage v: 75,000 v= (60 V) = 50 V 75,000 + 15,000 © 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. 3–6 CHAPTER 3. Simple Resistive Circuits [b] The meter resistance is a series combination of resistances: Rm = 149,950 + 50 = 150,000 Ω We can use voltage division to find v, but first we must calculate the equivalent resistance of the parallel combination of the 75 kΩ resistor and the voltmeter: (75,000)(150,000) 75,000 Ωk150,000 Ω = = 50 kΩ 75,000 + 150,000 Thus, vmeas = 50,000 (60 V) = 46.15 V 50,000 + 15,000 AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite resistors must be equal. Therefore, (1000)(150) = 1500 Ω = 1.5 kΩ 100 [b] When the bridge is balanced, there is no current flowing through the meter, so the meter acts like an open circuit. This places the following branches in parallel: The branch with the voltage source, the branch with the series combination R1 and R3 and the branch with the series combination of R2 and Rx . We can find the current in the latter two branches using Ohm’s law: 100Rx = (1000)(150) so Rx = 5V 5V = 20 mA; iR2 ,Rx = = 2 mA 100 Ω + 150 Ω 1000 + 1500 We can calculate the power dissipated by each resistor using the formula p = Ri2 : iR1 ,R3 = p100Ω = (100 Ω)(0.02 A)2 = 40 mW p150Ω = (150 Ω)(0.02 A)2 = 60 mW p1000Ω = (1000 Ω)(0.002 A)2 = 4 mW p1500Ω = (1500 Ω)(0.002 A)2 = 6 mW Since none of the power dissipation values exceeds 250 mW, the bridge can be left in the balanced state without exceeding the power-dissipating capacity of the resistors. © 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. Problems 3–7 AP 3.8 Convert the three Y-connected resistors, 20 Ω, 10 Ω, and 5 Ω to three ∆-connected resistors Ra , Rb, and Rc . To assist you the figure below has both the Y-connected resistors and the ∆-connected resistors (5)(10) + (5)(20) + (10)(20) = 17.5 Ω 20 (5)(10) + (5)(20) + (10)(20) Rb = = 35 Ω 10 (5)(10) + (5)(20) + (10)(20) = 70 Ω Rc = 5 Ra = The circuit with these new ∆-connected resistors is shown below: From this circuit we see that the 70 Ω resistor is parallel to the 28 Ω resistor: 70 Ωk28 Ω = (70)(28) = 20 Ω 70 + 28 Also, the 17.5 Ω resistor is parallel to the 105 Ω resistor: 17.5 Ωk105 Ω = (17.5)(105) = 15 Ω 17.5 + 105 Once the parallel combinations are made, we can see that the equivalent 20 Ω resistor is in series with the equivalent 15 Ω resistor, giving an equivalent resistance of 20 Ω + 15 Ω = 35 Ω. Finally, this equivalent 35 Ω resistor is in parallel with the other 35 Ω resistor: 35 Ωk35 Ω = (35)(35) = 17.5 Ω 35 + 35 © 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. 3–8 CHAPTER 3. Simple Resistive Circuits Thus, the resistance seen by the 2 A source is 17.5 Ω, and the voltage can be calculated using Ohm’s law: v = (17.5 Ω)(2 A) = 35 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. Problems 3–9 Problems P 3.1 [a] The 6 kΩ and 12 kΩ resistors are in series, as are the 9 kΩ and 7 kΩ resistors. The simplified circuit is shown below: [b] The 3 kΩ, 5 kΩ, and 7 kΩ resistors are in series. The simplified circuit is shown below: [c] The 300 Ω, 400 Ω, and 500 Ω resistors are in series. The simplified circuit is shown below: P 3.2 [a] The 10 Ω and 40 Ω resistors are in parallel, as are the 100 Ω and 25 Ω resistors. The simplified circuit is shown below: [b] The 9 kΩ, 18 kΩ, and 6 kΩ resistors are in parallel. The simplified circuit is shown below: © 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. 3–10 CHAPTER 3. Simple Resistive Circuits [c] The 600 Ω, 200 Ω, and 300 Ω resistors are in parallel. The simplified circuit is shown below: P 3.3 Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Req = 6 + 12 + [4k(9 + 7)] = 6 + 12 + 4k16 = 6 + 12 + 3.2 = 21.2 Ω [b] Req = 4 k + [10 kk(3 k + 5 k + 7 k)] = 4 k + 10 kk15 k = 4 k + 6 k = 10 kΩ [c] Req = 300 + 400 + 500 + (600k1200) = 300 + 400 + 500 + 400 = 1600 Ω P 3.4 Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Req = 18 + [100k25k(10k40 + 22)] = 18 + [100k25k(8 + 22)] = 18 + [100k25k30] = 18 + 12 = 30 Ω [b] Req = 10 kk[5 k + 2 k + (9 kk18 kk6 k)] = 10 kk[5 k + 2 k + 3 k] = 10 kk10 k = 5 kΩ [c] Req = 600k200k300k(250 + 150) = 600k200k300k400 = 80 Ω P 3.5 [a] Rab = 10 + (5k20) + 6 = 10 + 4 + 6 = 20 Ω [b] Rab = 30 kk60 kk[20 k + (200 kk50 k)] = 30 kk60 kk(20 k + 40 k) = 30 kk60 kk60 k = 15 kΩ © 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. Problems P 3.6 [a] 60k20 = 1200/80 = 15 Ω 15 + 8 + 7 = 30 Ω 3–11 12k24 = 288/36 = 8 Ω 30k120 = 3600/150 = 24 Ω Rab = 15 + 24 + 25 = 64 Ω [b] 35 + 40 = 75 Ω 75k50 = 3750/125 = 30 Ω 30 + 20 = 50 Ω 50k75 = 3750/125 = 30 Ω 30 + 10 = 40 Ω 40k60 + 9k18 = 24 + 6 = 30 Ω 30k30 = 15 Ω Rab = 10 + 15 + 5 = 30 Ω [c] 50 + 30 = 80 Ω P 3.7 80k20 = 16 Ω 16 + 14 = 30 Ω 30 + 24 = 54 Ω 54k27 = 18 Ω 18 + 12 = 30 Ω 30k30 = 15 Ω Rab = 3 + 15 + 2 = 20 Ω [a] For circuit (a) Rab = 4k(3 + 7 + 2) = 4k12 = 3 Ω For circuit (b) Rab = 6 + 2 + [8k(7 + 5k2.5k7.5k5k(9 + 6))] = 6 + 2 + 8k(7 + 1) = 6 + 2 + 4 = 12 Ω For circuit (c) 144k(4 + 12) = 14.4 Ω 14.4 + 5.6 = 20 Ω 20k12 = 7.5 Ω 7.5 + 2.5 = 10 Ω 10k15 = 6 Ω 14 + 6 + 10 = 30 Ω Rab = 30k60 = 20 Ω [b] Pa = Pb = 152 = 75 W 3 482 = 192 W 12 Pc = 52 (20) = 500 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. 3–12 P 3.8 CHAPTER 3. Simple Resistive Circuits [a] p4Ω = i2s 4 = (12)2 4 = 576 W p18Ω = (4)2 18 = 288 W p3Ω = (8)2 3 = 192 W p6Ω = (8)2 6 = 384 W [b] p120V (delivered) = 120is = 120(12) = 1440 W [c] pdiss = 576 + 288 + 192 + 384 = 1440 W P 3.9 [a] From Ex. 3-1: i1 = 4 A, i2 = 8 A, is = 12 A at node b: −12 + 4 + 8 = 0, at node d: 12 − 4 − 8 = 0 [b] v1 = 4is = 48 V v3 = 3i2 = 24 V v2 = 18i1 = 72 V v4 = 6i2 = 48 V loop abda: −120 + 48 + 72 = 0, loop bcdb: −72 + 24 + 48 = 0, loop abcda: −120 + 48 + 24 + 48 = 0 P 3.10 Req = 10k[6 + 5k(8 + 12)] = 10k(6 + 5k20) = 10k(6 + 4) = 5 Ω v10A = v10Ω = (10 A)(5 Ω) = 50 V Using voltage division: v5Ω = 5k(8 + 12) 4 (50) = (50) = 20 V 6 + 5k(8 + 12) 6+4 Thus, p5Ω = P 3.11 2 v5Ω 202 = = 80 W 5 5 [a] Req = (10 + 20)k[12 + (90k10)] = 30k15 = 10 Ω v2.4A = 10(2.4) = 24 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. Problems vo = v20Ω = v90Ω = io = 3–13 20 (24) = 16 V 10 + 20 90k10 9 (24) = (24) = 14.4 V 6 + (90k10) 15 14.4 = 0.16 A 90 (v2.4A − v90Ω)2 (24 − 14.4)2 = = 15.36 W 6 6 [c] p2.4A = −(2.4)(24) = −57.6 W Thus the power developed by the current source is 57.6 W. [b] p6Ω = P 3.12 [a] R + R = 2R [b] R + R + R + · · · + R = nR [c] R + R = 2R = 3000 so R = 1500 = 1.5 kΩ This is a resistor from Appendix H. [d] nR = 4000; so if n = 4, R = 1 kΩ This is a resistor from Appendix H. P 3.13 [a] Req R2 R = RkR = = 2R 2 [b] Req = = = RkRkRk · · · kR (n R’s) R Rk n−1 R2 /(n − 1) R2 R = = R + R/(n − 1) nR n R = 5000 so R = 10 kΩ 2 This is a resistor from Appendix H. R = 4000 so R = 4000n [d] n If n = 3 r = 4000(3) = 12 kΩ This is a resistor from Appendix H. So put three 12k resistors in parallel to get 4kΩ. [c] P 3.14 4= 20R2 R2 + 40 so R2 = 10 Ω 3= 20Re 40 + Re so Re = Thus, 120 10RL = 17 10 + RL 120 Ω 17 so RL = 24 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. 3–14 P 3.15 CHAPTER 3. Simple Resistive Circuits [a] vo = 160(3300) = 66 V (4700 + 3300) [b] i = 160/8000 = 20 mA PR1 = (400 × 10−6 )(4.7 × 103 ) = 1.88 W PR2 = (400 × 10−6 )(3.3 × 103 ) = 1.32 W [c] Since R1 and R2 carry the same current and R1 > R2 to satisfy the voltage requirement, first pick R1 to meet the 0.5 W specification iR1 160 − 66 = , R1 Thus, R1 ≥ 942 0.5 Therefore, or 94 R1 2 R1 ≤ 0.5 R1 ≥ 17,672 Ω Now use the voltage specification: R2 (160) = 66 R2 + 17,672 Thus, R2 = 12,408 Ω P 3.16 [a] vo = 40R2 =8 R1 + R2 Let Re = R2 kRL = vo = so R2 RL R2 + RL 40Re = 7.5 R1 + Re so Then, 4R2 = 4.33Re = Thus, R2 = 300 Ω R1 = 4R2 R1 = 4.33Re 4.33(3600R2 ) 3600 + R2 and R1 = 4(300) = 1200 Ω [b] The resistor that must dissipate the most power is R1 , as it has the largest resistance and carries the same current as the parallel combination of R2 and the load resistor. The power dissipated in R1 will be maximum when the voltage across R1 is maximum. This will occur when the voltage divider has a resistive load. Thus, vR1 = 40 − 7.5 = 32.5 V pR1 = 32.52 = 880.2 m W 1200 Thus the minimum power rating for all resistors should be 1 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. Problems P 3.17 3–15 Refer to the solution to Problem 3.16. The voltage divider will reach the maximum power it can safely dissipate when the power dissipated in R1 equals 1 W. Thus, vR2 1 =1 1200 so vR1 = 34.64 V vo = 40 − 34.64 = 5.36 V So, 40Re = 5.36 1200 + Re Thus, and (300)RL = 185.68 300 + RL Re = 185.68 Ω and RL = 487.26 Ω The minimum value for RL from Appendix H is 560 Ω. P 3.18 Begin by using the relationships among the branch currents to express all branch currents in terms of i4: i1 = 2i2 = 2(2i3 ) = 4(2i4 ) i2 = 2i3 = 2(2i4 ) i3 = 2i4 Now use KCL at the top node to relate the branch currents to the current supplied by the source. i1 + i2 + i3 + i4 = 1 mA Express the branch currents in terms of i4 and solve for i4: 1 mA = 8i4 + 4i4 + 2i4 + i4 = 15i4 so i4 = 0.001 A 15 Since the resistors are in parallel, the same voltage, 1 V appears across each of them. We know the current and the voltage for R4 so we can use Ohm’s law to calculate R4 : R4 = vg 1V = = 15 kΩ i4 (1/15) mA Calculate i3 from i4 and use Ohm’s law as above to find R3 : i3 = 2i4 = 0.002 A 15 vg 1V .·. R3 = = = 7.5 kΩ i3 (2/15) mA © 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. 3–16 CHAPTER 3. Simple Resistive Circuits Calculate i2 from i4 and use Ohm’s law as above to find R2 : i2 = 4i4 = 0.004 A 15 vg 1V .·. R2 = = = 3750 Ω i2 (4/15) mA Calculate i1 from i4 and use Ohm’s law as above to find R1 : i1 = 8i4 = 0.008 A 15 vg 1V .·. R1 = = = 1875 Ω i1 (8/15) mA The resulting circuit is shown below: P 3.19 [a ] 40 kΩ + 60 kΩ = 100 kΩ 25 kΩk100 kΩ = 20 kΩ vo1 = vo = 20,000 (380) = 80 V (75,000 + 20,000) 60,000 (vo1 ) = 48 V (100,000) [b ] i= 380 = 3.8 mA 100,000 25,000i = 95 V vo = 60,000 (95) = 57 V 100,000 © 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. Problems 3–17 [c] It removes loading effect of second voltage divider on the first voltage divider. Observe that the open circuit voltage of the first divider is 0 vo1 = 25,000 (380) = 95 V (100,000) Now note this is the input voltage to the second voltage divider when the current controlled voltage source is used. P 3.20 (24)2 = 80, R1 + R2 + R3 Therefore, R1 + R2 + R3 = 7.2 Ω (R1 + R2 )24 = 12 (R1 + R2 + R3 ) Therefore, 2(R1 + R2) = R1 + R2 + R3 Thus, R1 + R2 = R3 ; 2R3 = 7.2; R3 = 3.6 Ω R2 (24) =5 R1 + R2 + R3 4.8R2 = R1 + R2 + 3.6 = 7.2 Thus, R2 = 1.5 Ω; P 3.21 R1 = 7.2 − R2 − R3 = 2.1 Ω [a] Let vo be the voltage across the parallel branches, positive at the upper terminal, then ig = voG1 + voG2 + · · · + voGN = vo (G1 + G2 + · · · + GN ) It follows that vo = ig (G1 + G2 + · · · + GN ) The current in the k th branch is ik = [b] i5 = P 3.22 ik = vo Gk ; Thus, ig Gk [G1 + G2 + · · · + GN ] 40(0.2) = 3.2 A 2 + 0.2 + 0.125 + 0.1 + 0.05 + 0.025 [a] At no load: At full load: vo = kvs = R2 vs . R1 + R2 vo = αvs = Re vs , R1 + Re where Re = Ro R2 Ro + R2 © 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. 3–18 CHAPTER 3. Simple Resistive Circuits Therefore k = α = Thus 1−α α R2 R1 + R2 Re R1 + Re R1 = [b] R1 = R2 = and (1 − k) R2 k (1 − α) R1 = Re α R1 = R2 Ro (1 − k) = R2 Ro + R2 k Solving for R2 yields Also, and R2 = (1 − k) R2 k (k − α) Ro α(1 − k) .·. R1 = (k − α) Ro αk 0.05 Ro = 2.5 kΩ 0.68 0.05 Ro = 14.167 kΩ 0.12 [c ] Maximum dissipation in R2 occurs at no load, therefore, PR2 (max) [(60)(0.85)]2 = = 183.6 mW 14,167 Maximum dissipation in R1 occurs at full load. PR1 (max) = [60 − 0.80(60)]2 = 57.60 mW 2500 [d ] PR1 = PR2 = (60)2 = 1.44 W = 1440 mW 2500 (0)2 =0W 14,167 © 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. Problems P 3.23 3–19 [a] The equivalent resistance of the circuit to the right of the 18 Ω resistor is 100k25k[(40k10) + 22] = 100k25k30 = 12 Ω Thus by voltage division, v18 = 18 (60) = 36 V 18 + 12 [b] The current in the 18 Ω resistor can be found from its voltage using Ohm’s law: 36 =2A i18 = 18 [c] The current in the 18 Ω resistor divides among three branches – one containing 100 Ω, one containing 25 Ω and one containing (22 + 40k10) = 30 Ω. Using current division, 100k25k30 12 (i18) = (2) = 0.96 A 25 25 [d] The voltage drop across the 25 Ω resistor can be found using Ohm’s law: i25 = v25 = 25i25 = 25(0.96) = 24 V [e] The voltage v25 divides across the 22 Ω resistor and the equivalent resistance 40k10 = 8 Ω. Using voltage division, v10 = P 3.24 8 (24) = 6.4 V 8 + 22 [a] The equivalent resistance to the right of the 10 kΩ resistor is 5 k + 2 k + [9 kk18 kk6 k)] = 10 kΩ. Therefore, 10 kk10 k (0.050) = 25 mA 10 k [b] The voltage drop across the 10 kΩ resistor can be found using Ohm’s law: i10k = v10k = (10, 000)i10k = (10, 000)(0.025) = 250 V [c] The voltage v10k drops across the 5 kΩ resistor, the 2 kΩ resistor and the equivalent resistance of the 9 kΩ, 18 kΩ and 6 kΩ resistors in parallel. Thus, using voltage division, v6k = 2k 2 (250) = (250) = 50 V 5 k + 2 k + [9 kk18 kk6 k] 10 [d] The current through the 2 kΩ resistor can be found from its voltage using Ohm’s law: v2k 50 i2k = = = 25 mA 2000 2000 © 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. 3–20 CHAPTER 3. Simple Resistive Circuits [e] The current through the 2 kΩ resistor divides among the 9 kΩ, 18 kΩ, and 6 kΩ. Using current division, i18k = P 3.25 9 kk18 kk6 k 3 (0.025) = (0.025) = 4.167 mA 18 k 18 The equivalent resistance of the circuit to the right of the 90 Ω resistor is Req = [(150k75) + 40]k(30 + 60) = 90k90 = 45 Ω Use voltage division to find the voltage drop between the top and bottom nodes: vReq = 45 (3) = 1 V 45 + 90 Use voltage division again to find v1 from vReq: v1 = 150k75 50 5 (1) = (1) = V 150k75 + 40 90 9 Use voltage division one more time to find v2 from vReq: v2 = P 3.26 30 1 (1) = V 30 + 60 3 i10k = (18)(15 k) = 6.75 mA 40 k v15k = −(6.75 m)(15 k) = −101.25 V i3k = 18 m − 6.75 m = 11.25 mA v12k = −(12 k)(11.25 m) = −135 V vo = −101.25 − (−135) = 33.75 V P 3.27 [a] v6k = 6 (18) = 13.5 V 6+2 v3k = 3 (18) = 4.5 V 3+9 vx = v6k − v3k = 13.5 − 4.5 = 9 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. Problems 3–21 6 [b] v6k = (Vs ) = 0.75Vs 8 3 v3k = (Vs ) = 0.25Vs 12 vx = (0.75Vs ) − (0.25Vs ) = 0.5Vs P 3.28 5 Ωk20 Ω = 4 Ω; Therefore, ig = i6Ω = P 3.29 4 Ω + 6 Ω = 10 Ω; 10k(15 + 12 + 13) = 8 Ω; 125 = 12.5 A 2+8 8 (12.5) = 10 A; 6+4 io = 5k20 (10) = 2 A 20 [a] The equivalent resistance seen by the voltage source is 60k[8 + 30k(4 + 80k20)] = 60k[8 + 30k20] = 60k20 = 15 Ω Thus, 300 ig = = 20 A 15 [b] Use current division to find the current in the 8 Ω division: 15 (20) = 15 A 20 Use current division again to find the current in the 30 Ω resistor: 12 i30 = (15) = 6 A 30 Thus, p30 = (6)2 (30) = 1080 W P 3.30 [a] The voltage across the 9 Ω resistor is 1(12 + 6) = 18 V. The current in the 9 Ω resistor is 18/9 = 2 A. The current in the 2 Ω resistor is 1 + 2 = 3 A. Therefore, the voltage across the 24 Ω resistor is (2)(3) + 18 = 24 V. The current in the 24 Ω resistor is 1 A. The current in the 3 Ω resistor is 1 + 2 + 1 = 4 A. Therefore, the voltage across the 72 Ω resistor is 24 + 3(4) = 36 V. The current in the 72 Ω resistor is 36/72 = 0.5 A. The 20 Ωk5 Ω resistors are equivalent to a 4 Ω resistor. The current in this equivalent resistor is 0.5 + 1 + 3 = 4.5 A. Therefore the voltage across the 108 Ω resistor is 36 + 4.5(4) = 54 V. The current in the 108 Ω resistor is 54/108 = 0.5 A. The current in the 1.2 Ω resistor is 4.5 + 0.5 = 5 A. Therefore, vg = (1.2)(5) + 54 = 60 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. 3–22 CHAPTER 3. Simple Resistive Circuits [b] The current in the 20 Ω resistor is (4.5)(4) 18 = = 0.9 A 20 20 Thus, the power dissipated by the 20 Ω resistor is i20 = p20 = (0.9)2 (20) = 16.2 W P 3.31 For all full-scale readings the total resistance is RV + Rmovement = full-scale reading 10−3 We can calculate the resistance of the movement as follows: Rmovement = Therefore, 20 mV = 20 Ω 1 mA RV = 1000 (full-scale reading) − 20 [a] RV = 1000(50) − 20 = 49, 980 Ω [b] RV = 1000(5) − 20 = 4980 Ω [c] RV = 1000(0.25) − 20 = 230 Ω [d] RV = 1000(0.025) − 20 = 5 Ω P 3.32 [a] vmeas = (50 × 10−3 )[15k45k(4980 + 20)] = 0.5612 V [b] vtrue = (50 × 10−3 )(15k45) = 0.5625 V % error = P 3.33 0.5612 − 1 × 100 = −0.224% 0.5625 The measured value is ig = 50 = 1.995526 A; (15.05618 + 10) The true value is ig = 60k20.1 = 15.05618 Ω. 60 (1.996) = 1.494768 A 80.1 60k20 = 15 Ω. 50 = 2 A; (15 + 10) %error = imeas = itrue = 60 (2) = 1.5 A 80 1.494768 − 1 × 100 = −0.34878% ≈ −0.35% 1.5 © 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. Problems P 3.34 3–23 Begin by using current division to find the actual value of the current io : itrue = 15 (50 mA) = 12.5 mA 15 + 45 imeas = 15 (50 mA) = 12.4792 mA 15 + 45 + 0.1 12.4792 % error = − 1 100 = −0.166389% ≈ −0.17% 12.5 P 3.35 [a] The model of the ammeter is an ideal ammeter in parallel with a resistor whose resistance is given by 100 mV = 50 Ω. 2 mA We can calculate the current through the real meter using current division: (25/12) 25 1 im = (imeas ) = (imeas ) = imeas 50 + (25/12) 625 25 Rm = [b] At full scale, imeas = 5 A and im = 2 mA so 5 − 0.002 = 4998 mA flows throught the resistor RA : RA = im = 100 mV 100 = Ω 4998 mA 4998 (100/4998) 1 (imeas) = (imeas) 50 + (100/4998) 2500 [c] Yes P 3.36 Original meter: Re = 50 × 10−3 = 0.01 Ω 5 Modified meter: Re = (0.02)(0.01) = 0.00667 Ω 0.03 .·. (Ifs )(0.00667) = 50 × 10−3 .·. Ifs = 7.5 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. 3–24 P 3.37 CHAPTER 3. Simple Resistive Circuits [a ] 20 × 103 i1 + 80 × 103 (i1 − iB ) = 7.5 80 × 103 (i1 − iB) = 0.6 + 40iB (0.2 × 103 ) . ·. 100i1 − 80iB = 7.5 × 10−3 80i1 − 88iB = 0.6 × 10−3 Calculator solution yields iB = 225 µA [b] With the insertion of the ammeter the equations become 100i1 − 80iB = 7.5 × 10−3 (no change) 80 × 103 (i1 − iB) = 103 iB + 0.6 + 40iB (200) 80i1 − 89iB = 0.6 × 10−3 Calculator solution yields iB = 216 µA 216 [c] % error = − 1 100 = −4% 225 P 3.38 The current in the shunt resistor at full-scale deflection is iA = ifullscale = 2 × 10−3 A. The voltage across RA at full-scale deflection is always 50 mV; therefore, 50 × 10−3 50 RA = = −3 ifullscale − 2 × 10 1000ifullscale − 2 50 = 5.001 mΩ 10,000 − 2 50 [b] RA = = 50.1 mΩ 1000 − 2 50 [c] RA = = 1.042 mΩ 50 − 2 [a] RA = © 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. Problems [d] RA = P 3.39 50 =∞ 2−2 3–25 (open circuit) At full scale the voltage across the shunt resistor will be 50 mV; therefore the power dissipated will be PA = (50 × 10−3 )2 RA Therefore RA ≥ (50 × 10−3 )2 = 5 mΩ 0.5 Otherwise the power dissipated in RA will exceed its power rating of 0.5 W When RA = 5 mΩ, the shunt current will be iA = 50 × 10−3 = 10 A 5 × 10−3 The measured current will be imeas = 10 + 0.001 = 10.001 A .·. Full-scale reading for practical purposes is 10 A. P 3.40 Rmeter = Rm + Rmovement = 750 V = 500 kΩ 1.5 mA vmeas = (25 kΩk125 kΩk50 kΩ)(30 mA) = (20 kΩ)(30 mA) = 600 V vtrue = (25 kΩk125 kΩ)(30 mA) = (20.83 kΩ)(30 mA) = 625 V 600 % error = − 1 100 = −4% 625 P 3.41 [a] Since the unknown voltage is greater than either voltmeter’s maximum reading, the only possible way to use the voltmeters would be to connect them in series. [b ] Rm1 = (300)(900) = 270 kΩ; Rm2 = (150)(1200) = 180 kΩ .·. Rm1 + Rm2 = 450 kΩ © 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. 3–26 CHAPTER 3. Simple Resistive Circuits i1 max = 300 × 10−3 = 1.11 mA; 270 i2 max = 150 × 10−3 = 0.833 mA 180 .·. imax = 0.833 mA since meters are in series vmax = (0.833 × 10−3 )(270 + 180)103 = 375 V Thus the meters can be used to measure the voltage. 320 [c] im = = 0.711 mA 450 × 103 vm1 = (0.711)(270) = 192 V; P 3.42 vm2 = (0.711)(180) = 128 V The current in the series-connected voltmeters is 205.2 136.8 = = 0.76 mA 270,000 180,000 im = v50 kΩ = (0.76 × 10−3 )(50,000) = 38 V Vpower P 3.43 supply = 205.2 + 136.8 + 38 = 380 V [a] vmeter = 180 V [b] Rmeter = (100)(200) = 20 kΩ 20k70 = 15.555556 kΩ vmeter = 180 × 15.555556 = 78.75 V 35.555556 [c] 20k20 = 10 kΩ vmeter = 180 (10) = 22.5 V 80 [d] vmeter a = 180 V vmeter b + vmeter c = 101.26 V No, because of the loading effect. P 3.44 From the problem statement we have Vs (10) 50 = (1) Vs in mV; Rs in MΩ 10 + Rs 48.75 = Vs (6) 6 + Rs (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. Problems [a] From Eq (1) 3–27 10 + Rs = 0.2Vs .·. Rs = 0.2Vs − 10 Substituting into Eq (2) yields 48.75 = 6Vs 0.2Vs − 4 or Vs = 52 mV [b] From Eq (1) 50 = 520 10 + Rs or 50Rs = 20 So Rs = 400 kΩ P 3.45 [a] R1 = (100/2)103 = 50 kΩ R2 = (10/2)103 = 5 kΩ R3 = (1/2)103 = 500 Ω [b] Let ia = actual current in the movement id = design current in the movement ia Then % error = − 1 100 id For the 100 V scale: 100 100 = , ia = 50,000 + 25 50,025 ia 50,000 = = 0.9995 id 50,025 For the 10 V scale: ia 5000 = = 0.995 id 5025 For the 1 V scale: ia 500 = = 0.9524 id 525 P 3.46 id = 100 50,000 % error = (0.9995 − 1)100 = −0.05% % error = (0.995 − 1.0)100 = −0.4975% % error = (0.9524 − 1.0)100 = −4.76% [a] Rmovement = 50 Ω R1 + Rmovement = 30 = 30 kΩ 1 × 10−3 R2 + R1 + Rmovement = .·. R1 = 29,950 Ω 150 = 150 kΩ 1 × 10−3 .·. R2 = 120 kΩ © 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. 3–28 CHAPTER 3. Simple Resistive Circuits R3 + R2 + R1 + Rmovement = 300 = 300 kΩ 1 × 10−3 .·. R3 = 150 kΩ [b] v1 = (0.96 m)(150 k) = 144 V imove = i1 = 144 = 0.96 mA 120 + 29.95 + 0.05 144 = 0.192 mA 750 k i2 = imove + i1 = 0.96 m + 0.192 m = 1.152 mA vmeas = vx = 144 + 150i2 = 316.8 V [c] v1 = 150 V; i2 = 1 m + 0.20 m = 1.20 mA i1 = 150/750,000 = 0.20 mA .·. vmeas = vx = 150 + (150 k)(1.20 m) = 330 V P 3.47 [a] Rmeter = 300 kΩ + 600 kΩk200 kΩ = 450 kΩ 450k360 = 200 kΩ Vmeter = 200 (600) = 500 V 240 [b] What is the percent error in the measured voltage? True value = % error = 360 (600) = 540 V 400 500 − 1 100 = −7.41% 540 © 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. Problems P 3.48 3–29 Since the bridge is balanced, we can remove the detector without disturbing the voltages and currents in the circuit. It follows that i1 = i2 = ig (R2 + Rx ) ig (R2 + Rx ) X = R1 + R2 + R3 + Rx R ig (R1 + R3 ) ig (R1 + R3 ) X = R1 + R2 + R3 + Rx R v3 = R3 i1 = vx = i2Rx R3 ig (R2 + Rx ) Rx ig (R1 + R3 ) X X .·. = R R .·. R3 (R2 + Rx ) = Rx (R1 + R3 ) From which Rx = P 3.49 R2 R3 R1 Note the bridge structure is balanced, that is 15 × 5 = 3 × 25, hence there is no current in the 5 kΩ resistor. It follows that the equivalent resistance of the circuit is Req = 750 + (15,000 + 3000)k(25,000 + 5000) = 750 + 11,250 = 12 kΩ The source current is 192/12,000 = 16 mA. The current down through the branch containing the 15 kΩ and 3 kΩ resistors is i3k = 11,250 (0.016) = 10 mA 18,000 .·. p3k = 3000(0.01)2 = 0.3 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. 3–30 P 3.50 CHAPTER 3. Simple Resistive Circuits Redraw the circuit, replacing the detector branch with a short circuit. 6 kΩk30 kΩ = 5 kΩ 12 kΩk20 kΩ = 7.5 kΩ is = 75 = 6 mA 12,500 v1 = 0.006(5000) = 30 V v2 = 0.006(7500) = 45 V i1 = 30 = 5 mA 6000 i2 = 45 = 3.75 mA 12,000 id = i1 − i2 = 1.25 mA P 3.51 [a] The condition for a balanced bridge is that the product of the opposite resistors must be equal: (500)(Rx ) = (1000)(750) so Rx = (1000)(750) = 1500 Ω 500 © 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. Problems 3–31 [b] The source current is the sum of the two branch currents. Each branch current can be determined using Ohm’s law, since the resistors in each branch are in series and the voltage drop across each branch is 24 V: is = 24 V 24 V + = 28.8 mA 500 Ω + 750 Ω 1000 Ω + 1500 Ω [c] We can use Ohm’s law to find the current in each branch: ileft = 24 = 19.2 mA 500 + 750 iright = 24 = 9.6 mA 1000 + 1500 Now we can use the formula p = Ri2 to find the power dissipated by each resistor: p500 = (500)(0.0192)2 = 184.32 mW p1000 = (1000)(0.0096)2 = 92.16 mW p750 = (750)(0.0192)2 = 276.18 mW p1500 = (1500)(0.0096)2 = 138.24 mW Thus, the 750 Ω resistor absorbs the most power; it absorbs 276.48 mW of power. [d] From the analysis in part (c), the 1000 Ω resistor absorbs the least power; it absorbs 92.16 mW of power. P 3.52 In order that all four decades (1, 10, 100, 1000) that are used to set R3 contribute to the balance of the bridge, the ratio R2 /R1 should be set to 0.001. P 3.53 Begin by transforming the ∆-connected resistors (10 Ω, 40 Ω, 50 Ω) to Y-connected resistors. Both the Y-connected and ∆-connected resistors are shown below to assist in using Eqs. 3.44 – 3.46: Now use Eqs. 3.44 – 3.46 to calculate the values of the Y-connected resistors: R1 = (40)(10) = 4 Ω; 10 + 40 + 50 R2 = (10)(50) = 5 Ω; 10 + 40 + 50 R3 = (40)(50) = 20 Ω 10 + 40 + 50 © 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. 3–32 CHAPTER 3. Simple Resistive Circuits The transformed circuit is shown below: The equivalent resistance seen by the 24 V source can be calculated by making series and parallel combinations of the resistors to the right of the 24 V source: Req = (15 + 5)k(1 + 4) + 20 = 20k5 + 20 = 4 + 20 = 24 Ω Therefore, the current i in the 24 V source is given by i= 24 V =1A 24 Ω Use current division to calculate the currents i1 and i2 . Note that the current i1 flows in the branch containing the 15 Ω and 5 Ω series connected resistors, while the current i2 flows in the parallel branch that contains the series connection of the 1 Ω and 4 Ω resistors: i1 = 4 4 (i) = (1 A) = 0.2 A, 15 + 5 20 and i2 = 1 A − 0.2 A = 0.8 A Now use KVL and Ohm’s law to calculate v1. Note that v1 is the sum of the voltage drop across the 4 Ω resistor, 4i2 , and the voltage drop across the 20 Ω resistor, 20i: v1 = 4i2 + 20i = 4(0.8 A) + 20(1 A) = 3.2 + 20 = 23.2 V Finally, use KVL and Ohm’s law to calculate v2 . Note that v2 is the sum of the voltage drop across the 5 Ω resistor, 5i1 , and the voltage drop across the 20 Ω resistor, 20i: v2 = 5i1 + 20i = 5(0.2 A) + 20(1 A) = 1 + 20 = 21 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. Problems P 3.54 3–33 [a] After the 20 Ω—100 Ω—50 Ω wye is replaced by its equivalent delta, the circuit reduces to Now the circuit can be reduced to 96 (1000) = 240 mA 400 400 io = (240) = 96 mA 1000 80 [b] i1 = (240) = 48 mA 400 [c] Now that io and i1 are known return to the original circuit i= v2 = (50)(0.048) + (600)(0.096) = 60 V i2 = v2 60 = = 600 mA 100 100 [d] vg = v2 + 20(0.6 + 0.048) = 60 + 12.96 = 72.96 V pg = −(vg )(1) = −72.96 W Thus the current source delivers 72.96 W. P 3.55 The top of the pyramid can be replaced by a resistor equal to R1 = (18)(9) = 6 kΩ 27 © 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. 3–34 CHAPTER 3. Simple Resistive Circuits The lower left and right deltas can be replaced by wyes. Each resistance in the wye equals 3 kΩ. Thus our circuit can be reduced to Now the 12 kΩ in parallel with 6 kΩ reduces to 4 kΩ. .·. Rab = 3 k + 4 k + 3 k = 10 kΩ P 3.56 [a] Calculate the values of the Y-connected resistors that are equivalent to the 10 Ω, 40 Ω, and 50Ω ∆-connected resistors: RX = (10)(50) = 5 Ω; 10 + 40 + 50 RZ = (10)(40) = 4Ω 10 + 40 + 50 RY = (50)(40) = 20 Ω; 10 + 40 + 50 Replacing the R2 —R3—R4 delta with its equivalent Y gives Now calculate the equivalent resistance Rab by making series and parallel combinations of the resistors: Rab = 13 + 5 + [(8 + 4)k(20 + 4)] + 7 = 33 Ω [b] Calculate the values of the ∆-connected resistors that are equivalent to the 10 Ω, 8 Ω, and 40 Ω Y-connected resistors: (10)(8) + (8)(40) + (10)(40) 800 RX = = = 100 Ω 8 8 (10)(8) + (8)(40) + (10)(40) 800 RY = = = 80 Ω 10 10 (10)(8) + (8)(40) + (10)(40) 800 RZ = = = 20 Ω 40 40 © 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. Problems 3–35 Replacing the R2 , R4 , R5 wye with its equivalent ∆ gives Make series and parallel combinations of the resistors to find the equivalent resistance Rab: 100 Ωk50 Ω = 33.33 Ω; 80 Ωk4 Ω = 3.81 Ω .·. 20k(33.33 + 3.81) = 13 Ω .·. Rab = 13 + 13 + 7 = 33 Ω [c] Convert the delta connection R4 —R5—R6 to its equivalent wye. Convert the wye connection R3 —R4 —R6 to its equivalent delta. P 3.57 [a] Convert the upper delta to a wye. R1 = (50)(50) = 12.5 Ω 200 R2 = (50)(100) = 25 Ω 200 R3 = (100)(50) = 25 Ω 200 Convert the lower delta to a wye. R4 = (60)(80) = 24 Ω 200 R5 = (60)(60) = 18 Ω 200 R6 = (80)(60) = 24 Ω 200 Now redraw the circuit using the wye equivalents. © 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. 3–36 CHAPTER 3. Simple Resistive Circuits Rab = 1.5 + 12.5 + (120)(80) + 18 = 14 + 48 + 18 = 80 Ω 200 [b] When vab = 400 V 400 ig = =5A 80 48 i31 = (5) = 3 A 80 p31Ω = (31)(3)2 = 279 W P 3.58 Replace the upper and lower deltas with the equivalent wyes: R1U = (10)(50) (50)(40) (10)(40) = 5 Ω; R2U = = 20 Ω; R3U = = 4Ω 100 100 100 R1L = (10)(60) (60)(30) (10)(30) = 6 Ω; R2L = = 18 Ω; R3L = = 3Ω 100 100 100 The resulting circuit is shown below: Now make series and parallel combinations of the resistors: (4 + 6)k(20 + 32 + 20 + 18) = 10k90 = 9 Ω Rab = 33 + 5 + 9 + 3 + 40 = 90 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. Problems P 3.59 3–37 8 + 12 = 20 Ω 20k60 = 15 Ω 15 + 20 = 35 Ω 35k140 = 28 Ω 28 + 22 = 50 Ω 50k75 = 30 Ω 30 + 10 = 40 Ω ig = 240/40 = 6 A io = (6)(50)/125 = 2.4 A i140Ω = (6 − 2.4)(35)/175 = 0.72 A p140Ω = (0.72)2 (140) = 72.576 W P 3.60 [a] Replace the 60—120—20 Ω delta with a wye equivalent to get is = 750 750 = = 10 A 5 + (24 + 36)k(14 + 6) + 12 + 43 75 i1 = (24 + 36)k(14 + 6) 15 (10) = (10) = 2.5 A 24 + 36 60 [b] io = 10 − 2.5 = 7.5 A v = 36i1 − 6io = 36(2.5) − 6(7.5) = 45 V v 45 = 7.5 + = 8.25 A 60 60 [d] Psupplied = (750)(10) = 7500 W [c] i2 = io + © 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. 3–38 CHAPTER 3. Simple Resistive Circuits P 3.61 25k6.25 = 5 Ω i1 = 60k30 = 20 Ω (6)(15) = 2.25 A; (40) vx = 20i1 = 45 V vg = 25i1 = 56.25 V v6.25 = vg − vx = 11.25 V Pdevice = P 3.62 11.252 452 56.252 + + = 298.6875 W 6.25 30 15 [a] Subtracting Eq. 3.42 from Eq. 3.43 gives R1 − R2 = (Rc Rb − Rc Ra )/(Ra + Rb + Rc ). Adding this expression to Eq. 3.41 and solving for R1 gives R1 = Rc Rb/(Ra + Rb + Rc). To find R2, subtract Eq. 3.43 from Eq. 3.41 and add this result to Eq. 3.42. To find R3 , subtract Eq. 3.41 from Eq. 3.42 and add this result to Eq. 3.43. [b] Using the hint, Eq. 3.43 becomes R1 + R3 = Rb [(R2/R3 )Rb + (R2 /R1 )Rb ] Rb(R1 + R3 )R2 = (R2/R1 )Rb + Rb + (R2 /R3 )Rb (R1 R2 + R2 R3 + R3 R1 ) Solving for Rb gives Rb = (R1 R2 + R2R3 + R3 R1 )/R2 . To find Ra : First use Eqs. 3.44–3.46 to obtain the ratios (R1/R3 ) = (Rc /Ra ) or © 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. Problems 3–39 Rc = (R1 /R3 )Ra and (R1/R2 ) = (Rb /Ra ) or Rb = (R1/R2 )Ra. Now use these relationships to eliminate Rb and Rc from Eq. 3.42. To find Rc , use Eqs. 3.44–3.46 to obtain the ratios Rb = (R3 /R2 )Rc and Ra = (R3 /R1 )Rc . Now use the relationships to eliminate Rb and Ra from Eq. 3.41. P 3.63 P 3.64 1 R1 = Ra R1 R2 + R2 R3 + R3R1 1/G1 = (1/G1 )(1/G2 ) + (1/G2 )(1/G3 ) + (1/G3 )(1/G1 ) (1/G1 )(G1 G2 G3 ) G2 G3 = = G1 + G2 + G3 G1 + G2 + G3 Similar manipulations generate the expressions for Gb and Gc . Ga = [a] Rab = 2R1 + Therefore Thus R2 (2R1 + RL ) = RL 2R1 + R2 + RL 2R1 − RL + R2 (2R1 + RL ) =0 2R1 + R2 + RL R2L = 4R21 + 4R1 R2 = 4R1 (R1 + R2 ) When Rab = RL , the current into terminal a of the attenuator will be vi /RL Using current division, the current in the RL branch will be vi R2 · RL 2R1 + R2 + RL Therefore and vo = vi R2 · RL RL 2R1 + R2 + RL vo R2 = vi 2R1 + R2 + RL [b] (600)2 = 4(R1 + R2 )R1 9 × 104 = R21 + R1 R2 vo R2 = 0.6 = vi 2R1 + R2 + 600 .·. 1.2R1 + 0.6R2 + 360 = R2 0.4R2 = 1.2R1 + 360 R2 = 3R1 + 900 .·. 9 × 104 = R21 + R1 (3R1 + 900) = 4R21 + 900R1 .·. R21 + 225R1 − 22,500 = 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. 3–40 CHAPTER 3. Simple Resistive Circuits R1 = −112.5 ± .·. R1 = 75 Ω q (112.5)2 + 22,500 = −112.5 ± 187.5 .·. R2 = 3(75) + 900 = 1125 Ω [c] From Appendix H, choose R1 = 68 Ω and R2 = 1.2 kΩ. For these values, Rab = RL = q (4)(68)(68 + 1200) = 587.3 Ω 587.3 % error = − 1 100 = −2.1% 600 vo 1200 = = 0.624 vi 2(68) + 1200 + 587.3 0.624 % error = − 1 100 = 4% 0.6 P 3.65 [a] After making the Y-to-∆ transformation, the circuit reduces to Combining the parallel resistors reduces the circuit to Now note: Therefore 0.75R + Rab 2.25R2 + 3.75RRL 3R 3R + RL 3R(3R + 5RL ) ! = = 2 2.25R + 3.75RRL 15R + 9RL 3R + 3R + RL If R = RL , we have Therefore 3RRL 2.25R2 + 3.75RRL = 3R + RL 3R + RL ! Rab = 3RL (8RL ) = RL 24RL Rab = RL © 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. Problems 3–41 [b] When R = RL , the circuit reduces to io = ii(3RL ) 1 1 vi = ii = , 4.5RL 1.5 1.5 RL Therefore P 3.66 1 vo = 0.75RL io = vi , 2 vo = 0.5 vi [a] 3.5(3R − RL ) = 3R + RL 10.5R − 1050 = 3R + 300 7.5R = 1350, R = 180 Ω 2(180)(300)2 R2 = = 4500 Ω 3(180)2 − (300)2 [b ] vo = vi 42 = = 12 V 3.5 3.5 io = 12 = 40 mA 300 i1 = 42 − 12 30 = = 6.67 mA 4500 4500 ig = 42 = 140 mA 300 i2 = 140 − 6.67 = 133.33 mA i3 = 40 − 6.67 = 33.33 mA i4 = 133.33 − 33.33 = 100 mA © 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. 3–42 CHAPTER 3. Simple Resistive Circuits p4500 top = (6.67 × 10−3 )2(4500) = 0.2 W = (133.33 × 10−3 )2 (180) = 3.2 W p180 left p180 right p180 vertical p300 load = (33.33 × 10−3 )2 (180) = 0.2 W = (100 × 10−3 )2 (180) = 0.48 W = (40 × 10−3 )2 (300) = 0.48 W The 180 Ω resistor carrying i2 [c] p180 left = 3.2 W [d] Two resistors dissipate minimum power – the 4500 Ω resistor and the 180 Ω resistor carrying i3. [e] They both dissipate 0.2 W. P 3.67 [a ] va = vinR4 Ro + R4 + ∆R vb = R3 vin R2 + R3 vo = va − vb = R4 vin R3 − vin Ro + R4 + ∆R R2 + R3 When the bridge is balanced, R4 R3 vin = vin Ro + R4 R2 + R3 . ·. R4 R3 = Ro + R4 R2 + R3 Thus, vo = = = ≈ R4 vin R4 vin − Ro + R4 + ∆R Ro + R4 1 1 R4 vin − Ro + R4 + ∆R Ro + R4 R4 vin(−∆R) (Ro + R4 + ∆R)(Ro + R4) −(∆R)R4 vin , since ∆R << R4 (Ro + R4 )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. Problems 3–43 [b] ∆R = 0.03Ro Ro = R2 R4 (1000)(5000) = = 10,000 Ω R3 500 ∆R = (0.03)(104 ) = 300 Ω . ·. v o ≈ [c] vo = = = P 3.68 −300(5000)(6) = −40 mV (15,000)2 −(∆R)R4 vin (Ro + R4 + ∆R)(Ro + R4 ) −300(5000)(6) (15,300)(15,000) −39.2157 mV −(∆R)R4 vin (Ro + R4 )2 [a] approx value = true value = . ·. −(∆R)R4vin (Ro + R4 + ∆R)(Ro + R4 ) approx value (Ro + R4 + ∆R) = true value (Ro + R4 ) .·. % error = Ro + R4 −∆R − 1 × 100 = × 100 Ro + R4 + ∆R Ro + R4 Note that in the above expression, we take the ratio of the true value to the approximate value because both values are negative. But Ro = R2 R4 R3 .·. % error = [b] % error = P 3.69 −R3∆R R4 (R2 + R3 ) −(500)(300) × 100 = −2% (5000)(1500) ∆R(R3 )(100) = 0.5 (R2 + R3)R4 ∆R(500)(100) = 0.5 (1500)(5000) .·. ∆R = 75 Ω % change = 75 × 100 = 0.75% 10,000 © 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. 3–44 P 3.70 CHAPTER 3. Simple Resistive Circuits [a] From Eq 3.64 we have i1 i2 2 = R22 R21 (1 + 2σ)2 Substituting into Eq 3.63 yields R2 = R22 R1 R21 (1 + 2σ)2 Solving for R2 yields R2 = (1 + 2σ)2 R1 [b] From Eq 3.67 we have i1 R2 = ib R1 + R2 + 2Ra But R2 = (1 + 2σ)2 R1 and Ra = σR1 therefore (1 + 2σ)2 R1 (1 + 2σ)2 i1 = = ib R1 + (1 + 2σ)2 R1 + 2σR1 (1 + 2σ) + (1 + 2σ)2 = 1 + 2σ 2(1 + σ) It follows that i1 ib 2 = (1 + 2σ)2 4(1 + σ)2 Substituting into Eq 3.66 gives Rb = P 3.71 (1 + 2σ)2 Ra (1 + 2σ)2 σR1 = 4(1 + σ)2 4(1 + σ)2 From Eq 3.69 i1 R2 R3 = i3 D But D = (R1 + 2Ra )(R2 + 2Rb ) + 2Rb R2 where Ra = σR1 ; R2 = (1 + 2σ)2 R1 and Rb = (1 + 2σ)2σR1 4(1 + σ)2 Therefore D can be written as © 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. Problems D = = = = 3–45 2(1 + 2σ)2 σR1 (R1 + 2σR1 ) (1 + 2σ) R1 + + 4(1 + σ)2 " # (1 + 2σ)2 σR1 2 2(1 + 2σ) R1 4(1 + σ)2 " # (1 + 2σ)σ σ 3 2 (1 + 2σ) R1 1 + + 2(1 + σ)2 2(1 + σ)2 (1 + 2σ)3R21 {2(1 + σ)2 + σ + (1 + 2σ)σ} 2(1 + σ)2 (1 + 2σ)3R21 {1 + 3σ + 2σ 2} (1 + σ)2 " # 2 (1 + 2σ)4R21 D= (1 + σ) i1 .·. i3 R2 R3 (1 + σ) (1 + 2σ)4R21 (1 + 2σ)2R1 R3 (1 + σ) = (1 + 2σ)4R21 (1 + σ)R3 = (1 + 2σ)2R1 When this result is substituted into Eq 3.69 we get R3 = = (1 + σ)2R23 R1 (1 + 2σ)4 R21 Solving for R3 gives R3 = P 3.72 (1 + 2σ)4R1 (1 + σ)2 From the dimensional specifications, calculate σ and R3 : y 0.025 σ= = = 0.025; x 1 Vdc2 122 R3 = = = 1.2 Ω p 120 Calculate R1 from R3 and σ: R1 = (1 + σ)2 R3 = 1.0372 Ω (1 + 2σ)4 Calculate Ra , Rb , and R2 : Ra = σR1 = 0.0259 Ω Rb = (1 + 2σ)2σR1 = 0.0068 Ω 4(1 + σ)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. 3–46 CHAPTER 3. Simple Resistive Circuits R2 = (1 + 2σ)2R1 = 1.1435 Ω Using symmetry, R4 = R2 = 1.1435 Ω R5 = R1 = 1.0372 Ω Rc = Rb = 0.0068 Ω Rd = Ra = 0.0259 Ω Test the calculations by checking the power dissipated, which should be 120 W/m. Calculate D, then use Eqs. (3.58)-(3.60) to calculate ib , i1, and i2: D = (R1 + 2Ra )(R2 + 2Rb ) + 2R2 Rb = 1.2758 ib = Vdc (R1 + R2 + 2Ra ) = 21 A D i1 = Vdc R2 = 10.7561 A D i2 = Vdc (R1 + 2Ra ) = 10.2439 A D It follows that i2b Rb = 3 W and the power dissipation per meter is 3/0.025 = 120 W/m. The value of i21R1 = 120 W/m. The value of i22 R2 = 120 W/m. Finally, i21Ra = 3 W/m. P 3.73 From the solution to Problem 3.72 we have ib = 21 A and i3 = 10 A. By symmetry ic = 21 A thus the total current supplied by the 12 V source is 21 + 21 + 10 or 52 A. Therefore the total power delivered by the source is p12 V (del) = (12)(52) = 624 W. We also have from the solution that pa = pb = pc = pd = 3 W. Therefore the total power delivered to the vertical resistors is pV = (8)(3) = 24 W. The total power delivered to the five horizontal resistors is pH = 5(120) = 600 W. .·. P 3.74 X pdiss = pH + pV = 624 W = X pdel [a] σ = 0.03/1.5 = 0.02 Since the power dissipation is 200 W/m the power dissipated in R3 must be 200(1.5) or 300 W. Therefore 122 = 0.48 Ω 300 From Table 3.1 we have (1 + σ)2R3 R1 = = 0.4269 Ω (1 + 2σ)4 R3 = Ra = σR1 = 0.0085 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. Problems 3–47 R2 = (1 + 2σ)2 R1 = 0.4617 Ω Rb = (1 + 2σ)2σR1 = 0.0022 Ω 4(1 + σ)2 Therefore R4 = R2 = 0.4617 Ω R5 = R1 = 0.4269 Ω Rc = Rb = 0.0022 Ω Rd = Ra = 0.0085 Ω [b] D = [0.4269 + 2(0.0085)][0.4617 + 2(0.0022)] + 2(0.4617)(0.0022) = 0.2090 i1 = Vdc R2 = 26.51 A D i21 R1 = 300 W or 200 W/m i2 = R1 + 2Ra Vdc = 25.49 A D i22 R2 = 300 W or 200 W/m i21 Ra = 6 W or 200 W/m ib = R1 + R2 + 2Ra Vdc = 52 A D i2b Rb = 6 W or 200 W/m isource = 52 + 52 + 12 = 129 A 0.48 pdel = 12(129) = 1548 W pH = 5(300) = 1500 W pV = 8(6) = 48 W X pdel = X pdiss = 1548 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. 4 Techniques of Circuit Analysis Assessment Problems AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages: The two node voltage equations are v1 v1 − v2 v1 + + = 0 −15 + 60 15 5 v2 v2 − v1 5+ + = 0 2 5 Place in standard these equations form: 1 1 1 1 v1 + + + v2 − = 15 60 15 5 5 1 1 1 v1 − + v2 + = −5 5 2 5 Solving, v1 = 60 V and v2 = 10 V; Therefore, i1 = (v1 − v2)/5 = 10 A [b] p15A = −(15 A)v1 = −(15 A)(60 V) = −900 W = 900 W(delivered) [c] p5A = (5 A)v2 = (5 A)(10 V) = 50 W= −50 W(delivered) © 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 4–1 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. 4–2 CHAPTER 4. Techniques of Circuit Analysis AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown: The two node voltage equations are: v1 v1 − v2 −4.5 + + = 0 1 6+2 v2 v2 − v1 v2 − 30 + + = 0 12 6+2 4 Place form: theseequationsin standard 1 1 + v2 − = 4.5 v1 1 + 8 8 1 1 1 1 + v2 + + = 7.5 v1 − 8 12 8 4 Solving, v1 = 6 V v2 = 18 V To find the voltage v, first find the current i through the series-connected 6 Ω and 2 Ω resistors: i= v1 − v2 6 − 18 = = −1.5 A 6+2 8 Using a KVL equation, calculate v: v = 2i + v2 = 2(−1.5) + 18 = 15 V AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as shown: The node voltage equations are: v1 − 50 v1 v1 − v2 + + − 3i1 = 6 8 2 v2 v2 − v1 −5 + + + 3i1 = 4 2 0 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. Problems 4–3 The dependent source requires the following constraint equation: 50 − v1 i1 = 6 Place these equations in standard form: 1 50 1 1 1 v1 + + + v2 − + i1(−3) = 6 8 2 2 6 1 1 1 v1 − + v2 + + i1(3) = 5 2 4 2 50 1 v1 + v2(0) + i1(1) = 6 6 Solving, v1 = 32 V; v2 = 16 V; i1 = 3 A Using these values to calculate the power associated with each source: p50V = −50i1 = −150 W p5A = −5(v2) = −80 W p3i1 = 3i1 (v2 − v1) = −144 W [b] All three sources are delivering power to the circuit because the power computed in (a) for each of the sources is negative. AP 4.4 Redraw the circuit and label the reference node and the node at which the node voltage equation will be written: The node voltage equation is vo vo − 10 vo + 20i∆ + + =0 40 10 20 The constraint equation required by the dependent source is i∆ = i10 Ω + i30 Ω = 10 − vo 10 + 20i∆ + 10 30 Place these equations in standard form: 1 1 1 vo + + + i∆ (1) = 40 10 20 1 20 vo + i∆ 1 − = 10 30 Solving, i∆ = −3.2 A and 1 1+ 10 30 vo = 24 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. 4–4 CHAPTER 4. Techniques of Circuit Analysis AP 4.5 Redraw the circuit identifying the three node voltages and the reference node: Note that the dependent voltage source and the node voltages v and v2 form a supernode. The v1 node voltage equation is v1 v1 − v + − 4.8 = 0 7.5 2.5 The supernode equation is v − v1 v v2 v2 − 12 + + + =0 2.5 10 2.5 1 The constraint equation due to the dependent source is ix = v1 7.5 The constraint equation due to the supernode is v + ix = v2 Place this set of equations in standard form: 1 1 1 v1 + + v − + v2 (0) + 7.5 2.5 2.5 1 1 1 1 v1 − + v + + v2 +1 + 2.5 2.5 10 2.5 1 v1 − + v(0) + v2 (0) + 7.5 v1(0) + v(1) + v2 (−1) + ix (0) = 4.8 ix (0) = 12 ix (1) = 0 ix (1) = 0 Solving this set of equations gives v1 = 15 V, v2 = 10 V, ix = 2 A, and v = 8 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. Problems 4–5 AP 4.6 Redraw the circuit identifying the reference node and the two unknown node voltages. Note that the right-most node voltage is the sum of the 60 V source and the dependent source voltage. The node voltage equation at v1 is v1 − 60 v1 v1 − (60 + 6iφ ) + + =0 2 24 3 The constraint equation due to the dependent source is iφ = 60 + 6iφ − v1 3 Place these two equations in standard form: 1 1 1 + + + iφ(−2) = 30 + 20 v1 2 24 3 1 v1 + iφ(1 − 2) = 20 3 Solving, iφ = −4 A and v1 = 48 V AP 4.7 [a] Redraw the circuit identifying the three mesh currents: The mesh current equations are: −80 + 5(i1 − i2 ) + 26(i1 − i3) = 0 30i2 + 90(i2 − i3) + 5(i2 − i1) = 0 8i3 + 26(i3 − i1 ) + 90(i3 − i2) = 0 Place these equations in standard form: © 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. 4–6 CHAPTER 4. Techniques of Circuit Analysis 31i1 − 5i2 − 26i3 = 80 −5i1 + 125i2 − 90i3 = 0 −26i1 − 90i2 + 124i3 Solving, = 0 i1 = 5 A; i2 = 2 A; i3 = 2.5 A p80V = −(80)i1 = −(80)(5) = −400 W Therefore the 80 V source is delivering 400 W to the circuit. [b] p8Ω = (8)i23 = 8(2.5)2 = 50 W, so the 8 Ω resistor dissipates 50 W. AP 4.8 [a] b = 8, n = 6, b−n+1 = 3 [b] Redraw the circuit identifying the three mesh currents: The three mesh-current equations are −25 + 2(i1 − i2) + 5(i1 − i3) + 10 = 0 −(−3vφ ) + 14i2 + 3(i2 − i3 ) + 2(i2 − i1 ) = 0 1i3 − 10 + 5(i3 − i1 ) + 3(i3 − i2 ) = 0 The dependent source constraint equation is vφ = 3(i3 − i2) Place these four equations in standard form: 7i1 − 2i2 − 5i3 + 0vφ = 15 −2i1 + 19i2 − 3i3 + 3vφ = 0 −5i1 − 3i2 + 9i3 + 0vφ = 10 0i1 + 3i2 − 3i3 + 1vφ = 0 Solving i1 = 4 A; i2 = −1 A; i3 = 3 A; vφ = 12 V pds = −(−3vφ)i2 = 3(12)(−1) = −36 W Thus, the dependent source is delivering 36 W, or absorbing −36 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. Problems 4–7 AP 4.9 Redraw the circuit identifying the three mesh currents: The mesh current equations are: −25 + 6(ia − ib) + 8(ia − ic) = 0 2ib + 8(ib − ic) + 6(ib − ia) = 0 5iφ + 8(ic − ia) + 8(ic − ib) = 0 The dependent source constraint equation is iφ = ia. We can substitute this simple expression for iφ into the third mesh equation and place the equations in standard form: 14ia − 6ib − 8ic = 25 −6ia + 16ib − 8ic = 0 −3ia − 8ib + 16ic = 0 Solving, ia = 4 A; ib = 2.5 A; ic = 2 A Thus, vo = 8(ia − ic) = 8(4 − 2) = 16 V AP 4.10 Redraw the circuit identifying the mesh currents: Since there is a current source on the perimeter of the i3 mesh, we know that i3 = −16 A. The remaining two mesh equations are −30 + 3i1 + 2(i1 − i2) + 6i1 = 0 8i2 + 5(i2 + 16) + 4i2 + 2(i2 − 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. 4–8 CHAPTER 4. Techniques of Circuit Analysis Place these equations in standard form: 11i1 − 2i2 = 30 −2i1 + 19i2 = −80 Solving: i1 = 2 A, i2 = −4 A, i3 = −16 A The current in the 2 Ω resistor is i1 − i2 = 6 A .·. Thus, the 2 Ω resistors dissipates 72 W. p2 Ω = (6)2 (2) = 72 W AP 4.11 Redraw the circuit and identify the mesh currents: There are current sources on the perimeters of both the ib mesh and the ic mesh, so we know that ib = −10 A; ic = 2vφ 5 The remaining mesh current equation is −75 + 2(ia + 10) + 5(ia − 0.4vφ ) = 0 The dependent source requires the following constraint equation: vφ = 5(ia − ic) = 5(ia − 0.4vφ ) Place the mesh current equation and the dependent source equation is standard form: 7ia − 2vφ = 55 5ia − 3vφ = 0 Solving: ia = 15 A; Thus, ia = 15 A. ib = −10 A; ic = 10 A; vφ = 25 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. Problems 4–9 AP 4.12 Redraw the circuit and identify the mesh currents: The 2 A current source is shared by the meshes ia and ib. Thus we combine these meshes to form a supermesh and write the following equation: −10 + 2ib + 2(ib − ic) + 2(ia − ic) = 0 The other mesh current equation is −6 + 1ic + 2(ic − ia) + 2(ic − ib) = 0 The supermesh constraint equation is ia − ib = 2 Place these three equations in standard form: 2ia + 4ib − 4ic = 10 −2ia − 2ib + 5ic = 6 ia − ib + 0ic = 2 Solving, Thus, ia = 7 A; ib = 5 A; ic = 6 A p1 Ω = i2c(1) = (6)2 (1) = 36 W AP 4.13 Redraw the circuit and identify the reference node and the node voltage v1: The node voltage equation is v1 − 20 v1 − 25 −2+ =0 15 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. 4–10 CHAPTER 4. Techniques of Circuit Analysis Rearranging and solving, v1 1 1 20 25 + = 2+ + 15 10 15 10 .·. v1 = 35 V p2A = −35(2) = −70 W Thus the 2 A current source delivers 70 W. AP 4.14 Redraw the circuit and identify the mesh currents: There is a current source on the perimeter of the i3 mesh, so i3 = 4 A. The other two mesh current equations are −128 + 4(i1 − 4) + 6(i1 − i2 ) + 2i1 = 0 30ix + 5i2 + 6(i2 − i1 ) + 3(i2 − 4) = 0 The constraint equation due to the dependent source is ix = i1 − i3 = i1 − 4 Substitute the constraint equation into the second mesh equation and place the resulting two mesh equations in standard form: 12i1 − 6i2 = 144 24i1 + 14i2 = 132 Solving, i1 = 9 A; i2 = −6 A; i3 = 4 A; ix = 9 − 4 = 5 A .·. v4A = 3(i3 − i2 ) − 4ix = 10 V p4A = −v4A (4) = −(10)(4) = −40 W Thus, the 2 A current source delivers 40 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. Problems 4–11 AP 4.15 [a] Redraw the circuit with a helpful voltage and current labeled: Transform the 120 V source in series with the 20 Ω resistor into a 6 A source in parallel with the 20 Ω resistor. Also transform the −60 V source in series with the 5 Ω resistor into a −12 A source in parallel with the 5 Ω resistor. The result is the following circuit: Combine the three current sources into a single current source, using KCL, and combine the 20 Ω, 5 Ω, and 6 Ω resistors in parallel. The resulting circuit is shown on the left. To simplify the circuit further, transform the resulting 30 A source in parallel with the 2.4 Ω resistor into a 72 V source in series with the 2.4 Ω resistor. Combine the 2.4 Ω resistor in series with the 1.6 Ω resisor to get a very simple circuit that still maintains the voltage v. The resulting circuit is on the right. Use voltage division in the circuit on the right to calculate v as follows: 8 (72) = 48 V 12 [b] Calculate i in the circuit on the right using Ohm’s law: v= v 48 = =6A 8 8 Now use i to calculate va in the circuit on the left: i= va = 6(1.6 + 8) = 57.6 V Returning back to the original circuit, note that the voltage va is also the voltage drop across the series combination of the 120 V source and 20 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. 4–12 CHAPTER 4. Techniques of Circuit Analysis resistor. Use this fact to calculate the current in the 120 V source, ia: ia = 120 − va 120 − 57.6 = = 3.12 A 20 20 p120V = −(120)ia = −(120)(3.12) = −374.40 W Thus, the 120 V source delivers 374.4 W. AP 4.16 To find RTh , replace the 72 V source with a short circuit: Note that the 5 Ω and 20 Ω resistors are in parallel, with an equivalent resistance of 5k20 = 4 Ω. The equivalent 4 Ω resistance is in series with the 8 Ω resistor for an equivalent resistance of 4 + 8 = 12 Ω. Finally, the 12 Ω equivalent resistance is in parallel with the 12 Ω resistor, so RTh = 12k12 = 6 Ω. Use node voltage analysis to find vTh . Begin by redrawing the circuit and labeling the node voltages: The node voltage equations are v1 − 72 v1 v1 − vTh + + = 0 5 20 8 vTh − v1 vTh − 72 + = 0 8 12 Place these equations in standard form: 1 1 1 1 72 v1 + + + vTh − = 5 20 8 8 5 1 1 1 v1 − + vTh + = 6 8 8 12 Solving, v1 = 60 V and vTh = 64.8 V. Therefore, the Thévenin equivalent circuit is a 64.8 V source in series with a 6 Ω resistor. © 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. Problems 4–13 AP 4.17 We begin by performing a source transformation, turning the parallel combination of the 15 A source and 8 Ω resistor into a series combination of a 120 V source and an 8 Ω resistor, as shown in the figure on the left. Next, combine the 2 Ω, 8 Ω and 10 Ω resistors in series to give an equivalent 20 Ω resistance. Then transform the series combination of the 120 V source and the 20 Ω equivalent resistance into a parallel combination of a 6 A source and a 20 Ω resistor, as shown in the figure on the right. Finally, combine the 20 Ω and 12 Ω parallel resistors to give RN = 20k12 = 7.5 Ω. Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a 7.5 Ω resistor. AP 4.18 Find the Thévenin equivalent with respect to A, B using source transformations. To begin, convert the series combination of the −36 V source and 12 kΩ resistor into a parallel combination of a −3 mA source and 12 kΩ resistor. The resulting circuit is shown below: Now combine the two parallel current sources and the two parallel resistors to give a −3 + 18 = 15 mA source in parallel with a 12 kk60 k= 10 kΩ resistor. Then transform the 15 mA source in parallel with the 10 kΩ resistor into a 150 V source in series with a 10 kΩ resistor, and combine this 10 kΩ resistor in series with the 15 kΩ resistor. The Thévenin equivalent is thus a 150 V source in series with a 25 kΩ resistor, as seen to the left of the terminals A,B in the circuit below. Now attach the voltmeter, modeled as a 100 kΩ resistor, to the Thévenin equivalent and use voltage division to calculate the meter reading vAB : vAB = 100,000 (150) = 120 V 125,000 © 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. 4–14 CHAPTER 4. Techniques of Circuit Analysis AP 4.19 Begin by calculating the open circuit voltage, which is also vTh , from the circuit below: Summing the currents away from the node labeled vTh We have vTh vTh − 24 + 4 + 3ix + =0 8 2 Also, using Ohm’s law for the 8 Ω resistor, ix = vTh 8 Substituting the second equation into the first and solving for vTh yields vTh = 8 V. Now calculate RTh . To do this, we use the test source method. Replace the voltage source with a short circuit, the current source with an open circuit, and apply the test voltage vT , as shown in the circuit below: Write a KCL equation at the middle node: iT = ix + 3ix + vT /2 = 4ix + vT /2 Use Ohm’s law to determine ix as a function of vT : ix = vT /8 Substitute the second equation into the first equation: iT = 4(vT /8) + vT /2 = vT © 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. Problems 4–15 Thus, RTh = vT /iT = 1 Ω The Thévenin equivalent is an 8 V source in series with a 1 Ω resistor. AP 4.20 Begin by calculating the open circuit voltage, which is also vTh , using the node voltage method in the circuit below: The node voltage equations are v v − (vTh + 160i∆ ) + − 4 = 0, 60 20 vTh vTh vTh + 160i∆ − v + + = 0 40 80 20 The dependent source constraint equation is i∆ = vTh 40 Substitute the constraint equation into the node voltage equations and put the two equations in standard form: 1 1 5 v + + vTh − = 4 60 20 20 1 1 1 5 v − + vTh + + = 0 20 40 80 20 Solving, v = 172.5 V and vTh = 30 V. Now use the test source method to calculate the test current and thus RTh . Replace the current source with a short circuit and apply the test source to get the following circuit: Write a KCL equation at the rightmost node: iT = vT vT vT + 160i∆ + + 80 40 80 © 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. 4–16 CHAPTER 4. Techniques of Circuit Analysis The dependent source constraint equation is i∆ = vT 40 Substitute the constraint equation into the KCL equation and simplify the right-hand side: iT = vT 10 Therefore, RTh = vT = 10 Ω iT Thus, the Thévenin equivalent is a 30 V source in series with a 10 Ω resistor. AP 4.21 First find the Thévenin equivalent circuit. To find vTh , create an open circuit between nodes a and b and use the node voltage method with the circuit below: The node voltage equations are: vTh − (100 + vφ) vTh − v1 + = 0 4 4 v1 − 100 v1 − 20 v1 − vTh + + = 0 4 4 4 The dependent source constraint equation is vφ = v1 − 20 Place these three equations in standard form: 1 1 1 1 vTh + + v1 − + vφ − 4 4 4 4 1 1 1 1 vTh − + v1 + + + vφ (0) 4 4 4 4 vTh (0) + v1 (1) + vφ (−1) = 25 = 30 = 20 Solving, vTh = 120 V, v1 = 80 V, and vφ = 60 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. Problems 4–17 Now create a short circuit between nodes a and b and use the mesh current method with the circuit below: The mesh current equations are −100 + 4(i1 − i2) + vφ + 20 = 0 −vφ + 4i2 + 4(i2 − isc) + 4(i2 − i1) = 0 −20 − vφ + 4(isc − i2) = 0 The dependent source constraint equation is vφ = 4(i1 − isc) Place these four equations in standard form: 4i1 − 4i2 + 0isc + vφ = 80 −4i1 + 12i2 − 4isc − vφ = 0 0i1 − 4i2 + 4isc − vφ = 20 4i1 + 0i2 − 4isc − vφ = 0 Solving, i1 = 45 A, i2 = 30 A, isc = 40 A, and vφ = 20 V. Thus, RTh = vTh 120 = = 3Ω isc 40 [a] For maximum power transfer, R = RTh = 3 Ω [b] The Thévenin voltage, vTh = 120 V, splits equally between the Thévenin resistance and the load resistance, so 120 = 60 V 2 Therefore, vload = pmax = 2 vload 602 = = 1200 W Rload 3 © 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. 4–18 CHAPTER 4. Techniques of Circuit Analysis AP 4.22 Sustituting the value R = 3 Ω into the circuit and identifying three mesh currents we have the circuit below: The mesh current equations are: −100 + 4(i1 − i2 ) + vφ + 20 = 0 −vφ + 4i2 + 4(i2 − i3) + 4(i2 − i1) = 0 −20 − vφ + 4(i3 − i2) + 3i3 = 0 The dependent source constraint equation is vφ = 4(i1 − i3) Place these four equations in standard form: 4i1 − 4i2 + 0i3 + vφ = 80 −4i1 + 12i2 − 4i3 − vφ = 0 0i1 − 4i2 + 7i3 − vφ = 20 4i1 + 0i2 − 4i3 − vφ = 0 Solving, i1 = 30 A, i2 = 20 A, i3 = 20 A, and vφ = 40 V. [a] p100V = −(100)i1 = −(100)(30) = −3000 W. Thus, the 100 V source is delivering 3000 W. [b] pdepsource = −vφi2 = −(40)(20) = −800 W. Thus, the dependent source is delivering 800 W. [c] From Assessment Problem 4.21(b), the power delivered to the load resistor is 1200 W, so the load power is (1200/3800)100 = 31.58% of the combined power generated by the 100 V source and the dependent source. © 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. Problems 4–19 Problems P 4.1 [a] 11 branches, 8 branches with resistors, 2 branches with independent sources, 1 branch with a dependent source [b] The current is unknown in every branch except the one containing the 8 A current source, so the current is unknown in 10 branches. [c] 9 essential branches – R4 − R5 forms an essential branch as does R8 − 10 V. The remaining seven branches are essential branches that contain a single element. [d] The current is known only in the essential branch containing the current source, and is unknown in the remaining 8 essential branches [e] From the figure there are 6 nodes – three identified by rectangular boxes, two identified with single black dots, and one identified by a triangle. [f] There are 4 essential nodes, three identified with rectangular boxes and one identified with a triangle [g] A mesh is like a window pane, and as can be seen from the figure there are 6 window panes or meshes. P 4.2 [a] From Problem 4.1(d) there are 8 essential branches where the current is unknown, so we need 8 simultaneous equations to describe the circuit. [b] From Problem 4.1(f), there are 4 essential nodes, so we can apply KCL at (4 − 1) = 3 of these essential nodes. There would also be a dependent source constraint equation. [c] The remaining 4 equations needed to describe the circuit will be derived from KVL equations. © 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. 4–20 CHAPTER 4. Techniques of Circuit Analysis [d] We must avoid using the topmost mesh and the leftmost mesh. Each of these meshes contains a current source, and we have no way of determining the voltage drop across a current source. P 4.3 [a] As can be seen from the figure, the circuit has 2 separate parts. [b] There are 5 nodes – the four black dots and the node betweem the voltage source and the resistor R1. [c] There are 7 branches, each containing one of the seven circuit components. [d] When a conductor joins the lower nodes of the two separate parts, there is now only a single part in the circuit. There would now be 4 nodes, because the two lower nodes are now joined as a single node. The number of branches remains at 7, where each branch contains one of the seven individual circuit components. P 4.4 [a] There are six circuit components, five resistors and the current source. Since the current is known only in the current source, it is unknown in the five resistors. Therefore there are five unknown currents. [b] There are four essential nodes in this circuit, identified by the dark black dots in Fig. P4.4. At three of these nodes you can write KCL equations that will be independent of one another. A KCL equation at the fourth node would be dependent on the first three. Therefore there are three independent KCL equations. © 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. Problems 4–21 [c] Sum the currents at any three of the four essential nodes a, b, c, and d. Using nodes a, b, and c we get −ig + i1 + i2 = 0 −i1 + i4 + i3 = 0 i5 − i2 − i3 = 0 [d] There are three meshes in this circuit: one on the left with the components ig , R1 , and R4 ; one on the top right with components R1 , R2 , and R3 ; and one on the bottom right with components R3 , R4, and R5 . We cannot write a KVL equation for the left mesh because we don’t know the voltage drop across the current source. Therefore, we can write KVL equations for the two meshes on the right, giving a total of two independent KVL equations. [e] Sum the voltages around two independent closed paths, avoiding a path that contains the independent current source since the voltage across the current source is not known. Using the upper and lower meshes formed by the five resistors gives R1 i1 + R3i3 − R2i2 = 0 R3 i3 + R5i5 − R4i4 = 0 P 4.5 [a] At node a: − ig + i1 + i2 = 0 At node b: − i1 + i3 + i4 = 0 At node c: − i2 − i3 + i5 = 0 At node d: ig − i4 − i5 = 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. 4–22 CHAPTER 4. Techniques of Circuit Analysis [b] There are many possible solutions. For example, solve the equations at nodes a and d for ig : ig = i4 + i5 ig = i1 + i2 so i1 + i2 = i4 + i5 Solve this expression for i1 : i1 = i4 + i5 − i2 Substitute this expression for i1 into the equation for node b: −(i4 + i5 − i2) + i3 + i4 = 0 so − i2 − i3 + i5 = 0 The result above is the equation at node c. P 4.6 v1 − 144 v1 v1 − v2 + + =0 4 10 80 v2 − v1 v2 + =0 −3 + 80 5 Solving, v1 = 100 V; so 29v1 − v2 = 2880 so −v1 + 17v2 = 240 v2 = 20 V P 4.7 −2 + vo vo − 45 + =0 50 1+4 vo = 50 V p2A = −(50)(2) = −100 W (delivering) The 2 A source extracts −100 W from the circuit, because it delivers 100 W to the circuit. © 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. Problems P 4.8 −6 + 4–23 v1 v1 − v2 + =0 40 8 v2 − v1 v2 v2 + + +1 =0 8 80 120 Solving, v1 = 120 V; CHECK: (120)2 = 360 W 40 p40Ω = p8Ω = v2 = 96 V (120 − 96)2 = 72 W 8 p80Ω = (96)2 = 115.2 W 80 p120Ω = (96)2 = 76.8 W 120 p6A = −(6)(120) = −720 W p1A = (1)(96) = 96 W X pabs = 360 + 72 + 115.2 + 76.8 + 96 = 720 W X pdev = 720 W (CHECKS) P 4.9 Use the lower terminal of the 25 Ω resistor as the reference node. vo − 24 vo + + 0.04 = 0 20 + 80 25 Solving, vo = 4 V P 4.10 [a] From the solution to Problem 4.9 we know vo = 4 V, therefore p40mA = 0.04vo = 0.16 W .·. p40mA (developed) = −160 mW [b] The current into the negative terminal of the 24 V source is ig = 24 − 4 = 0.2 A 20 + 80 p24V = −24(0.2) = −4.8 W .·. p24V (developed) = 4800 mW © 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. 4–24 CHAPTER 4. Techniques of Circuit Analysis [c] p20Ω = (0.2)2 (20) = 800 mW p80Ω = (0.2)2 (80) = 3200 mW p25Ω = (4)2 /25 = 640 mW X X pdev = 4800 mW pdis = 160 + 800 + 3200 + 640 = 4800 mW v0 − 24 vo + + 0.04 = 0; vo = 4 V 20 + 80 25 [b] Let vx = voltage drop across 40 mA source P 4.11 [a] vx = vo − (50)(0.04) = 2 V p40mA = (2)(0.04) = 80 mW so p40mA (developed) = −80 mW [c] Let ig = be the current into the positive terminal of the 24 V source ig = (4 − 24)/100 = −0.2 A p24V = (−0.2)(24) = −4800 mW so p24V (developed) = 4800 mW [d] X pdis = (0.2)2 (20) + (0.2)2 (80) + (4)2 /25 + (0.04)2 (50) + 0.08 = 4800 mW [e] vo is independent of any finite resistance connected in series with the 40 mA current source P 4.12 [a] v1 − 125 v1 − v2 v1 − v3 + + 1 6 24 v2 − v1 v2 v2 − v3 + + 6 2 12 v3 + 125 v3 − v2 v3 − v1 + + 1 12 24 = 0 = 0 = 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. Problems 4–25 In standard form: 1 1 1 1 1 + + + v2 − + v3 − = 1 6 24 6 24 1 1 1 1 1 v1 − + v2 + + + v3 − = 6 6 2 12 12 1 1 1 1 1 v1 − + v2 − + v3 + + = 24 12 1 12 24 v1 Solving, v1 = 101.24 V; v2 = 10.66 V; 125 − v1 = 23.76 A 1 v2 i2 = = 5.33 A 2 v3 + 125 = 18.43 A i3 = 1 Thus, i1 = [b] P 4.13 [a] X 125 0 −125 v3 = −106.57 V v1 − v2 = 15.10 A 6 v2 − v3 i5 = = 9.77 A 12 v1 − v3 i6 = = 8.66 A 24 i4 = Pdev = 125i1 + 125i3 = 5273.09 W X Pdis = i21(1) + i22 (2) + i23 (1) + i24(6) + i25(12) + i26(24) = 5273.09 W v1 − 128 v1 v1 − v2 + + = 0 5 60 4 v2 − v1 v2 v2 − 320 + + = 0 4 80 10 In standard form, 1 1 1 1 v1 + + + v2 − = 5 60 4 4 1 1 1 1 v1 − + v2 + + = 4 4 80 10 Solving, v1 = 162 V; 128 5 320 10 v2 = 200 V ia = 128 − 162 = −6.8 A 5 ib = 162 = 2.7 A 60 ic = 162 − 200 = −9.5 A 4 © 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. 4–26 CHAPTER 4. Techniques of Circuit Analysis id = 200 = 2.5 A 80 ie = 200 − 320 = −12 A 10 [b] p128V = −(128)(−6.8) = 870.4 W (abs) p320V = (320)(−12) = −3840 W (dev) Therefore, the total power developed is 3840 W. P 4.14 v1 + 40 v1 v1 − v2 + + +5 =0 12 25 20 v2 − v1 v2 − v1 −5+ + −7.5 = 0 20 40 v3 v3 − v2 + + 7.5 = 0 40 40 Solving, v1 = −10 V; v2 = 132 V; v3 = −84 V; i40V = −10 + 40 = 2.5 A 12 p5A = 5(v1 − v2) = 5(−10 − 132) = −710 W (del) p7.5A = (−84 − 132)(7.5) = −1620 W (del) p40V = −(40)(2.5) = −100 W (del) p12Ω = (2.5)2 (12) = 75 W p25Ω = v12 102 = =4W 25 25 p20Ω = 1422 (v1 − v2)2 = = 1008.2 W 20 20 © 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. Problems p40Ω (lower) = (v3)2 842 = = 176.4 W 40 40 p40Ω (right) = (v2 − v3)2 2162 = = 1166.4 W 40 40 X pdiss = 75 + 4 + 1008.2 + 176.4 + 1166.4 = 2430 W X pdev = 710 + 1620 + 100 = 2430 W 4–27 (CHECKS) The total power dissipated in the circuit is 2430 W. P 4.15 [a] v1 v1 − 40 v1 − v2 + + =0 so 31v1 − 20v2 + 0v3 = 400 40 4 2 v2 − v1 v2 − v3 + − 28 = 0 so −2v1 + 3v2 − v3 = 112 2 4 v3 v3 − v2 + + 28 = 0 so 0v1 − v2 + 3v3 = −112 2 4 Solving, v1 = 60 V; v2 = 73 V; v3 = −13 V, [b] ig = 40 − 60 = −5 A 4 pg = (40)(−5) = −200 W Thus the 40 V source delivers 200 W of power. P 4.16 [a] vo − v1 vo − v2 vo − v3 vo − vn + + + ··· + =0 R R R R .·. nvo = v1 + v2 + v3 + · · · + vn . ·. v o = 1 1 Xn [v1 + v2 + v3 + · · · + vn ] = vk n n k=1 1 [b] vo = (100 + 80 − 60) = 40 V 3 © 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. 4–28 CHAPTER 4. Techniques of Circuit Analysis P 4.17 [a] −25 + v1 v1 v1 − v2 + + = 0 so 21v1 − 16v2 + 0i∆ = 4000 40 160 10 v2 − v1 v2 v2 − 84i∆ + + = 0 so − 16v1 + 44v2 − 1680i∆ = 0 10 20 8 v1 i∆ = so v1 + (0)v2 − 160i∆ = 0 160 Solving, v1 = 352 V; idepsource = v2 = 212 V; i∆ = 2.2 A; 212 − 84(2.2) = 3.4 A 8 p84i∆ = 84(2.2)(3.4) = 628.32 W(abs) p25A = −25(352) = −8800 W(del) .·. pdev = 8800 W [b] X pabs = . ·. P 4.18 −3 + (352)2 (352)2 (352 − 212)2 (212)2 + + + 40 160 10 20 +(3.4)2 (8) + 628.32 = 8800 W X pdev = X pabs = 8800 W vo vo + 5i∆ vo − 80 + + = 0; 200 10 20 i∆ = vo − 80 20 [a] Solving, vo = 50 V vo + 5i∆ [b] ids = 10 i∆ = (50 − 80)/20 = −1.5 A .·. ids = 4.25 A; 5i∆ = −7.5 V : pds = (−5i∆ )(ids) = 31.875 W [c] p3A = −3vo = −3(50) = −150 W (del) p80V = 80i∆ = 80(−1.5) = −120 W (del) X pdel = 150 + 120 = 270 W CHECK: p200Ω = 2500/200 = 12.5 W p20Ω = (80 − 50)2 /20 = 900/20 = 45 W p10Ω = (4.25)2 (10) = 180.625 W X pdiss = 31.875 + 180.625 + 12.5 + 45 = 270 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. Problems 4–29 P 4.19 vo − 160 vo vo − 150iσ + + = 0; 10 100 50 Solving, vo = 100 V; io = iσ = − vo 100 iσ = −1 A 100 − (150)(−1) =5A 50 p150iσ = 150iσ io = −750 W .·. The dependent voltage source delivers 750 W to the circuit. P 4.20 [a] io = v2 40 v1 v1 − v2 + =0 20 5 v2 − v1 v2 v2 − v3 + + 5 40 10 v3 − v2 v3 − 11.5io v3 − 96 + + =0 10 5 4 −5io + Solving, v1 = 156 V; [b] io = v2 = 120 V; so 10v1 − 13v2 + 0v3 = 0 so −8v1 + 13v2 − 4v3 = 0 so 0v1 − 63v2 + 220v3 = 9600 v3 = 78 V v2 120 = =3A 40 40 i3 = v3 − 11.5io 78 − 11.5(3) = = 8.7 A 5 5 © 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. 4–30 CHAPTER 4. Techniques of Circuit Analysis ig = 78 − 96 = −4.5 A 4 p5io = −5io v1 = −5(3)(156) = −2340 W(dev) p11.5io = 11.5io i3 = 11.5(3)(8.7) = 300.15 W(abs) p96V = 96(−4.5) = −432 W(dev) X pdev = 2340 + 432 = 2772 W CHECK X . ·. P 4.21 pdis = 1562 (156 − 120)2 1202 (120 − 78)2 + + + 20 5 40 50 +(8.7)2(5) + (4.5)2 (4) + 300.15 = 2772 W X pdev = X pdis = 2772 W v1 v1 − v2 v1 − 20 + + =0 30,000 5000 2000 v2 v2 − v1 v2 − 20 + + =0 1000 5000 5000 Solving, v1 = 15 V; Thus, io = so 22v1 − 6v2 = 300 so −v1 + 7v2 = 20 v2 = 5 V v1 − v2 = 2 mA 5000 P 4.22 [a] There is only one node voltage equation: © 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. Problems va + 30 va va − 80 + + + 0.01 = 0 5000 500 1000 Solving, va + 30 + 10va + 5va − 400 + 50 = 0 . ·. va = 20 V Calculate the currents: i1 = (−30 − 20)/5000 = −10 mA i2 = 20/500 = 40 mA i4 = 80/4000 = 20 mA i5 = (80 − 20)/1000 = 60 mA so 4–31 16va = 320 i3 + i4 + i5 − 10 mA = 0 so i3 = 0.01 − 0.02 − 0.06 = −0.07 = −70 mA [b] p30V = (30)(−0.01) = −0.3 W p10mA = (20 − 80)(0.01) = −0.6 W p80V = (80)(−0.07) = −5.6 W p5k = (−0.01)2 (5000) = 0.5 W p500Ω = (0.04)2 (500) = 0.8 W p1k = (80 − 20)2 /(1000) = 3.6 W p4k = (80)2 /(4000) = 1.6 W X P 4.23 [a] X pabs = 0.5 + 0.8 + 3.6 + 1.6 = 6.5 W pdel = 0.3 + 0.6 + 5.6 = 6.5 W (checks!) v2 − 230 v2 − v4 v2 − v3 + + =0 1 1 1 v3 − v2 v3 v3 − v5 + + =0 1 1 1 so 3v2 − 1v3 − 1v4 + 0v5 = 230 so −1v2 + 3v3 + 0v4 − 1v5 = 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. 4–32 CHAPTER 4. Techniques of Circuit Analysis v4 − v2 v4 − 230 v4 − v5 + + =0 1 6 2 v5 − v3 v5 v5 − v4 + + =0 1 6 2 Solving, v2 = 150 V; i2Ω = v3 = 80 V; so −12v2 + 0v3 + 20v4 − 6v5 = 460 so 0v2 − 12v3 − 6v4 + 20v5 = 0 v4 = 140 V; v5 = 90 V v4 − v5 140 − 90 = = 25 A 2 2 p2Ω = (25)2 (2) = 1250 W [b] i230V v1 − v2 v1 − v4 + 1 6 230 − 150 230 − 140 + = 80 + 15 = 95 A 1 6 = = p230V = (230)(95) = 21,850 W Check: X Pdis = (80)2 (1) + (70)2 (1) + (80)2 (1) + (15)2 (6) + (10)2 (1) +(10)2 (1) + (25)2 (2) + (15)2 (6) = 21,850 W P 4.24 © 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. Problems The two node voltage equations are: v1 − 50 v1 v1 − v2 + + = 0 80 50 40 v2 − v1 v2 v2 − 50 − 0.75 + + = 0 40 200 800 Place these equations in standard form: 1 1 1 1 v1 + + + v2 − = 80 50 40 40 1 1 1 1 + v2 + + = v1 − 40 40 200 800 4–33 50 80 0.75 + 50 800 Solving, v1 = 34 V; v2 = 53.2 V. Thus, vo = v2 − 50 = 53.2 − 50 = 3.2 V. POWER CHECK: ig = (50 − 34)/80 + (50 − 53.2)/800 = 196 m A p50V = −(50)(0.196) = −9.8 W p80Ω = (50 − 34)2 /80 = 3.2 W p800Ω = (50 − 53.2)2 /800 = 12.8 m W p40Ω = (53.2 − 34)2 /40 = 9.216 W p50Ω = 342 /50 = 23.12 W p200Ω = 53.22 /200 = 14.1512 W p0.75A = −(53.2)(0.75) = −39.9 W X pabs = 3.2 + .0128 + 9.216 + 23.12 + 14.1512 = 49.7 W = 9.8 + 39.9 = 49.7 X pdel = P 4.25 The two node voltage equations are: vb vb − vc 7+ + = 0 3 1 vc − vb vc − 4 −2vx + + = 0 1 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. 4–34 CHAPTER 4. Techniques of Circuit Analysis The constraint equation for the dependent source is: vx = vc − 4 Place these equations in standard form: 1 vb +1 + vc(−1) + vx (0) = 3 1 vb(−1) + vc 1 + + vx (−2) = 2 vb(0) + vc(1) + vx (−1) = −7 4 2 4 Solving, vc = 9 V, vx = 5 V, and vo = vb = 1.5 V P 4.26 [a] This circuit has a supernode includes the nodes v1, v2 and the 25 V source. The supernode equation is v2 v2 v1 + + =0 2+ 50 150 75 The supernode constraint equation is v1 − v2 = 25 Place these two equations in standard form: 1 1 1 v1 + v2 + = −2 50 150 75 v1(1) + v2(−1) = 25 Solving, v1 = −37.5 V and v2 = −62.5 V, so vo = v1 = −37.5 V. p2A = (2)vo = (2)(−37.5) = −75 W The 2 A source delivers 75 W. [b] © 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. Problems 4–35 This circuit now has only one non-reference essential node where the voltage is not known – note that it is not a supernode. The KCL equation at v1 is v1 v1 + 25 v1 + 25 + + =0 50 150 75 Solving, v1 = 37.5 V so v0 = −v1 = −37.5 V. −2 + p2A = (2)vo = (2)(−37.5) = −75 W The 2 A source delivers 75 W. [c] The choice of a reference node in part (b) resulted in one simple KCL equation, while the choice of a reference node in part (a) resulted in a supernode KCL equation and a second supernode constraint equation. Both methods give the same result but the choice of reference node in part (b) yielded fewer equations to solve, so is the preferred method. P 4.27 Place 5v∆ inside a supernode and use the lower node as a reference. Then v∆ − 15 v∆ v∆ − 5v∆ v∆ − 5v∆ + + + =0 10 2 20 40 12v∆ = 60; v∆ = 5 V vo = v∆ − 5v∆ = −4(5) = −20 V P 4.28 Node equations: v1 v1 − 20 v3 − v2 v3 + + + + 3.125v∆ = 0 20 2 4 80 v2 v2 − v3 v2 − 20 + + =0 40 4 1 Constraint equations: v∆ = 20 − v2 © 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. 4–36 CHAPTER 4. Techniques of Circuit Analysis v1 − 35iφ = v3 iφ = v2 /40 Solving, v1 = −20.25 V; v2 = 10 V; v3 = −29 V Let ig be the current delivered by the 20 V source, then ig = 20 − (20.25) 20 − 10 + = 30.125 A 2 1 pg (delivered) = 20(30.125) = 602.5 W P 4.29 For the given values of v3 and v4: v∆ = 120 − v3 = 120 − 108 = 12 V iφ = v4 − v3 81.6 − 108 = = −3.3 A 8 8 40 iφ = −44 V 3 v1 = v4 + 40 iφ = 81.6 − 44 = 37.6 V 3 Let ia be the current from right to left through the dependent voltage source: ia = v1 v1 − v2 + = 1.88 − 20.6 = −18.72 A 20 4 Let ib be the current supplied by the 120 V source: ib = 120 − 37.6 120 − 108 + = 20.6 + 6 = 26.6 A 4 2 Then p120V = −(120)(26.6) = −3192 W pCCVS = [(40/3)(−3.3)](−18.72) = −823.68 W pVCVS = (81.6)[1.75(12)] = 1713.6 W .·. X pdev = 3192 + 823.68 = 4015.68 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. Problems 4–37 The total power dissipated by the resistors is (37.6)2 (82.4)2 (12)2 (108)2 + + + 2 4 2 40 pR = = +(3.3)2 (8) + .·. X Thus, (81.6)2 = 2302.08 W 80 pdiss = 2302.08 + 1713.6 = 4015.68 W X pdev = X pdiss; Agree with analyst P 4.30 From Eq. 4.16, iB = vc /(1 + β)RE From Eq. 4.17, iB = (vb − Vo )/(1 + β)RE From Eq. 4.19, " # 1 VCC (1 + β)RE R2 + Vo R1 R2 iB = − Vo (1 + β)RE R1 R2 + (1 + β)RE (R1 + R2 ) = VCC R2 − Vo (R1 + R2 ) [VCC R2 /(R1 + R2 )] − Vo = R1 R2 + (1 + β)RE (R1 + R2) [R1R2 /(R1 + R2)] + (1 + β)RE P 4.31 [a] The three mesh current equations are: −125 + 1i1 + 6(i1 − i6) + 2(i1 − i3) = 0 24i6 + 12(i6 − i3) + 6(i6 − i1) = 0 −125 + 2(i3 − i1) + 12(i3 − i6) + 1i3 = 0 Place these equations in standard form: © 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. 4–38 CHAPTER 4. Techniques of Circuit Analysis i1 (1 + 6 + 2) + i3(−2) + i6 (−6) = 125 i1 (−6) + i3(−12) + i6(24 + 12 + 6) = 0 i1 (−2) + i3(2 + 12 + 1) + i6(−12) = 125 Solving, i1 = 23.76 A; i3 = 18.43 A; i6 = 8.66 A Now calculate the remaining branch currents: i2 = i1 − i3 = 5.33 A i4 = i1 − i6 = 15.10 A i5 = i3 − i6 = 9.77 A [b] psources = ptop + pbottom = −(125)(23.76) − (125)(18.43) = −2970 − 2304 = −5274 W Thus, the power developed in the circuit is 5274 W. Now calculate the power absorbed by the resistors: p1top = (23.76)2 (1) = 564.54 W p2 = (5.33)2 (2) = 56.82 W p1bot = (18.43)2 (1) = 339.66 W p6 = (15.10)2 (6) = 1368.06 W p12 = (9.77)2 (12) = 1145.43 W p24 = (8.66)2 (24) = 1799.89 W The power absorbed by the resistors is 564.54 + 56.82 + 339.66 + 1368.06 + 1145.43 + 1799.89 = 5274 W so the power balances. P 4.32 [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. Problems 4–39 The three mesh current equations are: −128 + 5ia + 60(ia − ic ) = 0 4ic + 80(ic − ie) + 60(ic − ia ) = 0 320 + 80(ie − ic ) + 10ie = 0 Place these equations in standard form: ia (5 + 60) + ic(−60) + ie(0) = 128 ia (−60) + ic(4 + 80 + 60) + ie(−80) = 0 ia (0) + ic(−80) + ie(80 + 10) = −320 Solving, ia = −6.8 A; ic = −9.5 A; ie = −12 A Now calculate the remaining branch currents: ib = ia − ic = 2.7 A id = ic − ie = 2.5 A [b] p128V = −(128)ia = −(128)(−6.8) = 870.4 W (abs) p320V = (320)ie = (320)(−12) = −3840 W (dev) Thus, the power developed in the circuit is 3840 W. Note that the resistors cannot develop power! P 4.33 [a] 60 = 15i1 − 10i2 −20 = −10i1 + 15i2 Solving, i1 = 5.6 A; ia = i1 = 5.6 A; i2 = 2.4 A ib = i1 − i2 = 3.2 A; ic = −i2 = −2.4 A [b] If the polarity of the 60 V source is reversed, we have −60 = 15i1 − 10i2 © 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. 4–40 CHAPTER 4. Techniques of Circuit Analysis −20 = −10i1 + 15i2 i1 = −8.8 A and i2 = −7.2 A ia = i1 = −8.8 A; ib = i1 − i2 = −1.6 A; ic = −i2 = 7.2 A P 4.34 [a] 230 − 115 = 7i1 − 1i2 − 2i3 0 = −1i1 + 10i2 − 3i3 115 − 460 = −2i1 − 3i2 + 10i3 Solving, i1 = 4.4 A; i2 = −10.6 A; i3 = −36.8 A p230 = −230i1 = −1012 W(del) p115 = 115(i1 − i3) = 4738 W(abs) p460 = 460i3 = −16,928 W(del) . ·. X pdev = 17,940 W [b] p6Ω = (10.6)2 (6) = 674.16 W p1Ω = (15)2 (1) = 225 W p3Ω = (26.2)2 (3) = 2059.32 W p2Ω = (41.2)2 (2) = 3394.88 W p4Ω = (4.4)2 (4) = 77.44 W p5Ω = (36.8)2 (5) = 6771.2 W . ·. X pabs = 4738 + 674.16 + 225 + 2059.32 + 3394.88 +77.44 + 6771.2 = 17,940 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. Problems 4–41 P 4.35 The three mesh current equations are: −20 + 2000(i1 − i2) + 30,000(i1 − i3) = 0 5000i2 + 5000(i2 − i3) + 2000(i2 − i1) = 0 1000i3 + 30,000(i3 − i1) + 5000(i3 − i2) = 0 Place these equations in standard form: i1(32,000) + i2(−2000) + i3 (−30,000) = 20 i1(−2000) + i2(12,000) + i3 (−5000) = 0 i1(−30,000) + i2 (−5000) + i3(36,000) = 0 Solving, i1 = 5.5 mA; i2 = 3 mA; Thus, io = i3 − i2 = 2 mA. i3 = 5 mA P 4.36 [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. 4–42 CHAPTER 4. Techniques of Circuit Analysis The four mesh current equations are: −230 + 1(i1 − i2) + 1(i1 − i3) + 1(i1 − i4) = 0 6i2 + 1(i2 − i3) + 1(i2 − i1) = 0 2i3 + 1(i3 − i4) + 1(i3 − i1) + 1(i3 − i2) = 0 6i4 + 1(i4 − i1) + 1(i4 − i3) = 0 Place these equations in standard form: i1 (3) + i2(−1) + i3 (−1) + i4(−1) = 230 i1 (−1) + i2(8) + i3 (−1) + i4(0) = 0 i1 (−1) + i2(−1) + i3 (5) + i4(−1) = 0 i1 (−1) + i2(0) + i3 (−1) + i4(8) = 0 Solving, i1 = 95 A; i2 = 15 A; i3 = 25 A; The power absorbed by the 5 Ω resistor is i4 = 15 A p5 = i23(2) = (25)2 (2) = 1250 W [b] p230 = −(230)i1 = −(230)(95) = −21,850 W P 4.37 [a] 25 = 30i1 − 20i2 + 0i∆ 0 = −20i1 + 44i2 + 6i∆ 21 = 0i1 + 0i2 + 14i∆ Solving, i1 = 1 A; i2 = 0.25 A; i∆ = 1.5 A vo = 20(i1 − i2) = 20(0.75) = 15 V [b] p6i∆ = 6i∆ i2 = (6)(1.5)(0.25) = 2.25 W (abs) .·. p6i∆ (deliver) = −2.25 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. Problems 4–43 P 4.38 −135 + 25i1 − 3i2 − 20i3 + 0iσ = 0 −3i1 + 12i2 − 4i3 + 0iσ = 0 −20i1 − 4i2 + 25i3 + 10iσ = 0 1i1 − 1i2 + 0i3 + 1iσ = 0 Solving, i1 = 64.8 A i2 = 39 A i3 = 68.4 A iσ = −25.8 A p20Ω = (68.4 − 64.8)2 (20) = 259.2 W P 4.39 660 = 30i1 − 10i2 − 15i3 20iφ = −10i1 + 60i2 − 50i3 0 = −15i1 − 50i2 + 90i3 iφ = i2 − i3 Solving, i1 = 42 A; i2 = 27 A; i3 = 22 A; iφ = 5 A 20iφ = 100 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. 4–44 CHAPTER 4. Techniques of Circuit Analysis p20iφ = −100i2 = −100(27) = −2700 W .·. p20iφ (developed) = 2700 W CHECK: p660V = −660(42) = −27,720 W (dev) .·. X Pdev X Pdis = 27,720 + 2700 = 30,420 W = (42)2 (5) + (22)2 (25) + (20)2 (15) + (5)2 (50)+ (15)2 (10) = 30,420 W P 4.40 Mesh equations: 53i∆ + 8i1 − 3i2 − 5i3 = 0 0i∆ − 3i1 + 30i2 − 20i3 = 30 0i∆ − 5i1 − 20i2 + 27i3 = 30 Constraint equations: i∆ = i2 − i3 Solving, i1 = 110 A; i2 = 52 A; i3 = 60 A; i∆ = −8 A pdepsource = 53i∆ i1 = (53)(−8)(110) = −46,640 W Therefore, the dependent source is developing 46,640 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. Problems 4–45 CHECK: p30V = −30i2 = −1560 W (left source) p30V = −30i3 = −1800 W (right source) X pdev = 46,640 + 1560 + 1800 = 50 k W p3Ω = (110 − 52)2 (3) = 10,092 W p5Ω = (110 − 60)2 (5) = 12,500 W p20Ω = (−8)2 (20) = 1280 W p7Ω = (52)2 (7) = 18,928 W p2Ω = (60)2 (2) = 7200 W X pdiss = 10,092 + 12,500 + 1280 + 18,928 + 7200 = 50 kW P 4.41 Mesh equations: 128i1 − 80i2 = 240 −80i1 + 200i2 = 120 Solving, i1 = 3 A; i2 = 1.8 A Therefore, v1 = 40(6 − 3) = 120 V; v2 = 120(1.8 − 1) = 96 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. 4–46 CHAPTER 4. Techniques of Circuit Analysis P 4.42 120 = 32i1 − 20i2 − 7i3 −80 = −20i1 + 25i2 − 1i3 −4 = 0i1 + 0i2 + 1i3 Solving, i1 = 1.55 A; i2 = −2.12 A; i3 = −4 A [a] v4A = 7(−4 − 1.55) + 1(−4 + 2.12) = −40.73 V p4A = 4v4A = 4(−40.73) = −162.92 W Therefore, the 4 A source delivers 162.92 W. [b] p120V = −120(1.55) = −186 W p80V = −80(−2.12) = 169.6 W Therefore, the total power delivered is 162.92 + 186 + 169.6 = 518.52 W [c] P 4.43 [a] X presistors = (1.55)2 (5) + (2.12)2 (4) + (3.67)2 (20) + (5.55)2 (7) + (1.88)2 (1) X = 518.52 W pabs = 518.52 W = X pdel (CHECKS) © 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. Problems 4–47 Mesh equations: 65i1 − 40i2 + 0i3 − 100io = 0 −40i1 + 55i2 − 5i3 + 11.5io = 0 0i1 − 5i2 + 9i3 − 11.5io = 0 −1i1 + 1i2 + 0i3 + 1io = 0 Solving, i1 = 7.2 A; i2 = 4.2 A; i3 = −4.5 A; io = 3 A Therefore, v1 = 20[5(3) − 7.2] = 156 V; v2 = 40(7.2 − 4.2) = 120 V v3 = 5(4.2 + 4.5) + 11.5(3) = 78 V [b] p5io = −5io v1 = −5(3)(156) = −2340 W p11.5io = 11.5io (i2 − i3 ) = 11.5(3)(4.2 + 4.5) = 300.15 W p96V = 96i3 = 96(−4.5) = −432 W Thus, the total power dissipated in the circuit, which equals the total power developed in the circuit is 2340 + 432 = 2772 W. P 4.44 [a] The mesh current equation for the right mesh is: 5400(i1 − 0.005) + 3700i1 − 150(0.005 − i1) = 0 Solving, 9250i1 = 27.75 .·. i1 = 3 mA Then, i∆ = 5 − 3 = 2 mA [b] vo = (0.005)(10,000) + (5400)(0.002) = 60.8 V p5mA = −(60.8)(0.005) = −304 mW Thus, the 5 mA source delivers 304 mW [c] pdep source = −150i∆ i1 = (−150)(0.002)(0.003) = −0.9 mW The dependent source delivers 0.9 mW. © 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. 4–48 CHAPTER 4. Techniques of Circuit Analysis P 4.45 Mesh equations: 10i∆ − 4i1 = 0 −4i∆ + 24i1 + 6.5i∆ = 400 Solving, i1 = 15 A; i∆ = 16 A v20A = 1i∆ + 6.5i∆ = 7.5(16) = 120 V p20A = −20v20A = −(20)(120) = −2400 W (del) p6.5i∆ = 6.5i∆ i1 = (6.5)(16)(15) = 1560 W (abs) Therefore, the independent source is developing 2400 W, all other elements are absorbing power, and the total power developed is thus 2400 W. CHECK: p1Ω = (16)2 (1) = 256 W p5Ω = (20 − 16)2 (5) = 80 W p4Ω = (1)2 (4) = 4 W p20Ω = (20 − 15)2 (20) = 500 W X pabs = 1560 + 256 + 80 + 4 + 500 = 2400 W (CHECKS) © 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. Problems 4–49 P 4.46 [a] Mesh equations: −50 + 6i1 − 4i2 + 9i∆ = 0 −9i∆ − 4i1 + 29i2 − 20i3 = 0 Constraint equations: i∆ = i2 ; i3 = −1.7v∆ ; Solving, i1 = −5 A; v∆ = 2i1 i2 = 16 A; i3 = 17 A; v∆ = −10 V 9i∆ = 9(16) = 144 V ia = i2 − i1 = 21 A ib = i2 − i3 = −1 A vb = 20ib = −20 V p50V = −50i1 = 250 W (absorbing) p9i∆ = −ia(9i∆ ) = −(21)(144) = −3024 W (delivering) p1.7V = −1.7v∆ vb = i3 vb = (17)(−20) = −340 W (delivering) [b] X Pdev = 3024 + 340 = 3364 W X Pdis = 250 + (−5)2 (2) + (21)2 (4) + (16)2 (5) + (−1)2 (20) = 3364 W P 4.47 [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. 4–50 CHAPTER 4. Techniques of Circuit Analysis Supermesh equations: 1000ib + 4000(ic − id ) + 500(ic − ia) = 0 ic − ib = 0.01 Two remaining mesh equations: 5500ia − 500ic = −30 4000id − 4000ic = −80 In standard form, −500ia + 1000ib + 4500ic − 4000id = 0 0ia − 1ib + 1ic + 0id = 0.01 5500ia + 0ib − 500ic + 0id = −30 0ia + 0ib − 4000ic + 4000id = −80 Solving: ia = −10 mA; ib = −60 mA; ic = −50 mA; id = −70 mA Then, i1 = ia = −10 mA; i2 = ia − ic = 40 mA; i3 = id = −70 mA [b] psources = 30(−0.01) + [1000(−0.06)](0.01) + 80(−0.07) = −6.5 W presistors = 1000(0.06)2 + 5000(0.01)2 + 500(0.04)2 +4000(−0.05 + 0.07)2 = 6.5 W P 4.48 −20 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0; Solving, i1 = 10 A; i1 − i2 = 6 i2 = 4 A p20V = −20i1 = −200 W (diss) p4 Ω = (10)2 (4) = 400 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. Problems 4–51 p1 Ω = (10)2 (1) = 100 W p9 Ω = (4)2 (9) = 144 W p6 Ω = (4)2 (6) = 96 W vo = 9(4) − 90 + 6(4) = −30 V p6A = 6vo = −180 W p90V = −90i2 = −360 W X pdev = 200 + 180 + 360 = 740 W X pdiss = 400 + 100 + 144 + 96 = 740 W Thus the total power dissipated is 740 W. P 4.49 [a] Summing around the supermesh used in the solution to Problem 4.48 gives −60 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0; .·. i1 = 12 A; i1 − i2 = 6 i2 = 6 A p60V = −60(12) = −720 W (del) vo = 9(6) − 90 + 6(6) = 0 V p6A = 6vo = 0 W p90V = −90i2 = −540 W (del) X X pdiss = (12)2 (4 + 1) + (6)2 (9 + 6) = 1260 W pdev = 720 + 0 + 540 = 1260 W = X pdiss [b] With 6 A current source replaced with a short circuit 5i1 = 60; 15i2 = 90 Solving, i1 = 12 A, . ·. X i2 = 6 A Psources = −(60)(12) − (90)(6) = −1260 W [c] A 6 A source with zero terminal voltage is equivalent to a short circuit carrying 6 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. 4–52 CHAPTER 4. Techniques of Circuit Analysis P 4.50 [a] −4id + 10(ie − id ) + 5(ie − ix ) = 0 5(ix − ie ) + 10(id − ie ) − 240 + 40(ix − 19) = 0 id − ix = 2ib = 2(ie − ix ) Solving, id = 10 A; ie = 18 A; ia = 19 − ix = −7 A; ix = 26 A ib = ie − ix = −8 A; [b] va = 40ia = −280 V; ic = ie − id = 8 A; vb = 5ib + 40ia = −320 V p19A = −19va = 5320 W p4id = −4id ie = −720 W p2ia = −2ib vb = −5120 W p240V = −240id = −2400 W p40Ω = (7)2 (40) = 1960 W = p5Ω = (8)2 (5) = 320 W p10Ω = (8)2 (10) = 640 W X X Pgen = 720 + 5120 + 2400 = 8240 W Pdiss = 5320 + 1960 + 320 + 640 = 8240 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. Problems 4–53 P 4.51 [a] 200 = 85i1 − 25i2 − 50i3 0 = −75i1 + 35i2 + 150i3 (supermesh) i3 − i2 = 4.3(i1 − i2) Solving, i1 = 4.6 A; ia = i2 = 5.7 A; i2 = 5.7 A; i3 = 0.97 A ib = i1 = 4.6 A ic = i3 = 0.97 A; id = i1 − i2 = −1.1 A ie = i1 − i3 = 3.63 A [b] 10i2 + vo + 25(i2 − i1) = 0 .·. vo = −57 − 27.5 = −84.5 V p4.3id = −vo (4.3id) = −(−84.5)(4.3)(−1.1) = −399.685 W(dev) p200V = −200(4.6) = −920 W(dev) X Pdev X Pdis = 1319.685 W = (5.7)2 10 + (1.1)2 (25) + (0.97)2 100 + (4.6)2 (10)+ (3.63)2 (50) = . ·. X Pdev = 1319.685 W X Pdis = 1319.685 W P 4.52 [a] The node voltage method requires summing the currents at two supernodes in terms of four node voltages and using two constraint equations to reduce the system of equations to two unknowns. If the connection at the bottom of the circuit is used as the reference node, then the voltages controlling the dependent sources are node voltages. This makes it easy to formulate the constraint equations. The current in the 20 V source is obtained by summing the currents at either terminal of the source. © 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. 4–54 CHAPTER 4. Techniques of Circuit Analysis The mesh current method requires summing the voltages around the two meshes not containing current sources in terms of four mesh currents. In addition the voltages controlling the dependent sources must be expressed in terms of the mesh currents. Thus the constraint equations are more complicated, and the reduction to two equations and two unknowns involves more algebraic manipulation. The current in the 20 V source is found by subtracting two mesh currents. Because the constraint equations are easier to formulate in the node voltage method, it is the preferred approach. [b] Node voltage equations: v1 v2 + 0.003v∆ + − 0.2 = 0 100 250 v3 v4 0.2 + + − 0.003v∆ = 0 100 200 Constraints: v2 = va ; v3 = v∆ ; v4 − v3 = 0.4va ; Solving, v1 = 24 V; v2 = 44 V; v2 io = 0.2 − = 24 m A 250 v2 − v1 = 20 v3 = −72 V; v4 = −54 V. p20V = 20(0.024) = 480 m W Thus, the 20 V source absorbs 480 mW. P 4.53 [a] There are three unknown node voltages and only two unknown mesh currents. Use the mesh current method to minimize the number of simultaneous equations. © 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. Problems 4–55 [b] The mesh current equations: 2500(i1 − 0.01) + 2000i1 + 1000(i1 − i2 ) = 0 5000(i2 − 0.01) + 1000(i2 − i1 ) + 1000i2 = 0 Place the equations in standard form: i1 (2500 + 2000 + 1000) + i2 (−1000) = 25 i1 (−1000) + i2(5000 + 1000 + 1000) = 50 Solving, i1 = 6 mA; i2 = 8 mA Find the power in the 1 kΩ resistor: i1k = i1 − i2 = −2 m A p1k = (−0.002)2 (1000) = 4 mW [c] No, the voltage across the 10 A current source is readily available from the mesh currents, and solving two simultaneous mesh-current equations is less work than solving three node voltage equations. [d] vg = 2000i1 + 1000i2 = 12 + 8 = 20 V p10mA = −(20)(0.01) = −200 m W Thus the 10 mA source develops 200 mW. P 4.54 [a] There are three unknown node voltages and three unknown mesh currents, so the number of simultaneous equations required is the same for both methods. The node voltage method has the advantage of having to solve the three simultaneous equations for one unknown voltage provided the connection at either the top or bottom of the circuit is used as the reference node. Therefore recommend the node voltage method. [b] © 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. 4–56 CHAPTER 4. Techniques of Circuit Analysis The node voltage equations are: v1 v1 − v2 v1 − v3 + + = 0 5000 2500 1000 v2 − v1 v2 − v3 v2 −0.01 + + + = 0 4000 2500 2000 v3 − v1 v3 − v2 v3 + + = 0 1000 2000 1000 Putthe equations in standard form: 1 1 1 1 1 v1 + + + v2 − + v3 − = 5000 2500 1000 2500 1000 1 1 1 1 1 v1 − + v2 + + + v3 − = 2500 4000 2500 2000 2000 1 1 1 1 1 v1 − + v2 − + v3 + + = 1000 2000 2000 1000 1000 Solving, v1 = 6.67 V; v2 = 13.33 V; v3 = 5.33 V p10m = −(13.33)(0.01) = −133.33 m W Therefore, the 10 mA source is developing 133.33 mW 0 0.01 0 P 4.55 [a] Both the mesh-current method and the node-voltage method require three equations. The mesh-current method is a bit more intuitive due to the presence of the voltage sources. We choose the mesh-current method, although technically it is a toss-up. [b] 125 = 10i1 − 0.5i2 − 9.2i3 125 = −0.5i1 + 20i2 − 19.2i3 0 = −9.2i1 − 19.2i2 + 40i3 Solving, i1 = 32.25 A; i2 = 26.29 A; v1 = 9.2(i1 − i3) = 112.35 V v2 = 19.2(i2 − i3) = 120.09 V v3 = 11.6i3 = 232.44 V i3 = 20.04 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. Problems [c] pR1 = (i1 − i3)2 (9.2) = 1371.93 W pR2 = (i2 − i3)2 (19.2) = 751.13 W pR3 = i23(11.6) = 4657.52 W [d] X 4–57 pdev = 125(i1 + i2) = 7317.72 W X pload = 6780.58 W % delivered = 6780.58 × 100 = 92.66% 7317.72 [e] 250 = 29i1 − 28.4i2 0 = −28.4i1 + 40i2 Solving, i1 = 28.29 A; i2 = 20.09 A i1 − i2 = 8.2 A v1 = (8.2)(9.2) = 75.44 V v2 = (8.2)(19.2) = 157.44 V Note v1 is low and v2 is high. Therefore, loads designed for 125 V would not function properly, and could be damaged. P 4.56 © 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. 4–58 CHAPTER 4. Techniques of Circuit Analysis The mesh current equations: 125 = (R + 0.8)ia − 0.5ib − Ric 125 = −0.5ia + (R + 0.8)ib − Ric .·. (R + 0.8)ia − 0.5ib − Ric = −0.5ia + (R + 0.8)ib − Ric .·. (R + 0.8)ia − 0.5ib = −0.5ia + (R + 0.8)ib .·. (R + 1.3)ia = (R + 1.3)ib Thus ia = ib so io = ib − ia = 0 P 4.57 [a] Write the mesh current equations. Note that if io = 0, then i1 = 0: −23 + 5(−i2) + 10(−i3 ) + 46 = 0 30i2 + 15(i2 − i3 ) + 5i2 = 0 Vdc + 25i3 − 46 + 10i3 + 15(i3 − i2) = 0 i2 (−5) + i3(−10) + Vdc(0) = −23 i2 (30 + 15 + 5) + i3(−15) + Vdc (0) = 0 i2 (−15) + i3(25 + 10 + 15) + Vdc (1) = 46 Place the equations in standard form: Solving, i2 = 0.6 A; i3 = 2 A; Vdc = −45 V Thus, the value of Vdc required to make io = 0 is −45 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. Problems 4–59 [b] Calculate the power: p23V = −(23)(0) = 0 W p46V = −(46)(2) = −92 W pVdc = (−45)(2) = −90 W p30Ω = (30)(0.6)2 = 10.8 W p5Ω = (5)(0.6)2 = 1.8 W p15Ω = (15)(2 − 0.6)2 = 29.4 W p10Ω = (10)(2)2 = 40 W p20Ω = (20)(0)2 = 0 W p25Ω = (25)(2)2 = 100 W X X pdev = 92 + 90 = 182 W pdis = 10.8 + 1.8 + 29.4 + 40 + 0 + 100 = 182 W(checks) P 4.58 Choose the reference node so that a node voltage is identical to the voltage across the 4 A source; thus: Since the 4 A source is developing 0 W, v1 must be 0 V. Since v1 is known, we can sum the currents away from node 1 to find v2; thus: 0 − (240 + v2 ) 0 − v2 0 + + −4=0 12 20 15 .·. v2 = −180 V Now that we know v2 we sum the currents away from node 2 to find v3; thus: v2 + 240 − 0 v2 − 0 v2 − v3 + + =0 12 20 40 .·. v3 = −340 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. 4–60 CHAPTER 4. Techniques of Circuit Analysis Now that we know v3 we sum the currents away from node 3 to find idc; thus: v3 v3 − v2 + = idc 50 40 .·. idc = −10.8 A P 4.59 [a] Apply source transformations to both current sources to get io = −(5.4 + 0.6) = −1 mA 2700 + 2300 + 1000 [b] The node voltage equations: v1 − v2 v1 + = 0 −2 × 10−3 + 2700 2300 v2 v2 − v1 + + 0.6 × 10−3 = 0 1000 2300 Place standard form: these equations in 1 1 1 v1 + + v2 − = 2 × 10−3 2700 2300 2300 1 1 1 v1 − + v2 + = −0.6 × 10−3 2300 1000 2300 Solving, v1 = 2.7 V; v2 = 0.4 V v2 − v1 . ·. i o = = −1 mA 2300 P 4.60 [a] Applying a source transformation to each current source yields © 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. Problems 4–61 Now combine the 12 V and 5 V sources into a single voltage source and the 6 Ω, 6 Ω and 5 Ω resistors into a single resistor to get Now use a source transformation on each voltage source, thus which can be reduced to . ·. i o = − 8.5 (1) = −0.85 A 10 [b] 34ia − 17ib = 12 + 5 + 34 = 51 −17ia + 18.5ib = −34 Solving, ib = −0.85 A = io P 4.61 [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. 4–62 CHAPTER 4. Techniques of Circuit Analysis io = 120 = 4 mA 30,000 [b] va = ia = ib = vb = ig = p100V = (15,000)(0.004) = 60 V va = 1 mA 60,000 12 − 1 − 4 = 7 mA 60 − (0.007)(4000) = 32 V 32 0.007 − = 6.6 mA 80,000 −(100)(6.6 × 10−3 ) = −660 mW © 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. Problems Check: p12mA X X 4–63 = −(60)(12 × 10−3 ) = −720 mW Pdev = 660 + 720 = 1380 mW Pdis = (20,000)(6.6 × 10−3 )2 + (80,000)(0.4 × 10−3 )2 + (4000)(7 × 10−3 )2 +(60,000)(1 × 10−3 )2 + (15,000)(4 × 10−3 )2 = 1380 mW P 4.62 [a] First remove the 16 Ω and 260 Ω resistors: Next use a source transformation to convert the 1 A current source and 40 Ω resistor: which simplifies to 250 (480) = 400 V 300 [b] Return to the original circuit with vo = 400 V: . ·. v o = ig = 520 + 1.6 = 3.6 A 260 © 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. 4–64 CHAPTER 4. Techniques of Circuit Analysis p520V = −(520)(3.6) = −1872 W Therefore, the 520 V source is developing 1872 W. [c] v1 = −520 + 1.6(4 + 250 + 6) = −104 V vg = v1 − 1(16) = −104 − 16 = −120 V p1A = (1)(−120) = −120 W Therefore the 1 A source is developing 120 W. [d] X pdev = 1872 + 120 = 1992 W X . ·. P 4.63 vTh = pdiss = (1)2 (16) + X pdiss = X (104)2 (520)2 + + (1.6)2 (260) = 1992 W 40 260 pdev 30 (80) = 60 V 40 RTh = 2.5 + (30)(10) = 10 Ω 40 P 4.64 First we make the observation that the 10 mA current source and the 10 kΩ resistor will have no influence on the behavior of the circuit with respect to the terminals a,b. This follows because they are in parallel with an ideal voltage source. Hence our circuit can be simplified to or © 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. Problems 4–65 Therefore the Norton equivalent is P 4.65 [a] Open circuit: v2 − 9 v2 + − 1.8 = 0 20 70 v2 = 35 V vTh = 60 v2 = 30 V 70 Short circuit: v2 − 9 v2 + − 1.8 = 0 20 10 .·. v2 = 15 V ia = 9 − 15 = −0.3 A 20 isc = 1.8 − 0.3 = 1.5 A RTh = 30 = 20 Ω 1.5 © 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. 4–66 CHAPTER 4. Techniques of Circuit Analysis [b] RTh = (20 + 10k60 = 20 Ω (CHECKS) P 4.66 After making a source transformation the circuit becomes 500 = 20i1 − 8i2 300 = −8i1 + 43.2i2 .·. i1 = 30 A and i2 = 12.5 A vTh = 12i1 + 5.2i2 = 425 V RTh = (8k12 + 5.2)k30 = 7.5 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. Problems 4–67 P 4.67 50i1 − 40isc = 60 + 40 −40i1 + 48iscs = 32 Solving, RTh = 8 + isc = 7 A (10)(40) = 16 Ω 50 P 4.68 First, find the Thévenin equivalent with respect to Ro . © 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. 4–68 CHAPTER 4. Techniques of Circuit Analysis Ro (Ω) io (A) vo (V) 10 1.2 12 15 1.067 16 22 0.923 20.31 33 0.762 25.14 47 0.623 29.30 68 0.490 33.31 P 4.69 12.5 = vTh − 2RTh 11.7 = vTh − 18RTh Solving the above equations for VTh and RTh yields vTh = 12.6 V, .·. IN = 252 A, RTh = 50 mΩ RN = 50 mΩ P 4.70 i1 = 100/20 = 5 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. Problems 100 = vTh − 5RTh , 4–69 vTh = 100 + 5RTh i2 = 200/50 = 4 A 200 = vTh − 4RTh , vTh = 200 + 4RTh .·. 100 + 5RTh = 200 + 4RTh so RTh = 100Ω vTh = 100 + 500 = 600 V P 4.71 [a] First, find the Thévenin equivalent with respect to a,b using a succession of source transformations. .·. vTh = 54 V RTh = 4.5 kΩ © 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. 4–70 CHAPTER 4. Techniques of Circuit Analysis 54 (85.5) = 51.3 V 90 51.3 − 54 [b] %error = × 100 = −5% 54 vmeas = P 4.72 v1 = 200 (18) = 5.143 V 700 v2 = 1200 (18) = 5.139 V 4203 vTh = v1 − v2 = 5.143 − 5.139 = 3.67 mV RTh = igal = (500)(200) (3003)(1200) + = 1000.24 Ω 700 4203 3.67 × 10−3 = 3.5 µA 1050.24 © 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. 4–71 Problems P 4.73 The node voltage equations are: v1 − 40 v1 v1 − v2 + + 2000 20,000 5000 v2 v2 − v3 v1 v2 − v1 + + + 30 5000 50,000 10,000 20,000 v3 − v2 v3 v1 + − 30 10,000 40,000 20,000 = 0 = 0 = 0 In standard form: ! 1 1 1 1 40 v1 + + + v2 − + v3(0) = 2000 20,000 5000 5000 2000 ! ! v1 1 30 1 1 1 1 + + v2 + + + v3 − − 5000 20,000 5000 50,000 10,000 10,000 v1 30 − 20,000 ! + v2 ! 1 1 1 − + v3 + 10,000 10,000 40,000 Solving, v1 = 24 V; VTh = v3 = 280 V v2 = −10 V; ! ! =0 =0 v3 = 280 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. 4–72 CHAPTER 4. Techniques of Circuit Analysis The mesh current equations are: −40 + 2000i1 + 20,000(i1 − i2) = 0 5000i2 + 50,000(i2 − isc) + 20,000(i2 − i1) = 0 50,000(isc − i2 ) + 10,000(isc − 30i∆ ) = 0 The constraint equation is: i∆ = i1 − i2 Put these equations in standard form: i1(22,000) + i2(−20,000) + isc (0) + i∆ (0) = 40 i1(−20,000) + i2 (75,000) + isc (−50,000) + i∆ (0) = 0 i1(0) + i2(−50,000) + isc (60,000) + i∆ (−300,000) = 0 i1(−1) + i2 (1) + isc(0) + i∆ (1) = 0 Solving, RTh i1 = 13.6 mA; 280 = = 20 kΩ 0.014 i2 = 12.96 mA; isc = 14 mA; i∆ = 640 µA P 4.74 OPEN CIRCUIT v2 = −80ib (50 × 103 ) = −40 × 105 ib 4 × 10−5 v2 = −160ib 1310ib + 4 × 10−5 v2 = 1310ib − 160ib = 1150ib So 1150ib is the voltage across the 100 Ω resistor. From KCL at the top left node, 500 µA = 1150ib + ib = 12.5ib 100 © 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. Problems 4–73 500 × 10−6 .·. ib = = 40 µA 12.5 vTh = −40 × 105 (40 × 10−6 ) = −160 V SHORT CIRCUIT v2 = 0; ib = isc = −80ib 100 (500 × 10−6 ) = 35.46 µA 100 + 1310 isc = −80(35.46) = −2837 µA RTh = −160 = 56.4 kΩ −2837 × 10−6 P 4.75 [a] Use source transformations to simplify the left side of the circuit. ib = Let . ·. 7.7 − 5.5 = 0.1 mA 22,000 Ro = Rmeter k1.3 kΩ = 5.5/4.4 = 1.25 kΩ (Rmeter)(1.3) = 1.25; Rmetere + 1.3 Rmeter = (1.25)(1.3) = 32.5 kΩ 0.05 [b] Actual value of ve : ib = 7.7 = 0.0972 mA 22 + (44)(1.3) ve = 44ib (1.3) = 5.56 V % error = 5.5 − 5.56 × 100 = −1.1% 5.56 © 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. 4–74 CHAPTER 4. Techniques of Circuit Analysis P 4.76 [a] Find the Thévenin equivalent with respect to the terminals of the ammeter. This is most easily done by first finding the Thévenin with respect to the terminals of the 4.8 Ω resistor. Thévenin voltage: note iφ is zero. vTh vTh vTh vTh − 16 + + + =0 100 25 20 2 Solving, vTh = 20 V. Short-circuit current: isc = 12 + 2isc , RTh = .·. isc = −12 A 20 = −(5/3) Ω −12 Rtotal = 20 = 3.33 Ω 6 Rmeter = 3.33 − 3.13 = 0.2 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. Problems 4–75 [b] Actual current: iactual = 20 = 6.38 A 3.13 % error = 6 − 6.38 × 100 = −6% 6.38 P 4.77 VTh = 0, since circuit contains no independent sources. v1 − 10i∆ v1 v1 − vT + + =0 10 2.5 12 vT − v1 vT − 10i∆ + −1=0 12 6 i∆ = vT − v1 12 In standard form: v1 1 1 1 1 + + + vT − + i∆ (−1) = 0 10 2.5 12 12 v1 − 1 1 1 10 + vT + + i∆ − =1 12 12 6 6 v1(1) + vT (−1) + i∆ (12) = 0 Solving, v1 = 2 V; vT = 8 V; i∆ = 0.5 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. 4–76 CHAPTER 4. Techniques of Circuit Analysis vT .·. RTh = = 8Ω 1 A P 4.78 VTh = 0 since there are no independent sources in the circuit. Thus we need only find RTh . iT = vT + ia 10 ia = i∆ − 21i∆ = −20i∆ i∆ = vT − 300i∆ , 700 1000i∆ = vT vT vT .·. iT = − 20 = 0.08vT 10 1000 vT = 1/0.08 = 12.5 Ω iT .·. RTh = 12.5 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. Problems 4–77 P 4.79 [a] v − 12 v − 10 v + + =0 12,000 20,000 12,500 Solving, v = 7.03125 V 10,000 v10k = (7.03125) = 5.625 V 12,500 .·. VTh = v − 10 = −4.375 V RTh = [(12,000k20,000) + 2500] = 5 kΩ Ro = RTh = 5 kΩ [b] pmax = (−437.5 × 10−6 )2(5000) = 957 µW [c] The resistor closest to 5 kΩ from Appendix H has a value of 4.7 kΩ. Use voltage division to find the voltage drop across this load resistor, and use the voltage to find the power delivered to it: 4700 v4.7k = (−4.375) = −2.12 V 4700 + 5000 (−2.12)2 p4.7k = = 956.12 µW 4700 The percent error between the maximum power and the power delivered to the best resistor from Appendix H is 956 % error = − 1 (100) = −0.1% 957 © 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. 4–78 CHAPTER 4. Techniques of Circuit Analysis P 4.80 Write KCL equations at each of the labeled nodes, place them in standard form, and solve: v1 − v2 v1 + =0 4000 8000 At v1 : − 3 × 10−3 + At v2 : v2 − v1 v2 − 10 v2 − v3 + + =0 8000 20,000 2500 At v3 : v3 − v2 v3 v3 − 10 + + =0 2500 10,000 5000 Standard form: v1 v1 1 1 1 + + v2 − + v3(0) = 0.003 4000 8000 8000 ! 1 1 1 1 1 − + v2 + + + v3 − 8000 8000 20,000 2500 2500 v1(0) + v2 1 1 1 1 − + v3 + + 2500 2500 10,000 5000 ! = = 10 20,000 10 5000 Calculator solution: v1 = 10.890625 V v2 = 8.671875 V v3 = 7.8125 V Calculate currents: i2 = 10 − v2 = 66.40625 µ A 20,000 i3 = 10 − v3 = 437.5 µ A 5000 Calculate power delivered by the sources: p3mA = (3 × 10−3 )v1 = (3 × 10−3 )(10.890625) = 32.671875 mW © 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. Problems 4–79 p10Vmiddle = i2(10) = (66.40625 × 10−6 )(10) = 0.6640625 mW p10Vtop = i3(10) = (437.5 × 10−6 )(10) = 4.375 mW pdeliveredtotal = 32.671875 + 0.6640625 + 4.375 = 37.7109375 mW Calculate power absorbed by the 5 kΩ resistor and the percentage power: p5k = i23 (5000) = (437.5 × 10−6 )2(5000) = 0.95703125 mW % delivered to Ro : 0.95793125 (100) = 2.54% 37.7109375 P 4.81 [a] Since 0 ≤ Ro ≤ ∞ maximum power will be delivered to the 6 Ω resistor when Ro = 0. 302 [b] P = = 150 W 6 P 4.82 [a] From the solution to Problem 4.68 we have Ro (Ω) po (W) 10 14.4 15 17.07 22 18.75 33 19.16 47 18.26 68 16.31 The 33 Ω resistor dissipates the most power, because its value is closest to the Thévenin equivalent resistance of the circuit. [b] © 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. 4–80 CHAPTER 4. Techniques of Circuit Analysis [c] Ro = 33 Ω, po = 19.16 W. Compare this to Ro = RTh = 30 Ω, which then gives the maximum power delivered to the load, po (max) = 19.2 W. P 4.83 We begin by finding the Thévenin equivalent with respect to Ro . After making a couple of source transformations the circuit simplifies to i∆ = 160 − 30i∆ ; 50 i∆ = 2 A VTh = 20i∆ + 30i∆ = 50i∆ = 100 V Using the test-source method to find the Thévenin resistance gives iT = vT vT − 30(−vT /30) + 30 20 iT 1 1 4 2 = + = = vT 30 10 30 15 RTh = vT 15 = = 7.5 Ω iT 2 Thus our problem is reduced to analyzing the circuit shown below. p= 100 7.5 + Ro 2 Ro = 250 © 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. Problems 4–81 104 Ro = 250 R2o + 15Ro + 56.25 104 Ro = R2o + 15Ro + 56.25 250 40Ro = R2o + 15Ro + 56.25 R2o − 25Ro + 56.25 = 0 Ro = 12.5 ± √ 156.25 − 56.25 = 12.5 ± 10 Ro = 22.5 Ω Ro = 2.5 Ω P 4.84 [a] From the solution of Problem 4.73 we have RTh = 20 kΩ and VTh = 280 V. Therefore Ro = RTh = 20 kΩ [b] p = (140)2 = 980 mW 20,000 [c] The node voltage equations are: v1 − 40 v1 v1 − v2 + + = 0 2000 20,000 5000 v2 − v1 v2 v2 − v3 + + + 30i∆ = 0 5000 50,000 10,000 v3 − v2 v3 v3 + − 30i∆ + = 0 10,000 40,000 20,000 The dependent source constraint equation is: v1 i∆ = 20,000 © 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. 4–82 CHAPTER 4. Techniques of Circuit Analysis Place these equations in standard form: ! 1 1 1 1 40 v1 + + + v2 − + v3 (0) + i∆ (0) = 2000 20,000 5000 5000 2000 v1 v1 (0) + v2 v1 ! ! 1 1 1 1 1 − + v2 + + + v3 − + i∆ (30) = 0 4000 4000 50,000 10,000 10,000 1 − 10,000 ! + v3 ! 1 1 1 + + + i∆ (−30) = 0 10,000 40,000 20,000 ! −1 + v2(0) + v3(0) + i∆ (1) = 0 20,000 Solving, v1 = 18.4 V; v2 = −31 V; v3 = 140 V; Calculate the power: 40 − 18.4 ig = = 10.8 mA 2000 p40V = −(40)(10.8 × 10−3 ) = −432 mW pX dep source = (v2 − v3 )(30i∆ ) = −4719.6 mW pdev = 432 + 4719.6 = 5151.6 mW i∆ = 920 µA 980 × 10−3 × 100 = 19.02% 5151.6 × 10−3 [d] There are two resistor values in Appendix H that fit the criterion – 18 kΩ and 22 kΩ. Let’s use the Thévenin equivalent circuit to calculate the power delivered to each in turn, first by calculating the current through the load resistor and then using the current to calculate to power delivered to the load: 280 i18k = = 7.368 m A 20,000 + 18,000 % delivered = p18k = (7.368)2 (18,000) = 977.17 m W i22k = 280 = 6.667 m A 20,000 + 22,000 p22k = (6.667)2 (22,000) = 977.88 m W We select the 22 kΩ resistor, as the power delivered to it is closer to the maximum power of 980 mW. © 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. Problems 4–83 [e] Now substitute the 22 kΩ resistor into the original circuit and calculate the power developed by the sources in this circuit: The node voltage equations are: v1 − 40 v1 v1 − v2 + + = 0 2000 20,000 5000 v2 − v1 v2 v2 − v3 + + + 30i∆ = 0 5000 50,000 10,000 v3 v3 v3 − v2 + − 30i∆ + = 0 10,000 40,000 22,000 The dependent source constraint equation is: v1 i∆ = 20,000 Place these equations in standard form: ! 1 1 1 1 40 v1 + + + v2 − + v3 (0) + i∆ (0) = 2000 20,000 5000 5000 2000 v1 v1 (0) + v2 v1 ! ! 1 1 1 1 1 − + v2 + + + v3 − + i∆ (30) = 0 5000 5000 50,000 10,000 10,000 1 − 10,000 ! + v3 ! 1 1 1 + + + i∆ (−30) = 0 10,000 40,000 22,000 ! −1 + v2(0) + v3(0) + i∆ (1) = 0 20,000 Solving, v1 = 18.67 V; v2 = −30 V; v3 = 146.67 V; Calculate the power: 40 − 18.67 ig = = 10.67 mA 2000 p40V = −(40)(10.67 × 10−3 ) = −426.67 mW pX dep source = (v2 − v3 )(30i∆ ) = −4946.67 mW pdev = 426.67 + 4946.67 = 5373.33 mW i∆ = 933.3 µA pL = (146.67)2 /22,000 = 977.78 mW 977.78 × 10−3 % delivered = × 100 = 18.20% 5373.33 × 10−3 © 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. 4–84 CHAPTER 4. Techniques of Circuit Analysis P 4.85 [a] Open circuit voltage Node voltage equations: v1 − 24 v1 − 6i∆ v1 − v2 + + =0 2 4 5 v2 − v1 − 0.85v∆ = 0 5 Constraint equations: v1 − v2 i∆ = ; v∆ = 24 − v1 5 Solving, v2 = 84 V = vTh Thévenin resistance using a test source: v1 v1 − 6i∆ v1 − vT + + =0 2 4 5 vT − v1 − 0.85v∆ − 1 = 0 5 i∆ = v1 − vT ; 5 v∆ = −v1 Solving, vT = 10 vT RTh = = 10 Ω 1 .·. Ro = RTh = 10 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. Problems 4–85 [b] pmax = (42)2 = 176.4 W 10 [c] v1 − 24 v1 − 6i∆ v1 − 42 + + =0 2 4 5 i∆ = v1 − 42 5 Solving, v1 = 12 V; i24V = i∆ = −6 A; v∆ = 24 − v1 = 24 − 12 = 12 V 24 − v1 =6A 2 p24V = −24i24V = −24(6) = −144 W iCCVS = v1 − 6i∆ = 12 A 4 pCCVS = [6(−6)](12) = −432 W pVCCS = −[0.85(12)](42) = −428.4 W X pdev = 144 + 432 + 428.4 = 1004.4 W % delivered = 176.4 × 100 = 17.56% 1004.4 © 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. 4–86 CHAPTER 4. Techniques of Circuit Analysis P 4.86 Find the Thévenin equivalent with respect to the terminals of Ro . Open circuit voltage: (440 − 220) = 5ia − 2ib − 3ic 0 = −2ia + 10ib − 1ic ic = 0.5v∆ ; v∆ = 2(ia − ib ) Solving, ib = 26.4 A .·. vTh = 7ib = 184.8 V Short circuit current: 440 − 220 = 5ia − 2isc − 3ic 0 = −2ia + 3isc − 1ic ic = 0.5v∆ ; v∆ = 2(ia − isc) © 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. Problems Solving, isc = 60 A; ia = 80 A; 4–87 ic = 20 A RTh = vTh /isc = 184.8/60 = 3.08 Ω Ro = 3.08 Ω Therefore, the Thévenin equivalent circuit configured for maximum power to the load is From this circuit, pmax = (92.4)2 = 2772 W 3.08 With Ro equal to 3.08 Ω the original circuit becomes 440 − 220 = 5ia − 2ib − 3ic ic = 0.5v∆ ; v∆ = 2(ia − ib ) −92.4 = −2ia + 3ib − 1ic Solving, ia = 88.4 A; ib = 43.2 A; ic = 45.2 A v∆ = 2(88.4 − 43.2) = 90.4 V p440V = −(440)(88.4) = −38,896 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. 4–88 CHAPTER 4. Techniques of Circuit Analysis p220V = (220)(88.4 − 45.2) = 9504 W pdep.source = (440 − 92.4)[0.5(90.4)] = 15,711.52 W Therefore, only the 440 V source supplies power to the circuit, and the power supplied is 38,896 W. % delivered = 2772 = 7.13% 38,896 P 4.87 [a] Find the Thévenin equivalent with respect to the terminals of RL . Open circuit voltage: The mesh current equations are: −240 + 3(i1 − i2) + 20(i1 − i3) + 2i1 = 0 2i2 + 4(i2 − i3) + 3(i2 − i1) = 0 10iβ + 1i3 + 20(i3 − i1) + 4(i3 − i2) = 0 The dependent source constraint equation is: iβ = i2 − i1 Place these equations in standard form: i1 (3 + 20 + 2) + i2 (−3) + i3(−20) + iβ (0) = 240 i1 (−3) + i2(2 + 4 + 3) + i3 (−4) + iβ (0) = 0 i1 (−20) + i2(−4) + i3(1 + 20 + 4) + iβ (10) = 0 i1 (−1) + i2(1) + i3 (0) + iβ (−1) = 0 Solving, i1 = 99.6 A; i2 = 78 A; VTh = 20(i1 − i3 ) = −24 V i3 = 100.8 A; iβ = 21.6 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. Problems 4–89 Short-circuit current: The mesh current equations are: −240 + 3(i1 − i2) + 2i1 = 0 2i2 + 4(i2 − i3) + 3(i2 − i1) = 0 10iβ + 1i3 + 4(i3 − i2) = 0 The dependent source constraint equation is: iβ = i2 − i1 Place these equations in standard form: i1 (3 + 2) + i2 (−3) + i3(0) + iβ (0) = 240 i1 (−3) + i2(2 + 4 + 3) + i3 (−4) + iβ (0) = 0 i1 (0) + i2(−4) + i3 (4 + 1) + iβ (10) = 0 i1 (−1) + i2(1) + i3 (0) + iβ (−1) = 0 Solving, i1 = 92 A; isc = i1 − i3 = −4 A; i2 = 73.33 A; i3 = 96 A; iβ = 18.67 A VTh −24 RTh = = = 6Ω isc −4 RL = RTh = 6 Ω [b] pmax = 122 = 24 W 6 © 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. 4–90 CHAPTER 4. Techniques of Circuit Analysis P 4.88 [a] First find the Thévenin equivalent with respect to Ro . Open circuit voltage: iφ = 0; 50iφ = 0 v1 − 280 v1 − 280 v1 v1 + + + + 0.5125v∆ = 0 100 10 25 400 (280 − v1) 5 = 56 − 0.2v1 25 v1 = 210 V; v∆ = 14 V v∆ = VTh = 280 − v∆ = 280 − 14 = 266 V Short circuit current v1 v1 − 280 v2 v2 + + + + 0.5125(280) = 0 100 10 20 400 v∆ = 280 V v2 + 50iφ = v1 280 v2 + = 56 + 0.05v2 5 20 v2 = −968 V; v1 = −588 V iφ = iφ = isc = 56 + 0.05(−968) = 7.6 A RTh = VTh /isc = 266/7.6 = 35 Ω .·. Ro = 35 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. Problems 4–91 [b] pmax = (133)2 /35 = 505.4 W [c] v1 v1 − 280 v2 − 133 v2 + + + + 0.5125(280 − 133) = 0 100 10 20 400 v2 + 50iφ = v1; iφ = 133/35 = 3.8 A Therefore, v1 = −189 V and v2 = −379 V; thus, ig = 280 − 133 280 + 189 + = 76.30 A 5 10 p280V (dev) = (280)(76.3) = 21,364 W P 4.89 [a] We begin by finding the Thévenin equivalent with respect to the terminals of Ro . Open circuit voltage © 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. 4–92 CHAPTER 4. Techniques of Circuit Analysis The mesh current equations are: −100 + 4(i1 − i2) + 80(i1 − i3) + 16i1 = 0 124i∆ + 8(i2 − i3) + 4(i2 − i1) = 0 50 + 12i3 + 80(i3 − i1 ) + 8(i3 − i2 ) = 0 The constraint equation is: i∆ = i3 − i1 Place these equations in standard form: i1 (4 + 80 + 16) + i2(−4) + i3(−80) + i∆ (0) = 100 i1 (−4) + i2(8 + 4) + i3(−8) + i∆ (124) = 0 i1 (−80) + i2(−8) + i3(12 + 80 + 8) + i∆ (0) = −50 i1 (1) + i2(0) + i3(−1) + i∆ (1) = 0 Solving, i1 = 4.7 A; i2 = 10.5 A; Also, VTh = vab = −80i∆ = 48 V Now find the short-circuit current. i3 = 4.1 A; i∆ = −0.6 A Note with the short circuit from a to b that i∆ is zero, hence 124i∆ is also zero. The mesh currents are: −100 + 4(i1 − i2) + 16i1 = 0 8(i2 − i3) + 4(i2 − i1) = 0 50 + 12i3 + 8(i3 − i2) = 0 Place these equations in standard form: i1 (4 + 16) + i2(−4) + i3 (0) = 100 i1 (−4) + i2(8 + 4) + i3(−8) = 0 i1 (0) + i2(−8) + i3 (12 + 8) = −50 Solving, i1 = 5 A; i2 = 0 A; Then, isc = i1 − i3 = 7.5 A RTh = 48/7.5 = 6.4 Ω i3 = −2.5 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. Problems 4–93 For maximum power transfer Ro = RTh = 6.4 Ω 242 [b] pmax = = 90 W 6.4 [c] The resistor from Appendix H that is closest to the Thévenin resistance is 10 Ω. To calculate the power delivered to a 10 Ω load resistor, calculate the current using the Thévenin circuit and use it to find the power delivered to the load resistor: 48 i10 = = 2.927 A 6.4 + 10 p10 = 10(2.927)2 = 85.7 W Thus, using a 10 Ω resistor selected from Appendix H will cause 85.7 W of power to be delivered to the load, compared to the maximum power of 90 W that will be delivered if a 6.4 Ω resistor is used. P 4.90 From the solution of Problem 4.89 we know that when Ro is 6.4 Ω, the voltage across Ro is 24 V, positive at the upper terminal. Therefore our problem reduces to the analysis of the following circuit. In constructing the circuit we have used the fact that i∆ is −0.3 A, and hence 124i∆ is −37.2 V. Using the node voltage method to find v1 and v2 yields 4.05 + 24 − v1 24 − v2 + =0 4 8 2v1 + v2 = 104.4; Solving, v1 = 22.4 V; v1 + 37.2 = v2 v2 = 59.6 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. 4–94 CHAPTER 4. Techniques of Circuit Analysis It follows that 22.4 − 100 i g1 = = −4.85 A 16 59.6 − 50 i g2 = = 0.8 A 12 59.6 − 24 i2 = = 4.45 A 8 ids = −4.45 − 0.8 = −5.25 A p100V = 100ig1 = −485 W p50V = 50ig2 = 40 W pds = 37.2ids = −195.3 W .·. X pdev = 485 + 195.3 = 680.3 W .·. % delivered = 90 (100) = 13.23% 680.3 .·. 13.23% of developed power is delivered to load P 4.91 [a] 110 V source acting alone: 10(14) 35 = Ω 24 6 110 132 i0 = = A 5 + 35/6 13 Re = 35 132 770 = V = 59.231 V 6 13 13 4 A source acting alone: v0 = 5 Ωk10 Ω = 50/15 = 10/3 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. Problems 4–95 10/3 + 2 = 16/3 Ω 16/3k12 = 48/13 Ω Hence our circuit reduces to: It follows that va00 = 4(48/13) = (192/13) V and v 00 = −va00 5 (10/3) = − va00 = −(120/13) V = −9.231 V (16/3) 8 . ·. v = v 0 + v 00 = [b] p = 770 120 − = 50 V 13 13 v2 = 250 W 10 P 4.92 Voltage source acting alone: io1 = 10 10 = = 0.2 A 45 + (5 + 5)k10 45 + 5 vo1 = 20 (−10) = −2.5 V 20 + 60 © 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. 4–96 CHAPTER 4. Techniques of Circuit Analysis Current source acting alone: v2 v2 − v3 +2+ =0 5 5 v3 v3 − v2 v3 + + =0 10 5 45 Solving, v2 = −7.25 V = vo2; io2 = − i20 = v3 = −4.5 V v3 = −0.1 A 45 60k20 (2) = 1.5 A 20 vo2 = −20i20 = −20(1.5) = −30 V .·. vo = vo1 + vo2 = −2.5 − 30 = −32.5 V io = io1 + io2 = 0.2 + 0.1 = 0.3 A P 4.93 240 V source acting alone: vo1 = 20k5 (240) = 60 V 5 + 7 + 20k5 © 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. Problems 4–97 84 V source acting alone: vo2 = 20k12 (−84) = −50.4 V 1 + 4 + 20k12 16 A current source acting alone: v1 − v2 v1 + − 16 = 0 5 7 v2 − v1 v2 v2 − v3 + + =0 5 20 4 v3 − v2 v3 + + 16 = 0 4 1 Solving, v2 = 18.4 V = vo3. Therefore, vo = vo1 + vo2 + vo3 = 60 − 50.4 + 18.4 = 28 V P 4.94 6 A source: 30 Ωk5 Ωk60 Ω = 4 Ω .·. io1 = 20 (6) = 4.8 A 20 + 5 © 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. 4–98 CHAPTER 4. Techniques of Circuit Analysis 10 A source: io2 = 4 (10) = 1.6 A 25 75 V source: io3 = − 4 (15) = −2.4 A 25 io = io1 + io2 + io3 = 4.8 + 1.6 − 2.4 = 4 A P 4.95 [a] By hypothesis i0o + i00o = 3 mA. i000 o = −5 (2) = −1.25 mA; (8) .·. io = 3.5 − 1.25 = 2.25 mA [b] With all three sources in the circuit write a single node voltage equation. vb vb − 8 + + 5 − 10 = 0 6 2 .·. vb = 13.5 V io = vb = 2.25 mA 6 © 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. Problems 4–99 P 4.96 70-V source acting alone: v 0 = 70 − 4i0b i0s = v0 vb0 + = i0a + i0b 2 10 70 = 20i0a + vb0 i0a = .·. 70 − vb0 20 i0b = vb0 v0 70 − vb0 11 v0 + − = vb0 + − 3.5 2 10 20 20 10 v 0 = vb0 + 2i0b .·. vb0 = v 0 − 2i0b 11 v0 .·. i0b = (v 0 − 2i0b ) + − 3.5 20 10 13 0 70 .·. v 0 = 70 − 4 v − 42 42 or or i0b = v0 = 13 0 70 v − 42 42 3220 1610 = V = 34.255 V 94 47 50-V source acting alone: v 00 = −4i00b © 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. 4–100 CHAPTER 4. Techniques of Circuit Analysis v 00 = vb00 + 2i00b v 00 = −50 + 10i00d v 00 + 50 .·. i00d = 10 i00s = vb00 v 00 + 50 + 2 10 i00b = vb00 v 00 v 00 v 00 + 50 11 v 00 + 50 + i00s = b + b + = vb00 + 20 20 2 10 20 10 vb00 = v 00 − 2i00b v 00 + 50 11 .·. i00b = (v 00 − 2i00b ) + 20 10 Thus, Hence, 13 00 100 v = −4 v + 42 42 00 v = v 0 + v 00 = 13 00 100 v + 42 42 or i00b = or v 00 = − 200 V = −4.255 V 47 1610 200 1410 − = = 30 V 47 47 47 P 4.97 Voltage source acting alone: vo1 − 25 vo1 vo1 − 25 + − 2.2 =0 4000 20,000 4000 Simplifying 5vo1 − 125 + vo1 − 11vo1 + 275 = 0 .·. vo1 = 30 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. Problems 4–101 Current source acting alone: vo2 vo2 vo2 + + 0.005 − 2.2 4000 20,000 4000 Simplifying =0 5vo2 + vo2 + 100 − 11vo2 = 0 .·. vo2 = 20 V vo = vo1 + vo2 = 30 + 20 = 50 V P 4.98 100 = 6ia − 1ib + 0ic − 2id − 2ie + 0if − 1ig 0 = −1ia + 4ib − 2ic + 0id + 0ie + 0if + 0ig 0 = 0ia − 2ib + 13ic − 3id + 0ie + 0if + 0ig 0 = −2ia + 0ib − 3ic + 9id − 4ie + 0if + 0ig 0 = −2ia + 0ib + 0ic − 4id + 9ie − 3if + 0ig 0 = 0ia + 0ib + 0ic + 0id − 3ie + 13if − 2ig 0 = −1ia + 0ib + 0ic + 0id + 0ie − 2if + 4ig © 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. 4–102 CHAPTER 4. Techniques of Circuit Analysis A computer solution yields ia = 30 A; ie = 15 A; ib = 10 A; if = 5 A; ic = 5 A; ig = 10 A; id = 15 A .·. i = id − ie = 0 A CHECK: p1T = p1B = (ib )2 = (ig )2 = 100 W p1L = (ia − ib )2 = (ia − ig )2 = 400 W p2C = 2(ib − ic )2 = (ig − if )2 = 50 W p3 = 3(ic − id )2 = 3(ie − if )2 = 300 W p4 = 4(id − ie )2 = 0 W p8 = 8(ic )2 = 8(if )2 = 200 W p2L = 2(ia − id )2 = 2(ia − ie )2 = 450 W X P 4.99 X pabs pgen = 100 + 400 + 50 + 200 + 300 + 450 + 0 + 450 + 300+ 200 + 50 + 400 + 100 = 3000 W = 100ia = 100(30) = 3000 W (CHECKS) The mesh equations are: −125 + 0.15ia + 18.4(ia − ic) + 0.25(ia − ib ) = 0 −125 + 0.25(ib − ia) + 38.4(ib − id) + 0.15ib = 0 0.15ic + 18.4(ic − ie) + 0.25(ic − id ) + 18.4(ic − ia) = 0 0.15id + 38.4(id − ib) + 0.25(id − ic) + 38.4(id − ie ) = 0 11.6ie + 38.4(ie − id) + 18.4(ie − ic ) = 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. Problems 4–103 Place these equations in standard form: ia(18.8) + ib(−0.25) + ic(−18.4) + id(0) + ie (0) = 125 ia(−0.25) + ib(38.8) + ic(0) + id(−38.4) + ie (0) = 125 ia(−18.4) + ib(0) + ic (37.2) + id(−0.25) + ie (−18.4) = 0 ia(0) + ib(−38.4) + ic (−0.25) + id (77.2) + ie (−38.4) = 0 ia(0) + ib(0) + ic (−18.4) + id(−38.4) + ie(68.4) = 0 Solving, ia = 32.77 A; ib = 26.46 A; Find the requested voltages: v1 = 18.4(ic − ie) = 113.90 V v2 = 38.4(id − ie) = 120.19 V v3 = 11.6ie = 233.62 V ic = 26.33 A; id = 23.27 A; ie = 20.14 A P 4.100 KCL equations at nodes B, D, and E: vB − vA vB − vE + − 0.1 = 0 4 7 0.1 + vD vD + 13v∆ + −5=0 2 3 vE − vB vE − vA vE + + +5 =0 7 5 6 Multiply the first equation by 28, the second by 6, and the third by 42 to get −7vA + 11vB − 4vE = 2.8 5vD + 26v∆ = 29.4 © 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. 4–104 CHAPTER 4. Techniques of Circuit Analysis −8.4vA − 6vB + 21.4vE = −210 Constraint equations: vA = 3vx ; vσ = vx = vE − vC − 0.9; vA − vB = 0.25vA − 0.25vB ; 4 v∆ = vB − vE 5iσ = vB = vC Use the constraint equations to solve for vA , vB and v∆ in terms of vC and vE : vA = 3vE − 3vC − 2.7 vB = 15 11 vE − vC − 1.5 9 9 6 11 v∆ = vE − vC − 1.5 9 9 Substitute these three expressions into the previous three equations to yield: 68vC + 0vD − 60vE = 3.6 −286vC + 45vD + 156vE = 615.6 292.8vC + 0vD − 124.2vE = −2175.12 Solving, vC = −14.3552 V; vD = −20.9474 V; vC = 16.3293 V From the circuit diagram, p5A = 5v5A = 5(vE − vD) = 23.09 W Therefore the 5 A source is absorbing 23.09 W of power. P 4.101 [a] In studying the circuit in Fig. P4.101 we note it contains six meshes and six essential nodes. Further study shows that by replacing the parallel resistors with their equivalent values the circuit reduces to four meshes and four essential nodes as shown in the following diagram. The node Voltage approach will require solving three node Voltage equations along with equations involving v∆ and iβ . The mesh-current approach will require writing one supermesh equation plus three constraint equations involving the three current sources. Thus © 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. Problems 4–105 at the outset we know the supermesh equation can be reduced to a single unknown current. Since we are interested in the power developed by the 1 V source, we will retain the mesh current ib and eliminate the mesh currents ia, ic And id. The supermesh is denoted by the dashed line in the following figure. [b] Summing the voltages around the supermesh yields 4 −9iβ + ia + 0.75ib + 1 + 5ib + 7(ic − id ) + 8ic = 0 3 Note that iβ = ib; make that substitution and multiply the equation by 12: −108ib + 16ia + 9ib + 12 + 60ib + 84(ic − id) + 96ic = 0 or 16ia − 39ib + 180ic − 84id = −12 Use the following constraints: ia − ic = −2; . ·. ib − ic = 3ib ia = −2 + ic = −2 − 2ib Therefore, 16(−2 − 2ib ) − 39ib + 180(−2ib ) − 84id = −12 so −431ib − 84id = 20 Finally use the following constraint: 4 id = −6v∆ = −6 − ia = 8ia = −16 − 16ib 3 Thus, −431ib − 84(−16 − 16ib ) = 20 © 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. 4–106 CHAPTER 4. Techniques of Circuit Analysis so 913ib = −1324 and ib = −1.45 A Finally, p1V = 1ib = −1.45 W The 1 V source delivers 1.45 W of power. P 4.102 [a] v − v1 v v − v2 + + =0 2xr R 2r(L − x) " # 1 1 1 v1 v2 v + + = + 2xr R 2r(L − x) 2xr 2r(L − x) v= v1RL + xR(v2 − v1) RL + 2rLx − 2rx2 [b] Let D = RL + 2rLx − 2rx2 dv (RL + 2rLx − 2rx2 )R(v2 − v1) − [v1RL + xR(v2 − v1)]2r(L − 2x) = dx D2 dv = 0 when numerator is zero. dx The numerator simplifies to x2 + 2Lv1 RL(v2 − v1) − 2rv1 L2 x+ =0 (v2 − v1) 2r(v2 − v1) Solving for the roots of the quadratic yields L x= −v1 ± v2 − v1 L −v1 ± [c] x = v2 − v1 v2 = 1200 V, s s R v1 v2 − (v2 − v1)2 2rL R (v1 − v2 )2 v1 v2 − 2rL v1 = 1000 V, r = 5 × 10−5 Ω/m; L = 16 km R = 3.9 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. Problems L 16,000 = = 80; v2 − v1 1200 − 1000 4–107 v1v2 = 1.2 × 106 R 3.9(−200)2 2 (v1 − v2 ) = = 0.975 × 105 −5 3 2rL (10 × 10 )(16 × 10 ) √ x = 80{−1000 ± 1.2 × 106 − 0.0975 × 106 } = 80{−1000 ± 1050} = 80(50) = 4000 m [d] vmin = v1RL + R(v2 − v1)x RL + 2rLx − 2rx2 = (1000)(3.9)(16 × 103 ) + 3.9(200)(4000) (3.9)(16,000) + 10 × 10−5 (16,000)(4000) − 10 × 10−5 (16 × 106 ) = 975 V P 4.103 [a] voc = VTh = 75 V; Therefore [b] iL = 60 = 3 A; 20 iL = 75 − 60 15 = RTh RTh 15 = 5Ω 3 vo VTh − vo = RL RTh Therefore P 4.104 RTh = iL = RTh VTh − vo VTh = = − 1 RL vo /RL vo dv1 −R1 [R2(R3 + R4 ) + R3 R4 ] = dIg1 (R1 + R2 )(R3 + R4) + R3R4 dv1 R1 R3 R4 = dIg2 (R1 + R2 )(R3 + R4) + R3R4 dv2 −R1 R3 R4 + dIg1 (R1 + R2 )(R3 + R4 ) + R3 R4 dv2 R3 R4 (R1 + R2) = dIg2 (R1 + R2 )(R3 + R4) + R3R4 © 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. 4–108 CHAPTER 4. Techniques of Circuit Analysis P 4.105 From the solution to Problem 4.104 we have dv1 −25[5(125) + 3750] 175 = =− V/A = −14.5833 V/A dIg1 30(125) + 3750 12 and −(25)(50)(75) dv2 = = −12.5 V/A dIg1 30(125) + 3750 By hypothesis, ∆Ig1 = 11 − 12 = −1 A 175 175 .·. ∆v1 = (− )(−1) = = 14.583 V 12 12 Thus, v1 = 25 + 14.583 = 39.583 V Also, ∆v2 = (−12.5)(−1) = 12.5 V Thus, v2 = 90 + 12.5 = 102.5 V The PSpice solution is v1 = 39.583 V and v2 = 102.5 V These values are in agreement with our predicted values. P 4.106 From the solution to Problem 4.104 we have dv1 (25)(50)(75) = = 12.5 V/A dIg2 30(125) + 3750 and dv2 (50)(75)(30) = = 15 V/A dIg2 30(125) + 3750 By hypothesis, ∆Ig2 = 17 − 16 = 1 A .·. ∆v1 = (12.5)(1) = 12.5 V Thus, v1 = 25 + 12.5 = 37.5 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. Problems 4–109 Also, ∆v2 = (15)(1) = 15 V Thus, v2 = 90 + 15 = 105 V The PSpice solution is v1 = 37.5 V and v2 = 105 V These values are in agreement with our predicted values. P 4.107 From the solutions to Problems 4.104 — 4.106 we have dv1 175 =− V/A; dIg1 12 dv1 = 12.5 V/A dIg2 dv2 = −12.5 V/A; dIg1 dv2 = 15 V/A dIg2 By hypothesis, ∆Ig1 = 11 − 12 = −1 A ∆Ig2 = 17 − 16 = 1 A Therefore, ∆v1 = 175 + 12.5 = 27.0833 V 12 ∆v2 = 12.5 + 15 = 27.5 V Hence v1 = 25 + 27.0833 = 52.0833 V v2 = 90 + 27.5 = 117.5 V The PSpice solution is v1 = 52.0830 V and v2 = 117.5 V These values are in agreement with our predicted values. © 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. 4–110 CHAPTER 4. Techniques of Circuit Analysis P 4.108 By hypothesis, ∆R1 = 27.5 − 25 = 2.5 Ω ∆R2 = 4.5 − 5 = −0.5 Ω ∆R3 = 55 − 50 = 5 Ω ∆R4 = 67.5 − 75 = −7.5 Ω So ∆v1 = 0.5833(2.5) − 5.417(−0.5) + 0.45(5) + 0.2(−7.5) = 4.9168 V .·. v1 = 25 + 4.9168 = 29.9168 V ∆v2 = 0.5(2.5) + 6.5(−0.5) + 0.54(5) + 0.24(−7.5) = −1.1 V .·. v2 = 90 − 1.1 = 88.9 V The PSpice solution is v1 = 29.6710 V and v2 = 88.5260 V Note our predicted values are within a fraction of a volt of the actual values. © 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. 5 The Operational Amplifier Assessment Problems AP 5.1 [a] This is an inverting amplifier, so vo = (−Rf /Ri )vs = (−80/16)vs , vs ( V) 0.4 2.0 vo ( V) −2.0 −10.0 so vo = −5vs 3.5 −0.6 −1.6 −2.4 −15.0 3.0 8.0 10.0 Two of the values, 3.5 V and −2.4 V, cause the op amp to saturate. [b] Use the negative power supply value to determine the largest input voltage: −15 = −5vs , vs = 3 V Use the positive power supply value to determine the smallest input voltage: 10 = −5vs , Therefore vs = −2 V − 2 ≤ vs ≤ 3 V AP 5.2 From Assessment Problem 5.1 vo = (−Rf /Ri )vs = (−Rx /16,000)vs = (−Rx /16,000)(−0.640) = 0.64Rx /16,000 = 4 × 10−5 Rx Use the negative power supply value to determine one limit on the value of Rx : 4 × 10−5 Rx = −15 so Rx = −15/4 × 10−5 = −375 kΩ © 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 5–1 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. 5–2 CHAPTER 5. The Operational Amplifier Since we cannot have negative resistor values, the lower limit for Rx is 0. Now use the positive power supply value to determine the upper limit on the value of Rx : 4 × 10−5 Rx = 10 so Rx = 10/4 × 10−5 = 250 kΩ Therefore, 0 ≤ Rx ≤ 250 kΩ AP 5.3 [a] This is an inverting summing amplifier so vo = (−Rf /Ra )va + (−Rf /Rb )vb = −(250/5)va − (250/25)vb = −50va − 10vb Substituting the values for va and vb: vo = −50(0.1) − 10(0.25) = −5 − 2.5 = −7.5 V [b] Substitute the value for vb into the equation for vo from part (a) and use the negative power supply value: vo = −50va − 10(0.25) = −50va − 2.5 = −10 V Therefore 50va = 7.5, so va = 0.15 V [c] Substitute the value for va into the equation for vo from part (a) and use the negative power supply value: vo = −50(0.10) − 10vb = −5 − 10vb = −10 V; Therefore 10vb = 5, so vb = 0.5 V [d] The effect of reversing polarity is to change the sign on the vb term in each equation from negative to positive. Repeat part (a): vo = −50va + 10vb = −5 + 2.5 = −2.5 V Repeat part (b): vo = −50va + 2.5 = −10 V; 50va = 12.5, va = 0.25 V Repeat part (c), using the value of the positive power supply: vo = −5 + 10vb = 15 V; 10vb = 20; vb = 2.0 V AP 5.4 [a] Write a node voltage equation at vn ; remember that for an ideal op amp, the current into the op amp at the inputs is zero: vn vn − vo + =0 4500 63,000 © 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. Problems 5–3 Solve for vo in terms of vn by multiplying both sides by 63,000 and collecting terms: 14vn + vn − vo = 0 so vo = 15vn Now use voltage division to calculate vp. We can use voltage division because the op amp is ideal, so no current flows into the non-inverting input terminal and the 400 mV divides between the 15 kΩ resistor and the Rx resistor: vp = Rx (0.400) 15,000 + Rx Now substitute the value Rx = 60 kΩ: vp = 60,000 (0.400) = 0.32 V 15,000 + 60,000 Finally, remember that for an ideal op amp, vn = vp , so substitute the value of vp into the equation for v0 vo = 15vn = 15vp = 15(0.32) = 4.8 V [b] Substitute the expression for vp into the equation for vo and set the resulting equation equal to the positive power supply value: 0.4Rx vo = 15 15,000 + Rx ! =5 15(0.4Rx ) = 5(15,000 + Rx ) so Rx = 75 kΩ AP 5.5 [a] Since this is a difference amplifier, we can use the expression for the output voltage in terms of the input voltages and the resistor values given in Eq. 5.22: vo = 20(60) 50 vb − va 10(24) 10 Simplify this expression and subsitute in the value for vb: vo = 5(vb − va) = 20 − 5va Set this expression for vo to the positive power supply value: 20 − 5va = 10 V so va = 2 V Now set the expression for vo to the negative power supply value: 20 − 5va = −10 V so va = 6 V Therefore 2 ≤ va ≤ 6 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. 5–4 CHAPTER 5. The Operational Amplifier [b] Begin as before by substituting the appropriate values into Eq. 5.22: vo = 8(60) vb − 5va = 4vb − 5va 10(12) Now substitute the value for vb: vo = 4(4) − 5va = 16 − 5va Set this expression for vo to the positive power supply value: 16 − 5va = 10 V so va = 1.2 V Now set the expression for vo to the negative power supply value: 16 − 5va = −10 V so va = 5.2 V Therefore 1.2 ≤ va ≤ 5.2 V AP 5.6 [a] Replace the op amp with the more realistic model of the op amp from Fig. 5.15: Write the node voltage equation at the left hand node: vn vn − vg vn − vo + + =0 500,000 5000 100,000 Multiply both sides by 500,000 and simplify: vn + 100vn − 100vg + 5vn − 5v0 = 0 so 21.2vn − vo = 20vg Write the node voltage equation at the right hand node: vo − 300,000(−vn ) vo − vn + =0 5000 100,000 Multiply through by 100,000 and simplify: 20vo + 6 × 106 vn + vo − vn = 0 so 6 × 106 vn + 21vo = 0 Use Cramer’s method to solve for vo : ∆= 21.2 −1 6 × 106 21 = 6,000,445.2 © 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. Problems No = vo = 21.2 20vg 6 6 × 10 0 5–5 = −120 × 106 vg No = −19.9985vg ; ∆ so vo = −19.9985 vg [b] Use Cramer’s method again to solve for vn : N1 = vn = 20vg −1 0 21 = 420vg N1 = 6.9995 × 10−5 vg ∆ vg = 1 V, vn = 69.995 µ V [c] The resistance seen at the input to the op amp is the ratio of the input voltage to the input current, so calculate the input current as a function of the input voltage: vg − vn vg − 6.9995 × 10−5 vg ig = = 5000 5000 Solve for the ratio of vg to ig to get the input resistance: Rg = vg 5000 = = 5000.35 Ω ig 1 − 6.9995 × 10−5 [d] This is a simple inverting amplifier configuration, so the voltage gain is the ratio of the feedback resistance to the input resistance: vo 100,000 =− = −20 vg 5000 Since this is now an ideal op amp, the voltage difference between the two input terminals is zero; since vp = 0, vn = 0 Since there is no current into the inputs of an ideal op amp, the resistance seen by the input voltage source is the input resistance: Rg = 5000 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. 5–6 CHAPTER 5. The Operational Amplifier Problems P 5.1 [a] The five terminals of the op amp are identified as follows: [b] The input resistance of an ideal op amp is infinite, which constrains the value of the input currents to 0. Thus, in = 0 A. [c] The open-loop voltage gain of an ideal op amp is infinite, which constrains the difference between the voltage at the two input terminals to 0. Thus, (vp − vn ) = 0. [d] Write a node voltage equation at vn : vn + 3 vn − vo + =0 5000 15,000 But vp = 0 and vn = vp = 0. Thus, 3 vo − = 0 so vo = 9 V 5000 15,000 P 5.2 vo = −(0.5 × 10−3 )(10,000) = −5 V vo −5 .·. io = = = −1 mA 5000 5000 P 5.3 vb − va vb − vo + = 0, 20,000 100,000 [a] va = 4 V, vb = 0 V, [b] va = 2 V, vb = 0 V, [c] va = 2 V, vb = 1 V, [d] va = 1 V, vb = 2 V, [e] va = 1.5 V, [f] If vb = 1.6 V, therefore vo = 6vb − 5va vb = 4 V, vo = −15 V (sat) vo = −10 V vo = −4 V vo = 7 V vo = 15 V (sat) vo = 9.6 − 5va = ±15 .·. −1.08 V ≤ va ≤ 4.92 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. Problems P 5.4 vp = 5–7 3000 (3) = 1 V = vn 3000 + 6000 vn − 5 vn − vo + =0 10,000 5000 (1 − 5) + 2(1 − vo ) = 0 vo = −1.0 V iL = vo 1 =− = −250 × 10−6 4000 4000 iL = −250 µA P 5.5 Since the current into the inverting input terminal of an ideal op-amp is zero, the voltage across the 2.2 MΩ resistor is (2.2 × 106 )(3.5 × 10−6 ) or 7.7 V. Therefore the voltmeter reads 7.7 V. P 5.6 [a] i2 = 150 × 10−3 = 75 µA 2000 v1 = −40 × 103 i2 = −3 V v1 v1 v1 − vo [b] + + =0 20,000 40,000 50,000 .·. vo = 4.75v1 = −14.25 V [c] i2 = 75 µA, (from part [a]) v1 − vo −vo + = 795 µ A [d] io = 25,000 50,000 P 5.7 [a] First, note that vn = vp = 2.5 V Let vo1 equal the voltage output of the op-amp. Then 2.5 − vg 2.5 − vo1 + = 0, 5000 10,000 .·. vo1 = 7.5 − 2vg Also note that vo1 − 2.5 = vo , .·. vo = 5 − 2vg © 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. 5–8 CHAPTER 5. The Operational Amplifier [b] Yes, the circuit designer is correct! P 5.8 [a] The gain of an inverting amplifier is the ratio of the feedback resistor to the input resistor. If the gain of the inverting amplifier is to be 3, the feedback resistor must be 3 times as large as the input resistor. There are many possible designs that use a resistor value chosen from Appendix H. We present two here that use 3.3 kΩ resistors. Use a single 3.3 kΩ resistor as the input resistor, and use three 3.3 kΩ resistors in series as the feedback resistor to give a total of 9.9 kΩ. Alternately, use a single 3.3 kΩ resistor as the feedback resistor and use three 3.3 kΩ resistors in parallel as the input resistor to give a total of 1.1 kΩ. [b] To amplify a 5 V signal without saturating the op amp, the power supply voltages must be greater than or equal to the product of the input voltage and the amplifier gain. Thus, the power supplies should have a magnitude of (5)(3) = 15 V. P 5.9 [a] Replace the combination of vg , 1.6 kΩ, and the 6.4 kΩ resistors with its Thévenin equivalent. −[12 + σ50] (0.20) 1.28 At saturation vo = −5 V; Then vo = − 12 + σ50 (0.2) = −5, 1.28 or therefore σ = 0.4 © 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. Problems 5–9 Thus for 0 ≤ σ ≤ 0.40 the operational amplifier will not saturate. −(12 + 13.6) [b] When σ = 0.272, vo = (0.20) = −4 V 1.28 vo vo Also + + io = 0 10 25.6 . ·. i o = − P 5.10 vo 4 4 vo − = + mA = 556.25 µA 10 25.6 10 25.6 [a] Let v∆ be the voltage from the potentiometer contact to ground. Then 0 − vg 0 − v∆ + =0 2000 50,000 .·. v∆ = −25(40 × 10−3 ) = −1 V −25vg − v∆ = 0, v∆ − 0 v∆ − vo v∆ + + =0 αR∆ 50,000 (1 − α)R∆ v∆ v∆ − vo + 2v∆ + =0 α 1−α v∆ 1 1 +2+ α 1−α = vo 1−α " (1 − α) .·. vo = −1 1 + 2(1 − α) + α When α = 0.2, When α = 1, # vo = −1(1 + 1.6 + 4) = −6.6 V vo = −1(1 + 0 + 0) = −1 V .·. −6.6 V ≤ vo ≤ −1 V " # (1 − α) [b] −1 1 + 2(1 − α) + = −7 α α + 2α(1 − α) + (1 − α) = 7α α + 2α − 2α2 + 1 − α = 7α .·. 2α2 + 5α − 1 = 0 P 5.11 α∼ = 0.186 " Rf Rf Rf vo = − (0.2) + (0.15) + (0.4) 4000 5000 20,000 −6 = −0.1 × 10−3 Rf ; P 5.12 so Rf = 60 kΩ; # .·. 0 ≤ Rf ≤ 60 kΩ [a] This circuit is an example of an inverting summing amplifier. © 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. 5–10 CHAPTER 5. The Operational Amplifier 220 220 220 va − vb − vc = −5 − 12 + 11 = −6 V 44 27.5 80 [c] vo = −6 − 8vb = ±10 [b] vo = − .·. vb = −0.5 V when vo = 10 V; vb = 2 V when vo = −10 V .·. −0.5 V ≤ vb ≤ 2 V P 5.13 We want the following expression for the output voltage: vo = −(3va + 5vb + 4vc + 2vd ) This is an inverting summing amplifier, so each input voltage is amplified by a gain that is the ratio of the feedback resistance to the resistance in the forward path for the input voltage. Pick a feedback resistor with divisors of 3, 5, 4, and 2 – say 60 kΩ: " 60k 60k 60k 60k vo = − va + vb + vc + vd Ra Rb Rc Rd # Solve for each input resistance value to yield the desired gain: .·. Ra = 60,000/3 = 20 kΩ Rc = 60,000/4 = 15 kΩ Rb = 60,000/5 = 12 kΩ Rd = 60,000/2 = 30 kΩ Now create the 5 resistor values needed from the realistic resistor values in Appendix H. Note that Rb = 12 kΩ and Rc = 15 kΩ are already values from Appendix H. Create Rf = 60 kΩ by combining 27 kΩ and 33 kΩ in series. Create Ra = 20 kΩ by combining two 10 kΩ resistors in series. Create Rd = 30 kΩ by combining 18 kΩ and 12 kΩ in series. Of course there are many other acceptable possibilities. The final circuit is shown here: P 5.14 [a] Write a KCL equation at the inverting input to the op amp: vd − va vd − vb vd − vc vd vd − vo + + + + =0 40,000 22,000 100,000 352,000 220,000 © 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. Problems 5–11 Multiply through by 220,000, plug in the values of input voltages, and rearrange to solve for vo: vo = 220,000 4 −1 −5 + + 40,000 22,000 100,000 8 8 + + 352,000 220,000 ! = 14 V [b] Write a KCL equation at the inverting input to the op amp. Use the given values of input voltages in the equation: 8 − va 8−9 8 − 13 8 8 − vo + + + + =0 40,000 22,000 100,000 352,000 220,000 Simplify and solve for vo : 44 − 5.5va − 10 − 11 + 5 + 8 − vo = 0 so vo = 36 − 5.5va Set vo to the positive power supply voltage and solve for va : 36 − 5.5va = 15 .·. va = 3.818 V Set vo to the negative power supply voltage and solve for va : 36 − 5.5va = −15 . ·. va = 9.273 V Therefore, 3.818 V ≤ va ≤ 9.273 V P 5.15 [a] 8−4 8−9 8 − 13 8 8 − v0 + + + + =0 40,000 22,000 100,000 352,000 Rf 8 − vo = −2.7272 × 10−5 Rf For vo = 15 V, For vo = −15 V, so Rf = 8 − vo −2.727 × 10−5 Rf = 256.7 kΩ Rf < 0 " so this solution is not possible. # 15 − 8 15 [b] io = −(if + i10k) = − + = −1527 µA 3 256.7 × 10 10,000 P 5.16 [a] The circuit shown is a non-inverting amplifier. [b] We assume the op amp to be ideal, so vn = vp = 3 V. Write a KCL equation at vn : 3 3 − vo + =0 40,000 80,000 Solving, vo = 9 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. 5–12 P 5.17 CHAPTER 5. The Operational Amplifier [a] This circuit is an example of the non-inverting amplifier. [b] Use voltage division to calculate vp: vp = 10,000 vs vs = 10,000 + 30,000 4 Write a KCL equation at vn = vp = vs /4: vs /4 vs /4 − vo + =0 4000 28,000 Solving, vo = 7vs /4 + vs /4 = 2vs [c] 2vs = 8 so 2vs = −12 vs = 4 V so vs = −6 V Thus, −6 V ≤ vs ≤ 4 V. P 5.18 [a] vp = vn = . ·. 68 vg = 0.85vg 80 0.85vg 0.85vg − vo + = 0; 30,000 63,000 .·. vo = 2.635vg = 2.635(4), vo = 10.54 V [b] vo = 2.635vg = ±12 vg = ±4.55 V, [c] −4.55 ≤ vg ≤ 4.55 V 0.85vg 0.85vg − vo + =0 30,000 Rf ! 0.85Rf + 0.85 vg = vo = ±12 30,000 .·. 1.7Rf + 51 = ±360; P 5.19 1.7Rf = 360 − 51; Rf = 181.76 kΩ [a] From the equation for the non-inverting amplifier, Rs + Rf =4 Rs so Rs + Rf = 4Rs and therefore Rf = 3Rs Choose Rf = 30 kΩ and implement this choice from components in Appendix H by combining two 15 kΩ resistors in series. Choose Rs = Rg = 10 kΩ, which is a component in Appendix H. The resulting non-inverting amplifier circuit is shown here: © 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. Problems [b] vo = 4vg = 12 so vo = 4vg = −12 5–13 vg = 3 V so vg = −3 V Therefore, −3 V ≤ vg ≤ 3 V P 5.20 [a] This circuit is an example of a non-inverting summing amplifier. [b] Write a KCL equation at vp and solve for vp in terms of vs : vp − vs vp − 6 + =0 15,000 30,000 2vp − 2vs + vp − 6 = 0 so vp = 2vs /3 + 2 Now write a KCL equation at vn and solve for vo : vn vn − vo + =0 so vo = 4vn 20,000 60,000 Since we assume the op amp is ideal, vn = vp. Thus, vo = 4(2vs /3 + 2) = 8vs /3 + 8 [c] 8vs /3 + 8 = 16 8vs /3 + 8 = −12 so vs = 3 V so vs = −7.5 V Thus, −7.5 V ≤ vs ≤ 3 V. P 5.21 [a] This is a non-inverting summing amplifier. vp − va vp − vb [b] + =0 3 13 × 10 27 × 103 .·. 40vp = 27va + 13vb so vp = 0.675va + 0.325vb vn vn − vo + =0 11,000 110,000 .·. vo = 11vn = 11vp = 11(0.675va + 0.325vb ) = 11[0.675(0.8) + 0.325(0.4)] = 7.37 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. 5–14 CHAPTER 5. The Operational Amplifier [c] vp = vn = ia = ib = vo = 0.667 V 11 va − vp = 10 µA 13 × 103 vb − vp = −10 µA 27 × 103 [d] 7.425 for va ; P 5.22 [a] 3.575 for vb vp − va vp − vb vp − vc + + =0 Ra Rb Rc . ·. v p = RbRc Ra Rc Ra Rb va + vb + vc D D D where D = Rb Rc + Ra Rc + Ra Rb vn − vo vn + =0 20,000 100,000 ! 100,000 + 1 vn = 6vn = vo 20,000 . ·. v o = 6Rb Rc 6Ra Rc 6Ra Rb va + vb + vc D D D By hypothesis, 6Rb Rc = 1; D Then 6Ra Rb /D 3 = 6Ra Rc /D 2 6Ra Rc = 2; D so 6Ra Rb =3 D Rb = 1.5Rc But from the circuit Rb = 15 kΩ so Rc = 10 kΩ Similarly, 6Rb Rc /D 1 = 6Ra Rb /D 3 so 3Rc = Ra Thus, Ra = 30 kΩ [b] vo = 1(0.7) + 2(0.4) + 3(1.1) = 4.8 V vn = vo /6 = 0.8 V = vp ia = va − vp 0.7 − 0.8 = = −3.33 µA 30,000 30,000 © 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. Problems ib = vb − vp 0.4 − 0.8 = = −26.67 µA 15,000 15,000 ic = vc − vp 1.1 − 0.8 = = 30 µA 10,000 10,000 5–15 Check: ia + ib + ic = 0? P 5.23 [a] − 3.33 − 26.67 + 30 = 0 (checks) vp vp − va vp − vb vp − vc + + + =0 Ra Rb Rc Rg . ·. v p = RbRc Rg Ra Rc Rg Ra Rb Rg va + vb + vc D D D where D = Rb Rc Rg + Ra Rc Rg + Ra Rb Rg + Ra Rb Rc vn vn − vo + =0 Rs Rf vn 1 1 + Rs Rf . ·. v o = 1 + = vo Rf Rf vn = kvn Rs Rf where k = 1 + Rs vp = vn .·. vo = kvp or kRg Rb Rc kRg Ra Rc kRg Ra Rb va + vb + vc D D D vo = kRg Rb Rc =6 D . ·. kRg Ra Rc =3 D Rb 6 = =2 Ra 3 Since Rc 3 = = 0.75 Rb 4 Ra = 1 kΩ Rb = 2 kΩ kRg Ra Rb =4 D Rc 6 = = 1.5 Ra 4 Rc = 1.5 kΩ .·. D = [(2)(1.5)(3) + (1)(1.5)(3) + (1)(2)(3) + (1)(2)(1.5)] × 109 = 22.5 × 109 k(3)(2)(1.5) × 109 =6 22.5 × 109 k= 135 × 109 = 15 9 × 109 © 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. 5–16 CHAPTER 5. The Operational Amplifier .·. 15 = 1 + Rf Rs Rf = 14 Rs Rf = (14)(15,000) = 210 kΩ [b] vo = 6(0.5) + 3(2.5) + 4(1) = 14 V vn = vp = P 5.24 14.5 = 0.967 V 15 ia = 0.5 − 0.967 = −466.67 µA 1000 ib = 2.5 − 0.967 = 766.67 µA 2000 ic = 1 − 0.967 = 22.22 µA 1500 ig = 0.967 = 322.22 µA 3000 is = vn 0.967 = = 64.44 µA 15,000 15,000 [a] Assume va is acting alone. Replacing vb with a short circuit yields vp = 0, therefore vn = 0 and we have 0 − va 0 − vo0 + + in = 0, Ra Rb Therefore vo0 va =− , Rb Ra vo0 = − in = 0 Rb va Ra Assume vb is acting alone. Replace va with a short circuit. Now vp = vn = vbRd Rc + Rd vn vn − vo00 + + in = 0, Ra Rb 1 1 + Ra Rb vo00 Rd v 00 vb − o = 0 Rc + Rd Rb Rb = +1 Ra in = 0 vo = vo0 + vo00 = Rd Rd vb = Rc + Rd Ra Rd Ra Ra + Rb vb Rc + Rd Ra + Rb Rb vb − va Rc + Rd Ra © 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. Problems Rd [b] Ra Ra + Rb Rc + Rd = Rb , Ra Rd Ra = Rb Rc , Rd When Ra therefore Rd(Ra + Rb ) = Rb(Rc + Rd ) therefore Ra + Rb Rc + Rd = [a] vo = Ra Rc = Rb Rd Rb Ra Eq. (5.22) reduces to vo = P 5.25 5–17 Rb Rb Rb vb − va = (vb − va ). Ra Ra Ra Rd (Ra + Rb) Rb 120(24 + 75) 75 vb − va = (5) − (8) Ra (Rc + Rd ) Ra 24(130 + 120) 24 vo = 9.9 − 25 = −15.1 V [b] v1 − 8 v1 − 15.1 + =0 24,000 75,000 ia = so v1 = 2.4 V 8 − 2.4 = 233 µ A 24,000 Rina = va 8 = = 34.3 kΩ ia 233 × 10−6 [c] Rin b = Rc + Rd = 250 kΩ P 5.26 Use voltage division to find vp : vp = 2000 (5) = 1 V 2000 + 8000 Write a KCL equation at vn and solve it for vo : vn − va vn − vo + =0 5000 Rf so Rf Rf + 1 vn − va = vo 5000 5000 Since the op amp is ideal, vn = vp = 1V, so vo = Rf Rf +1 − va 5000 5000 To satisfy the equation, Rf +1 =5 5000 and Rf =4 5000 Thus, Rf = 20 kΩ. © 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. 5–18 P 5.27 CHAPTER 5. The Operational Amplifier vbRb = vn Ra + Rb vp = vn − va vn − vo + =0 4700 Rf vn Rf vaRf +1 − = vo 4700 4700 Rf Rb Rf +1 vb − va = vo 4700 Ra + Rb 4700 .·. .·. Rf = 10; 4700 Rf = 47 kΩ (a value from Appendix H) Ra + Rb = 220 kΩ Thus, 47 1+ 4700 .·. Rb 220,000 Rb = 200 kΩ ! = 10 and Ra = 220 − 200 = 20 kΩ Use two 100 kΩ resistors in series for Rb and use two 10 kΩ resistors in series for Ra . P 5.28 [a] vp vp − vc vp − vd + + =0 20,000 30,000 20,000 .·. 8vp = 2vc + 3vd = 8vn © 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. Problems 5–19 vn − va vn − vb vn − vo + + =0 20,000 18,000 180,000 . ·. v o = 20vn − 9va − 10vb = 20[(1/4)vc + (3/8)vd ] − 9va − 10vb = 20(0.75 + 1.5) − 9(1) − 10(2) = 16 V [b] vo = 5vc + 30 − 9 − 20 = 5vc + 1 ±20 = 5vc + 1 .·. vb = −4.2 V and vb = 3.8 V .·. −4.2 V ≤ vb ≤ 3.8 V P 5.29 vp = 1000ib 1000ib 1000ib − vo + − ia = 0 Ra Rf .·. .·. 1000ib 1000ib 1 1 + Ra Rf ! Rf 1+ Ra − Rf ia = vo − ia = vo Rf By hypopthesis, vo = 5000(ib − ia). Therefore, Rf = 5 kΩ 1000 1 + (use two 10 kΩ resistors in parallel) Rf Ra = 5000 so Ra = 1250 Ω To construct the 1250 Ω resistor, combine a 1.2 kΩ resistor in series with a parallel combination of two 100 Ω resistors. P 5.30 vo = Rd (Ra + Rb ) Rb vb − va Ra (Rc + Rd ) Ra By hypothesis: Rb /Ra = 4; Rc + Rd = 470 kΩ; Rd (Ra + 4Ra ) .·. =3 Ra 470,000 Rd = 282 kΩ; so Rd (Ra + Rb) =3 Ra (Rc + Rd) Rc = 188 kΩ © 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. 5–20 CHAPTER 5. The Operational Amplifier Create Rd = 282 kΩ by combining a 270 kΩ resistor and a 12 kΩ resistor in series. Create Rc = 188 kΩ by combining a 120 kΩ resistor and a 68 kΩ resistor in series. Also, when vo = 0 we have vn − va vn + =0 Ra Rb Ra .·. vn 1 + Rb ia = = va ; vn = 0.8va va − 0.8va va = 0.2 ; Ra Ra .·. Ra = 4.4 kΩ; Rin = va = 5Ra = 22 kΩ ia Rb = 17.6 kΩ Create Ra = 4.4 kΩ by combining two 2.2 kΩ resistors in series. Create Rb = 17.6 kΩ by combining a 12 kΩ resistor and a 5.6 kΩ resistor in series. P 5.31 vp = 1500 (−18) = −3 V = vn 9000 −3 + 18 −3 − vo + =0 1600 Rf .·. vo = 0.009375Rf − 3 vo = 9 V; Rf = 1280 Ω vo = −9 V; But P 5.32 Rf = −640 Ω Rf ≥ 0, .·. Rf = 1.28 kΩ αRg [a] vp = vg αRg + (Rg − αRg ) vn = vp = αvg vn − vg vn − vo + =0 R1 Rf (vn − vg ) Rf + vn − vo = 0 R1 vo = ! Rf Rf 1+ αvg − vg Rg R1 = (αvg − vg )4 + αvg = [(α − 1)4 + α]vg = (5α − 4)vg = (5α − 4)(2) = 10α − 8 © 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. Problems α vo α vo α 5–21 vo 0.0 −8 V 0.4 −4 V 0.8 0 V 0.1 −7 V 0.5 −3 V 0.9 1 V 0.2 −6 V 0.6 −2 V 1.0 2 V 0.3 −5 V 0.7 −1 V [b] Rearranging the equation for vo from (a) gives Rf Rf vo = + 1 vg α + − vg R1 R1 Therefore, Rf slope = + 1 vg ; R1 Rf intercept = − vg R1 [c] Using the equations from (b), −6 = Rf + 1 vg ; R1 4=− Rf vg R1 Solving, vg = −2 V; P 5.33 Rf =2 R1 Acm = (20)(50) − (50)Rx 20(50 + Rx ) Adm = 50(20 + 50) + 50(50 + Rx ) 2(20)(50 + Rx ) Adm Rx + 120 = Acm 2(20 − Rx ) .·. Rx + 120 = ±1000 for the limits on the value of Rx 2(20 − Rx ) If we use +1000 Rx = 19.93 kΩ © 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. 5–22 CHAPTER 5. The Operational Amplifier If we use −1000 Rx = 20.07 kΩ 19.93 kΩ ≤ Rx ≤ 20.07 kΩ P 5.34 [a] Adm = (24)(26) + (25)(25) = 24.98 (2)(1)(25) (1)(24) − 25(1) = −0.04 1(25) 24.98 [c] CMRR = = 624.50 0.04 [b] Acm = P 5.35 [a] vp = vs , vn = R1 vo , R1 + R2 Therefore vo = vn = vp R2 R1 + R2 vs = 1 + vs R1 R1 [b] vo = vs [c] Because vo = vs , thus the output voltage follows the signal voltage. P 5.36 It follows directly from the circuit that vo = −(120/7.5)vg = −16vg From the plot of vg we have vg = 0, t < 0 vg = t 0 ≤ t ≤ 0.5 vg = 1−t 0.5 ≤ t ≤ 1.5 vg = t−2 1.5 ≤ t ≤ 2.5 vg = 3−t 2.5 ≤ t ≤ 3.5 vg = t − 4 Therefore vo = −16t 3.5 ≤ t ≤ 4.5, etc. 0 ≤ t ≤ 0.5 vo = 16t − 16 0.5 ≤ t ≤ 1.5 vo = 32 − 16t 1.5 ≤ t ≤ 2.5 vo = 16t − 48 2.5 ≤ t ≤ 3.5 vo = 64 − 16t 3.5 ≤ t ≤ 4.5, etc. These expressions for vo are valid as long as the op amp is not saturated. Since the peak values of vo are ±5, the output is clipped at ±5. The plot is shown below. © 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. Problems P 5.37 vp = 5–23 5.6 vg = 0.7vg = 7 sin(π/3)t V 8.0 vn − vo vn + =0 15,000 75,000 6vn = vo ; vn = vp .·. vo = 42 sin(π/3)t V vo = 0 0≤t≤∞ t≤0 At saturation 42 sin π t = ±21; 3 π π .·. t= , 3 6 t = 0.50 s, 5π , 6 2.50 s, π sin t = ±0.5 3 7π , 6 3.50 s, 11π , 6 etc. 5.50 s, etc. © 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. 5–24 P 5.38 CHAPTER 5. The Operational Amplifier [a] vn − va vn − vo + =0 R R 2vn − va = vo va va − vn va − vo + + =0 Ra R R va 1 2 vn vo + − = Ra R R R va 2 + R − vn = vo Ra vn = vp = va + vg .·. 2vn − va = 2va + 2vg − va = va + 2vg .·. va − vo = −2vg 2va + va (1) R − va − vg = vo Ra R . ·. v a 1 + − vo = vg Ra (2) Now combining equations (1) and (2) yields −va R = −3vg Ra or va = 3vg Ra R Hence ia = va 3vg = Ra R Q.E.D. © 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. Problems 5–25 [b] At saturation vo = ± Vcc .·. va = ± Vcc − 2vg (3) and R . ·. v a 1 + Ra = ± Vcc + vg (4) Dividing Eq (4) by Eq (3) gives 1+ R ± Vcc + vg = Ra ± Vcc − 2vg . ·. R ± Vcc + vg 3vg = −1= Ra ± Vcc − 2vg ± Vcc − 2vg or Ra = P 5.39 (± Vcc − 2vg ) R 3vg Q.E.D. (320 × 10−3 )2 = 6.4 µW (16 × 103 ) 16 [b] v16 kΩ = (320) = 80 mV 64 [a] p16 kΩ = p16 kΩ = (80 × 10−3 )2 = 0.4 µW (16 × 103 ) pa 6.4 = = 16 pb 0.4 [d] Yes, the operational amplifier serves several useful purposes: [c] • First, it enables the source to control 16 times as much power delivered to the load resistor. When a small amount of power controls a larger amount of power, we refer to it as power amplification. • Second, it allows the full source voltage to appear across the load resistor, no matter what the source resistance. This is the voltage follower function of the operational amplifier. • Third, it allows the load resistor voltage (and thus its current) to be set without drawing any current from the input voltage source. This is the current amplification function of the circuit. P 5.40 [a] Assume the op-amp is operating within its linear range, then iL = 8 = 2 mA 4000 For RL = 4 kΩ vo = (4 + 4)(2) = 16 V Now since vo < 20 V our assumption of linear operation is correct, therefore iL = 2 mA © 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. 5–26 CHAPTER 5. The Operational Amplifier [b] 20 = 2(4 + RL ); RL = 6 kΩ [c] As long as the op-amp is operating in its linear region iL is independent of RL . From (b) we found the op-amp is operating in its linear region as long as RL ≤ 6 kΩ. Therefore when RL = 6 kΩ the op-amp is saturated. We can estimate the value of iL by assuming ip = in iL . Then iL = 20/(4000 + 16,000) = 1 mA. To justify neglecting the current into the op-amp assume the drop across the 50 kΩ resistor is negligible, since the input resistance to the op-amp is at least 500 kΩ. Then ip = in = (8 − 4)/(500 × 103 ) = 8 µA. But 8 µA 1 mA, hence our assumption is reasonable. © 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. Problems 5–27 [d] P 5.41 i1 = 15 − 10 = 1 mA 5000 i2 + i1 + 0 = 10 mA; i2 = 9 mA vo2 = 10 + (400)(9) × 10−3 = 13.6 V i3 = 15 − 13.6 = 0.7 mA 2000 i4 = i3 + i1 = 1.7 mA © 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. 5–28 CHAPTER 5. The Operational Amplifier vo1 = 15 + 1.7(0.5) = 15.85 V P 5.42 [a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output voltage of the amplifier on the right. Then vo1 = ia = −47 (1) = −4.7 V; 10 −220 (−0.15) = 1.0 V 33 vo2 − vo1 = 5.7 mA 1000 [b] ia = 0 when vo1 = vo2 Thus −47 (vL) = 1 10 vL = − P 5.43 vo2 = so from (a) vo2 = 1 V 10 = −212.77 mV 47 [a] Replace the op amp with the model from Fig. 5.15: Write two node voltage equations, one at the left node, the other at the right node: vn − vg vn − vo vn + + =0 5000 100,000 500,000 vo + 3 × 105 vn vo − vn vo + + =0 5000 100,000 500 Simplify and place in standard form: 106vn − 5vo = 100vg (6 × 106 − 1)vn + 221vo = 0 Let vg = 1 V and solve the two simultaneous equations: vo = −19.9844 V; vn = 736.1 µV Thus the voltage gain is vo/vg = −19.9844. [b] From the solution in part (a), vn = 736.1 µV. © 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. Problems [c] ig = 5–29 vg − vn vg − 736.1 × 10−6 vg = 5000 5000 Rg = vg 5000 = = 5003.68 Ω ig 1 − 736.1 × 10−6 [d] For an ideal op amp, the voltage gain is the ratio between the feedback resistor and the input resistor: 100,000 vo =− = −20 vg 5000 For an ideal op amp, the difference between the voltages at the input terminals is zero, and the input resistance of the op amp is infinite. Therefore, vn = vp = 0 V; P 5.44 Rg = 5000 Ω [a] vn − vg vn − vo + =0 2000 10,000 .·. vo = 6vn − 5vg Also vo = A(vp − vn ) = −Avn . ·. v n = −vo A 6 . ·. v o 1 + = −5vg A vo = −5A vg (6 + A) −5(194)(1) = −4.85 V 200 −5 [c] vo = (1) = −5 V 1 + (6/A) [b] vo = © 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. 5–30 CHAPTER 5. The Operational Amplifier [d] P 5.45 [a] −5A (1) = −0.99(5) A+6 .·. −0.05A = −29.7 so − 5A = −4.95(A + 6) so A = 594 vn vn − vg vn − vo + + =0 16,000 800,000 200,000 or 55vn − 4vo = vg Eq (1) vo − vn vo − 50,000(vp − vn ) vo + + =0 20,000 200,000 8000 36vo − vn − 125 × 104 (vp − vn ) = 0 vp = vg + (vn − vg )(240) = (0.7)vg + (0.3)vn 800 36vo − vn − 125 × 104 [(0.7)vg − (0.7)vn ] = 0 36vo + 874,999vn = 875,000vg Eq (2) Let vg = 1 V and solve Eqs. (1) and (2) simultaneously: vn = 999.446 mV and vo . ·. = 13.49 vg vo = 13.49 V [b] From part (a), vn = 999.446 mV. vp = (0.7)(1000) + (0.3)(999.446) = 999.834 mV [c] vp − vn = 387.78 µV (1000 − 999.83)10−3 = 692.47 pA [d] ig = 24 × 103 vg vg − vo [e] + = 0, since vn = vp = vg 16,000 200,000 vo = 13.5 .·. vo = 13.5vg , vg vn = vp = 1 V; P 5.46 vp − vn = 0 V; ig = 0 A [a] vn − 0.88 vn vn − vTh + + =0 1600 500,000 24,000 © 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. Problems 5–31 vTh + 105 vn vTh − vn + =0 2000 24,000 Solving, vTh = −13.198 V Short-circuit current calculation: vn vn − 0.88 vn − 0 + + =0 500,000 1600 24,000 .·. vn = 0.8225 V isc = vn 105 − vn = −41.13 A 24,000 2000 RTh = vTh = 320.9 mΩ isc [b] The output resistance of the inverting amplifier is the same as the Thévenin resistance, i.e., Ro = RTh = 320.9 mΩ [c] 330 vo = (−13.2) = −13.18 V 330.3209 © 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. 5–32 CHAPTER 5. The Operational Amplifier vn − 0.88 vn vn + 13.18 + + =0 1600 500,000 24,000 .·. vn = 942 µV 0.88 − 942 × 10−6 = 549.41 µA ig = 1600 Rg = P 5.47 0.88 = 1601.71 Ω ig [a] vTh = − 24,000 (0.88) = −13.2 V 1600 RTh = 0, since op-amp is ideal [b] Ro = RTh = 0 Ω [c] Rg = 1.6 kΩ P 5.48 since vn = 0 From Eq. 5.57, vref 1 1 1 = vn + + R + ∆R R + ∆R R − ∆R Rf ! − vo Rf Substituting Eq. 5.59 for vp = vn : 1 1 vref R+∆R + R−∆R + R1f vref = 1 1 R + ∆R (R − ∆R) R+∆R + R−∆R + 1 Rf − vo Rf Rearranging, vo 1 1 = vref − Rf R − ∆R R + ∆R © 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. Problems 5–33 Thus, vo = vref P 5.49 2∆R Rf R2 − ∆R2 [a] Use Eq. 5.61 to solve for Rf ; note that since we are using 1% strain gages, ∆ = 0.01: (5)(120) vo R Rf = = = 2 kΩ 2∆vref (2)(0.01)(15) [b] Now solve for ∆ given vo = 50 mV: ∆= vo R (0.05)(120) = = 100 × 10−6 2Rf vref 2(2000)(15) The change in strain gage resistance that corresponds to a 50 mV change in output voltage is thus ∆R = ∆R = (100 × 10−6 )(120) = 12 mΩ P 5.50 [a] Let R1 = R + ∆R vp vp vp − vin + + =0 Rf R R1 . ·. v p " # 1 1 1 vin + + = Rf R R1 R1 . ·. v p = RRf vin = vn RR1 + Rf R1 + Rf R vn vn − vin vn − vo + + =0 R R Rf vn " # 1 1 1 vo vin + + − = R R Rf Rf R . ·. v n " # R + 2Rf vin vo − = RRf R Rf © 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. 5–34 CHAPTER 5. The Operational Amplifier . ·. " vo R + 2Rf = Rf RRf #" # RRf vin vin − RR1 + Rf R1 + Rf R R " # vo R + 2Rf 1 . ·. = − vin Rf RR1 + Rf R1 + Rf R R [R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf · . . vo = vin R[R1(R + Rf ) + RRf ] Now substitute R1 = R + ∆R and get vo = −∆R(R + Rf )Rf vin R[(R + ∆R)(R + Rf ) + RRf ] If ∆R R (R + Rf )Rf (−∆R)vin vo ≈ R2 (R + 2Rf ) [b] vo ≈ P 5.51 47 × 104 (48 × 104 )(−95)15 ≈ −3.384 V 108 (95 × 104 ) [c] vo = −95(48 × 104 )(47 × 104 )15 = −3.368 V 104 [(1.0095)104 (48 × 104 ) + 47 × 108 ] [a] vo ≈ (R + Rf )Rf (−∆R)vin R2 (R + 2Rf ) vo = . ·. (R + Rf )(−∆R)Rf vin R[(R + ∆R)(R + Rf ) + RRf ] approx value R[(R + ∆R)(R + Rf ) + RRf ] = true value R2 (R + 2Rf ) .·. Error = R[(R + ∆R)(R + Rf ) + RRf ] − R2 (R + 2Rf ) R2 (R + 2Rf ) = .·. % error = [b] % error = P 5.52 1= ∆R (R + Rf ) R (R + 2Rf ) ∆R(R + Rf ) × 100 R(R + 2Rf ) 95(48 × 104 ) × 100 = 0.48% 104 (95 × 104 ) ∆R(48 × 104 ) × 100 104 (95 × 104 ) 9500 .·. ∆R = = 197.91667 Ω 48 © 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. Problems 5–35 197.19667 .·. % change in R = × 100 ≈ 1.98% 104 P 5.53 [a] It follows directly from the solution to Problem 5.50 that vo = [R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf vin R[R1 (R + Rf ) + RRf ] Now R1 = R − ∆R. Substituting into the expression gives vo = (R + Rf )Rf (∆R)vin R[(R − ∆R)(R + Rf ) + RRf ] Now let ∆R R and get vo ≈ (R + Rf )Rf ∆Rvin R2 (R + 2Rf ) [b] It follows directly from the solution to Problem 5.50 that . ·. approx value R[(R − ∆R)(R + Rf ) + RRf ] = true value R2 (R + 2Rf ) .·. Error = = (R − ∆R)(R + Rf ) + RRf − R(R + 2Rf ) R(R + 2Rf ) −∆R(R + Rf ) R(R + 2Rf ) .·. % error = −∆R(R + Rf ) × 100 R(R + 2Rf ) [c] R − ∆R = 9810 Ω .·. ∆R = 10,000 − 9810 = 190 Ω (48 × 104 )(47 × 104 )(190)(15) · . . vo ≈ ≈ 6.768 V 108 (95 × 104 ) −190(48 × 104 )(100) [d] % error = = −0.96% 104 (95 × 104 ) © 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. 6 Inductance, Capacitance, and Mutual Inductance Assessment Problems AP 6.1 [a] ig = 8e−300t − 8e−1200tA v=L dig = −9.6e−300t + 38.4e−1200tV, dt t > 0+ v(0+ ) = −9.6 + 38.4 = 28.8 V [b] v = 0 when 38.4e−1200t = 9.6e−300t or t = (ln 4)/900 = 1.54 ms [c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W dp = 0 when e1800t − 12.5e900t + 16 = 0 [d] dt Let x = e900t and solve the quadratic x2 − 12.5x + 16 = 0 x = 1.44766, t= ln 1.45 = 411.05 µs 900 x = 11.0523, t= ln 11.05 = 2.67 ms 900 p is maximum at t = 411.05 µs [e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W [f] W is max when i is max, i is max when di/dt is zero. When di/dt = 0, v = 0, therefore t = 1.54 ms. [g] imax = 8[e−0.3(1.54) − e−1.2(1.54)] = 3.78 A wmax = (1/2)(4 × 10−3 )(3.78)2 = 28.6 mJ © 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 6–1 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. 6–2 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance AP 6.2 [a] i = C dv d = 24 × 10−6 [e−15,000t sin 30,000t] dt dt = [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A, π [b] i ms = −31.66 mA, 80 i(0+ ) = 0.72 A π v ms = 20.505 V, 80 p = vi = −649.23 mW [c] w = 1 Cv 2 = 126.13 µJ 2 1 AP 6.3 [a] v = C = t Z i dx + v(0− ) 0− 1 0.6 × 10−6 Z t 0− 3 cos 50,000x dx = 100 sin 50,000t V [b] p(t) = vi = [300 cos 50,000t] sin 50,000t = 150 sin 100,000t W, [c] w(max) = p(max) = 150 W 1 2 Cvmax = 0.30(100)2 = 3000 µJ = 3 mJ 2 60(240) = 48 mH 300 [b] i(0+ ) = 3 + −5 = −2 A Z 125 t [c] i = (−0.03e−5x ) dx − 2 = 0.125e−5t − 2.125 A 6 0+ Z 50 t [d] i1 = (−0.03e−5x ) dx + 3 = 0.1e−5t + 2.9 A + 3 0 AP 6.4 [a] Leq = i2 = 25 6 Z t 0+ (−0.03e−5x ) dx − 5 = 0.025e−5t − 5.025 A i1 + i2 = i AP 6.5 v1 = 0.5 × 106 Z 240 × 10−6 e−10x dx − 10 = −12e−10t + 2 V 0+ v2 = 0.125 × 106 v1(∞) = 2 V, t Z t 0+ 240 × 10−6 e−10x dx − 5 = −3e−10t − 2 V v2 (∞) = −2 V 1 1 W = (2)(4) + (8)(4) × 10−6 = 20 µJ 2 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. Problems 6–3 AP 6.6 [a] Summing the voltages around mesh 1 yields di1 d(i2 + ig ) +8 + 20(i1 − i2 ) + 5(i1 + ig ) = 0 dt dt or 4 di2 dig di1 + 25i1 + 8 − 20i2 = − 5ig + 8 4 dt dt dt ! Summing the voltages around mesh 2 yields 16 di1 d(i2 + ig ) +8 + 20(i2 − i1) + 780i2 = 0 dt dt or di1 di2 dig 8 − 20i1 + 16 + 800i2 = −16 dt dt dt [b] From the solutions given in part (b) i1 (0) = −0.4 − 11.6 + 12 = 0; i2 (0) = −0.01 − 0.99 + 1 = 0 These values agree with zero initial energy in the circuit. At infinity, i1 (∞) = −0.4A; i2(∞) = −0.01A When t = ∞ the circuit reduces to 7.8 7.8 7.8 + = −0.4A; i2(∞) = − = −0.01A 20 780 780 From the solutions for i1 and i2 we have .·. i1(∞) = − di1 = 46.40e−4t − 60e−5t dt di2 = 3.96e−4t − 5e−5t dt Also, dig = 7.84e−4t dt Thus di1 4 = 185.60e−4t − 240e−5t dt 25i1 = −10 − 290e−4t + 300e−5t © 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. 6–4 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 8 di2 = 31.68e−4t − 40e−5t dt 20i2 = −0.20 − 19.80e−4t + 20e−5t 5ig = 9.8 − 9.8e−4t 8 dig = 62.72e−4t dt Test: 185.60e−4t − 240e−5t − 10 − 290e−4t + 300e−5t + 31.68e−4t − 40e−5t ? +0.20 + 19.80e−4t − 20e−5t = −[9.8 − 9.8e−4t + 62.72e−4t ] −9.8 + (300 − 240 − 40 − 20)e−5t ? +(185.60 − 290 + 31.68 + 19.80)e−4t = −(9.8 + 52.92e−4t ) ? −9.8 + 0e−5t + (237.08 − 290)e−4t = −9.8 − 52.92e−4t −9.8 − 52.92e−4t = −9.8 − 52.92e−4t (OK) Also, 8 di1 = 371.20e−4t − 480e−5t dt 20i1 = −8 − 232e−4t + 240e−5t 16 di2 = 63.36e−4t − 80e−5t dt 800i2 = −8 − 792e−4t + 800e−5t 16 dig = 125.44e−4t dt Test: 371.20e−4t − 480e−5t + 8 + 232e−4t − 240e−5t + 63.36e−4t − 80e−5t ? −8 − 792e−4t + 800e−5t = −125.44e−4t (8 − 8) + (800 − 480 − 240 − 80)e−5t ? +(371.20 + 232 + 63.36 − 792)e−4t = −125.44e−4t ? (800 − 800)e−5t + (666.56 − 792)e−4t = −125.44e−4t −125.44e−4t = −125.44e−4t (OK) © 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. Problems 6–5 Problems P 6.1 0 ≤ t ≤ 2s : 103 iL = 2.5 Z t 3 × 10−3 e−4x dx + 1 = 1.2 0 = −0.3e−4t + 1.3 A, e−4x −4 t +1 0 0 ≤ t ≤ 2s iL (2) = −0.3e−8 + 1.3 = 1.3 A t ≥ 2s : 103 iL = 2.5 Z t 2 −3 −4(x−2) −3 × 10 e = 0.3e−4(t−2) + 1 A, P 6.2 [a] v = L e−4(x−2) dx + 1.3 = −1.2 −4 t + 1.3 2 t ≥ 2s di dt = (50 × 10−6 )(18)[e−10t − 10te−10t ] = 900e−10t (1 − 10t) µV © 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. 6–6 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [b] i(200 ms) = 18(0.2)(e−2 ) = 487.21 mA v(200 ms) = 900(e−2 )(1 − 2) = −121.8 µV p(200 ms) = vi = (487.21 × 10−3 )(−121.8 × 10−6 ) = −59.34 µW [c] delivering 59.34 µW [d] i(200 ms) = 487.21 mA (from part [b]) 1 1 w = Li2 = (50 × 10−6 )(0.48721)2 = 5.93 µJ 2 2 [e] The energy is a maximum where the current is a maximum: diL = 0 when 1 − 10t = 0 dt or t = 0.1 s imax = 18(0.1)e−1 = 662.18 mA 1 wmax = (50 × 10−6 )(0.66218)2 = 10.96 µJ 2 P 6.3 [a] 0 ≤ t ≤ 2 ms : 1 i= L = Z 0 t 106 vs dx + i(0) = 200 Z 0 t 5 × 10−3 dx + 0 5000 t x = 25t A 200 0 2 ms ≤ t < ∞ : [b] i = 25t mA, 106 i= 200 0 ≤ t ≤ 2 ms; Z t 2×10−3 (0) dx + 2 × 10−3 = 50 mA i = 50 mA, t ≥ 2 ms © 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. Problems P 6.4 [a] i = 0 t<0 i = 50t A 0 ≤ t ≤ 5 ms i = 0.5 − 50t A 5 ≤ t ≤ 10 ms i = 0 10 ms < t [b] v = L di = 20 × 10−3 (50) = 1 V dt 6–7 0 ≤ t ≤ 5 ms v = 20 × 10−3 (−50) = −1 V 5 ≤ t ≤ 10 ms v = 0 t<0 v = 1V 0 < t < 5 ms v = −1 V 5 < t < 10 ms v = 0 10 ms < t p = vi p = 0 t<0 p = (50t)(1) = 50t W 0 < t < 5 ms p = (0.5 − 50t)(−1) = 50t − 0.5 W 5 < t < 10 ms p = 0 10 ms < t w w w w P 6.5 = 0 t = Z t<0 t = Z = 25x2 − 0.5x 2 (50x) dx = 50 0 0.005 x 2 t = 25t2 J 0 < t < 5 ms 0 (50x − 0.5) dx + 0.625 × 10−3 t 0.005 +0.625 × 10−3 = 25t2 − 0.5t + 2.5 × 10−3 J 5 < t < 10 ms = 0 10 ms < t [a] 0 ≤ t ≤ 1 s : v = −100t i= 1 5 Z 0 t −100x dx + 0 = −20 x2 2 t 0 2 i = −10t A © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 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Inductance, Capacitance, and Mutual Inductance 1s ≤ t ≤ 3s : v = 100t − 200 i(1) = −10 A . ·. i = = 1 5 t Z 1 Z 20 (100x − 200) dx − 10 t 1 (x − 2) dx − 10 = 10t2 − 40t + 20 A 3s ≤ t ≤ 5s : v = 100 V i(3) = 90 − 120 + 20 = −10 A i = = 1 5 Z t 3 (100) dx − 10 20(t − 3) − 10 = 20t − 70 A 5s ≤ t ≤ 6s : v = 600t − 100 i(5) = 100 − 70 = 30 A i = 1 5 Z t 5 (600x − 100) dx + 30 t Z = 20 = 120t − 600 − 10t2 + 250 + 30 = t ≥ 6s : 5 (6 − x) dx + 30 −10t2 + 120t − 320 A v=0 i(6) = 720 − 360 − 320 = 40 A i = = 1 5 Z t 6 0 dx + 40 40 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. Problems 6–9 [b] v = 0 at t = 2 s and t = 6 s i(2) = 10(4) − 40(2) + 20 = −20 A i(6) = 40 A [c] P 6.6 [a] i(0) = A1 + A2 = 0.04 di = −10,000A1 e−10,000t − 40,000A2 e−40,000t dt v = −200A1 e−10,000t − 800A2 e−40,000t V v(0) = −200A1 − 800A2 = 28 Solving, A1 = 0.1 and A2 = −0.06 Thus, i1 = (100e−10,000t − 60e−40,000t) mA v = −20e−10,000t + 48e−40,000t V, t≥0 t≥0 [b] i = 0 when 100e−10,000t = 60e−40,000t Therefore e30,000t = 0.6 so t = −17.03 µs which is not possible! v = 0 when 20e−10,000t = 48e−40,000t Therefore e30,000t = 2.4 so t = 29.18 µs Thus the power is zero at t = 29.18 µs. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 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Inductance, Capacitance, and Mutual Inductance [a] From Problem 6.6 we have i = A1 e−10,000t + A2e−40,000t A v = −20A1 e−10,000t + 48A2 e−40,000t V i(0) = A1 + A2 = 0.04 v(0) = −200A1 − 800A2 = −68 Solving, Thus, A1 = −0.06; A2 = 0.1 i = −60e−10,000t + 100e−40,000t mA t ≥ 0 v = 12e−10,000t − 80e−40,000t V t ≥ 0 [b] i = 0 when 60e−10,000t = 100e−40,000t .·. e30,000t = 5/3 so t = 17.03 µs Thus, i > 0 for 0 ≤ t ≤ 17.03 µs and i < 0 for 17.03 µs ≤ t < ∞ v = 0 when 12e−10,000t = 80e−40,000t .·. e30,000t = 20/3 so t = 63.24 µs Thus, v < 0 for 0 ≤ t ≤ 63.24 µs and v > 0 for 63.24 µs ≤ t < ∞ and 63.24 µs ≤ t < ∞ Therefore, p < 0 for 0 ≤ t ≤ 17.03 µs (inductor delivers energy) p > 0 for 17.03 µs ≤ t ≤ 63.24 µs (inductor stores energy) [c] The energy stored at t = 0 is 1 1 w(0) = L[i(0)]2 = (0.02)(0.04)2 = 16 µJ 2 2 p = vi = 6e−50,000t − 8e−80,000t − 0.72e−20,000t W For t > 0: w= Z ∞ 0 6e−50,000t dt − 6e−50,000t = −50,000 ∞ 0 Z ∞ 0 8e−80,000t dt − 8e−80,000t − −80,000 ∞ 0 Z ∞ 0 0.72e−20,000t dt 0.72e−20,000t − −20,000 ∞ 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. Problems 6–11 = (1.2 − 1 − 0.36) × 10−4 = −16 µJ Thus, the energy stored equals the energy extracted. P 6.8 [a] v = L di dt v = −25 × 10−3 d [10 cos 400t + 5 sin 400t]e−200t dt = −25 × 10−3 (−200e−200t [10 cos 400t + 5 sin 400t] +e−200t[−4000 sin 400t + 2000 cos 400t]) v = −25 × 10−3 e−200t(−1000 sin 400t − 4000 sin 400t) = −25 × 10−3 e−200t(−5000 sin 400t) = 125e−200t sin 400t V dv = 125(e−200t(400) cos 400t − 200e−200t sin 400t) dt = 25,000e−200t (2 cos 400t − sin 400t) V/s dv =0 when dt .·. tan 400t = 2, 2 cos 400t = sin 400t 400t = 1.11; t = 2.77 ms [b] v(2.77 ms) = 125e−0.55 sin 1.11 = 64.27 V P 6.9 [a] i = = = [b] p 1000 20 Z −2500 t 0 − 50 sin 250x dx + 10 − cos 250x 250 t + 10 0 10 cos 250t A = vi = (−50 sin 250t)(10 cos 250t) = −500 sin 250t cos 250t p w = −250 sin 500t W 1 2 = Li 2 = 1 (20 × 10−3 )(10 cos 250t)2 2 = 1000 cos 2 250t mJ w = (500 + 500 cos 500t) mJ © 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. 6–12 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [c] Absorbing power: P 6.10 Delivering power: 2π ≤ t ≤ 4π ms 0 ≤ t ≤ 2π ms 6π ≤ t ≤ 8π ms 4π ≤ t ≤ 6π ms i = (B1 cos 4t + B2 sin 4t)e−t/2 i(0) = B1 = 10 A di = (B1 cos 4t + B2 sin 4t)(−0.5e−t/2 ) + e−t/2(−4B1 sin 4t + 4B2 cos 4t) dt = [(4B2 − 0.5B1 ) cos 4t − (4B1 + 0.5B2 ) sin 4t]e−t/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. Problems v=4 6–13 di = [(16B2 − 2B1 ) cos 4t − (16B1 + 2B2 ) sin 4t]e−t/2 dt v(0) = 60 = 16B2 − 2B1 = 16B2 − 20 .·. B2 = 5 A Thus, i = (10 cos 4t + 5 sin 4t)e−t/2, A, t≥0 v = (60 cos 4t − 170 sin 4t)e−t/2 V, i(1) = −26, A; t≥0 v(1) = 54.25 V p(1) = (−26)(54.25) = −339.57 W delivering P 6.11 p = vi = 40t[e−10t − 10te−20t − e−20t] W= Z 0 ∞ p dx = Z 0 ∞ 40x[e−10x − 10xe−20x − e−20x ] dx = 0.2 J This is energy stored in the inductor at t = ∞. P 6.12 [a] v(20 µs) = v(20 µs) = = v(40 µs) = = 12.5 × 109 (20 × 10−6 )2 = 5 V (end of first interval) 106 (20 × 10−6 ) − (12.5)(400) × 10−3 − 10 5 V (start of second interval) 106 (40 × 10−6 ) − (12.5)(1600) × 10−3 − 10 10 V (end of second interval) [b] p(10µs) = 62.5 × 1012 (10−5 )3 = 62.5 mW, i(10µs) = 50 mA, v(10 µs) = 1.25 V, p(10 µs) = vi = (1.25)(50 m) = 62.5 mW (checks) p(30 µs) = 437.50 mW, v(30 µs) = 8.75 V, i(30 µs) = 0.05 A p(30 µs) = vi = (8.75)(0.05) = 62.5 mW (checks) [c] w(10 µs) = 15.625 × 1012 (10 × 10−6 )4 = 0.15625 µJ w = 0.5Cv 2 = 0.5(0.2 × 10−6 )(1.25)2 = 0.15625 µJ w(30 µs) = 7.65625 µJ w(30 µs) = 0.5(0.2 × 10−6 )(8.75)2 = 7.65625 µJ © 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. 6–14 P 6.13 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance For 0 ≤ t ≤ 1.6 s: iL = 1 5 t Z 0 3 × 10−3 dx + 0 = 0.6 × 10−3 t iL (1.6 s) = (0.6 × 10−3 )(1.6) = 0.96 mA Rm = (20)(1000) = 20 kΩ vm (1.6 s) = (0.96 × 10−3 )(20 × 103 ) = 19.2 V P 6.14 [a] i = 400 × 10−3 t = 80 × 103 t −6 5 × 10 i = 400 × 10−3 i= q 0 ≤ t ≤ 5 µs 5 ≤ t ≤ 20 µs 300 × 10−3 t − 0.5 = 104 t − 0.5 30 × 10−6 5×10−6 = Z = 8 × 104 0 4 8 × 10 t dt + t2 2 Z 15×10−6 5×10−6 20 µs ≤ t ≤ 50 µs 0.4 dt 5×10−6 +0.4(10 × 10−6 ) 0 = 4 × 104 (25 × 10−12 ) + 4 × 10−6 = 5 µC [b] v = 4 × 106 Z 5×10−6 0 6 + 4 × 10 = 4 × 106 Z 8 × 104 x dx + 4 × 106 30×10−6 20×10−6 20×10−6 5×10−6 0.4 dx (104 x − 0.5) dx x2 8 × 104 2 " Z 5×10−6 20×10−6 +0.4x 0 5×10−6 x2 +104 2 30×10−6 20×10−6 30×10−6 −0.5x 20×10−6 # = 4 × 106 [4 × 104 (25 × 10−12 ) + 0.4(15 × 10−6 ) + 5000(900 × 10−12 400 × 10−12 ) − 0.5(10 × 10−6 )] = 18 V v(30 µs) = 18 V © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 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Problems 6–15 [c] v(50 µs) = 4 × 106 [10−6 + 6 × 10−6 + 5000(2500 × 10−12 − 400 × 10−12 ) − 0.5(30 × 10−6 )] = 10 V 1 1 w = Cv 2 = (0.25 × 10−6 )(10)2 = 12.5 µJ 2 2 P 6.15 [a] v = 1 0.5 × 10−6 Z 3e = 100 × 10 500×10−6 0 −2000t −2000 50 × 10−3 e−2000t dt − 20 500×10−6 −20 0 = 50(1 − e−1 ) − 20 = 11.61 V w = 1 Cv 2 2 = 12 (0.5)(10−6 )(11.61)2 = 33.7 µJ [b] v(∞) = 50 − 20 = 30V 1 w(∞) = (0.5 × 10−6 )(30)2 = 225 µJ 2 P 6.16 [a] 0 ≤ t ≤ 10 µs 1 = 10 × 106 C C = 0.1 µF v = 10 × 106 t Z 0 − 0.05 dx + 15 v = −50 × 104 t + 15 V 0 ≤ t ≤ 10 µs v(10 µs) = −5 + 15 = 10 V [b] 10 µs ≤ t ≤ 20 µs 6 v = 10 × 10 Z t 10×10−6 v = 106 t V 0.1 dx + 10 = 106 t − 10 + 10 10 ≤ t ≤ 20 µs v(20 µs) = 106 (20 × 10−6 ) = 20 V [c] 20 µs ≤ t ≤ 40 µs 6 v = 10 × 10 Z t 20×10−6 1.6 dx + 20 = 1.6 × 106 t − 32 + 20 v = 1.6 × 106 t − 12 V, 20 µs ≤ t ≤ 40 µs © 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. 6–16 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [d] 40 µs ≤ t < ∞ v(40 µs) = 64 − 12 = 52 V P 6.17 P 6.18 40 µs ≤ t < ∞ iC = C(dv/dt) 0 < t < 0.5 : iC = 20 × 10−6 (60)t = 1.2t mA 0.5 < t < 1 : iC = 20 × 10−6 (60)(t − 1) = 1.2(t − 1) mA 1 1 [a] w(0) = C[v(0)]2 = (0.20) × 10−6 (150)2 = 2.25 mJ 2 2 −5000t [b] v = (A1t + A2)e v(0) = A2 = 150 V dv dt = −5000e−5000t (A1t + A2) + e−5000t(A1) = (−5000A1 t − 5000A2 + A1 )e−5000t dv (0) = A1 − 5000A2 dt i=C dv , dt i(0) = C dv(0) dt © 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. Problems 6–17 dv(0) i(0) 250 × 10−3 = = = 1250 × 103 dt C 0.2 × 10−6 . ·. .·. 1.25 × 106 = A1 − 5000(150) Thus, A1 = 1.25 × 106 + 75 × 104 = 2 × 106 V s [c] v = (2 × 106 t + 150)e−5000t dv d = 0.2 × 10−6 (2 × 106 t + 150)e−5000t dt dt d = [(0.4t + 10 × 30−6 )e−5000t] dt i=C i = (0.4t + 30 × 10−6 )(−5000)e−5000t + e−5000t(0.4) = (−2000t − 150 × 10−3 + 0.4)e−5000t = (0.25 − 2000t)e−5000t A, P 6.19 t≥0 dv = 0, t < 0 dt dv d [b] i = C = 4 × 10−6 [100 − 40e−2000t (3 cos 1000t + sin 1000t)] dt dt [a] i = C = 4 × 10−6 [−40(−2000)e−2000t (3 cos 1000t + sin 1000t) −40(1000)e−2000t (−3 sin 1000t + cos 1000t)] = 0.32e−2000t(3 cos 1000t + sin 1000t) − 0.16(−3 sin 1000t + cos 1000t) = 0.8e−2000t[cos 1000t + sin 1000t] A, [c] no, [d] yes, t≥0 v(0− ) = −20 V v(0+ ) = 100 − 40(1)(3) = −20 V i(0− ) = 0 A i(0+ ) = 0.8 A [e] v(∞) = 100 V 1 1 w = Cv 2 = (4 × 10−6 )(100)2 = 20 mJ 2 2 P 6.20 30k20 = 12 H 80k(8 + 12) = 16 H 60k(14 + 16) = 20 H 15k(20 + 10) = 20 H Lab = 5 + 10 = 15 H © 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. 6–18 P 6.21 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 5k(12 + 8) = 4 H 4k4 = 2 H 15k(8 + 2) = 6 H 3k6 = 2 H 6 + 2 = 8H P 6.22 [a] Combine three 1 mH inductors in series to get a 3 mH equivalent inductor. [b] Combine two 100 µH inductors in parallel to get a 50 µH inductor. Then combine this parallel pair in series with two more 100 µH inductors: 100 µk100 µ + 100 µ + 100 µ = 50 µ + 100 µ + 100 µ = 250 µH [c] Combine two 100 µH inductors in parallel to get a 50 µH inductor. Then combine this parallel pair with a 10 µH inductor in series: 100 µk100 µ + 10 µ = 50 µ + 10 µ = 60 µH P 6.23 [a] io(0) = −i1 (0) − i2(0) = 6 − 1 = 5 A [b] io = − 1 4 Z t 0 2000e−100x dx + 5 = −500 = 5(e−100t − 1) + 5 = 5e−100t A, e−100x −100 t +5 0 t≥0 [c] © 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. Problems va = 3.2(−500e−100t ) = −1600e−100t V vc = va + vb = −1600e−100t + 2000e−100t = = 400e−100t V Z 1 t 400e−100x dx − 6 1 0 = −4e−100t + 4 − 6 = −4e−100t − 2 A i1 i1 [d] i2 1 4 = Z 6–19 t≥0 t 0 400e−100x dx + 1 = −e−100t + 2 A, t≥0 1 1 1 [e] w(0) = (1)(6)2 + (4)(1)2 + (3.2)(5)2 = 60 J 2 2 2 1 [f] wdel = (4)(5)2 = 50 J 2 [g] wtrapped = 60 − 50 = 10 J or P 6.24 1 1 wtrapped = (1)(2)2 + (4)(2)2 + 10 J (check) 2 2 vb = 2000e−100t V io = 5e−100t A p = 10,000e−200t W w= Z 0 t 104 e−200x dx = 10,000 e−200x −200 t 0 = 50(1 − e−200t) W wtotal = 50 J 80%wtotal = 40 J Thus, 50 − 50e−200t = 40; e200t = 5; .·. t = 8.05 ms © 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. 6–20 P 6.25 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [a] 3.2 di = 64e−4t dt Z i(t) = 20 = 20 i(t) = [b] 4 di = 20e−4t dt so t 0 e−4x dx − 5 t e−4x −4 −5 0 −5e−4t A di1 = 64e−4t dt Z t i1 (t) = 16 e−4x dx − 10 0 = i1 (t) = [c] 16 16 0 −10 −4e−4t − 6 A di2 = 64e−4t dt Z i2 (t) = 4 = 4 i2 (t) = t e−4x −4 di2 = 4e−4t dt so t 0 e−4x dx + 5 t e−4x −4 +5 0 −e−4t + 6 A [d] p = −vi = (−64e−4t )(−5e−4t ) = 320e−8t W w ∞ = Z = 320 = 0 p dt = e−8t −8 Z ∞ 0 320e−8t dt ∞ 0 40 J 1 1 [e] w = (4)(−10)2 + (16)(5)2 = 400 J 2 2 [f] wtrapped = winitial − wdelivered = 400 − 40 = 360 J © 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. Problems 1 1 [g] wtrapped = (4)(−6)2 + (16)(6)2 = 360 J 2 2 P 6.26 1 1 1 1 = + = ; C1 48 16 12 6–21 checks C1 = 12 µF C2 = 3 + 12 = 15 µF 1 1 1 1 = + = ; C3 30 15 10 C3 = 10 µF C4 = 10 + 10 = 20 µF 1 1 1 1 1 = + + = ; C5 5 20 4 2 C5 = 2 µF Equivalent capacitance is 2 µF with an initial voltage drop of +25 V. P 6.27 1 1 5 + = 4 6 12 .·. Ceq = 2.4 µF 1 1 4 + = 4 12 12 .·. Ceq = 3 µF © 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. 6–22 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 1 1 4 + = 24 8 24 P 6.28 .·. Ceq = 6 µF [a] Combine two 220 µF capacitors in series to get a 110 µF capacitor. Then combine the series pair in parallel with another 220 µF capacitor to get 330 µF: (220 µ + 220 µ)k220 µ = 110 µk220 µ = 330 µF [b] Create a 1500 nF capacitor as follows: (1 µ + 1 µ)k1 µ = 500 nk1000 n = 1500 nF Create a second 1500 nF capacitor using the same three resistors. Place these two 1500 nF in series: 1500 n + 1500 n = 750 nF [c] Combine two 100 pF capacitors in series to get a 50 pF capacitor. Then combine the series pair in parallel with another 100 pF capacitor to get 150 pF: (100 p + 100 p)k100 p = 50 pk100 p = 150 pF P 6.29 1 1 1 1 10 = + + = =2 Ce 1 5 1.25 5 .·. C2 = 0.5 µF vb = 20 − 250 + 30 = −200 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. Problems 6–23 [a] vb Z − = 10,000 = [b] va 106 0.5 = t 0 − 5 × 10−3 e−50x dx − 200 e−50x −50 t 0 −200 −200e−50t V = − 106 0.5 Z t 0 − 5 × 10−3 e−50x dx − 20 = 20(e−50t − 1) − 20 = 20e−50t − 40 V [c] vc [d] vd = 106 Z t − 5 × 10−3 e−50x dx − 30 1.25 0 = 80(e−50t − 1) − 30 = 80e−50t − 110 V 6 = 10 Z t 0 − 5 × 10−3 e−50x dx + 250 = 100(e−50t − 1) + 250 = 100e−50t + 150 V CHECK: vb [e] i1 [f] i2 = −vc − vd − va = −200e−50t V (checks) d [100e−50t + 150] dt = 0.2 × 10−6 = 0.2 × 10−6 (−5000e−50t ) = −e−50t mA = 0.8 × 10−6 = −4e−50t mA d [100e−50t + 150] dt © 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. 6–24 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance CHECK: ib = i1 + i2 = −5e−50t mA P 6.30 [a] w(0) 1 (0.2 2 = = × 10−6 )(250)2 + 12 (0.8 × 10−6 )(250)2 + 21 (5 × 10−6 )(20)2 1 (1.25 2 + (OK) × 10−6 )(30)2 32,812.5 µJ [b] w(∞) = 21 (5 × 10−6 )(40)2 + 12 (1.25 × 10−6 )(110)2 + 12 (0.2 × 10−6 )(150)2 + 21 (0.8 × 10−6 )(150)2 = 22,812.5 µJ 1 [c] w = (0.5 × 10−6 )(200)2 = 10,000 µJ 2 CHECK: 32,812.5 − 22,812.5 = 10,000 µJ 10,000 × 100 = 30.48% [d] % delivered = 32,812.5 [e] w t = Z = 10(1 − e−100t) mJ 0 (−0.005e−50x )(−200e−50x ) dx = .·. 10−2 (1 − e−100t) = 7.5 × 10−3 ; Thus, t = P 6.31 Z t 0 e−100x dx e−100t = 0.25 ln 4 = 13.86 ms. 100 [a] vo = 106 1.6 Z t 0 800 × 10−6 e−25x dx − 20 e−25x = 500 −25 t 0 −20 = −20e−25t V, [b] v1 = t≥0 106 e−25x (800 × 10−6 ) 2 −25 = −16e−25t + 21 V, t +5 0 t≥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. Problems [c] v2 106 e−25x (800 × 10−6 ) 8 −25 = [d] p t 0 −25 = −4e−25t − 21 V, = −vi = −(−20e−25t)(800 × 10−6 )e−25t = 16 × 10−3 e−50t w [e] w t≥0 ∞ = Z = 16 × 10−3 = 6–25 0 16 × 10−3 e−50t dt e−50t −50 ∞ 0 −0.32 × 10−3 (0 − 1) = 320 µJ × 10−6 )(5)2 + 12 (8 × 10−6 )(25)2 = 1 (2 2 = 2525 µJ [f] wtrapped = winitial − wdelivered = 2525 − 320 = 2205 µJ [g] wtrapped = 1 (2 2 × 10−6 )(21)2 + 12 (8 × 10−6 )(−21)2 = 2205 µJ P 6.32 From Figure 6.17(a) we have v= 1 C1 v= Z 0 i dx + v1(0) + 1 1 + + ··· C1 C2 Therefore P 6.33 t Z 1 C2 Z 0 t i dx + v2(0) + · · · t 0 i dx + v1(0) + v2(0) + · · · 1 1 1 = + + ··· , Ceq C1 C2 veq(0) = v1(0) + v2(0) + · · · From Fig. 6.18(a) i = C1 dv dv dv + C2 + · · · = [C1 + C2 + · · ·] dt dt dt Therefore Ceq = C1 + C2 + · · ·. Because the capacitors are in parallel, the initial voltage on every capacitor must be the same. This initial voltage would appear on Ceq. © 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. 6–26 P 6.34 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance dio dt = (5){e−2000t[−8000 sin 4000t + 4000 cos 4000t] +(−2000e−2000t )[2 cos 4000t + sin 4000t]} = e−2000t{− 50,000 sin 4000t} V dio + (0 ) = (1)[sin(0)] = 0 dt .·. 10 × 10−3 dio + (0 ) = 0 dt so v2 (0+ ) = 0 v1(0+ ) = 40io (0+ ) + v2(0+ ) = 40(10) = 0 = 400 V P 6.35 vc vL vo P 6.36 1 0.625 × 10−6 Z t Z t = − = 150(e−16,000t − 1) − 200(e−4000t − 1) − 50 = 150e−16,000t − 200e−4000t V dio 25 × 10−3 dt = 0 1.5e−16,000x dx − 0 0.5e−4000x dx − 50 = 25 × 10−3 (−24,000e−16,000t + 2000e−4000t ) = −600e−16,000t + 50e−4000t V = vc − vL = (150e−16,000t − 200e−4000t ) − (−600e−16,000t + 50e−4000t) = 750e−16,000t − 250e−4000t V, t > 0 [a] Rearrange by organizing the equations by di1/dt, i1, di2/dt, i2 and transfer the ig terms to the right hand side of the equations. We get 4 di1 di2 dig + 25i1 − 8 − 20i2 = 5ig − 8 dt dt dt −8 di1 di2 dig − 20i1 + 16 + 80i2 = 16 dt dt dt [b] From the given solutions we have di1 = −320e−5t + 272e−4t dt di2 = 260e−5t − 204e−4t dt © 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. Problems 6–27 Thus, 4 di1 = −1280e−5t + 1088e−4t dt 25i1 = 100 + 1600e−5t − 1700e−4t 8 di2 = 2080e−5t − 1632e−4t dt 20i2 = 20 − 1040e−5t + 1020e−4t 5ig = 80 − 80e−5t 8 dig = 640e−5t dt Thus, −1280e−5t + 1088e−4t + 100 + 1600e−5t − 1700e−4t − 2080e−5t ? +1632e−4t − 20 + 1040e−5t − 1020e−4t = 80 − 80e−5t − 640e−5t 80 + (1088 − 1700 + 1632 − 1020)e−4t ? +(1600 − 1280 − 2080 + 1040)e−5t = 80 − 720e−5t 80 + (2720 − 2720)e−4t + (2640 − 3360)e−5t = 80 − 720e−5t 8 (OK) di1 = −2560e−5t + 2176e−4t dt 20i1 = 80 + 1280e−5t − 1360e−4t 16 di2 = 4160e−5t − 3264e−4t dt 80i2 = 80 − 4160e−5t + 4080e−4t 16 dig = 1280e−5t dt 2560e−5t − 2176e−4t − 80 − 1280e−5t + 1360e−4t + 4160e−5t − 3264e−4t ? +80 − 4160e−5t + 4080e−4t = 1280e−5t (−80 + 80) + (2560 − 1280 + 4160 − 4160)e−5t ? +(1360 − 2176 − 3264 + 4080)e−4t = 1280e−5t 0 + 1280e−5t + 0e−4t = 1280e−5t P 6.37 (OK) [a] Yes, using KVL around the lower right loop vo = v20Ω + v60Ω = 20(i2 − i1) + 60i2 © 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. 6–28 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [b] vo = 20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t )+ 60(1 − 52e−5t + 51e−4t ) = 20(−3 − 116e−5t + 119e−4t ) + 60 − 3120e−5t + 3060e−4t vo = −5440e−5t + 5440e−4t V [c] vo = L2 d d (15 + 36e−5t − 51e−4t ) + 8 (4 + 64e−5t − 68e−4t ) dt dt −5t −4t −5t −2880e + 3264e − 2560e + 2176e−4t = 16 = P 6.38 d di1 (ig − i2) + M dt dt vo = −5440e−5t + 5440e−4t V [a] vg = 5(ig − i1) + 20(i2 − i1) + 60i2 = 5(16 − 16e−5t − 4 − 64e−5t + 68e−4t )+ 20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t )+ 60(1 − 52e−5t + 51e−4t ) = 60 + 5780e−4t − 5840e−5t V [b] vg (0) = 60 + 5780 − 5840 = 0 V [c] pdev = vg ig = 960 + 92,480e−4t − 94,400e−5t − 92,480e−9t + 93,440e−10t W [d] pdev (∞) = 960 W [e] i1(∞) = 4 A; i2(∞) = 1 A; ig (∞) = 16 A; p5Ω = (16 − 4)2 (5) = 720 W p20Ω = 32 (20) = 180 W p60Ω = 12 (60) = 60 W X . ·. P 6.39 [a] −2 pabs = 720 + 180 + 60 = 960 W X pdev = X pabs = 960 W dig di2 + 16 + 32i2 = 0 dt dt 16 di2 dig + 32i2 = 2 dt dt © 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. Problems 6–29 [b] i2 = e−t − e−2t A di2 = −e−t + 2e−2t A/s dt ig = 8 − 8e−t A dig = 8e−t A/s dt .·. −16e−t + 32e−2t + 32e−t − 32e−2t = 16e−t [c] v1 dig di2 −2 dt dt = 4 = 4(8e−t ) − 2(−e−t + 2e−2t ) = 34e−t − 4e−2t V, t>0 [d] v1(0) = 34 − 4 = 30 V; Also di2 dig v1(0) = 4 (0) − 2 (0) dt dt = 4(8) − 2(−1 + 2) = 32 − 2 = 30 V Yes, the initial value of v1 is consistent with known circuit behavior. P 6.40 [a] vab = L1 di di di di di + L2 + M + M = (L1 + L2 + 2M) dt dt dt dt dt It follows that Lab = (L1 + L2 + 2M) [b] vab = L1 di di di di di − M + L2 − M = (L1 + L2 − 2M) dt dt dt dt dt Therefore Lab = (L1 + L2 − 2M) P 6.41 [a] vab = L1 0 = L1 d(i1 − i2) di2 +M dt dt d(i2 − i1) di2 d(i1 − i2) di2 −M +M + L2 dt dt dt dt Collecting coefficients of [di1/dt] and [di2/dt], the two mesh-current equations become vab = L1 di1 di2 + (M − L1 ) dt dt and di1 di2 + (L1 + L2 − 2M) dt dt Solving for [di1/dt] gives 0 = (M − L1 ) di1 L1 + L2 − 2M = vab dt L1 L2 − M 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. 6–30 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance from which we have vab = L1 L2 − M 2 L1 + L2 − 2M .·. Lab = ! di1 dt ! L1 L2 − M 2 L1 + L2 − 2M [b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses, therefore Lab = P 6.42 L1 L2 − M 2 L1 + L2 + 2M When the switch is opened the induced voltage is negative at the dotted terminal. Since the voltmeter kicks upscale, the induced voltage across the voltmeter must be positive at its positive terminal. Therefore, the voltage is negative at the negative terminal of the voltmeter. Thus, the lower terminal of the unmarked coil has the same instantaneous polarity as the dotted terminal. Therefore, place a dot on the lower terminal of the unmarked coil. P 6.43 [a] Dot terminal 1; the flux is up in coil 1-2, and down in coil 3-4. Assign the current into terminal 4; the flux is down in coil 3-4. Therefore, dot terminal 4. Hence, 1 and 4 or 2 and 3. [b] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4. Assign the current into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3. [c] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4. Assign the current into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3. [d] Dot terminal 1; the flux is down in coil 1-2, and down in coil 3-4. Assign the current into terminal 4; the flux is down in coil 3-4. Therefore, dot terminal 4. Hence, 1 and 4 or 2 and 3. P 6.44 [a] W = (0.5)L1 i21 + (0.5)L2 i22 + Mi1 i2 q M = 0.85 (18)(32) = 20.4 mH W = [9(36) + 16(81) + 20.4(54)] = 2721.6 mJ [b] W = [324 + 1296 + 1101.6] = 2721.6 mJ [c] W = [324 + 1296 − 1101.6] = 518.4 mJ [d] W = [324 + 1296 − 1101.6] = 518.4 mJ © 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. Problems P 6.45 q [a] M = 1.0 (18)(32) = 24 mH, i1 = 6 A Therefore 16i22 + 144i2 + 324 = 0, 9 Therefore i2 = − ± 2 6–31 s 2 9 2 i22 + 9i2 + 20.25 = 0 − 20.25 = −4.5 ± √ 0 Therefore i2 = −4.5 A [b] No, setting W equal to a negative value will make the quantity under the square root sign negative. P 6.46 M 22.8 =√ = 0.95 L1 L2 576 √ = 576 = 24 mH [a] k = √ [b] Mmax [c] N 2 P1 N1 L1 = 12 = L2 N2 P2 N2 2 N1 2 60 = = 6.25 N2 9.6 N1 √ = 6.25 = 2.5 N2 . ·. P 6.47 [a] L1 = N12 P1 ; P1 = dφ11 P11 = = 0.2; dφ21 P21 72 × 10−3 = 1152 nWb/A 6.25 × 104 P21 = 2P11 .·. 1152 × 10−9 = P11 + P21 = 3P11 P11 = 192 nWb/A; q P21 = 960 nWb/A q M = k L1 L2 = (2/3) (0.072)(0.0405) = 36 mH N2 = M 36 × 10−3 = = 150 turns N1P21 (250)(960 × 10−9 ) L2 40.5 × 10−3 = = 1800 nWb/A N22 (150)2 [c] P11 = 192 nWb/A [see part (a)] φ22 P22 P2 − P12 P2 [d] = = = −1 φ12 P12 P12 P12 [b] P2 = P21 = P21 = 960 nWb/A; P2 = 1800 nWb/A φ22 1800 = − 1 = 0.875 φ12 960 © 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. 6–32 P 6.48 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance M2 k 2 L1 [a] L2 = s N1 = N2 [b] P1 = P1 = = (0.09)2 = 50 mH (0.75)2 (0.288) s L1 = L2 288 = 2.4 50 0.288 L1 = = 0.2 × 10−6 Wb/A 2 N1 (1200)2 P2 = P 6.49 ! L2 0.05 = = 0.2 × 10−6 Wb/A 2 N2 (500)2 L1 = 2 nWb/A; N12 P12 = P21 = P2 = L2 = 2 nWb/A; N22 q M = k L1 L2 = 180 µH M = 1.2 nWb/A N1 N2 P11 = P1 − P21 = 0.8 nWb/A P 6.50 1 P11 P22 P11 [a] 2 = 1 + 1+ = 1+ k P12 P12 P21 Therefore P12P21 k2 = (P21 + P11)(P12 + P22) P22 1+ P12 Now note that φ1 = φ11 + φ21 = P11N1i1 + P21N1 i1 = N1i1 (P11 + P21) and similarly φ2 = N2 i2 (P22 + P12) It follows that (P11 + P21) = φ1 N1 i1 and (P22 + P12) = φ2 N2i2 ! Therefore (φ12/N2 i2 )(φ21/N1 i1 ) φ12φ21 k2 = = (φ1 /N1 i1 )(φ2/N2 i2) φ1 φ2 or v ! u u φ21 k=t φ1 φ12 φ2 ! [b] The fractions (φ21/φ1 ) and (φ12/φ2 ) are by definition less than 1.0, therefore k < 1. © 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. Problems P 6.51 6–33 When the button is not pressed we have C2 dv d = C1 (vs − v) dt dt or (C1 + C2) dvs dv = C1 dt dt dv C1 dvs = dt (C1 + C2) dt Assuming C1 = C2 = C dv dvs = 0.5 dt dt or v = 0.5vs (t) + v(0) When the button is pressed we have C1 dv dv d(v − vs ) + C3 + C2 =0 dt dt dt dv C2 dvs .·. = dt C1 + C2 + C3 dt Assuming C1 = C2 = C3 = C dv 1 dvs = dt 3 dt © 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. 6–34 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 1 v = vs (t) + v(0) 3 Therefore interchanging the fixed capacitor and the button has no effect on the change in v(t). P 6.52 With no finger touching and equal 10 pF capacitors v(t) = 10 (vs (t)) + 0 = 0.5vs (t) 20 With a finger touching Let Ce = equivalent capacitance of person touching lamp Ce = Then (10)(100) = 9.091 pF 110 C + Ce = 10 + 9.091 = 19.091 pF .·. v(t) = 10 vs = 0.344vs 29.091 .·. ∆v(t) = (0.5 − 0.344)vs = 0.156vs P 6.53 With no finger on the button the circuit is C1 d d (v − vs ) + C2 (v + vs ) = 0 dt dt when C1 = C2 = C (2C) dv =0 dt © 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. Problems 6–35 With a finger on the button C1 d(v + vs ) dv d(v − vs ) + C2 + C3 =0 dt dt dt (C1 + C2 + C3 ) when dvs dvs dv + C2 − C1 =0 dt dt dt C1 = C2 = C3 = C (3C) dv =0 dt .·. there is no change in the output voltage of this circuit. © 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. 7 Response of First-Order RL and RC Circuits Assessment Problems AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 Ω resistor from the circuit. First combine the 30 Ω and 6 Ω resistors in parallel: 30k6 = 5 Ω Use voltage division to find the voltage drop across the parallel resistors: 5 v= (120) = 75 V 5+3 Now find the current using Ohm’s law: v 75 i(0− ) = − = − = −12.5 A 6 6 1 1 [b] w(0) = Li2 (0) = (8 × 10−3 )(12.5)2 = 625 mJ 2 2 [c] To find the time constant, we need to find the equivalent resistance seen by the inductor for t > 0. When the switch opens, only the 2 Ω resistor remains connected to the inductor. Thus, L 8 × 10−3 τ= = = 4 ms R 2 [d] i(t) = i(0− )et/τ = −12.5e−t/0.004 = −12.5e−250t A, t≥0 [e] i(5 ms) = −12.5e−250(0.005) = −12.5e−1.25 = −3.58 A So w (5 ms) = 21 Li2 (5 ms) = 12 (8) × 10−3 (3.58)2 = 51.3 mJ © 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 7–1 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. 7–2 CHAPTER 7. Response of First-Order RL and RC Circuits w (dis) = 625 − 51.3 = 573.7 mJ 573.7 % dissipated = 100 = 91.8% 625 AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor: Using current division, 10 i(0− ) = (6.4) = 4 A 10 + 6 Now use the circuit for t > 0 to find the equivalent resistance seen by the inductor, and use this value to find the time constant: L 0.32 = = 0.1 s Req 3.2 Use the initial inductor current and the time constant to find the current in the inductor: i(t) = i(0− )e−t/τ = 4e−t/0.1 = 4e−10t A, t ≥ 0 Use current division to find the current in the 10 Ω resistor: Req = 4k(6 + 10) = 3.2 Ω, io (t) = .·. τ= 4 4 (−i) = (−4e−10t ) = −0.8e−10t A, 4 + 10 + 6 20 t ≥ 0+ Finally, use Ohm’s law to find the voltage drop across the 10 Ω resistor: vo (t) = 10io = 10(−0.8e−10t ) = −8e−10t V, t ≥ 0+ [b] The initial energy stored in the inductor is 1 1 w(0) = Li2(0− ) = (0.32)(4)2 = 2.56 J 2 2 Find the energy dissipated in the 4 Ω resistor by integrating the power over all time: di v4Ω (t) = L = 0.32(−10)(4e−10t ) = −12.8e−10t V, t ≥ 0+ dt p4Ω (t) = 2 v4Ω = 40.96e−20t W, 4 t ≥ 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. Problems w4Ω (t) = Z 7–3 ∞ 0 40.96e−20t dt = 2.048 J Find the percentage of the initial energy in the inductor dissipated in the 4 Ω resistor: 2.048 % dissipated = 100 = 80% 2.56 AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like an open circuit. Find the voltage drop across the open circuit by finding the voltage drop across the 50 kΩ resistor. First use current division to find the current through the 50 kΩ resistor: 80 × 103 i50k = (7.5 × 10−3 ) = 4 mA 3 3 3 80 × 10 + 20 × 10 + 50 × 10 Use Ohm’s law to find the voltage drop: v(0− ) = (50 × 103 )i50k = (50 × 103 )(0.004) = 200 V [b] To find the time constant, we need to find the equivalent resistance seen by the capacitor for t > 0. When the switch opens, only the 50 kΩ resistor remains connected to the capacitor. Thus, τ = RC = (50 × 103 )(0.4 × 10−6 ) = 20 ms [c] v(t) = v(0− )e−t/τ = 200e−t/0.02 = 200e−50t V, t ≥ 0 1 1 [d] w(0) = Cv 2 = (0.4 × 10−6 )(200)2 = 8 mJ 2 2 1 1 2 [e] w(t) = Cv (t) = (0.4 × 10−6 )(200e−50t )2 = 8e−100t mJ 2 2 The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains: 8 × 10−3 e−100t = 2 × 10−3 , e100t = 4, t = (ln 4)/100 = 13.86 ms AP 7.4 [a] This circuit is actually two RC circuits in series, and the requested voltage, vo , is the sum of the voltage drops for the two RC circuits. The circuit for t < 0 is shown below: © 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. 7–4 CHAPTER 7. Response of First-Order RL and RC Circuits Find the current in the loop and use it to find the initial voltage drops across the two RC circuits: 15 i= = 0.2 mA, v5(0− ) = 4 V, v1(0− ) = 8 V 75,000 There are two time constants in the circuit, one for each RC subcircuit. τ5 is the time constant for the 5 µF – 20 kΩ subcircuit, and τ1 is the time constant for the 1 µF – 40 kΩ subcircuit: τ5 = (20 × 103 )(5 × 10−6 ) = 100 ms; τ1 = (40 × 103 )(1 × 10−6 ) = 40 ms Therefore, v5 (t) = v5(0− )e−t/τ5 = 4e−t/0.1 = 4e−10t V, t ≥ 0 v1 (t) = v1(0− )e−t/τ1 = 8e−t/0.04 = 8e−25t V, t ≥ 0 Finally, vo (t) = v1(t) + v5(t) = [8e−25t + 4e−10t] V, t≥0 [b] Find the value of the voltage at 60 ms for each subcircuit and use the voltage to find the energy at 60 ms: v1 (60 ms) = 8e−25(0.06) ∼ v5 (60 ms) = 4e−10(0.06) ∼ = 1.79 V, = 2.20 V w1 (60 ms) = 21 Cv12(60 ms) = 12 (1 × 10−6 )(1.79)2 ∼ = 1.59 µJ w5 (60 ms) = 21 Cv52(60 ms) = 12 (5 × 10−6 )(2.20)2 ∼ = 12.05 µJ w(60 ms) = 1.59 + 12.05 = 13.64 µJ Find the initial energy from the initial voltage: w(0) = w1 (0) + w2 (0) = 21 (1 × 10−6 )(8)2 + 12 (5 × 10−6 )(4)2 = 72 µJ Now calculate the energy dissipated at 60 ms and compare it to the initial energy: wdiss = w(0) − w(60 ms) = 72 − 13.64 = 58.36 µJ % dissipated = (58.36 × 10−6 /72 × 10−6 )(100) = 81.05 % AP 7.5 [a] Use the circuit at t < 0, shown below, to calculate the initial current in the inductor: i(0− ) = 24/2 = 12 A = i(0+ ) Note that i(0− ) = i(0+ ) because the current in an inductor is continuous. [b] Use the circuit at t = 0+ , shown below, to calculate the voltage drop across the inductor at 0+ . Note that this is the same as the voltage drop across the 10 Ω resistor, which has current from two sources — 8 A from the current source and 12 A from the initial current through the inductor. © 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. Problems 7–5 v(0+ ) = −10(8 + 12) = −200 V [c] To calculate the time constant we need the equivalent resistance seen by the inductor for t > 0. Only the 10 Ω resistor is connected to the inductor for t > 0. Thus, τ = L/R = (200 × 10−3 /10) = 20 ms [d] To find i(t), we need to find the final value of the current in the inductor. When the switch has been in position a for a long time, the circuit reduces to the one below: Note that the inductor behaves as a short circuit and all of the current from the 8 A source flows through the short circuit. Thus, if = −8 A Now, i(t) = if + [i(0+ ) − if ]e−t/τ = −8 + [12 − (−8)]e−t/0.02 = −8 + 20e−50t A, t ≥ 0 [e] To find v(t), use the relationship between voltage and current for an inductor: di(t) v(t) = L = (200 × 10−3 )(−50)(20e−50t ) = −200e−50t V, t ≥ 0+ dt AP 7.6 [a] From Example 7.6, vo (t) = −60 + 90e−100t V Write a KCL equation at the top node and use it to find the relationship between vo and vA : vA − vo vA vA + 75 + + =0 8000 160,000 40,000 20vA − 20vo + vA + 4vA + 300 = 0 25vA = 20vo − 300 vA = 0.8vo − 12 Use the above equation for vA in terms of vo to find the expression for vA : vA (t) = 0.8(−60 + 90e−100t ) − 12 = −60 + 72e−100t V, t ≥ 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. 7–6 CHAPTER 7. Response of First-Order RL and RC Circuits [b] t ≥ 0+ , since there is no requirement that the voltage be continuous in a resistor. AP 7.7 [a] Use the circuit shown below, for t < 0, to calculate the initial voltage drop across the capacitor: i= 40 × 103 (10 × 10−3 ) = 3.2 mA 125 × 103 ! vc (0− ) = (3.2 × 10−3 )(25 × 103 ) = 80 V so vc (0+ ) = 80 V Now use the next circuit, valid for 0 ≤ t ≤ 10 ms, to calculate vc (t) for that interval: For 0 ≤ t ≤ 100 ms: τ = RC = (25 × 103 )(1 × 10−6 ) = 25 ms vc (t) = vc (0− )et/τ = 80e−40t V 0 ≤ t ≤ 10 ms [b] Calculate the starting capacitor voltage in the interval t ≥ 10 ms, using the capacitor voltage from the previous interval: vc (0.01) = 80e−40(0.01) = 53.63 V Now use the next circuit, valid for t ≥ 10 ms, to calculate vc (t) for that interval: For t ≥ 10 ms : Req = 25 kΩk100 kΩ = 20 kΩ τ = ReqC = (20 × 103 )(1 × 10−6 ) = 0.02 s Therefore vc (t) = vc (0.01+ )e−(t−0.01)/τ = 53.63e−50(t−0.01) V, t ≥ 0.01 s © 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. Problems 7–7 [c] To calculate the energy dissipated in the 25 kΩ resistor, integrate the power absorbed by the resistor over all time. Use the expression p = v 2/R to calculate the power absorbed by the resistor. w25 k = Z 0.01 0 [80e−40t ]2 dt + 25,000 [53.63e−50(t−0.01)]2 dt = 2.91 mJ 25,000 0.01 Z ∞ [d] Repeat the process in part (c), but recognize that the voltage across this resistor is non-zero only for the second interval: w100 kΩ = [53.63e−50(t−0.01)]2 dt = 0.29 mJ 100,000 0.01 Z ∞ We can check our answers by calculating the initial energy stored in the capacitor. All of this energy must eventually be dissipated by the 25 kΩ resistor and the 100 kΩ resistor. Check: wstored = (1/2)(1 × 10−6 )(80)2 = 3.2 mJ wdiss = 2.91 + 0.29 = 3.2 mJ AP 7.8 [a] Prior to switch a closing at t = 0, there are no sources connected to the inductor; thus, i(0− ) = 0. At the instant A is closed, i(0+ ) = 0. For 0 ≤ t ≤ 1 s, The equivalent resistance seen by the 10 V source is 2 + (3k0.8). The current leaving the 10 V source is 10 = 3.8 A 2 + (3k0.8) The final current in the inductor, which is equal to the current in the 0.8 Ω resistor is 3 IF = (3.8) = 3 A 3 + 0.8 The resistance seen by the inductor is calculated to find the time constant: L 2 [(2k3) + 0.8]k3k6 = 1 Ω τ= = = 2s R 1 Therefore, i = iF + [i(0+ ) − iF ]e−t/τ = 3 − 3e−0.5t A, 0 ≤ t ≤ 1s © 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. 7–8 CHAPTER 7. Response of First-Order RL and RC Circuits For part (b) we need the value of i(t) at t = 1 s: i(1) = 3 − 3e−0.5 = 1.18 A . [b] For t > 1 s Use current division to find the final value of the current: 9 i= (−8) = −4.8 A 9+6 The equivalent resistance seen by the inductor is used to calculate the time constant: L 2 3k(9 + 6) = 2.5 Ω τ= = = 0.8 s R 2.5 Therefore, i = iF + [i(1+ ) − iF ]e−(t−1)/τ = −4.8 + 5.98e−1.25(t−1) A, t ≥ 1s AP 7.9 0 ≤ t ≤ 32 ms: 1 vo = − RCf Z 0 32×10−3 1 −10 dt + 0 = − (−10t) RCf RCf = (200 × 103 )(0.2 × 10−6 ) = 40 × 10−3 so 32×10−3 =− 0 1 (−320 × 10−3 ) RCf 1 = 25 RCf vo = −25(−320 × 10−3 ) = 8 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. Problems 7–9 t ≥ 32 ms: 1 Zt 1 vo = − 5 dy + 8 = − (5y) −3 RCf 32×10 RCf t +8 = − 32×10−3 RCf = (250 × 103 )(0.2 × 10−6 ) = 50 × 10−3 so 1 5(t − 32 × 10−3 ) + 8 RCf 1 = 20 RCf vo = −20(5)(t − 32 × 10−3 ) + 8 = −100t + 11.2 The output will saturate at the negative power supply value: −15 = −100t + 11.2 .·. t = 262 ms AP 7.10 [a] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (0 + 2)e−t/τ τ = (160 × 103 )(10 × 10−9 ) = 10−3 ; vp = −2 + 2e−625t V; 1/τ = 625 vn = vp Write a KVL equation at the inverting input, and use it to determine vo: vn vn − vo + =0 10,000 40,000 .·. vo = 5vn = 5vp = −10 + 10e−625t V The output will saturate at the negative power supply value: −10 + 10e−625t = −5; e−625t = 1/2; t = ln 2/625 = 1.11 ms [b] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (1 + 2)e−625t = −2 + 3e−625t V The analysis for vo is the same as in part (a): vo = 5vp = −10 + 15e−625t V The output will saturate at the negative power supply value: −10 + 15e−625t = −5; e−625t = 1/3; t = ln 3/625 = 1.76 ms © 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. 7–10 CHAPTER 7. Response of First-Order RL and RC Circuits Problems P 7.1 v = 25 Ω i 1 [b] τ = = 100 ms 10 L [c] τ = = 0.1 R L = (0.1)(25) = 2.5 H [a] R = 1 1 [d] w(0) = L[i(0)]2 = (2.5)(6.4)2 = 51.2 J 2 2 Z t e−20x t = 51.2(1 − e−20t) J [e] wdiss = 1024e−20x dx = 1024 −20 0 0 % dissipated = 51.2(1 − e−20t) (100) = 100(1 − e−20t) 51.2 .·. 100(1 − e−20t) = 60 Therefore t = P 7.2 so e−20t = 0.4 1 ln 2.5 = 45.81 ms 20 [a] Note that there are several different possible solutions to this problem, and the answer to part (c) depends on the value of inductance chosen. L τ Choose a 10 mH inductor from Appendix H. Then, 0.01 R= = 10 Ω which is a resistor value from Appendix H. 0.001 R= [b] i(t) = Io e−t/τ = 10e−1000t mA, t≥0 1 1 [c] w(0) = LIo2 = (0.01)(0.01)2 = 0.5 µJ 2 2 1 w(t) = (0.01)(0.01e−1000t )2 = 0.5 × 10−6 e−2000t 2 1 So 0.5 × 10−6 e−2000t = w(0) = 0.25 × 10−6 2 e−2000t = 0.5 then e2000t = 2 ln 2 . ·. t = = 346.57 µs (for a 10 mH inductor) 2000 © 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. Problems P 7.3 [a] iL(0) = 7–11 125 = 2.5 A 50 io (0+ ) = 125 − 2.5 = 5 − 2.5 = 2.5 A 25 io (∞) = 125 = 5A 25 [b] iL = 2.5e−t/τ ; τ= L 50 × 10−3 = = 2 ms R 25 iL = 2.5e−500t A t ≥ 0+ io = 5 − iL = 5 − 2.5e−500t A, [c] 5 − 2.5e−500t = 3 2 = 2.5e−500t e500t = 1.25 P 7.4 .·. t = 446.29 µs [a] t < 0 2 kΩk6 kΩ = 1.5 kΩ Find the current from the voltage source by combining the resistors in series and parallel and using Ohm’s law: ig (0− ) = 40 = 20 mA (1500 + 500) Find the branch currents using current division: i1 (0− ) = 2000 (0.02) = 5 mA 8000 i2 (0− ) = 6000 (0.02) = 15 mA 8000 [b] The current in an inductor is continuous. Therefore, i1 (0+ ) = i1(0− ) = 5 mA i2 (0+ ) = −i1(0+ ) = −5 mA (when switch is open) © 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. 7–12 CHAPTER 7. Response of First-Order RL and RC Circuits [c] τ = L 0.4 × 10−3 = = 5 × 10−5 s; R 8 × 103 i1 (t) = i1(0+ )e−t/τ = 5e−20,000t mA, 1 = 20,000 τ t≥0 when t ≥ 0+ [d] i2(t) = −i1(t) .·. i2(t) = −5e−20,000t mA, t ≥ 0+ [e] The current in a resistor can change instantaneously. The switching operation forces i2 (0− ) to equal 15 mA and i2(0+ ) = −5 mA. P 7.5 [a] io(0− ) = 0 since the switch is open for t < 0. [b] For t = 0− the circuit is: 120 Ωk60 Ω = 40 Ω . ·. i g = 12 = 0.24 A = 240 mA 10 + 40 iL (0− ) = 120 ig = 160 mA 180 [c] For t = 0+ the circuit is: 120 Ωk40 Ω = 30 Ω . ·. i g = ia = 12 = 0.30 A = 300 mA 10 + 30 120 300 = 225 mA 160 .·. io(0+ ) = 225 − 160 = 65 mA © 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. Problems 7–13 [d] iL(0+ ) = iL(0− ) = 160 mA [e] io (∞) = ia = 225 mA [f] iL(∞) = 0, since the switch short circuits the branch containing the 20 Ω resistor and the 100 mH inductor. L 100 × 10−3 1 [g] τ = = = 5 ms; = 200 R 20 τ .·. iL = 0 + (160 − 0)e−200t = 160e−200t mA, [h] vL(0− ) = 0 t≥0 since for t < 0 the current in the inductor is constant [i] Refer to the circuit at t = 0+ and note: 20(0.16) + vL(0+ ) = 0; [j] vL(∞) = 0, .·. vL (0+ ) = −3.2 V since the current in the inductor is a constant at t = ∞. [k] vL (t) = 0 + (−3.2 − 0)e−200t = −3.2e−200t V, [l] io (t) = ia − iL = 225 − 160e−200t mA, P 7.6 t ≥ 0+ t ≥ 0+ For t < 0 ig = −48 = −6.5 A 6 + (18k1.5) iL (0− ) = 18 (−6.5) = −6 A = iL (0+ ) 18 + 1.5 For t > 0 iL (t) = iL (0+ )e−t/τ A, τ= t≥0 L 0.5 = = 0.0125 s; R 10 + 12.45 + (54k26) 1 = 80 τ © 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. 7–14 CHAPTER 7. Response of First-Order RL and RC Circuits iL (t) = −6e−80t A, io (t) = P 7.7 t≥0 54 54 (−iL(t)) = (6e−80t) = 4.05e−80t V, 80 80 t ≥ 0+ 24 = 2A 12 1.6 L = = 20 ms [b] τ = R 80 [c] i = 2e−50t A, t≥0 [a] i(0) = v1 = L d (2e−50t ) = −160e−50t V dt v2 = −72i = −144e−50t V t ≥ 0+ t≥0 1 [d] w(0) = (1.6)(2)2 = 3.2 J 2 w72Ω = Z 0 t 72(4e−100x ) dx = 288 e−100x −100 t = 2.88(1 − e−100t) J 0 w72Ω (15 ms) = 2.88(1 − e−1.5 ) = 2.24 J % dissipated = P 7.8 2.24 (100) = 69.92% 3.2 1 w(0) = (10 × 10−3 )(5)2 = 125 mJ 2 0.9w(0) = 112.5 mJ 1 w(t) = (10 × 10−3 )i(t)2, 2 i(t) = 5e−t/τ A .·. w(t) = 0.005(25e−2t/τ ) = 125e−2t/τ ) mJ w(10 µs) = 125e−20×10 −6 /τ mJ −6 .·. 125e−20×10 /τ = 112.5 so τ= 20 × 10−6 L = ln(10/9) R R= 10 × 10−3 ln(10/9) = 52.68 Ω 20 × 10−6 e20×10 −6/τ = 10 9 © 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. Problems P 7.9 7–15 1 [a] w(0) = LIg2 2 e−2t/τ to (−2/τ ) 0 0 1 1 = Ig2Rτ (1 − e−2to /τ ) = Ig2L(1 − e−2to /τ ) 2 2 to Z wdiss = Ig2Re−2t/τ dt = Ig2R wdiss = σw(0) 1 2 1 2 . ·. LIg (1 − e−2to/τ ) = σ LI 2 2 g 1 − e−2to/τ = σ; " e2to/τ = # 2to 1 = ln ; τ (1 − σ) R= [b] R = 1 (1 − σ) R(2to ) = ln[1/(1 − σ)] L L ln[1/(1 − σ)] 2to (10 × 10−3 ) ln[1/0.9] 20 × 10−6 R = 52.68 Ω P 7.10 [a] vo(t) = vo(0+ )e−t/τ .·. vo(0+ )e−10 .·. e10 −3 /τ . ·. τ = −3 /τ = 0.5vo (0+ ) =2 L 10−3 = R ln 2 10 × 10−3 · .. L= = 14.43 mH ln 2 [b] vo(0+ ) = −10iL (0+ ) = −10(1/10)(30 × 10−3 ) = −30 mV vo (t) = −0.03e−t/τ V p10Ω = vo2 = 9 × 10−5 e−2t/τ 10 w10Ω = Z τ= 10−3 0 1 1000 ln 2 9 × 10−5 e−2t/τ dt = 4.5τ × 10−5 (1 − e−2×10 . ·. −3 /τ ) w10Ω = 48.69 nJ © 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. 7–16 CHAPTER 7. Response of First-Order RL and RC Circuits 1 1 wL (0) = Li2L (0) = (14.43 × 10−3 )(3 × 10−3 )2 = 64.92 nJ 2 2 48.69 × 100 = 75% 64.92 % diss in 1 ms = P 7.11 [a] t < 0 iL (0− ) = 150 (12) = 10 A 180 t≥0 τ= 1.6 × 10−3 = 200 × 10−6 ; 8 1/τ = 5000 io = −10e−5000t A t ≥ 0 1 [b] wdel = (1.6 × 10−3 )(10)2 = 80 mJ 2 [c] 0.95wdel = 76 mJ .·. 76 × 10−3 = Z 0 to 8(100e−10,000t) dt .·. 76 × 10−3 = −80 × 10−3 e−10,000t to = 80 × 10−3 (1 − e−10,000to ) 0 .·. e−10,000to = 0.05 so to = 299.57 µs . ·. to 299.57 × 10−6 = = 1.498 τ 200 × 10−6 so to ≈ 1.498τ © 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. Problems P 7.12 7–17 t < 0: 240 = 10 A; 16 + 8 iL (0+ ) = iL(0− ) = 10 40 = 8A 50 t > 0: Re = τ= (10)(40) + 10 = 18 Ω 50 L 72 × 10−3 = = 4 ms; Re 18 1 = 250 τ .·. iL = 8e−250t A vo = 8io = 64e−250t V, P 7.13 t ≥ 0+ p40Ω = vo2 (64)2 −500t = e = 102.4e−500t W 40 40 w40Ω = Z 0 ∞ 102.4e−500t dt = 102.4 e−500t −500 ∞ = 204.8 mJ 0 1 w(0) = (72 × 10−3 )(8)2 = 2304 mJ 2 % diss = 204.8 (100) = 8.89% 2304 © 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. 7–18 P 7.14 CHAPTER 7. Response of First-Order RL and RC Circuits [a] t < 0 : iL (0) = − 72 = −2.4 A 24 + 6 t > 0: i∆ = − 100 5 iT = − iT 160 8 vT = 20i∆ + iT (100)(60) = −12.5iT + 37.5iT 160 vT = RTh = −12.5 + 37.5 = 25 Ω iT L 250 × 10−3 = R 25 τ= iL = −2.4e−100t A, 1 = 100 τ t≥0 [b] vL = 250 × 10−3 (240e−100t ) = 60e−100t V, [c] i∆ = 0.625iL = −1.5e−100t A P 7.15 t ≥ 0+ t ≥ 0+ 1 w(0) = (250 × 10−3 )(−2.4)2 = 720 mJ 2 p60Ω = 60(−1.5e−100t)2 = 135e−200t W w60Ω = Z 0 ∞ 135e−200t dt = 135 % dissipated = e−200t −200 ∞ = 675 mJ 0 675 (100) = 93.75% 720 © 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. Problems P 7.16 7–19 t<0 iL (0− ) = iL (0+ ) = 4 A t>0 Find Thévenin resistance seen by inductor: iT = 4vT ; vT 1 = RTh = = 0.25 Ω iT 4 L 5 × 10−3 τ= = = 20 ms; R 0.25 io = 4e−50t A, vo = L 1/τ = 50 t≥0 dio = (5 × 10−3 )(−200e−50t ) = −e−50t V, dt t ≥ 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. 7–20 P 7.17 CHAPTER 7. Response of First-Order RL and RC Circuits [a] t < 0 : t = 0+ : 120 = iab + 18 + 12, iab = 90 A, t = 0+ [b] At t = ∞: iab = 240/2 = 120 A, [c] i1 (0) = 18, i2 (0) = 12, τ1 = t=∞ 2 × 10−3 = 0.2 ms 10 τ2 = 6 × 10−3 = 0.4 ms 15 i1 (t) = 18e−5000t A, t≥0 i2 (t) = 12e−2500t A, t≥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. Problems iab = 120 − 18e−5000t − 12e−2500t A, 7–21 t≥0 120 − 18e−5000t − 12e−2500t = 114 6 = 18e−5000t + 12e−2500t x = e−2500t Let P 7.18 so Solving x = 1 = e−2500t 3 .·. e2500t = 3 and 6 = 18x2 + 12x t= ln 3 = 439.44 µs 2500 [a] t < 0 1 kΩk4 kΩ = 0.8 kΩ 20 kΩk80 kΩ = 16 kΩ (105 × 10−3 )(0.8 × 103 ) = 84 V iL (0− ) = 84 = 5 mA 16,800 t>0 τ= L 6 = × 10−3 = 250 µs; R 24 1 = 4000 τ © 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. 7–22 CHAPTER 7. Response of First-Order RL and RC Circuits iL (t) = 5e−4000t mA, t≥0 p4k = 25 × 10−6 e−8000t(4000) = 0.10e−8000t W wdiss = t Z 0.10e−8000x dx = 12.5 × 10−6 [1 − e−8000t] J 0 1 w(0) = (6)(25 × 10−6 ) = 75 µJ 2 0.10w(0) = 7.5 µJ .·. e8000t = 2.5 12.5(1 − e−8000t) = 7.5; t= ln 2.5 = 114.54 µs 8000 [b] wdiss(total) = 75(1 − e−8000t) µJ wdiss(114.54 µs) = 45 µJ % = (45/75)(100) = 60% P 7.19 [a] t > 0: Leq = 1.25 + 60 = 5H 16 iL (t) = iL (0)e−t/τ mA; iL (t) = 2e−1500t A, iL (0) = 2 A; 1 R 7500 = = = 1500 τ L 5 t≥0 vR (t) = RiL (t) = (7500)(2e−1500t ) = 15,000e−1500t V, vo = −3.75 [b] io = P 7.20 −1 6 Z 0 diL = 11,250e−1500t V, dt t ≥ 0+ t ≥ 0+ t 11,250e−1500x dx + 0 = 1.25e−1500t − 1.25 A [a] From the solution to Problem 7.19, 1 1 w(0) = Leq[iL (0)]2 = (5)(2)2 = 10 J 2 2 1 1 [b] wtrapped = (10)(1.25)2 + (6)(1.25)2 = 12.5 J 2 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. Problems P 7.21 v = 8 kΩ i 1 1 1 = = 500; C= = 0.25 µF τ RC (500)(8000) 1 τ= = 2 ms 500 1 w(0) = (0.25 × 10−6 )(72)2 = 648 µJ 2 Z to (72)2 e−1000t wdiss = dt (800) 0 [a] R = [b] [c] [d] [e] = 0.648 e−1000t −1000 to = 648(1 − e−1000to ) µJ 0 %diss = 100(1 − e−1000to ) = 68 . ·. t = P 7.22 so e1000to = 3.125 ln 3.125 = 1139 µs 1000 [a] Note that there are many different possible correct solutions to this problem. τ R= C Choose a 100 µF capacitor from Appendix H. Then, 0.05 R= = 500 Ω 100 × 10−6 Construct a 500 Ω resistor by combining two 1 kΩ resistors in parallel: [b] v(t) = Vo e−t/τ = 50e−20t V, [c] 50e−20t = 10 . ·. t = P 7.23 7–23 so t≥0 e20t = 5 ln 5 = 80.47 ms 20 [a] v1(0− ) = v1 (0+ ) = 40 V v2(0+ ) = 0 Ceq = (1)(4)/5 = 0.8 µF τ = (25 × 103 )(0.8 × 10−6 ) = 20ms; 1 = 50 τ © 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. 7–24 CHAPTER 7. Response of First-Order RL and RC Circuits i= 40 −50t e = 1.6e−50t mA, 25,000 t ≥ 0+ −1 Z t 1.6 × 10−3 e−50x dx + 40 = 32e−50t + 8 V, t≥0 10−6 0 Z t 1 v2 = 1.6 × 10−3 e−50x dx + 0 = −8e−50t + 8 V, t≥0 −6 4 × 10 0 v1 = 1 [b] w(0) = (10−6 )(40)2 = 800 µJ 2 1 1 [c] wtrapped = (10−6 )(8)2 + (4 × 10−6 )(8)2 = 160 µJ. 2 2 The energy dissipated by the 25 kΩ resistor is equal to the energy dissipated by the two capacitors; it is easier to calculate the energy dissipated by the capacitors: 1 wdiss = (0.8 × 10−6 )(40)2 = 640 µJ. 2 Check: wtrapped + wdiss = 160 + 640 = 800 µJ; w(0) = 800 µJ. P 7.24 [a] t < 0: i1 (0− ) = i2(0− ) = 3 = 100 mA 30 [b] t > 0: i1 (0+ ) = 0.2 = 100 mA 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. Problems i2 (0+ ) = 7–25 −0.2 = −25 mA 8 [c] Capacitor voltage cannot change instantaneously, therefore, i1 (0− ) = i1(0+ ) = 100 mA [d] Switching can cause an instantaneous change in the current in a resistive branch. In this circuit i2 (0− ) = 100 mA and i2(0+ ) = 25 mA [e] vc = 0.2e−t/τ V, t≥0 τ = Re C = 1.6(2 × 10−6 ) = 3.2 µs; vc = 0.2e−312,000t V, [f] i2 = P 7.25 t≥0 vc = 0.1e−312,000t A, 2 i1 = 1 = 312,500 τ t≥0 −vc = −25e−312,000t mA, 8 t ≥ 0+ [a] t < 0: Req = 12 kk8 k = 10.2 kΩ vo (0) = 10,200 (−120) = −102 V 10,200 + 1800 t > 0: τ = [(10/3) × 10−6 )(12,000) = 40 ms; vo = −102e−25t V, p= 1 = 25 τ t≥0 vo2 = 867 × 10−3 e−50t W 12,000 © 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. 7–26 CHAPTER 7. Response of First-Order RL and RC Circuits wdiss = 12×10−3 Z 0 867 × 10−3 e−50t dt = 17.34 × 10−3 (1 − e−50(12×10 1 [b] w(0) = 2 −3) ) = 7824 µJ 10 (102)2 × 10−6 = 17.34 mJ 3 0.75w(0) = 13 mJ Z 0 to 867 × 10−3 e−50x dx = 13 × 10−3 .·. 1 − e−50to = 0.75; P 7.26 e50to = 4; so to = 27.73 ms [a] t < 0: io (0− ) = 6000 (40 m) = 24 mA 6000 + 4000 vo (0− ) = (3000)(24 m) = 72 V i2 (0− ) = 40 − 24 = 16 mA v2 (0− ) = (6000)(16 m) = 96 V t>0 τ = RC = (1000)(0.2 × 10−6 ) = 200 µs; io (t) = 24 e−t/τ = 24e−5000t mA, 1 × 103 1 = 5000 τ t ≥ 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. Problems 7–27 [b] Z t 1 24 × 10−3 e−5000x dx + 72 0.6 × 10−6 0 e−5000x t = (40,000) +72 −5000 0 = −8e−5000t + 8 + 72 vo = [−8e−5000t + 80] V, t≥0 vo = [c] wtrapped = (1/2)(0.3 × 10−6 )(80)2 + (1/2)(0.6 × 10−6 )(80)2 wtrapped = 2880 µJ. Check: 1 wdiss = (0.2 × 10−6 )(24)2 = 57.6 µJ 2 1 1 w(0) = (0.3 × 10−6 )(96)2 + (0.6 × 10−6 )(72)2 = 2937.6 µJ. 2 2 wtrapped + wdiss = w(0) 2880 + 57.6 = 2937.6 P 7.27 OK. [a] At t = 0− the voltage on each capacitor will be 150 V(5 × 30), positive at the upper terminal. Hence at t ≥ 0+ we have .·. isd (0+ ) = 5 + 150 150 + = 1055 A 0.2 0.5 At t = ∞, both capacitors will have completely discharged. .·. isd (∞) = 5 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. 7–28 CHAPTER 7. Response of First-Order RL and RC Circuits [b] isd (t) = 5 + i1 (t) + i2 (t) τ1 = 0.2(10−6 ) = 0.2 µs τ2 = 0.5(100 × 10−6 ) = 50 µs .·. i1(t) = 750e−5×10 t A, 6 i2(t) = 300e−20,000t A, t ≥ 0+ t≥0 .·. isd = 5 + 750e−5×10 t + 300e−20,000t A, 6 P 7.28 t ≥ 0+ [a] vT = 20 × 103 (iT + αv∆ ) + 5 × 103 iT v∆ = 5 × 103 iT vT = 25 × 103 iT + 20 × 103 α(5 × 103 iT ) RTh = 25,000 + 100 × 106 α τ = RTh C = 40 × 10−3 = RTh (0.8 × 10−6 ) RTh = 50 kΩ = 25,000 + 100 × 106 α α= 25,000 = 2.5 × 10−4 A/V 6 100 × 10 [b] vo(0) = (−5 × 10−3 )(3600) = −18 V t > 0: vo = −18e−25t V, t<0 t≥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. Problems 7–29 v∆ v∆ − vo + + 2.5 × 10−4 v∆ = 0 5000 20,000 4v∆ + v∆ − vo + 5v∆ = 0 vo = −1.8e−25t V, 10 . ·. v ∆ = P 7.29 t ≥ 0+ [a] pds = (16.2e−25t )(−450 × 10−6 e−25t ) = −7290 × 10−6 e−50t W wds = Z ∞ pds dt = −145.8 µJ. 0 .·. dependent source is delivering 145.8 µJ. [b] w5k = Z ∞ (5000)(0.36 × 10−3 e−25t)2 dt = 648 × 10−6 0 w20k = Z 0 ∞ (16.2e−25t )2 dt = 13,122 × 10−6 20,000 Z 0 Z 0 ∞ e−50t dt = 12.96 µJ ∞ e−50t dt = 262.44 µJ 1 wc (0) = (0.8 × 10−6 )(18)2 = 129.6 µJ 2 X X wdiss = 12.96 + 262.44 = 275.4 µJ wdev = 145.8 + 129.6 = 275.4 µJ. © 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. 7–30 P 7.30 CHAPTER 7. Response of First-Order RL and RC Circuits t<0 t>0 .·. vT = −5io − 15io = −20io = 20iT τ = RC = 40 µs; vo = 15e−25,000t V, io = − P 7.31 RTh = vT = 20 Ω iT 1 = 25,000 τ t≥0 vo = −0.75e−25,000t A, 20 t ≥ 0+ [a] The equivalent circuit for t > 0: τ = 2 ms; 1/τ = 500 © 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. Problems vo = 10e−500t V, t≥0 io = e−500t mA, t ≥ 0+ i24kΩ = e−500t 16 = 0.4e−500t mA, 40 7–31 t ≥ 0+ p24kΩ = (0.16 × 10−6 e−1000t)(24,000) = 3.84e−1000t mW w24kΩ = ∞ Z 3.84 × 10−3 e−1000t dt = −3.84 × 10−6 (0 − 1) = 3.84 µJ 0 1 1 w(0) = (0.25 × 10−6 )(40)2 + (1 × 10−6 )(50)2 = 1.45 mJ 2 2 % diss (24 kΩ) = 3.84 × 10−6 × 100 = 0.26% 1.45 × 10−3 [b] p400Ω = 400(1 × 10−3 e−500t)2 = 0.4 × 10−3 e−1000t w400Ω = ∞ Z p400 dt = 0.40 µJ 0 % diss (400 Ω) = i16kΩ = e −500t 0.4 × 10−6 × 100 = 0.03% 1.45 × 10−3 24 = 0.6e−500t mA, 40 t ≥ 0+ p16kΩ = (0.6 × 10−3 e−500t)2 (16,000) = 5.76 × 10−3 e−1000t W w16kΩ = Z 0 ∞ 5.76 × 10−3 e−1000t dt = 5.76 µJ % diss (16 kΩ) = 0.4% [c] X wdiss = 3.84 + 5.76 + 0.4 = 10 µJ wtrapped = w(0) − % trapped = X wdiss = 1.45 × 10−3 − 10 × 10−6 = 1.44 mJ 1.44 × 100 = 99.31% 1.45 Check: 0.26 + 0.03 + 0.4 + 99.31 = 100% P 7.32 [a] Ce = (2 + 1)6 = 2 µF 2+1+6 vo (0) = −5 + 30 = 25 V τ = (2 × 10−6 )(250 × 103 ) = 0.5 s; 1 =2 τ © 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. 7–32 CHAPTER 7. Response of First-Order RL and RC Circuits t ≥ 0+ vo = 25e−2t V, 1 1 [b] wo = (3 × 10−6 )(30)2 + (6 × 10−6 )(5)2 = 1425 µJ 2 2 1 wdiss = (2 × 10−6 )(25)2 = 625 µJ 2 625 × 100 = 43.86% % diss = 1425 vo [c] io = = 100e−2t µA 250 × 10−3 1 6 × 10−6 e−2x = −16.67 −2 Z v1 = − t 0 t Z 100 × 10−6 e−2x dx − 5 = −16.67 −5 = 8.33e−2t − 13.33 V t 0 e−2x dx − 5 t≥0 0 [d] v1 + v2 = vo v2 = vo − v1 = 25e−2t − 8.33e−2t + 13.33 = 16.67e−2t + 13.33 V t ≥ 0 1 1 [e] wtrapped = (6 × 10−6 )(13.33)2 + (3 × 10−6 )(13.33)2 = 800 µJ 2 2 wdiss + wtrapped = 625 + 800 = 1425 µJ (check) P 7.33 [a] From Eqs. (7.35) and (7.42) Vs Vs −(R/L)t i= + Io − e R R v = (Vs − Io R)e−(R/L)t . ·. Vs = 4; R Io − Vs − Io R = −80; Vs =4 R R = 40 L Vs = 8A R Now since Vs = 4R we have . ·. I o = 4 + 4R − 8R = −80; Vs = 80 V; L= R = 20 Ω R = 0.5 H 40 © 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. Problems [b] i = 4 + 4e−40t ; 7–33 i2 = 16 + 32e−40t + 16e−80t 1 1 w = Li2 = (0.5)[16 + 32e−40t + 16e−80t] = 4 + 8e−40t + 4e−80t 2 2 .·. 4 + 8e−40t + 4e−80t = 9 or e−80t + 2e−40t − 1.25 = 0 Let x = e−40t: x2 + 2x − 1.25 = 0; Solving, x = 0.5; x = −2.5 But x ≥ 0 for all t. Thus, e−40t = 0.5; P 7.34 e40t = 2; t = 25 ln 2 = 17.33 ms [a] Note that there are many different possible solutions to this problem. L τ Choose a 1 mH inductor from Appendix H. Then, R= 0.001 = 125 Ω 8 × 10−6 Construct the resistance needed by combining 100 Ω, 10 Ω, and 15 Ω resistors in series: R= [b] i(t) = If + (Io − If )e−t/τ Io = 0 A; If = 25 Vf = = 200 mA R 125 .·. i(t) = 200 + (0 − 200)e−125,000t mA = 200 − 200e−125,000t mA, t≥0 [c] i(t) = 0.2 − 0.2e−125,000t = (0.75)(0.2) = 0.15 e−125,000t = 0.25 . ·. t = so e125,000t = 4 ln 4 = 11.09 µs 125,000 © 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. 7–34 P 7.35 CHAPTER 7. Response of First-Order RL and RC Circuits [a] t < 0 iL (0− ) = −5 A t>0 iL (∞) = τ= 40 − 80 = −2 A 4 + 16 L 4 × 10−3 = = 200 µs; R 4 + 16 1 = 5000 τ iL = iL(∞) + [iL(0+ ) − iL (∞)]e−t/τ = −2 + (−5 + 2)e−5000t = −2 − 3e−5000t A, t≥0 vo = 16iL + 80 = 16(−2 − 3e−5000t) + 80 = 48 − 48e−5000t V, [b] vL = L diL = 4 × 10−3 (−5000)[−3e−5000t] = 60e−5000t V, dt t≥0 t ≥ 0+ vL (0+ ) = 60 V From part (a) vo (0+ ) = 0 V Check: at t = 0+ the circuit is: vL (0+ ) = 40 + (5 A)(4 Ω) = 60 V, vo (0+ ) = 80 − (16 Ω)(5 A) = 0 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. Problems P 7.36 7–35 [a] For t < 0, calculate the Thévenin equivalent for the circuit to the left and right of the 75 mH inductor. We get i(0− ) = 5 − 120 = −5 mA 15 k + 8 k i(0− ) = i(0+ ) = −5 mA [b] For t > 0, the circuit reduces to Therefore i(∞) = 5/15,000 = 0.333 mA L 75 × 10−3 = = 5 µs [c] τ = R 15,000 [d] i(t) = i(∞) + [i(0+ ) − i(∞)]e−t/τ = 0.333 + [−5 − 0.333]e−200,000t = 0.333 − 5.333e−200,000t mA, P 7.37 t≥0 [a] t < 0 KVL equation at the top node: vo vo vo 50 = + + 8 40 10 Multiply by 40 and solve: 2000 = (5 + 1 + 4)vo ; .·. io(0− ) = vo = 200 V vo = 200/10 = 20 A 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. 7–36 CHAPTER 7. Response of First-Order RL and RC Circuits t>0 Use voltage division to find the Thévenin voltage: 40 (800) = 200 V VTh = vo = 40 + 120 Remove the voltage source and make series and parallel combinations of resistors to find the equivalent resistance: RTh = 10 + 120k40 = 10 + 30 = 40 Ω The simplified circuit is: L 40 × 10−3 1 = = 1 ms; = 1000 R 40 τ 200 io (∞) = = 5A 40 .·. io = io (∞) + [io(0+ ) − io(∞)]e−t/τ τ= = 5 + (20 − 5)e−1000t = 5 + 15e−1000t A, [b] vo vo P 7.38 t≥0 = dio dt 10(5 + 15e−1000t) + 0.04(−1000)(15e−1000t ) = 50 + 150e−1000t − 600e−1000t = 50 − 450e−1000t V, = 10io + L t ≥ 0+ [a] − Vs v 1 + + R R L Z t 0 v dt + Io = 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. Problems 7–37 Differentiating both sides, 1 dv 1 + v=0 R dt L dv R + v=0 dt L . ·. [b] R dv =− v dt L R dv dt = − v dt dt L R dv = − v dt L so dv R = − dt v L v(t) dx Z Vo ln . ·. P 7.39 x =− RZ t dy L 0 v(t) R =− t Vo L v(t) = Vo e−(R/L)t = (Vs − RIo )e−(R/L)t [a] vo(0+ ) = −Ig R2 ; τ= L R1 + R2 vo (∞) = 0 vo (t) = −Ig R2 e−[(R1 +R2 )/L]t V, t ≥ 0+ [b] vo(0+ ) → ∞, and the duration of vo(t) → zero L [c] vsw = R2 io ; τ= R1 + R2 io (0+ ) = Ig ; Therefore Therefore [d] |vsw (0+ )| → ∞; P 7.40 io (∞) = Ig R1 R1 + R2 io (t) = Ig R1 R1 +R2 io (t) = R1 Ig (R1 +R2 ) vsw = h + Ig − + R1Ig (1+R1 /R2) Ig R1 R1 +R2 i e−[(R1+R2 )/L]t R2Ig e−[(R1 +R2 )/L]t (R1+R2 ) + R2 Ig e−[(R1+R2 )/L]t, (1+R1 /R2) t ≥ 0+ duration → 0 Opening the inductive circuit causes a very large voltage to be induced across the inductor L. This voltage also appears across the switch (part [d] of Problem 7.39), causing the switch to arc over. At the same time, the large voltage across L damages the meter movement. © 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. 7–38 P 7.41 CHAPTER 7. Response of First-Order RL and RC Circuits t > 0; calculate vo (0+ ) va va − vo (0+ ) + = 20 × 10−3 15 5 .·. va = 0.75vo (0+ ) + 75 × 10−3 15 × 10−3 + vo(0+ ) − va vo (0+ ) + − 9i∆ + 50 × 10−3 = 0 5 8 13vo (0+ ) − 8va − 360i∆ = −2600 × 10−3 i∆ = vo(0+ ) − 9i∆ + 50 × 10−3 8 vo (0+ ) · . . i∆ = + 5 × 10−3 80 .·. 360i∆ = 4.5vo (0+ ) + 1800 × 10−3 8va = 6vo (0+ ) + 600 × 10−3 .·. 13vo (0+ ) − 6vo (0+ ) − 600 × 10−3 − 4.5vo (0+ )− 1800 × 10−3 = −2600 × 10−3 2.5vo (0+ ) = −200 × 10−3 ; vo(0+ ) = −80 mV vo (∞) = 0 Find the Thévenin resistance seen by the 4 mH inductor: iT = vT vT + − 9i∆ 20 8 © 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. Problems vT .·. 10i∆ = ; 8 i∆ = vT − 9i∆ 8 iT = vT 10vT 9vT + − 20 80 80 i∆ = 7–39 vT 80 iT 1 1 5 1 = + = = S vT 20 80 80 16 .·. RTh = 16Ω τ= 4 × 10−3 = 0.25 ms; 16 1/τ = 4000 .·. vo = 0 + (−80 − 0)e−4000t = −80e−4000t mV, P 7.42 t ≥ 0+ For t < 0 vx vx − 480 − 0.8vφ + =0 15 21 vφ = vx − 480 21 vx vx − 480 vx − 480 − 0.8 + 15 21 21 = .·. vx vs − 480 + 0.2 15 21 = 21vx + 3(vx − 480) = 0 24vx = 1440 so vx = 60 V io (0− ) = vx = 4A 15 t>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. 7–40 CHAPTER 7. Response of First-Order RL and RC Circuits Find Thévenin equivalent with respect to a, b VTh − 320 VTh − 320 − 0.8 5 5 =0 vT = (iT + 0.8vφ )(5) = iT + 0.8 vT = 5iT + 0.8vT vT 5 VTh = 320 V (5) .·. 0.2vT = 5iT vT = RTh = 25 Ω iT io (∞) = 320/40 = 8 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. Problems τ= 80 × 10−3 = 2 ms; 40 1/τ = 500 io = 8 + (4 − 8)e−500t = 8 − 4e−500t A, P 7.43 7–41 t≥0 For t < 0, i80mH (0) = 50 V/10 Ω = 5 A For t > 0, after making a Thévenin equivalent we have Vs Vs −t/τ + Io − e i= R R 1 R 8 = = = 80 τ L 100 × 10−3 Io = 5 A; If = Vs −80 = = −10 A R 8 i = −10 + (5 + 10)e−80t = −10 + 15e−80t A, vo = 0.08 P 7.44 di = 0.08(−1200e−80t ) = −96e−80t V, dt t≥0 t ≥ 0+ [a] Let v be the voltage drop across the parallel branches, positive at the top node, then −Ig + v 1 + Rg L1 Z v 1 1 + + Rg L1 L2 v 1 + Rg Le Z 0 0 t v dx + Z 0 1 L2 Z 0 t v dx = 0 t v dx = Ig t v dx = Ig 1 dv v + =0 Rg dt Le © 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. 7–42 CHAPTER 7. Response of First-Order RL and RC Circuits dv Rg + v=0 dt Le Therefore v = Ig Rg e−t/τ ; τ = Le /Rg Thus Z 1 t Ig Rg e−x/τ t Ig Le i1 = Ig Rg e−x/τ dx = = (1 − e−t/τ ) L1 0 L1 (−1/τ ) 0 L1 i1 = Ig L1 Ig L2 (1 − e−t/τ ) and i2 = (1 − e−t/τ ) L1 + L2 L1 + L2 [b] i1(∞) = P 7.45 L2 Ig ; L1 + L2 i2(∞) = L1 Ig L1 + L2 [a] t < 0 t>0 iL (0− ) = iL (0+ ) = 25 mA; τ= 24 × 10−3 = 0.2 ms; 120 1 = 5000 τ iL (∞) = −50 mA iL = −50 + (25 + 50)e−5000t = −50 + 75e−5000t mA, vo = −120[75 × 10−3 e−5000t] = −9e−5000t V, t≥0 t ≥ 0+ t 1 [b] i1 = −9e−5000x dx + 10 × 10−3 = (30e−5000t − 20) mA, 60 × 10−3 0 Z t 1 [c] i2 = −9e−5000x dx + 15 × 10−3 = (45e−5000t − 30) mA, 40 × 10−3 0 Z P 7.46 t≥0 t≥0 t>0 τ= 1 40 © 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. Problems io = 5e−40t A, 7–43 t≥0 t > 0+ vo = 40io = 200e−40t V, 200e−40t = 100; e40t = 2 1 .·. t = ln 2 = 17.33 ms 40 P 7.47 1 1 [a] wdiss = Le i2 (0) = (1)(5)2 = 12.5 J 2 2 Z 1 t [b] i3H = (200)e−40x dx − 5 3 0 = 1.67(1 − e−40t) − 5 = −1.67e−40t − 3.33 A i1.5H = 1 Zt (200)e−40x dx + 0 1.5 0 = −3.33e−40t + 3.33 A 1 wtrapped = (4.5)(3.33)2 = 25 J 2 1 [c] w(0) = (3)(5)2 = 37.5 J 2 P 7.48 [a] v = Is R + (Vo − Is R)e−t/RC .·. Is R = 40, i = Is − Vo −t/RC e R Vo − Is R = −24 .·. Vo = 16 V Is − Vo = 3 × 10−3 ; R Is − .·. Is − 0.4Is = 3 × 10−3 ; R= 16 = 3 × 10−3 ; R R= 40 Is Is = 5 mA 40 × 103 = 8 kΩ 5 1 = 2500; RC C= 1 10−3 = = 50 nF; 2500R 20 × 103 τ = RC = 1 = 400 µs 2500 [b] v(∞) = 40 V 1 w(∞) = (50 × 10−9 )(1600) = 40 µJ 2 0.81w(∞) = 32.4 µJ v 2(to ) = 32.4 × 10−6 = 1296; 25 × 10−9 40 − 24e−2500to = 36; v(to) = 36 V e2500to = 6; .·. to = 716.70 µs © 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. 7–44 P 7.49 CHAPTER 7. Response of First-Order RL and RC Circuits [a] Note that there are many different possible solutions to this problem. τ R= C Choose a 10 µH capacitor from Appendix H. Then, 0.25 = 25 kΩ 10 × 10−6 Construct the resistance needed by combining 10 kΩ and 15 kΩ resistors in series: R= [b] v(t) = Vf + (Vo − Vf )e−t/τ Vo = 100 V; Vf = (If )(R) = (1 × 10−3 )(25 × 103 ) = 25 V .·. v(t) = 25 + (100 − 25)e−4t V = 25 + 75e−4t V, [c] v(t) = 25 + 75e−4t = 50 . ·. t = P 7.50 so e−4t = t≥0 1 3 ln 3 = 274.65 ms 4 [a] io (0+ ) = −36 = −7.2 mA 5000 [b] io(∞) = 0 [c] τ = RC = (5000)(0.8 × 10−6 ) = 4 ms [d] io = 0 + (−7.2)e−250t = −7.2e−250t mA, t ≥ 0+ [e] vo = −[36 + 1800(−7.2 × 10−3 e−250t)] = −36 + 12.96e−250t V, t ≥ 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. Problems P 7.51 7–45 [a] Simplify the circuit for t > 0 using source transformation: Since there is no source connected to the capacitor for t < 0 vo (0− ) = vo (0+ ) = 0 V From the simplified circuit, vo (∞) = 60 V τ = RC = (20 × 103 )(0.5 × 10−6 ) = 10 ms 1/τ = 100 vo = vo(∞) + [vo(0+ ) − vo (∞)]e−t/τ = (60 − 60e−100t ) V, t≥0 dvo dt ic = 0.5 × 10−6 (−100)(−60e−100t ) = 3e−100t mA [b] ic = C v1 = 8000ic + vo = (8000)(3 × 10−3 )e−100t + (60 − 60e−100t ) = 60 − 36e−100t V v1 io = = 1 − 0.6e−100t mA, t ≥ 0+ 3 60 × 10 [c] i1 (t) = io + ic = 1 + 2.4e−100t mA, t ≥ 0+ v1 [d] i2(t) = = 4 − 2.4e−100t mA, t ≥ 0+ 15 × 103 [e] i1(0+ ) = 1 + 2.4 = 3.4 mA At t = 0+ : Re = 15 kk60 kk8 k = 4800 Ω v1 (0+ ) = (5 × 10−3 )(4800) = 24 V i1 (0+ ) = P 7.52 v1 (0+ ) v1(0+ ) + = 0.4 m + 3 m = 3.4 mA (checks) 60,000 8000 [a] vo(0− ) = vo (0+ ) = 120 V vo (∞) = −150 V; τ = 2 ms; 1 = 500 τ © 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. 7–46 CHAPTER 7. Response of First-Order RL and RC Circuits vo = −150 + (120 − (−150))e−500t vo = −150 + 270e−500t V, t≥0 [b] io = −0.04 × 10−6 (−500)[270e−500t ] = 5.4e−500t mA, t ≥ 0+ [c] vg = vo − 12.5 × 103 io = −150 + 202.5e−500t V [d] vg (0+ ) = −150 + 202.5 = 52.5 V Checks: vg (0+ ) = io (0+ )[37.5 × 103 ] − 150 = 202.5 − 150 = 52.5 V vg = −3 + 4.05e−500t mA 50k vg = −1 + 1.35e−500t mA = 150k i50k = i150k -io + i50k + i150k + 4 = 0 P 7.53 (ok) For t < 0 Simplify the circuit: 80/10,000 = 8 mA, 10 kΩk40 kΩk24 kΩ = 6 kΩ 8 mA − 3 mA = 5 mA 5 mA × 6 kΩ = 30 V Thus, for t < 0 .·. vo (0− ) = vo(0+ ) = 30 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. Problems 7–47 t>0 Simplify the circuit: 8 mA + 2 mA = 10 mA 10 kk40 kk24 k = 6 kΩ (10 mA)(6 kΩ) = 60 V Thus, for t > 0 vo (∞) = −10 × 10−3 (6 × 103 ) = −60 V τ = RC = (10 k)(0.05 µ) = 0.5 ms; 1 = 2000 τ vo = vo (∞) + [vo(0+ ) − vo(∞)]e−t/τ = −60 + [30 − (−60)]e−2000t = −60 + 90e−2000t V P 7.54 t≥0 t < 0: io (0− ) = 20 (10 × 10−3 ) = 2 mA; 100 vo(0− ) = (2 × 10−3 )(50,000) = 100 V t = ∞: io (∞) = −5 × 10 −3 20 = −1 mA; 100 vo (∞) = io (∞)(50,000) = −50 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. 7–48 CHAPTER 7. Response of First-Order RL and RC Circuits RTh = 50 kΩk50 kΩ = 25 kΩ; C = 16 nF 1 = 2500 τ τ = (25,000)(16 × 10−9 ) = 0.4 ms; .·. vo (t) = −50 + 150e−2500t V, t≥0 ic = C dvo = −6e−2500t mA, dt i50k = vo = −1 + 3e−2500t mA, 50,000 t ≥ 0+ io = ic + i50k = −(1 + 3e−2500t) mA, P 7.55 t ≥ 0+ t ≥ 0+ [a] vc (0+ ) = 50 V [b] Use voltage division to find the final value of voltage: vc (∞) = 20 (−30) = −24 V 20 + 5 [c] Find the Thévenin equivalent with respect to the terminals of the capacitor: VTh = −24 V, RTh = 20k5 = 4 Ω, Therefore τ = ReqC = 4(25 × 10−9 ) = 0.1 µs The simplified circuit for t > 0 is: [d] i(0+ ) = −24 − 50 = −18.5 A 4 © 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. Problems 7–49 [e] vc = vc (∞) + [vc (0+ ) − vc(∞)]e−t/τ 7 = −24 + [50 − (−24)]e−t/τ = −24 + 74e−10 t V, [f] i = C P 7.56 t≥0 dvc 7 7 = (25 × 10−9 )(−107 )(74e−10 t ) = −18.5e−10 t A, dt t ≥ 0+ [a] Use voltage division to find the initial value of the voltage: vc (0+ ) = v9k = 9k (120) = 90 V 9k + 3k [b] Use Ohm’s law to find the final value of voltage: vc (∞) = v40k = −(1.5 × 10−3 )(40 × 103 ) = −60 V [c] Find the Thévenin equivalent with respect to the terminals of the capacitor: VTh = −60 V, RTh = 10 k + 40 k = 50 kΩ τ = RTh C = 1 ms = 1000 µs [d] vc = vc (∞) + [vc(0+ ) − vc (∞)]e−t/τ = −60 + (90 + 60)e−1000t = −60 + 150e−1000t V, t≥0 We want vc = −60 + 150e−1000t = 0: Therefore t = P 7.57 ln(150/60) = 916.3 µs 1000 Use voltage division to find the initial voltage: vo (0) = 60 (50) = 30 V 40 + 60 Use Ohm’s law to find the final value of voltage: vo (∞) = (−5 mA)(20 kΩ) = −100 V τ = RC = (20 × 103 )(250 × 10−9 ) = 5 ms; 1 = 200 τ vo = vo (∞) + [vo(0+ ) − vo(∞)]e−t/τ = −100 + (30 + 100)e−200t = −100 + 130e−200t V, t≥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. 7–50 P 7.58 CHAPTER 7. Response of First-Order RL and RC Circuits For t < 0, t > 0: vo (0) = 80 V 3 3 vTh = 30 × 10 i∆ + 0.8(100) = 30 × 10 −100 + 80 = 50 V 100 × 103 vT = 30 × 103 i∆ + 16 × 103 iT = 30 × 103 (0.8)iT + 16 × 103 iT = 40 × 103 iT RTh = vT = 40 kΩ iT t>0 vo = 50 + (80 − 50)e−t/τ τ = RC = (40 × 103 )(5 × 10−9 ) = 200 × 10−6 ; vo = 50 + 30e−5000t V, P 7.59 vo (0) = 50 V; 1 = 5000 τ t≥0 vo(∞) = 80 V RTh = 16 kΩ τ = (16)(5 × 10−6 ) = 80 × 10−6 ; 1 = 12,500 τ v = 80 + (50 − 80)e−12,500t = 80 − 30e−12,500t V, t≥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. Problems P 7.60 7–51 For t > 0 VTh = (−25)(16,000)ib = −400 × 103 ib ib = 33,000 (120 × 10−6 ) = 49.5 µA 80,000 VTh = −400 × 103 (49.5 × 10−6 ) = −19.8 V RTh = 16 kΩ vo(0+ ) = 0 vo (∞) = −19.8 V; τ = (16, 000)(0.25 × 10−6 ) = 4 ms; vo = −19.8 + 19.8e−250t V, 1/τ = 250 t≥0 1 w(t) = (0.25 × 10−6 )vo2 = w(∞)(1 − e−250t)2 J 2 (1 − e−250t)2 = 0.36w(∞) = 0.36 w(∞) 1 − e−250t = 0.6 e−250t = 0.4 P 7.61 . ·. t = 3.67 ms [a] 1 t Is R = Ri + i dx + Vo C 0+ di i 0=R + +0 dt C di i . ·. + =0 dt RC Z © 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. 7–52 CHAPTER 7. Response of First-Order RL and RC Circuits [b] di i =− ; dt RC Z i(t) i(0+ ) ln di dt =− i RC dy 1 =− y RC t Z 0+ dx i(t) −t = + i(0 ) RC i(t) = i(0+ )e−t/RC ; i(0+ ) = Vo Is R − Vo = Is − R R Vo −t/RC .·. i(t) = Is − e R P 7.62 [a] Let i be the current Zin the clockwise direction around the circuit. Then Z 1 t 1 t Vg = iRg + i dx + i dx C1 0 C2 0 1 1 iRg + + C1 C2 = Z 1 i dx = iRg + Ce 0 t Z t 0 i dx Now differentiate the equation 0 = Rg di i + dt Ce Therefore i = 1 v1 (t) = C1 0 t di 1 + i=0 dt Rg Ce Vg −t/Rg Ce Vg −t/τ e = e ; Rg Rg Vg −x/τ Vg e−x/τ e dx = Rg Rg C1 −1/τ τ = Rg Ce t =− 0 v1 (t) = Vg C2 (1 − e−t/τ ); C1 + C2 τ = Rg Ce v2 (t) = Vg C1 (1 − e−t/τ ); C1 + C2 τ = Rg Ce [b] v1(∞) = P 7.63 Z or C2 Vg ; C1 + C2 v2 (∞) = Vg Ce −t/τ (e − 1) C1 C1 Vg C1 + C2 [a] For t > 0: τ = RC = 250 × 103 × 8 × 10−9 = 2 ms; 1 = 500 τ © 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. Problems 7–53 t ≥ 0+ vo = 50e−500t V, vo 50e−500t [b] io = = = 200e−500t µA 250,000 250,000 −1 v1 = × 200 × 10−6 40 × 10−9 P 7.64 Z t 0 e−500x dx + 50 = 10e−500t + 40 V, t≥0 [a] t < 0 t>0 vo (0− ) = vo (0+ ) = 40 V vo (∞) = 80 V τ = (0.16 × 10−6 )(6.25 × 103 ) = 1 ms; vo = 80 − 40e−1000t V, [b] io = −C t≥0 dvo = −0.16 × 10−6 [40,000e−1000t ] dt = −6.4e−1000t mA; −1 [c] v1 = 0.2 × 10−6 Z 0 t ≥ 0+ t −6.4 × 10−3 e−1000x dx + 32 = 64 − 32e−1000t V, [d] v2 = 1/τ = 1000 t≥0 Z t −1 −6.4 × 10−3 e−1000x dx + 8 0.8 × 10−6 0 = 16 − 8e−1000t V, t≥0 1 1 [e] wtrapped = (0.2 × 10−6 )(64)2 + (0.8 × 10−6 )(16)2 = 512 µJ. 2 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. 7–54 P 7.65 CHAPTER 7. Response of First-Order RL and RC Circuits [a] Leq = τ= (3)(15) = 2.5 H 3 + 15 Leq 2.5 1 = = s R 7.5 3 io (0) = 0; 120 = 16 A 7.5 io (∞) = .·. io = 16 − 16e−3t A, t≥0 t ≥ 0+ vo = 120 − 7.5io = 120e−3t V, i1 = 1 3 Z 0 40 40 −3t − e A, 3 3 t 120e−3x dx = i2 = io − i1 = 8 8 −3t − e A, 3 3 t≥0 t≥0 [b] io(0) = i1 (0) = i2(0) = 0, consistent with initial conditions. vo (0+ ) = 120 V, consistent with io (0) = 0. vo = 3 di1 = 120e−3t V, dt t ≥ 0+ or di2 = 120e−3t V, t ≥ 0+ dt The voltage solution is consistent with the current solutions. vo = 15 λ1 = 3i1 = 40 − 40e−3t Wb-turns λ2 = 15i2 = 40 − 40e−3t Wb-turns .·. λ1 = λ2 as it must, since vo = dλ1 dλ2 = dt dt λ1 (∞) = λ2 (∞) = 40 Wb-turns λ1 (∞) = 3i1 (∞) = 3(40/3) = 40 Wb-turns λ2 (∞) = 15i2 (∞) = 15(8/3) = 40 Wb-turns .·. i1(∞) and i2(∞) are consistent with λ1 (∞) and λ2 (∞). P 7.66 [a] Leq = 5 + 10 − 2.5(2) = 10 H τ= L 10 1 = = ; R 40 4 i = 2 − 2e−4t A, 1 =4 τ t≥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. Problems 7–55 di1 di di − 2.5 = 2.5 = 2.5(8e−4t ) = 20e−4t V, t ≥ 0+ dt dt dt di1 di di [c] v2 (t) = 10 − 2.5 = 7.5 = 7.5(8e−4t ) = 60e−4t V, t ≥ 0+ dt dt dt [d] i(0) = 2 − 2 = 0, which agrees with initial conditions. [b] v1(t) = 5 80 = 40i1 + v1 + v2 = 40(2 − 2e−4t ) + 20e−4t + 60e−4t = 80 V Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior. P 7.67 [a] Leq = 5 + 10 + 2.5(2) = 20 H τ= L 20 1 = = ; R 40 2 i = 2 − 2e−2t A, 1 =2 τ t≥0 di1 di di + 2.5 = 7.5 = 7.5(4e−2t ) = 30e−2t V, t ≥ 0+ dt dt dt di1 di di [c] v2 (t) = 10 + 2.5 = 12.5 = 12.5(4e−2t ) = 50e−2t V, t ≥ 0+ dt dt dt [d] i(0) = 0, which agrees with initial conditions. [b] v1(t) = 5 80 = 40i1 + v1 + v2 = 40(2 − 2e−2t ) + 30e−2t + 50e−2t = 80 V Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior. P 7.68 [a] From Example 7.10, Leq = τ= L1 L2 − M 2 50 − 25 = = 1H L1 + L2 + 2M 15 + 10 1 L = ; R 20 1 = 20 τ .·. io(t) = 4 − 4e−20t A, t≥0 [b] vo = 80 − 20io = 80 − 80 + 80e−20t = 80e−20t V, di1 di2 [c] vo = 5 −5 = 80e−20t V dt dt t ≥ 0+ io = i1 + i2 dio di1 di2 = + = 80e−20t A/s dt dt dt . ·. di2 di1 = 80e−20t − dt dt © 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. 7–56 CHAPTER 7. Response of First-Order RL and RC Circuits .·. 80e−20t = 5 .·. 10 Z di1 = 480e−20t ; dt t1 dx = 0 i1 = di1 di1 − 400e−20t + 5 dt dt Z di1 = 48e−20t dt t 0 48e−20y dy 48 −20y t e = 2.4 − 2.4e−20t A, −20 0 t≥0 [d] i2 = io − i1 = 4 − 4e−20t − 2.4 + 2.4e−20t = 1.6 − 1.6e−20t A, t≥0 [e] io (0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy. vo = Leq dio = 1(80)e−20t = 80e−20t V, dt t ≥ 0+ (checks) Also, vo = 5 di1 di2 −5 = 80e−20t V, dt dt vo = 10 di2 di1 −5 = 80e−20t V, dt dt t ≥ 0+ (checks) t ≥ 0+ (checks) vo (0+ ) = 80 V, which agrees with io (0+ ) = 0 A io (∞) = 4 A; io (∞)Leq = (4)(1) = 4 Wb-turns i1 (∞)L1 + i2 (∞)M = (2.4)(5) + (1.6)(−5) = 4 Wb-turns (ok) i2 (∞)L2 + i1 (∞)M = (1.6)(10) + (2.4)(−5) = 4 Wb-turns (ok) Therefore, the final values of io , i1 , and i2 are consistent with conservation of flux linkage. Hence, the answers make sense in terms of known circuit behavior. P 7.69 [a] From Example 7.10, Leq = τ= L1 L2 − M 2 0.125 − 0.0625 = = 50 mH L1 + L2 + 2M 0.75 + 0.5 L 1 = ; R 5000 1 = 5000 τ .·. io(t) = 40 − 40e−5000t mA, t≥0 [b] vo = 10 − 250io = 10 − 250(0.04 + 0.04e−5000t = 10e−5000t V, t ≥ 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. Problems [c] vo = 0.5 7–57 di1 di2 − 0.25 = 10e−5000t V dt dt io = i1 + i2 dio di1 di2 = + = 200e−5000t A/s dt dt dt di2 di1 = 200e−5000t − dt dt . ·. .·. 10e−5000t = 0.5 .·. 0.75 Z di1 = 60e−5000t; dt t1 dx = 0 i1 = di1 di1 − 50e−5000t + 0.25 dt dt Z 0 di1 = 80e−5000t dt t 80e−5000y dy 80 −5000y t e = 16 − 16e−5000t mA, −5000 0 t≥0 [d] i2 = io − i1 = 40 − 40e−5000t − 16 + 16e−5000t = 24 − 24e−5000t mA, t≥0 [e] io (0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy. vo = Leq dio = (0.05)(200)e−5000t = 10e−5000t V, dt t ≥ 0+ (checks) Also, vo = 0.5 di1 di2 − 0.25 = 10e−5000t V, dt dt vo = 0.25 di2 di1 − 0.25 = 10e−5000t V, dt dt t ≥ 0+ (checks) t ≥ 0+ (checks) vo (0+ ) = 10 V, which agrees with io (0+ ) = 0 A io (∞) = 40 mA; io (∞)Leq = (0.04)(0.05) = 2 mWb-turns i1 (∞)L1 + i2 (∞)M = (16 m)(500) + (24 m)(−250) = 2 mWb-turns (ok) i2 (∞)L2 + i1 (∞)M = (24 m)(250) + (16 m)(−250) = 2 mWb-turns (ok) Therefore, the final values of io , i1 , and i2 are consistent with conservation of flux linkage. Hence, the answers make sense in terms of known circuit behavior. © 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. 7–58 P 7.70 CHAPTER 7. Response of First-Order RL and RC Circuits t < 0: iL (0− ) = 10 V/5 Ω = 2 A = iL(0+ ) 0 ≤ t ≤ 5: τ = 5/0 = ∞ iL (t) = 2e−t/∞ = 2e−0 = 2 iL (t) = 2 A 0 ≤ t ≤ 5 s 5 ≤ t < ∞: τ= 5 = 5 s; 1 1/τ = 0.2 iL (t) = 2e−0.2(t −5) A, P 7.71 t ≥ 5s For t < 0: i(0) = 10 (15) = 10 A 15 © 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. Problems 7–59 0 ≤ t ≤ 10 ms: i = 10e−100t A i(10 ms) = 10e−1 = 3.68 A 10 ms ≤ t ≤ 20 ms: Req = (5)(20) = 4Ω 25 1 R 4 = = = 80 τ L 50 × 10−3 i = 3.68e−80(t−0.01) A 20 ms ≤ t < ∞: i(20 ms) = 3.68e−80(0.02−0.01) = 1.65 A i = 1.65e−100(t−0.02) A vo = L di ; dt L = 50 mH di = 1.65(−100)e−100(t−0.02) = −165e−100(t−0.02) dt vo = (50 × 10−3 )(−165)e−100(t−0.02) = −8.26e−100(t−0.02) V, t > 20+ ms vo (25 ms) = −8.26e−100(0.025−0.02) = −5.013 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. 7–60 P 7.72 CHAPTER 7. Response of First-Order RL and RC Circuits From the solution to Problem 7.71, the initial energy is 1 w(0) = (50 mH)(10 A)2 = 2.5 J 2 0.04w(0) = 0.1 J 1 .·. (50 × 10−3 )i2L = 0.1 so iL = 2 A 2 Again, from the solution to Problem 7.73, t must be between 10 ms and 20 ms since i(10 ms) = 3.68 A and i(20 ms) = 1.65 A For 10 ms ≤ t ≤ 20 ms: i = 3.68e−80(t−0.01) = 2 e80(t−0.01) = P 7.73 3.68 2 so t − 0.01 = 0.0076 .·. t = 17.6 ms [a] t < 0: Using Ohm’s law, 800 ig = = 12.5 A 40 + 60k40 Using current division, 60 i(0− ) = (12.5) = 7.5 A = i(0+ ) 60 + 40 [b] 0 ≤ t ≤ 1 ms: i = i(0+ )e−t/τ = 7.5e−t/τ 1 R 40 + 120k60 = = = 1000 τ L 80 × 10−3 i = 7.5e−1000t i(200µs) = 7.5e−10 3 (200×10−6 ) = 7.5e−0.2 = 6.14 A [c] i(1 ms) = 7.5e−1 = 2.7591 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. Problems 7–61 1 ms ≤ t < ∞: 1 R 40 = = = 500 τ L 80 × 10−3 i = i(1 ms)e−(t−1 ms)/τ = 2.7591e−500(t−0.001) A i(6ms) = 2.7591e−500(0.005) = 2.7591e−2.5 = 226.48 mA [d] 0 ≤ t ≤ 1 ms: i = 7.5e−1000t v=L di = (80 × 10−3 )(−1000)(7.5e−1000t ) = −600e−1000t V dt v(1− ms) = −600e−1 = −220.73 V [e] 1 ms ≤ t ≤ ∞: i = 2.759e−500(t−0.001) v=L di = (80 × 10−3 )(−500)(2.759e−500(t−0.001)) dt = −110.4e−500(t−0.001) V v(1+ ms) = −110.4 V P 7.74 0 ≤ t ≤ 10 µs: τ = RC = (4 × 103 )(20 × 10−9 ) = 80 µs; vo (0) = 0 V; 1/τ = 12,500 vo (∞) = −20 V vo = −20 + 20e−12,500t V 0 ≤ t ≤ 10 µs © 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. 7–62 CHAPTER 7. Response of First-Order RL and RC Circuits 10 µs ≤ t < ∞: t → ∞: i= −50 V = −2.5 mA 20 kΩ vo (∞) = (−2.5 × 10−3 )(16,000) + 30 = −10 V vo (10 µs) = −20 + 20−0.125 = −2.35 V vo = −10 + (−2.35 + 10)e−(t − 10×10 −6 )/τ RTh = 4 kΩk16 kΩ = 3.2 kΩ τ = (3200)(20 × 10−9 ) = 64 µs; vo = −10 + 7.65e−15,625(t − 10×10 P 7.75 1/τ = 15,625 −6 ) 10 µs ≤ t < ∞ 0 ≤ t ≤ 200 µs; Re = 150k100 = 60 kΩ; 10 τ= × 10−9 (60,000) = 200 µs 3 vc = 300e−5000t V vc (200 µs) = 300e−1 = 110.36 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. Problems 7–63 200 µs ≤ t < ∞: Re = 30k60 + 120k40 = 20 + 30 = 50 kΩ 10 τ= × 10−9 (50,000) = 166.67 µs; 3 1 = 6000 τ vc = 110.36e−6000(t − 200 µs) V vc (300 µs) = 110.36e−6000(100 µs) = 60.57 V io (300 µs) = i1 = 60.57 = 1.21 mA 50,000 60 2 io = io ; 90 3 i2 = 40 1 io = io 160 4 2 1 5 5 isw = i1 − i2 = io − io = io = (1.21 × 10−3 ) = 0.50 mA 3 4 12 12 P 7.76 Note that for t > 0, vo = (4/6)vc , where vc is the voltage across the 0.5 µF capacitor. Thus we will find vc first. t<0 vc(0) = 3 (−75) = −15 V 15 © 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. 7–64 CHAPTER 7. Response of First-Order RL and RC Circuits 0 ≤ t ≤ 800 µs: τ = Re C, Re = (6000)(3000) = 2 kΩ 9000 1 = 1000 τ τ = (2 × 103 )(0.5 × 10−6 ) = 1 ms, vc = −15e−1000t V, t≥0 vc(800 µs) = −15e−0.8 = −6.74 V 800 µs ≤ t ≤ 1.1 ms: 1 = 333.33 τ τ = (6 × 103 )(0.5 × 10−6 ) = 3 ms, vc = −6.74e−333.33(t−800×10 −6) V 1.1 ms ≤ t < ∞: τ = 1 ms, 1 = 1000 τ −6 vc(1.1ms) = −6.74e−333.33(1100−800)10 vc = −6.1e−1000(t−1.1×10 −3) = −6.74e−0.1 = −6.1 V V vc(1.5ms) = −6.1e−1000(1.5−1.1)10 −3 = −6.1e−0.4 = −4.09 V vo = (4/6)(−4.09) = −2.73 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. Problems P 7.77 7–65 1 w(0) = (0.5 × 10−6 )(−15)2 = 56.25 µJ 2 0 ≤ t ≤ 800 µs: vc2 = 225e−2000t vc = −15e−1000t; p3k = 75e−2000t mW w3k = Z 800×10−6 75 × 10−3 e−2000t dt 0 −3 e 800×10−6 −2000t = 75 × 10 −2000 0 = −37.5 × 10−6 (e−1.6 − 1) = 29.93 µJ 1.1 ms ≤ t ≤ ∞: vc = −6.1e−1000(t−1.1×10 −3) V; −3) mW p3k = 12.4e−2000(t−1.1×10 w3k = Z vc2 = 37.19e−2000(t−1.1×10 −3 ) ∞ 1.1×10−3 12.4 × 10−3 e−2000(t−1.1×10 e−2000(t−1.1×10 ) −2000 −6 = −6.2 × 10 (0 − 1) = 6.2 µJ −3 −3) dt ∞ = 12.4 × 10−3 1.1×10−3 w3k = 29.93 + 6.2 = 36.13 µJ %= P 7.78 36.13 (100) = 64.23% 56.25 t < 0: vc (0− ) = −(5)(1000) × 10−3 = −5 V = vc (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. 7–66 CHAPTER 7. Response of First-Order RL and RC Circuits 0 ≤ t ≤ 5 s: τ = ∞; 1/τ = 0; vo = −5e−0 = −5 V 5 s ≤ t < ∞: τ = (100)(0.1) = 10 s; vo = −5e−0.1(t − 5) V 1/τ = 0.1; Summary: vo = −5 V, 0 ≤ t ≤ 5s vo = −5e−0.1(t − 5) V, P 7.79 5s ≤ t < ∞ [a] 0 ≤ t ≤ 2.5 ms vo (0+ ) = 80 V; vo (∞) = 0 L = 2 ms; R 1/τ = 500 τ= vo (t) = 80e−500t V, 0+ ≤ t ≤ 2.5− ms vo (2.5− ms) = 80e−1.25 = 22.92 V io (2.5− ms) = (80 − 22.92) = 2.85 A 20 vo (2.5+ ms) = −20(2.85) = −57.08 V vo (∞) = 0; τ = 2 ms; vo = −57.08e−500(t − 0.0025) V 1/τ = 500 t ≥ 2.5+ ms © 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. Problems 7–67 [b] [c] vo (5 ms) = −16.35 V io = P 7.80 +16.35 = 817.68 mA 20 [a] io(0) = 0; io (∞) = 25 mA 1 R 2000 = = × 103 = 8000 τ L 250 io = (25 − 25e−8000t) mA, vo = 0.25 0 ≤ t ≤ 75 µs dio = 50e−8000t V, dt 0 ≤ t ≤ 75 µs 75 µs ≤ t < ∞: io (75µs) = 25 − 25e−0.6 = 11.28 mA; io = 11.28e−8000(t−75×10 vo = (0.25) −6 ) io (∞) = 0 mA dio = −22.56e−8000(t−75µs) dt . ·. t < 0 : vo = 0 0 ≤ t ≤ 75 µs : vo = 50e−8000t V 75 µs ≤ t < ∞ : vo = −22.56e−8000(t−75µs) [b] vo(75− µs) = 50e−0.6 = 27.44 V vo (75+ µs) = −22.56 V [c] io (75− µs) = io(75+ µs) = 11.28 mA © 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. 7–68 P 7.81 CHAPTER 7. Response of First-Order RL and RC Circuits [a] 0 ≤ t ≤ 1 ms: vc (0+ ) = 0; vc (∞) = 50 V; RC = 400 × 103 (0.01 × 10−6 ) = 4 ms; 1/RC = 250 vc = 50 − 50e−250t vo = 50 − 50 + 50e−250t = 50e−250t V, 0 ≤ t ≤ 1 ms 1 ms ≤ t < ∞: vc (1 ms) = 50 − 50e−0.25 = 11.06 V vc (∞) = 0 V τ = 4 ms; 1/τ = 250 vc = 11.06e−250(t − 0.001) V vo = −vc = −11.06e−250(t − 0.001) V, t ≥ 1 ms [b] P 7.82 [a] t < 0; vo = 0 0 ≤ t ≤ 4 ms: τ = (200 × 103 )(0.025 × 10−6 ) = 5 ms; vo = 100 − 100e−200t V, 1/τ = 200 0 ≤ t ≤ 4 ms vo (4 ms) = 100(1 − e−0.8 ) = 55.07 V 4 ms ≤ t ≤ 8 ms: vo = −100 + 155.07e−200(t−0.004) V, 4 ms ≤ t ≤ 8 ms vo (8 ms) = −100 + 155.07e−0.8 = −30.32 V t ≥ 8 ms: vo = −30.32e−200(t−0.008) V, t ≥ 8 ms © 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. Problems 7–69 [b] [c] t ≤ 0 : vo = 0 0 ≤ t ≤ 4 ms: τ = (50 × 103 )(0.025 × 10−6 ) = 1.25 ms 1/τ = 800 vo = 100 − 100e−800t V, 0 ≤ t ≤ 4 ms vo (4 ms) = 100 − 100e−3.2 = 95.92 V 4 ms ≤ t ≤ 8 ms: vo = −100 + 195.92e−800(t−0.004) V, 4 ms ≤ t ≤ 8 ms vo (8 ms) = −100 + 195.92e−3.2 = −92.01 V t ≥ 8 ms: vo = −92.01e−800(t−0.008) V, P 7.83 t ≥ 8 ms [a] τ = RC = (20,000)(0.2 × 10−6 ) = 4 ms; io = vo = 0 t<0 16 io (0 ) = 20 = 16 mA, 20 + 1/τ = 250 .·. io = 16e−250t mA io(∞) = 0 0+ ≤ t ≤ 2− ms © 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. 7–70 CHAPTER 7. Response of First-Order RL and RC Circuits i16kΩ = 20 − 16e−250t mA .·. vo = 320 − 256e−250t V 0+ ≤ t ≤ 2− ms vc = vo − 4 × 103 io = 320 − 320e−250t V 0 ≤ t ≤ 2 ms vc (2 ms) = 320 − 320e−0.5 = 125.91 V .·. io(2+ ms) = 16e−0.5 = 9.7 mA io (∞) = 0 vc = 125.91e−250(t−0.002), io = C t ≥ 2 ms dvc = (0.2 × 10−6 )(−250)(125.91)e−250(t−0.002) dt = −6.3e−250(t−0.002) mA, t ≥ 2+ ms vo = 4000io + vc = 100.73e−250(t −0.002) V t ≥ 2+ ms Summary part (a) io = 0 t<0 io = 16e−250t mA (0+ ≤ t ≤ 2− ms) t ≥ 2+ ms io = −6.3e−250(t −0.002) mA vo = 0 t<0 vo = 320 − 256e−250t V, 0+ ≤ t ≤ 2− ms vo = 100.73e−250(t −0.002) V, t ≥ 2+ ms [b] io(0− ) = 0 io (0+ ) = 16 mA io (2− ms) = 16e−0.5 = 9.7 mA io (2+ ms) = −6.3 mA [c] vo (0− ) = 0 vo (0+ ) = 64 V vo (2− ms) = 320 − 256e−0.5 = 164.73 V vo (2+ ms) = 100.73 © 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. Problems 7–71 [d] [e] P 7.84 t > 0: vT = 12 × 104 i∆ + 16 × 103 iT i∆ = − 20 iT = −0.2iT 100 .·. vT = −24 × 103 iT + 16 × 103 iT RTh = vT = −8 kΩ iT τ = RC = (−8 × 103 )(2.5 × 10−6 ) = −0.02 1/τ = −50 © 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. 7–72 CHAPTER 7. Response of First-Order RL and RC Circuits vc = 20e50t V; 50t = ln 1000 P 7.85 20e50t = 20,000 .·. t = 138.16 ms Find the Thévenin equivalent with respect to the terminals of the capacitor. RTh calculation: iT = .·. vT vT vT + −4 2000 5000 5000 iT 5+2−8 1 = =− vT 10,000 10,000 vT 10,000 =− = −10 kΩ iT 1 Open circuit voltage calculation: The node voltage equations: voc voc − v1 + − 4i∆ = 0 2000 1000 v1 − voc v1 + − 5 × 10−3 = 0 1000 4000 The constraint equation: i∆ = v1 4000 Solving, voc = −80 V, v1 = −60 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. Problems vc(0) = 0; 7–73 vc(∞) = −80 V τ = RC = (−10,000)(1.6 × 10−6 ) = −16 ms; 1 = −62.5 τ vc = vc (∞) + [vc(0+ ) − vc (∞)]e−t/τ = −80 + 80e62.5t = 14,400 Solve for the time of the maximum voltage rating: e62.5t = 181; 62.5t = ln 181; t = 83.09 ms P 7.86 vT = 2000iT + 4000(iT − 2 × 10−3 vφ ) = 6000iT − 8vφ = 6000iT − 8(2000iT ) vT = −10,000 iT τ= 10 = −1 ms; −10,000 1/τ = −1000 i = 25e1000t mA .·. 25e1000t × 10−3 = 5; t= ln 200 = 5.3 ms 1000 © 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. 7–74 P 7.87 CHAPTER 7. Response of First-Order RL and RC Circuits [a] Using Ohm’s law, vT = 5000iσ Using current division, iσ = 20,000 (iT + βiσ ) = 0.8iT + 0.8βiσ 20,000 + 5000 Solve for iσ : iσ (1 − 0.8β) = 0.8iT iσ = 0.8iT ; 1 − 0.8β vT = 5000iσ = 4000iT (1 − 0.8β) Find β such that RTh = −5 kΩ: RTh = 4000 vT = = −5000 iT 1 − 0.8β 1 − 0.8β = −0.8 .·. β = 2.25 [b] Find VTh ; Write a KCL equation at the top node: VTh − 40 VTh + − 2.25iσ = 0 5000 20,000 The constraint equation is: (VTh − 40) =0 5000 Solving, iσ = VTh = 50 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. Problems 7–75 Write a KVL equation around the loop: 50 = −5000i + 0.2 di dt Rearranging: di = 250 + 25,000i = 25,000(i + 0.01) dt Separate the variables and integrate to find i; di = 25,000 dt i + 0.01 Z 0 i dx = x + 0.01 Z 0 t 25,000 dx .·. i = −10 + 10e25,000t mA di = (10 × 10−3 )(25,000)e25,000t = 250e25,000t dt Solve for the arc time: di v = 0.2 = 50e25,000t = 45,000; dt . ·. t = P 7.88 e25,000t = 900 ln 900 = 272.1 µs 25,000 [a] τ = (25)(2) × 10−3 = 50 ms; vc (0+ ) = 80 V; 1/τ = 20 vc (∞) = 0 vc = 80e−20t V .·. 80e−20t = 5; e20t = 16; t= ln 16 = 138.63 ms 20 © 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. 7–76 CHAPTER 7. Response of First-Order RL and RC Circuits [b] 0+ ≤ t ≤ 138.63− ms: i = (2 × 10−6 )(−1600e−20t ) = −3.2e−20t mA t ≥ 138.63+ ms: τ = (2)(4) × 10−3 = 8 ms; vc (138.63+ ms) = 5 V; 1/τ = 125 vc (∞) = 80 V vc = 80 − 75e−125(t−0.13863) V, t ≥ 138.63 ms i = 2 × 10−6 (9375)e−125(t−0.13863) = 18.75e−125(t−0.13863) mA, t ≥ 138.63+ ms [c] 80 − 75e−125∆t = 0.85(80) = 68 80 − 68 = 75e−125∆t = 12 e125∆t = 6.25; P 7.89 ∆t = ln 6.25 ∼ = 14.66 ms 125 [a] RC = (25 × 103 )(0.4 × 10−6 ) = 10 ms; vo = 0, 1 = 100 RC t<0 [b] 0 ≤ t ≤ 250 ms : vo = −100 Z t 0 −0.20 dx = 20t V [c] 250 ms ≤ t ≤ 500 ms; vo (0.25) = 20(0.25) = 5 V vo (t) = −100 Z t 0.25 0.20 dx + 5 = −20(t − 0.25) + 5 = −20t + 10 V [d] t ≥ 500 ms : vo (0.5) = −10 + 10 = 0 V vo (t) = 0 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. Problems P 7.90 [a] vo = 0, t<0 1 = 100 RC RC = (25 × 103 )(0.4 × 10−6 ) = 10 ms [b] Rf Cf = (5 × 106 )(0.4 × 10−6 ) = 2; vo = 7–77 1 = 0.5 Rf Cf −5 × 106 (−0.2)[1 − e−0.5t] = 40(1 − e−0.5t) V, 25 × 103 0 ≤ t ≤ 250 ms [c] vo (0.25) = 40(1 − e−0.125) ∼ = 4.70 V −Vm Rf Vm Rf + (2 − e−0.125)e−0.5(t−0.25) Rs Rs = −40 + 40(2 − e−0.125)e−0.5(t−0.25) = −40 + 44.70e−0.5(t−0.25) V, 250 ms ≤ t ≤ 500 ms vo = [d] vo(0.5) = −40 + 44.70e−0.125 ∼ = −0.55 V vo = −0.55e−0.5(t−0.5) V, t ≥ 500 ms © 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. 7–78 P 7.91 CHAPTER 7. Response of First-Order RL and RC Circuits 1 vo = − R(0.5 × 10−6 ) Z t 4 dx + 0 = 0 −4t R(0.5 × 10−6 ) −4(15 × 10−3 ) = −10 R(0.5 × 10−6 ) .·. P 7.92 vo = .·. P 7.93 R= −4(15 × 10−3 ) = 12 kΩ −10(0.5 × 10−6 ) −4(40 × 10−3 ) −4t + 6 = + 6 = −10 R(0.5 × 10−6 ) R(0.5 × 10−6 ) R= −4(40 × 10−3 ) = 20 kΩ −16(0.5 × 10−6 ) 1 = 1,250,000 RC [a] RC = (1000)(800 × 10−12 ) = 800 × 10−9 ; 0 ≤ t ≤ 1 µs: vg = 2 × 106 t 6 vo = −1.25 × 10 = −2.5 × 1012 t Z 2 × 106 x dx + 0 0 x2 t 2 = −125 × 1010 t2 V, 0 ≤ t ≤ 1 µs 0 10 vo (1 µs) = −125 × 10 (1 × 10−6 )2 = −1.25 V 1 µs ≤ t ≤ 3 µs: vg = 4 − 2 × 106 t 4 vo = −125 × 10 4 = −125 × 10 Z t 1×10−6 " t (4 − 2 × 106 x) dx − 1.25 6x 4x −2 × 10 2 t # − 1.25 2 1×10−6 = −5 × 106 t + 5 + 125 × 1010 t2 − 1.25 − 1.25 = 125 × 1010 t2 − 5 × 106 t + 2.5 V, 1 µs ≤ t ≤ 3 µs 1×10−6 vo (3 µs) = 125 × 1010 (3 × 10−6 )2 − 5 × 106 (3 × 10−6 ) + 2.5 = −1.25 3 µs ≤ t ≤ 4 µs: vg = −8 + 2 × 106 t 4 vo = −125 × 10 Z t (−8 + 2 × 106 x) dx − 1.25 3×10−6 " t x2 t = −125 × 104 −8x +2 × 106 − 1.25 2 3×10−6 3×10−6 = 107 t − 30 − 125 × 1010 t2 + 11.25 − 1.25 = −125 × 1010 t2 + 107 t − 20 V, 3 µs ≤ t ≤ 4 µs # © 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. Problems 7–79 vo (4 µs) = −125 × 1010 (4 × 10−6 )2 + 107 (4 × 10−6 ) − 20 = 0 [b] [c] The output voltage will also repeat. This follows from the observation that at t = 4 µs the output voltage is zero, hence there is no energy stored in the capacitor. This means the circuit is in the same state at t = 4 µs as it was at t = 0, thus as vg repeats itself, so will vo . P 7.94 [a] Cdvp vp − vb + = 0; dt R therefore dvp 1 vb + vp = dt RC RC vn − va d(vn − vo ) +C = 0; R dt dvo dvn vn va = + − dt dt RC RC therefore But vn = vp Therefore dvn vn dvp vp vb + = + = dt RC dt RC RC Therefore dvo 1 = (vb − va ); dt RC 1 vo = RC Z 0 t (vb − va) dy [b] The output is the integral of the difference between vb and va and then scaled by a factor of 1/RC. Z t 1 [c] vo = (vb − va ) dx RC 0 RC = (50 × 103 )(10 × 10−9 ) = 0.5 ms vb − va = −25 mV vo = 1 0.0005 Z 0 −50tsat = −6; t −25 × 10−3 dx = −50t tsat = 120 ms © 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. 7–80 P 7.95 CHAPTER 7. Response of First-Order RL and RC Circuits The equation for an integrating amplifier: 1 vo = RC t Z (vb − va) dy + vo(0) 0 Find the values and substitute them into the equation: RC = (100 × 103 )(0.05 × 10−6 ) = 5 ms 1 = 200; RC vb − va = −15 − (−7) = −8 V vo (0) = −4 + 12 = 8 V vo = 200 Z t 0 −8 dx + 8 = (−1600t + 8) V, 0 ≤ t ≤ tsat RC circuit analysis for v2: v2(0+ ) = −4 V; v2(∞) = −15 V; τ = RC = (100 k)(0.05 µ) = 5 ms v2 = v2(∞) + [v2(0+ ) − v2(∞)]e−t/τ = −15 + (−4 + 15)e−200t = −15 + 11e−200t V, vf + v2 = vo .·. 0 ≤ t ≤ tsat vf = vo − v2 = 23 − 1600t − 11e−200t V, 0 ≤ t ≤ tsat Note that −1600tsat + 8 = −20 . ·. tsat = −28 = 17.5 ms −1600 so the op amp operates in its linear region until it saturates at 17.5 ms. P 7.96 Use voltage division to find the voltage at the non-inverting terminal: vp = 80 (−45) = −36 V = vn 100 Write a KCL equation at the inverting terminal: −36 − 14 d + 2.5 × 10−6 (−36 − vo) = 0 80,000 dt .·. 2.5 × 10−6 dvo −50 = dt 80,000 © 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. Problems 7–81 Separate the variables and integrate: dvo = −250 dt Z vo (t) vo (0) .·. dx = −250 dvo = −250dt Z t 0 . ·. dy vo (t) − vo(0) = −250t vo (0) = −36 + 56 = 20 V vo (t) = −250t + 20 Find the time when the voltage reaches 0: 0 = −250t + 20 P 7.97 .·. t= 20 = 80 ms 250 [a] T2 is normally ON since its base current ib2 is greater than zero, i.e., ib2 = VCC /R when T2 is ON. When T2 is ON, vce2 = 0, therefore ib1 = 0. When ib1 = 0, T1 is OFF. When T1 is OFF and T2 is ON, the capacitor C is charged to VCC , positive at the left terminal. This is a stable state; there is nothing to disturb this condition if the circuit is left to itself. [b] When S is closed momentarily, vbe2 is changed to −VCC and T2 snaps OFF. The instant T2 turns OFF, vce2 jumps to VCC R1 /(R1 + RL ) and ib1 jumps to VCC /(R1 + RL ), which turns T1 ON. [c] As soon as T1 turns ON, the charge on C starts to reverse polarity. Since vbe2 is the same as the voltage across C, it starts to increase from −VCC toward +VCC . However, T2 turns ON as soon as vbe2 = 0. The equation for vbe2 is vbe2 = VCC − 2VCC e−t/RC . vbe2 = 0 when t = RC ln 2, therefore T2 stays OFF for RC ln 2 seconds. P 7.98 [a] For t < 0, vce2 = 0. When the switch is momentarily closed, vce2 jumps to vce2 = VCC 6(5) R1 = = 1.2 V R1 + RL 25 T2 remains open for (23,083)(250) × 10−12 ln 2 ∼ = 4 µs. © 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. 7–82 CHAPTER 7. Response of First-Order RL and RC Circuits [b] ib2 = VCC = 259.93 µA, R ib2 = 0, ib2 P 7.99 −5 ≤ t ≤ 0 µs 0 < t < RC ln 2 = VCC VCC −(t−RC ln 2)/RLC + e R RL = 259.93 + 300e−0.2×10 6 (t−4×10−6 ) µA, RC ln 2 < t [a] While T2 has been ON, C2 is charged to VCC , positive on the left terminal. At the instant T1 turns ON the capacitor C2 is connected across b2 − e2 , thus vbe2 = −VCC . This negative voltage snaps T2 OFF. Now the polarity of the voltage on C2 starts to reverse, that is, the right-hand terminal of C2 starts to charge toward +VCC . At the same time, C1 is charging toward VCC , positive on the right. At the instant the charge on C2 reaches zero, vbe2 is zero, T2 turns ON. This makes vbe1 = −VCC and T1 snaps OFF. Now the capacitors C1 and C2 start to charge with the polarities to turn T1 ON and T2 OFF. This switching action repeats itself over and over as long as the circuit is energized. At the instant T1 turns ON, the voltage controlling the state of T2 is governed by the following circuit: It follows that vbe2 = VCC − 2VCC e−t/R2 C2 . [b] While T2 is OFF and T1 is ON, the output voltage vce2 is the same as the voltage across C1 , thus © 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. Problems 7–83 It follows that vce2 = VCC − VCC e−t/RLC1 . [c] T2 will be OFF until vbe2 reaches zero. As soon as vbe2 is zero, ib2 will become positive and turn T2 ON. vbe2 = 0 when VCC − 2VCC e−t/R2C2 = 0, or when t = R2 C2 ln 2. [d] When t = R2C2 ln 2, we have vce2 = VCC − VCC e−[(R2 C2 ln 2)/(RLC1 )] = VCC − VCC e−10 ln 2 ∼ = VCC [e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have ib1 = VCC VCC −t/RLC1 + e R1 RL [f] At the instant T2 turns back ON, t = R2 C2 ln 2; therefore ib1 = VCC VCC −10 ln 2 ∼ VCC + e = R1 RL R1 [g] [h] © 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. 7–84 CHAPTER 7. Response of First-Order RL and RC Circuits P 7.100 [a] tOFF2 = R2 C2 ln 2 = 18 × 103 (2 × 10−9 ) ln 2 ∼ = 25 µs [b] tON2 = R1 C1 ln 2 ∼ = 25 µs [c] tOFF1 = R1 C1 ln 2 ∼ = 25 µs [d] tON1 = R2 C2 ln 2 ∼ = 25 µs 9 9 + = 3.5 mA 3 18 9 9 + e−6 ln 2 ∼ [f] ib1 = = 0.5469 mA 18 3 [g] vce2 = 9 − 9e−6 ln 2 ∼ = 8.86 V [e] ib1 = P 7.101 [a] tOFF2 = R2 C2 ln 2 = (18 × 103 )(2.8 × 10−9 ) ln 2 ∼ = 35 µs [b] tON2 = R1 C1 ln 2 ∼ = 37.4 µs [c] tOFF1 = R1 C1 ln 2 ∼ = 37.4 µs [d] tON1 = R2 C2 ln 2 = 35 µs [e] ib1 = 3.5 mA 9 [f] ib1 = + 3e−5.6 ln 2 ∼ = 0.562 mA 18 [g] vce2 = 9 − 9e−5.6 ln 2 ∼ = 8.81 V Note in this circuit T2 is OFF 35 µs and ON 37.4 µs of every cycle, whereas T1 is ON 35 µs and OFF 37.4 µs every cycle. P 7.102 If R1 = R2 = 50RL = 100 kΩ, C1 = 48 × 10−6 = 692.49 pF; 100 × 103 ln 2 If R1 = R2 = 6RL = 12 kΩ, C1 = then C2 = 36 × 10−6 = 519.37 pF 100 × 103 ln 2 then 48 × 10−6 = 5.77 nF; 12 × 103 ln 2 C2 = 36 × 10−6 = 4.33 nF 12 × 103 ln 2 Therefore 692.49 pF ≤ C1 ≤ 5.77 nF and 519.37 pF ≤ C2 ≤ 4.33 nF P 7.103 [a] We want the lamp to be in its nonconducting state for no more than 10 s, the value of to : 10 = R(10 × 10−6 ) ln 1−6 4−6 and R = 1.091 MΩ [b] When the lamp is conducting VTh = 20 × 103 (6) = 0.108 V 20 × 103 + 1.091 × 106 RTh = 20 kk1.091 M = 19,640 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. Problems 7–85 So, (tc − to ) = (19,640)(10 × 10−6 ) ln 4 − 0.108 = 0.289 s 1 − 0.108 The flash lasts for 0.289 s. P 7.104 [a] At t = 0 we have τ = (800)(25) × 10−3 = 20 sec; vc (∞) = 40 V; 1/τ = 0.05 vc (0) = 5 V vc = 40 − 35e−0.05t V, 0 ≤ t ≤ to .·. e0.05to = 1.4 40 − 35e−0.05to = 15; to = 20 ln 1.4 s = 6.73 s At t = to we have The Thévenin equivalent with respect to the capacitor is 800 20 τ= (25) × 10−3 = s; 81 81 vc (to ) = 15 V; vc(∞) = 1 81 = = 4.05 τ 20 40 V 81 40 40 −4.05(t−to) 40 1175 −4.05(t−to) vc (t) = + 15 − e V= + e 81 81 81 81 . ·. 40 1175 −4.05(t−to) + e =5 81 81 © 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. 7–86 CHAPTER 7. Response of First-Order RL and RC Circuits 1175 −4.05(t−to) 365 e = 81 81 e4.05(t−to) = t − to = 1175 = 3.22 365 1 ln 3.22 ∼ = 0.29 s 4.05 One cycle = 7.02 seconds. N = 60/7.02 = 8.55 flashes per minute [b] At t = 0 we have τ = 25R × 10−3 ; 1/τ = 40/R vc = 40 − 35e−(40/R)t 40 − 35e−(40/R)to = 15 . ·. t o = R ln 1.4, 40 R in kΩ At t = to : vTh = τ= 10 400 (40) = ; R + 10 R + 10 RTh = (25)(10R) × 10−3 0.25R = ; R + 10 R + 10 10R kΩ R + 10 1 4(R + 10) = τ R 4(R+10) 400 400 vc = + 15 − e− R (t−to ) R + 10 R + 10 . ·. or 400 15R − 250 − 4(R+10) (t−to ) R + e =5 R + 10 R + 10 15R − 250 − 4(R+10) (t−to ) 5R − 350 R e = R + 10 (R + 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. Problems . ·. e 4(R+10) (t−to ) R . ·. t − t o = = 7–87 3R − 50 R − 70 R 3R − 50 ln 4(R + 10) R − 70 At 12 flashes per minute to + (t − to ) = 5 s . ·. 3R − 50 R R ln 1.4 + ln =5 R − 70 |40 {z } 4(R + 10) dominant term Start the trial-and-error procedure by setting (R/40) ln 1.4 = 5, then R = 200/(ln 1.4) or 594.40 kΩ. If R = 594.40 kΩ then t − to ∼ = 0.29 s. Second trial set (R/40) ln 1.4 = 4.7 s or R = 558.74 kΩ. t − to ∼ = 0.30 s With R = 558.74 kΩ, This procedure converges to R = 559.3 kΩ. P 7.105 [a] to = RC ln Vmin − Vs Vmax − Vs = (3700)(250 × 10−6 ) ln −700 −100 = 1.80 s tc − to = RCRL Vmax − VTh ln R + RL Vmin − VTh RL 1.3 = = 0.26; R + RL 1.3 + 3.7 VTh = 1000(1.3) = 260 V; 1.3 + 3.7 RC = (3700)(2501̧0−6 ) = 0.925 s RTh = 3.7 kk1.3 k = 962 Ω .·. tc − to = (0.925)(0.26) ln(640/40) = 0.67 s .·. tc = 1.8 + 0.67 = 2.47 s flashes/min = 60 = 24.32 2.47 [b] 0 ≤ t ≤ to : vL = 1000 − 700e−t/τ1 τ1 = RC = 0.925 s to ≤ t ≤ tc : vL = 260 + 640e−(t−to )/τ2 τ2 = RTh C = 962(250) × 10−6 = 0.2405 s © 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. 7–88 CHAPTER 7. Response of First-Order RL and RC Circuits 0 ≤ t ≤ to : i= 1000 − vL 7 = e−t/0.925 A 3700 37 to ≤ t ≤ tc : i= 1000 − vL 74 64 −(t−to )/0.2405 = − e 3700 370 370 Graphically, i versus t is The average value of i will equal the areas (A1 + A2) divided by tc . .·. iavg = A1 + A2 tc 7 to −t/0.925 e dt 37 0 6.475 = (1 − e− ln 7 ) = 0.15 A–s 37 Z tc 74 − 64e−(t−to )/0.2405 A2 = dt 370 to 74 15.392 − ln 16 = (tc − to ) + (e − 1) 370 370 17.797 15.392 = ln 16 − (1 − e− ln 16) 370 370 = 0.09436 A–s A1 = iavg = Z (0.15 + 0.09436) (1000) = 99.06 mA 0.925 ln 7 + 0.2405 ln 16 [c] Pavg = (1000)(99.06 × 10−3 ) = 99.06 W No. of kw hrs/yr = (99.06)(24)(365) = 867.77 1000 Cost/year = (867.77)(0.05) = 43.39 dollars/year P 7.106 [a] Replace the circuit attached to the capacitor with its Thévenin equivalent, where the equivalent resistance is the parallel combination of the two resistors, and the open-circuit voltage is obtained by voltage division across the lamp resistance. The resulting circuit is © 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. Problems RTh = RkRL = RRL ; R + RL VTh = 7–89 RL Vs R + RL From this circuit, vC (∞) = VTh ; vC (0) = Vmax ; τ = RTh C Thus, vC (t) = VTh + (Vmax − VTh )e−(t−to )/τ where RRL C τ= R + RL [b] Now, set vC (tc) = Vmin and solve for (tc − to ): VTh + (Vmax − VTh )e−(tc−to )/τ = Vmin e−(tc −to )/τ = Vmin − VTh Vmax − VTh −(tc − to ) Vmin − VTh = ln τ Vmax − VTh (tc − to ) = − RRL C Vmin − VTh RRL C Vmax − VTh ln = ln R + RL Vmax − VTh R + RL Vmin − VTh P 7.107 [a] 0 ≤ t ≤ 0.5: 21 30 21 −t/τ i= + − e 60 60 60 where τ = L/R. i = 0.35 + 0.15e−60t/L i(0.5) = 0.35 + 0.15e−30/L = 0.40 .·. e30/L = 3; L= 30 = 27.31 H ln 3 [b] 0 ≤ t ≤ tr , where tr is the time the relay releases: 30 i=0+ − 0 e−60t/L = 0.5e−60t/L 60 .·. 0.4 = 0.5e−60tr /L ; e60tr /L = 1.25 tr = 27.31 ln 1.25 ∼ = 0.10 s 60 © 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. 8 Natural and Step Responses of RLC Circuits Assessment Problems 1 1 = , therefore C = 500 nF (2RC)2 LC 1 [b] α = 5000 = , therefore C = 1 µF 2RC AP 8.1 [a] s1,2 = −5000 ± [c] √ s 25 × 106 − 1 = 20,000, LC s1,2 = −40 ± therefore C = 125 nF q (40)2 − 202 103 , s1 = −5.36 krad/s, AP 8.2 iL = = 1 50 × 10−3 (103 )(106 ) = (−5000 ± j5000) rad/s 20 Z t 0 s2 = −74.64 krad/s [−14e−5000x + 26e−20,000x] dx + 30 × 10−3 ( −14e−5000x 20 −5000 t 26e−20,000t + −20,000 0 t 0 ) + 30 × 10−3 = 56 × 10−3 (e−5000t − 1) − 26 × 10−3 (e−20,000t − 1) + 30 × 10−3 = [56e−5000t − 56 − 26e−20,000t + 26 + 30] mA = 56e−5000t − 26e−20,000t mA, t≥0 AP 8.3 From the given values of R, L, and C, s1 = −10 krad/s and s2 = −40 krad/s. [a] v(0− ) = v(0+ ) = 0, therefore iR (0+ ) = 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 8–1 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. 8–2 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] iC (0+ ) = −(iL (0+ ) + iR(0+ )) = −(−4 + 0) = 4 A dvc (0+ ) dvc (0+ ) 4 = ic (0+ ) = 4, therefore = = 4 × 108 V/s dt dt C [d] v = [A1e−10,000t + A2 e−40,000t] V, t ≥ 0+ [c] C dv(0+ ) = −10,000A1 − 40,000A2 dt v(0+ ) = A1 + A2 , Therefore A1 + A2 = 0, −A1 − 4A2 = 40,000; [e] A2 = −40,000/3 V [f] v = [40,000/3][e−10,000t − e−40,000t] V, A1 = 40,000/3 V t≥0 1 = 8000, therefore R = 62.5 Ω 2RC 10 V [b] iR(0+ ) = = 160 mA 62.5 Ω AP 8.4 [a] iC (0+ ) = −(iL(0+ ) + iR (0+ )) = −80 − 160 = −240 mA = C Therefore dv(0+ ) −240 m = = −240 kV/s dt C dvc (0+ ) = ωd B2 − αB1 dt [c] B1 = v(0+ ) = 10 V, Therefore 6000B2 − 8000B1 = −240,000, [d] iL = −(iR + iC ); iR = v/R; v = e−8000t[10 cos 6000t − iC = C iC = e−8000t[−240 cos 6000t + iL = 10e−8000t[8 cos 6000t + 1 2 1 106 AP 8.5 [a] = = , 2RC LC 4 [b] 0.5CV02 = 12.5 × 10−3 , [c] 0.5LI02 = 12.5 × 10−3 , B2 = (−80/3) V dv dt 80 sin 6000t] V 3 Therefore iR = e−8000t[160 cos 6000t − dv(0+ ) dt 1280 sin 6000t] mA 3 460 sin 6000t] mA 3 82 sin 6000t] mA, 3 t≥0 1 = 500, 2RC therefore V0 = 50 V therefore R = 100 Ω I0 = 250 mA © 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. 8–3 Problems [d] D2 = v(0+ ) = 50, iR (0+ ) = dv(0+ ) = D1 − αD2 dt 50 = 500 mA 100 Therefore iC (0+ ) = −(500 + 250) = −750 mA 10−3 dv(0+ ) = −750 × = −75,000 V/s dt C 1 Therefore D1 − αD2 = −75,000; α= = 500, 2RC Therefore D1 = −50,000 V/s [e] v = [50e−500t − 50,000te−500t ] V iR = v = [0.5e−500t − 500te−500t ] A, R t ≥ 0+ V0 40 = = 0.08 A R 500 iC (0+ ) = I − iR(0+ ) − iL (0+ ) = −1 − 0.08 − 0.5 = −1.58 A diL (0+ ) Vo 40 = = = 62.5 A/s dt L 0.64 1 1 α= = 1000; = 1,562,500; s1,2 = −1000 ± j750 rad/s 2RC LC iL = if + B10 e−αt cos ωd t + B20 e−αt sin ωd t, if = I = −1 A AP 8.6 [a] iR(0+ ) = [b] [c] [d] [e] iL (0+ ) = 0.5 = if + B10 , therefore B10 = 1.5 A diL (0+ ) = 62.5 = −αB10 + ωd B20 , dt therefore B20 = (25/12) A Therefore iL(t) = −1 + e−1000t[1.5 cos 750t + (25/12) sin 750t] A, [f] v(t) = LdiL = 40e−1000t [cos 750t − (154/3) sin 750t]V dt t≥0 t≥0 AP 8.7 [a] i(0+ ) = 0, since there is no source connected to L for t < 0. + − [b] vc (0 ) = vC (0 ) = ! 15 k (80) = 50 V 15 k + 9 k di(0+ ) di(0+ ) = 100, = 10,000 A/s dt dt 1 [d] α = 8000; = 100 × 106 ; s1,2 = −8000 ± j6000 rad/s LC [e] i = if + e−αt[B10 cos ωd t + B20 sin ωd t]; if = 0, i(0+ ) = 0 [c] 50 + 80i(0+ ) + L Therefore B10 = 0; Therefore B20 = 1.67 A; di(0+ ) = 10,000 = −αB10 + ωd B20 dt i = 1.67e−8000t sin 6000t A, t≥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. 8–4 CHAPTER 8. Natural and Step Responses of RLC Circuits AP 8.8 vc (t) = vf + e−αt [B10 cos ωd t + B20 sin ωd t], dvc (0+ ) = 0; dt + vc (0 ) = 50 V; B10 = −50 V; B20 Therefore vf = 100 V therefore 50 = 100 + B10 0 = −αB10 + ωd B20 α 8000 = B10 = (−50) = −66.67 V ωd 6000 Therefore vc (t) = 100 − e−8000t[50 cos 6000t + 66.67 sin 6000t] V, t≥0 Problems P 8.1 [a] α = 1 1012 = = 25,000 2RC (4000)(10) 1012 1 = = 4 × 108 LC (250)(10) √ = −25,000 ± 625 × 106 − 400 × 106 = −25,000 ± 15,000 ωo2 = s1,2 s1 = −10,000 rad/s s2 = −40,000 rad/s [b] overdamped [c] ωd = q ωo2 − α2 .·. α2 = ωo2 − ωd2 = 4 × 108 − 144 × 106 = 256 × 106 α = 16 × 103 = 16,000 1 = 16,000; 2RC .·. R = 109 = 3125 Ω (32,000)(10) [d] s1 = −16,000 + j12,000 rad/s; s2 = −16,000 − j12,000 rad/s 1 1 [e] α = 4 × 104 = ; .·. R = = 2500 Ω 2RC 2C(4 × 104 ) © 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. Problems P 8.2 [a] iR(0) = 8–5 15 = 75 mA 200 iL (0) = −45 mA iC (0) = −iL (0) − iR (0) = 45 − 75 = −30 mA [b] α = 1 1 = = 12,500 2RC 2(200)(0.2 × 10−6 ) 1 1 = = 108 −3 LC (50 × 10 )(0.2 × 10−6 ) √ = −12,500 ± 1.5625 × 108 − 108 = −12,500 ± 7500 ωo2 = s1,2 s1 = −5000 rad/s; s2 = −20,000 rad/s v = A1 e−5000t + A2e−20,000t v(0) = A1 + A2 = 15 dv −30 × 10−3 (0) = −5000A1 − 20,000A2 = = −15 × 104 V/s dt 0.2 × 10−6 Solving, A1 = 10; A2 = 5 v = 10e−5000t + 5e−20,000t V, [c] iC t≥0 dv dt = C = 0.2 × 10−6 [−50,000e−5000t − 100,000e−20,000t] = −10e−5000t − 20e−20,000t mA iR = 50e−5000t + 25e−20,000t mA iL = −iC − iR = −40e−5000t − 5e−20,000t mA, P 8.3 t≥0 1 1 = = 8000 2RC 2(312.5)(0.2 × 10−6 ) 1 1 = = 108 −3 −6 LC (50 × 10 )(0.2 × 10 ) s1,2 = −8000 ± √ 80002 − 108 = −8000 ± j6000 rad/s .·. response is underdamped v(t) = B1 e−8000t cos 6000t + B2 e−8000t sin 6000t © 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. 8–6 CHAPTER 8. Natural and Step Responses of RLC Circuits v(0+ ) = 15 V = B1 ; iR (0+ ) = 15 = 48 mA 312.5 iC (0+ ) = [−iL(0+ ) + iR(0+ )] = −[−45 + 48] = −3 mA dv(0+ ) −3 × 10−3 = = −15,000 V/s dt 0.2 × 10−6 dv(0) = −8000B1 + 6000B2 = −15,000 dt 6000B2 = 8000(15) − 15,000; .·. B2 = 17.5 V v(t) = 15e−8000t cos 6000t + 17.5e−8000t sin 6000t V, P 8.4 α= t≥0 1 1 = = 104 −6 2RC 2(250)(0.2 × 10 ) α2 = 108 ; .·. α2 = ωo2 Critical damping: v = D1 te−αt + D2 e−αt iR(0+ ) = 15 = 60 mA 250 iC (0+ ) = −[iL(0+ ) + iR(0+ )] = −[−45 + 60] = −15 mA v(0) = D2 = 15 dv = D1 [t(−αe−αt) + e−αt ] − αD2 e−αt dt dv iC (0) −15 × 10−3 (0) = D1 − αD2 = = = −75,000 dt C 0.2 × 10−6 D1 = αD2 − 75,000 = (104 )(15) − 75,000 = 75,000 v = (75,000t + 15)e−10,000t V, t≥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. Problems P 8.5 [a] 8–7 1 = 50002 LC There are many possible solutions. This one begins by choosing L = 10 mH. Then, C= 1 (10 × 10−3 )(5000)2 = 4 µF We can achieve this capacitor value using components from Appendix H by combining four 1 µF capacitors in parallel. Critically damped: . ·. R = α = ω0 = 5000 1 2(4 × 10−6 )(5000) so 1 = 5000 2RC = 25 Ω We can create this resistor value using components from Appendix H by combining a 10 Ω resistor and a 15 Ω resistor in series. The final circuit: q [b] s1,2 = −α ± α2 − ω02 = −5000 ± 0 Therefore there are two repeated real roots at −5000 rad/s. P 8.6 [a] Underdamped response: α < ω0 so α < 5000 Therefore we choose a larger resistor value than the one used in Problem 8.5. Choose R = 100 Ω: 1 α= = 1250 2(100)(4 × 10−6 ) √ s1,2 = −1250 ± 12502 − 50002 = −1250 ± j4841.23 rad/s [b] Overdamped response: α > ω0 so α > 5000 Therefore we choose a smaller resistor value than the one used in Problem 8.5. Choose R = 20 Ω: 1 α= = 6250 2(20)(4 × 10−6 ) √ s1,2 = −1250 ± 62502 − 50002 = −1250 ± 3750 = −2500 rad/s; and − 10,000 rad/s © 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. 8–8 P 8.7 CHAPTER 8. Natural and Step Responses of RLC Circuits [a] α = 8000; ωd = q ωd = 6000 ωo2 − α2 .·. ωo2 = ωd2 + α2 = 36 × 106 + 64 × 106 = 100 × 106 1 = 100 × 106 LC C= [b] α = 1 = 25 nF (100 × 106 )(0.4) 1 2RC . ·. R = 1 1 = = 2500 Ω 2αC (16,000)(25 × 10−9 ) [c] Vo = v(0) = 75 V [d] Io = iL (0) = −iR(0) − iC (0) 75 = 30 mA 2500 iR (0) = iC (0) = C dv (0) = 25 × 10−9 [6000(−300) − 8000(75)] = −60 mA dt .·. Io = −30 + 60 = 30 mA [e] iC (t) = 25 × 10−9 iR (t) = dv(t) = e−8000t(48.75 sin 6000t − 60 cos 6000t) mA dt v(t) = e−8000t(30 cos 6000t − 120 sin 6000t) mA 2500 iL (t) = −iR (t) − iC (t) = e−8000t(30 cos 6000t + 71.25 sin 6000t) mA, t≥0 Check: diL L = 0.4 × 10−3 e−8000t[187,000 cos 6000t − 750,000 sin 6000t] dt v(t) = e−8000t[75 cos 6000t − 300 sin 6000t] V P 8.8 [a] −α + q −α − α2 − ωo2 = −250 q α2 − ωo2 = −1000 Adding the above equations, − 2α = −1250 α = 625 rad/s © 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. Problems 8–9 1 1 = = 625 2RC 2R(0.1 × 10−6 ) R = 8 kΩ q 2 α2 − ωo2 = 750 4(α2 − ωo2) = 562,500 .·. ωo = 500 rad/s 1 LC 1 . ·. L = = 40 H 4 (25 × 10 )(0.1 × 10−6 ) ωo2 = 25 × 104 = [b] iR = v(t) = −1e−250t + 4e−1000t mA, R iC = C t ≥ 0+ dv(t) = 0.2e−250t − 3.2e−1000t mA, dt iL = −(iR + iC ) = 0.8e−250t − 0.8e−1000t mA, P 8.9 [a] 1 2RC 2 . ·. C = = t ≥ 0+ t≥0 1 = (500)2 LC 1 = 1 µF (500)2 (4) 1 = 500 2RC . ·. R = 1 = 1 kΩ 2(500)(10−6 ) v(0) = D2 = 8 V iR (0) = 8 = 8mA 1000 iC (0) = −8 + 10 = 2 mA dv 2 × 10−3 (0) = D1 − 500D2 = = 2000 V/s dt 10−6 .·. D1 = 2000 + 500(8) = 6000 V/s [b] v = 6000te−500t + 8e−500t V, t≥0 dv = [−3 × 106 t + 2000]e−500t dt dv iC = C = (−3000t + 2)e−500t mA, dt t ≥ 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. 8–10 P 8.10 CHAPTER 8. Natural and Step Responses of RLC Circuits α = 500/2 = 250 R= 1 106 = = 1000 Ω 2αC (500)(18) v(0+ ) = −11 + 20 = 9 V 9 = 9 mA 1000 iR (0+ ) = dv = 1100e−100t − 8000e−400t dt dv(0+ ) = 1100 − 8000 = −6900 V/s dt iC (0+ ) = 2 × 10−6 (−6900) = −13.8 mA iL (0+ ) = −[iR(0+ ) + iC (0+ )] = −[9 − 13.8] = 4.8 mA P 8.11 [a] 2α = 1000; q α = 500 rad/s 2 α2 − ωo2 = 600; C= L= ωo = 400 rad/s 1 1 = = 4 µF 2αR 2(500)(250) 1 ωo2 C = 1 = 1.5625 H × 10−6 ) (400)2 (4 iC (0+ ) = A1 + A2 = 45 mA diC diL diR + + =0 dt dt dt diC (0) diL (0) diR (0) =− − dt dt dt diL (0) 0 = = 0 A/s dt 1.5625 diR (0) 1 dv(0) 1 iC (0) 45 × 10−3 = = = = 45 A/s dt R dt R C (250)(4 × 10−6 ) . ·. diC (0) = 0 − 45 = −45 A/s dt .·. 200A1 + 800A2 = 45; A1 + A2 = 0.045 Solving, A1 = −15 mA; A2 = 60 mA .·. iC = −15e−200t + 60e−800t mA, t ≥ 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. Problems 8–11 [b] By hypothesis v = A3 e−200t + A4 e−800t, t≥0 v(0) = A3 + A4 = 0 dv(0) 45 × 10−3 = = 11,250 V/s dt 4 × 10−6 −200A3 − 800A4 = 11,250; .·. A3 = 18.75 V; v = 18.75e−200t − 18.75e−800t V, t≥0 v = 75e−200t − 75e−800t mA, 250 [d] iL = −iR − iC [c] iR (t) = iL = −60e−200t + 15e−800t mA, P 8.12 A4 = −18.75 V t ≥ 0+ t≥0 From the form of the solution we have v(0) = A1 + A2 dv(0+ ) = −α(A1 + A2 ) + jωd (A1 − A2) dt We know both v(0) and dv(0+ )/dt will be real numbers. To facilitate the algebra we let these numbers be K1 and K2 , respectively. Then our two simultaneous equations are K1 = A1 + A2 K2 = (−α + jωd )A1 + (−α − jωd )A2 The characteristic determinant is 1 ∆= 1 (−α + jωd ) (−α − jωd ) = −j2ωd The numerator determinants are N1 = K1 1 K2 (−α − jωd ) and N2 = 1 = −(α + jωd )K1 − K2 K1 (−α + jωd ) K2 = K2 + (α − jωd )K1 © 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. 8–12 CHAPTER 8. Natural and Step Responses of RLC Circuits It follows that A1 = and A2 = N1 ωd K1 − j(αK1 + K2 ) = ∆ 2ωd N2 ωd K1 + j(αK1 + K2 ) = ∆ 2ωd We see from these expressions that P 8.13 A1 = A∗2 . By definition, B1 = A1 + A2. From the solution to Problem 8.12 we have A1 + A2 = 2ωd K1 = K1 2ωd But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. (8.30). By definition, B2 = j(A1 − A2 ). From Problem 8.12 we have B2 = j(A1 − A2 ) = j[−2j(αK1 + K2 )] αK1 + K2 = 2ωd ωd It follows that K2 = −αK1 + ωd B2 , but K2 = dv(0+ ) dt and K1 = B1 . Thus we have dv + (0 ) = −αB1 + ωd B2, dt which is identical to Eq. (8.31). P 8.14 1 = 800 rad/s 2RC 1 ωo2 = = 106 LC √ ωd = 106 − 8002 = 600 rad/s [a] α = .·. v = B1e−800t cos 600t + B2 e−800t sin 600t v(0) = B1 = 30 30 = 6 mA; iC (0+ ) = −12 mA 5000 dv + −0.012 . ·. (0 ) = = −96,000 V/s dt 125 × 10−9 iR (0+ ) = −96,000 = −αB1 + ωd B2 = −(800)(30) + 600B2 .·. B2 = −120 .·. v = 30e−800t cos 600t − 120e−800t sin 600t V, t≥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. Problems [b] 8–13 dv = 6000e−800t (13 sin 600t − 16 cos 600t) dt dv = 0 when 16 cos 600t = 13 sin 600t or dt .·. 600t1 = 0.8885, t1 = 1.48 ms 600t2 = 0.8885 + π, 600t3 = 0.8885 + 2π, tan 600t = 16 13 t2 = 6.72 ms t3 = 11.95 ms 2π 2π = = 10.47 ms ωd 600 10.48 Td [d] t2 − t1 = 5.24 ms; = = 5.24 ms 2 2 [e] v(t1) = 30e−(1.184)(cos 0.8885 − 4 sin 0.8885) = −22.7 V [c] t3 − t1 = 10.47 ms; Td = v(t2) = 30e−(5.376)(cos 4.032 − 4 sin 4.032) = 0.334 V v(t3) = 30e−(9.56) (cos 7.17 − 4 sin 7.17) = −5.22 mV [f] P 8.15 [a] α = 0; ωd = ωo = √ 106 = 1000 rad/s v = B1 cos ωo t + B2 sin ωo t; C v(0) = B1 = 30 dv (0) = −iL (0) = −0.006 dt −48,000 = −αB1 + ωd B2 = −0 + 1000B2 .·. B2 = −48,000 = −48 V 1000 v = 30 cos 1000t − 48 sin 1000t V, [b] 2πf = 1000; f = t≥0 1000 ∼ = 159.15 Hz 2π © 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. 8–14 P 8.16 CHAPTER 8. Natural and Step Responses of RLC Circuits [c] √ [a] ωo2 302 + 482 = 56.6 V 1 109 = = = 4 × 106 LC (2.5)(100) ωo = 2000 rad/s 1 = 2000; 2RC R= 1 = 2500 Ω 4000C [b] v(t) = D1 te−5000t + D2 e−5000t v(0) = −15 V = D2 iC (0) = 5 + 15 = 11 mA 2.5 iC (0) 11 × 10−3 dv (0) = = = 110,000 dt C 100 × 10−9 D1 − 2000(−15) = 110,000 so D1 = 80,000 V/s .·. v(t) = (80,000t − 15)e−2000t V, [c] iC (t) = 0 when t≥0 dv (t) = 0 dt dv = (110,000 − 160 × 106 t))e−2000t dt dv = 0 when 160 × 106 t1 = 110,000; dt .·. t1 = 687.5 µs v(687.5µs) = (55 − 15)e−1.375 = 10.1136 V 1 1 [d] w(0) = (100 × 10−9 )(15)2 + (2.5)(0.005)2 = 42.5 µJ 2 2 1 1 10.1136 w(687.5 µs) = (100 × 10−9 )(10.1136)2 + (2.5) 2 2 2500 % remaining = P 8.17 [a] α = = 25.571 µJ 25.571 (100) = 60.17% 42.5 1 = 1250, 2RC s1 = −500, 2 ωo = 103 , therefore overdamped s2 = −2000 therefore v = A1 e−500t + A2 e−2000t + v(0 ) = 0 = A1 + A2 ; " dv(0+ ) iC (0+ ) = = 98,000 V/s dt C # © 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. Problems Therefore A1 = − 500A1 − 2000A2 = 98,000 +980 , 15 A2 = −980 15 980 −500t v(t) = [e − e−2000t] V, 15 8–15 t≥0 [b] Example 8.4: vmax ∼ = 74.1 V at 1.4 ms Example 8.5: vmax ∼ = 36.1 V at 1.0 ms Problem 8.17: vmax ∼ = 30.9 at 0.92 ms P 8.18 t<0: Vo = 15 V, Io = −60 mA t > 0: iR(0) = 15 = 150 mA; 100 iL (0) = −60 mA iC (0) = −150 − (−60) = −90 mA © 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. 8–16 CHAPTER 8. Natural and Step Responses of RLC Circuits α= 1 1 = = 5000 rad/s 2RC 2(100)(10−6 ) ωo2 = 1 1 = = 16 × 106 LC (62.5 × 10−3 )(10−6 ) s1,2 = −5000 ± √ 25 × 106 − 16 × 106 = −5000 ± 3000 s1 = −2000 rad/s; s2 = −8000 rad/s .·. vo = A1e−2000t + A2 e−8000t A1 + A2 = vo (0) = 15 dvo −90 × 10−3 (0) = −2000A1 − 8000A2 = = −90,000 dt 10−6 Solving, A1 = 5 V, A2 = 10 V .·. vo = 5e−2000t + 10e−8000t V, P 8.19 ωo2 = α= t≥0 1 1 = = 16 × 106 LC (62.5 × 10−3 )(10−6 ) 1 1 = = 2500 2RC 2(200)(10−6 ) s1,2 = −2500 ± √ 25002 − 16 × 106 = −2500 ± j3122.5rad/s vo (t) = B1 e−2500t cos 3122.5t + B2 e−2500t sin 3122.5t vo (0) = B1 = 15 V iR(0) = 15 = 75 mA 200 iL (0) = −60 mA iC (0) = −iR(0) − iL(0) = −15 mA .·. iC (0) = −15,000 C dvo (0) = −2500B1 + 3122.5B2 = −15,000 dt .·. B2 = 7.21 vo (t) = 15e−2500t cos 3122.5t + 7.21e−2500t sin 3122.5t V, t≥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. Problems P 8.20 ωo2 = α= 8–17 1 1 = = 16 × 106 −3 −6 LC (62.5 × 10 )(10 ) 1 1 = = 4000 2RC 2(125)(10−6 ) .·. α2 = ωo2 (critical damping) vo (t) = D1 te−4000t + D2 e−4000t vo (0) = D2 = 15 V iR(0) = 15 = 120 mA 125 iL (0) = −60 mA iC (0) = −60 mA dvo (0) = −4000D2 + D1 dt iC (0) −60 × 10−3 = = −60,000 C 10−6 D1 − 4000D2 = −60,000; vo (t) = 15e−4000t V, D1 = 0 t≥0 P 8.21 vT = −16,000iφ + iT (15,000) = −16,000 −iT (40) + it(15,000) 64 vT = 10,000 + 15,000 = 25 kΩ iT © 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. 8–18 CHAPTER 8. Natural and Step Responses of RLC Circuits 4000 (7.5) = 6 V; 5000 Vo = Io = 0 iC (0) = −iR(0) − iL(0) = − 6 = −240 µA 25,000 −240 × 10−6 iC (0) = = −60,000 C 4 × 10−9 ωo2 = α= 1 109 = = 16 × 106 ; LC (4)(15.625) ωo = 4000 rad/s 109 1 = = 5000 rad/s 2RC (2)(4)(25 × 103 ) α2 > ω02 so the response is overdamped s1,2 = −5000 ± √ 50002 − 40002 = −5000 ± 3000 rad/s vo = A1 e−2000t + A2e−8000t vo (0) = A1 + A2 = 6 V dvo (0) = −2000A1 − 8000A2 = −60,000 dt .·. A1 = −2 V; A2 = 8 V vo = 8e−8000t − 2e−2000t V, t≥0 P 8.22 vT = −16,000iφ + iT (15,000) = −16,000 −iT (40) + it(15,000 64 vT = 10,000 + 15,000 = 25 kΩ iT © 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. Problems 4000 (7.5) = 6 V; 5000 Vo = 8–19 Io = 0 iC (0) = −iR(0) − iL(0) = − 6 = −240 µA 25,000 iC (0) −240 × 10−6 = = −60,000 C 4 × 10−9 ωo2 = α= 109 1 = = 25 × 106 ; LC (4)(10) ωo = 5000 rad/s 1 109 = = 5000 rad/s 2RC (2)(4)(25 × 103 ) α2 = ω02 so the response is critically damped vo = D1 te−5000t + D2 e−5000t vo (0) = D2 = 6 V dvo (0) = D1 − αD2 = −60,000 dt .·. D1 = −60,000 + (5000)(6) = −30,000 V/s vo = −30,000te−5000t + 6e−5000t V, t≥0 P 8.23 vT = −16,000iφ + iT (15,000) = −16,000 −iT (40) + it(15,000 64 vT = 10,000 + 15,000 = 25 kΩ iT Vo = 4000 (7.5) = 6 V; 5000 Io = 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. 8–20 CHAPTER 8. Natural and Step Responses of RLC Circuits iC (0) = −iR(0) − iL(0) = − 6 = −240 µA 25,000 iC (0) −240 × 10−6 = = −60,000 C 4 × 10−9 ωo2 = α= 1 109 = = 62502 ; LC (4)(6.4) 1 109 = = 5000 rad/s 2RC (2)(4)(25 × 103 ) α2 < ω02 ωd = ωo = 6250 rad/s √ so the response is underdamped 62502 − 50002 = 3750 rad/s vo = B1 e−5000t cos 3750t + B2e−5000t sin 3750t vo (0) = B1 = 6 V dvo (0) = −5000B1 + 3750B2 = −60,000 dt .·. B2 = −8 V vo = e−5000t(6 cos 3750t − 8 sin 3750t) V, P 8.24 P 8.25 t≥0 ! diL [a] v = L = 16[e−20,000t − e−80,000t] V, t≥0 dt v [b] iR = = 40[e−20,000t − e−80,000t] mA, t ≥ 0+ R [c] iC = I − iL − iR = [−8e−20,000t + 32e−80,000t] mA, diL [a] v = L dt ! = 40e−32,000t sin 24,000t V, [b] iC (t) = I − iR − iL = 24 × 10−3 − t ≥ 0+ t≥0 v − iL 625 = [24e−32,000t cos 24,000t − 32e−32,000t sin 24,000t] mA, P 8.26 diL v=L dt ! = 960,000te−40,000t V, t ≥ 0+ t≥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. Problems P 8.27 8–21 t < 0: vo (0− ) = vo (0+ ) = 625 (25) = 20 V 781.25 iL (0− ) = iL (0+ ) = 0 t > 0: −160 × 10−3 + 20 + iC (0+ ) + 0 = 0; 125 .·. iC (0+ ) = 0 1 1 = = 800 rad/s 2RC 2(125)(5 × 10−6 ) ωo2 = 1 1 = = 64 × 104 −3 LC (312.5 × 10 )(5 × 10−6 ) .·. α2 = ωo2 critically damped [a] vo = Vf + D10 te−800t + D20 e−800t Vf = 0 dvo (0) = −800D20 + D10 = 0 dt vo (0+ ) = 20 = D20 D10 = 800D20 = 16,000 V/s .·. vo = 16,000te−800t + 20e−800t V, t ≥ 0+ [b] iL = If + D30 te−800t + D40 e−800t iL (0+ ) = 0; If = 160 mA; .·. 0 = 160 + D40 ; −800D40 + D30 = 64; diL (0+ ) 20 = = 64 A/s dt 312.5 × 10−3 D40 = −160 mA; D30 = −64 A/s .·. iL = 160 − 64,000te−800t − 160e−800t mA t≥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. 8–22 P 8.28 CHAPTER 8. Natural and Step Responses of RLC Circuits [a] wL = Z ∞ 0 ∞ Z pdt = voiL dt 0 vo = 16,000te−800t + 20e−800t V iL = 0.16 − 64te−800t − 0.16e−800t A p = 3.2e−800t + 2560te−800t − 3840te−1600t −1,024,000t2 e−1600t − 3.2e−1600t W Z wL = 3.2 ∞ Z e−800t dt + 2560 0 Z − 1,024,000 e−800t = 3.2 −800 ∞ 0 ∞ 0 Z te−800t dt − 3480 ∞ 0 Z t2 e−1600t dt − 3.2 ∞ 0 te−1600t dt ∞ 0 e−1600t dt ∞ 2560 −800t + e (−2560t − 1) (800)2 0 3840 −1600t − e (−1600t − 1) (1600)2 ∞ 0 ∞ 1,024,000 −1600t − e (16002 t2 + 3200t + 2) 3 (−1600) 0 − 3.2 e−1600t (−1600) ∞ 0 All the upper limits evaluate to zero hence 2560 3840 (1,024,000)(2) 3.2 3.2 + − − − = 4 mJ 2 2 3 800 800 1600 1600 1600 Note this value corresponds to the final energy stored in the inductor, i.e. 1 wL (∞) = (312.5 × 10−3 )(0.16)2 = 4 mJ. 2 [b] v = 16,000te−800t + 20e−800t V v iR = = 128te−800t + 0.16e−800t A 125 wL = pR = viR = 2,048,000t2 e−1600t + 5120te−1600t + 3.2e−1600t wR = Z 0 ∞ pR dt Z = 2,048,000 = ∞ 0 Z t2 e−1600t dt + 5120 ∞ 0 2,048,000e−1600t [16002 t2 + 3200t + 2] −16003 5120e−1600t (−1600t − 1) 16002 ∞ + 0 Z te−1600t dt + 3.2 ∞ 0 e−1600t dt ∞ + 0 3.2e−1600t (−1600) ∞ 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. Problems 8–23 Since all the upper limits evaluate to zero we have 2,048,000(2) 5120 3.2 + + = 5 mJ 3 2 1600 1600 1600 [c] 160 = iR + iC + iL (mA) wR = iR + iL = 160 + 64,000te−800t mA .·. iC = 160 − (iR + iL ) = −64,000te−800t mA = −64te−800t A pC = viC = [16,000te−800t + 20e−800t][−64te−800t] = −1,024,000t2 e−1600t − 1280e−1600t wC = −1,024,000 wC = Z ∞ 2 −1600t t e 0 dt − 1280 Z ∞ 0 te−1600t dt −1,024,000e−1600t [16002 t2 + 3200t + 2] −16003 ∞ 0 − 1280e−1600t (−1600t − 1) 16002 ∞ 0 Since all upper limits evaluate to zero we have −1,024,000(2) 1280(1) − = −1 mJ 16003 16002 Note this 1 mJ corresponds to the initial energy stored in the capacitor, i.e., wC = 1 wC (0) = (5 × 10−6 )(20)2 = 1 mJ. 2 Thus wC (∞) = 0 mJ which agrees with the final value of v = 0. [d] is = 160 mA ps (del) = 160v mW = 0.16[16,000te−800t + 20e−800t ] = 3.2e−800t + 2560te−800t W Z ws = 3.2 ∞ 0 e −800t 3.2e−800t = −800 = ∞ 0 dt + Z ∞ 0 2560te−800t dt 2560e−800t + (−800t − 1) 8002 ∞ 0 3.2 2560 + = 8 mJ 800 800 [e] wL = 4 mJ (absorbed) wR = 5 mJ (absorbed) wC = 1 mJ (delivered) © 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. 8–24 CHAPTER 8. Natural and Step Responses of RLC Circuits wS = 8 mJ (delivered) X P 8.29 ωo2 = α= wdel = wabs = 9 mJ. 1 1 = = 108 ; −3 LC (50 × 10 )(0.2 × 10−6 ) ωo = 104 rad/s 1 1 = = 12,500 rad/s 2RC 2(200)(0.2 × 10−6 ) s1,2 = −12,500 ± .·. overdamped q s1 = −5000 rad/s; (12,500)2 − 108 = −12,500 ± 7500 rad/s s2 = −20,000 rad/s If = 60 mA iL = 60 × 10−3 + A01e−5000t + A02e−20,000t .·. −45 × 10−3 = 60 × 10−3 + A01 + A02 ; A01 + A02 = −105 × 10−3 diL 15 = −5000A01 − 20,000A02 = = 300 dt 0.05 Solving, A01 = −120 mA; A02 = 15 mA iL = 60 − 120e−5000t + 15e−20,000t mA, P 8.30 α= t≥0 1 1 = = 8000; 2RC 2(312.5)(0.2 × 10−6 ) ωo = 104 α2 = 64 × 106 underdamped √ s1,2 = −8000 ± j 80002 − 108 = −8000 ± j6000 rad/s iL = 60 × 10−3 + B10 e−8000t cos 6000t + B20 e−8000t sin 6000t −45 × 10−3 = 60 × 10−3 + B10 .·. B10 = −105 mA diL (0) = −8000B10 + 6000B20 = 300 dt .·. B20 = −90 mA iL = 60 − 105e−8000t cos 6000t − 90e−8000t sin 6000t mA, t≥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. Problems P 8.31 α= 8–25 1 1 = = 104 −6 2RC 2(250)(0.2 × 10 ) α2 = 104 = ωo2 critical damping 4 4 4 iL = If + D10 te−10 t + D20 e−10 t = 60 × 10−3 + D10 te−10 t + D20 e−10 iL (0) = −45 × 10−3 = 60 × 10−3 + D20 ; 4t .·. D20 = −105 mA diL (0) = −104 D20 + D10 = 300 A/s dt .·. D10 = 300 + 104 (−105 × 10−3 ) = −750 A/s 4 4 iL = 60 − 750,000te−10 t − 105e−10 t mA, P 8.32 −15 t<0: iL (0− ) = = −5 mA; 3000 The circuit reduces to: t≥0 vC (0− ) = 0 V iL (∞) = 4 mA ωo2 = 1 106 = = 6400; LC (62.5)(2.5) ωo = 80 rad/s 1 106 α= = = 100 2RC (4000)(2.5) s1,2 = −100 ± √ 1002 − 802 = −100 ± 60 s1 = −40 rad/s; s2 = −160 rad/s iL = If + A01e−40t + A02e−160t iL (∞) = If = 4mA iL (0) = A01 + A02 + If = −5 mA © 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. 8–26 CHAPTER 8. Natural and Step Responses of RLC Circuits .·. A01 + A02 + 4 = −5 so A01 + A02 = −9 mA diL (0) = 0 = −40A1 − 160A02 dt A01 = −12 mA, Solving, A02 = 3 mA iL = 4 − 12e−40t + 3e−160t mA, P 8.33 t≥0 1 vC (0+ ) = (240) = 120 V 2 iL (0+ ) = 60 mA; α= ωo2 iL (∞) = 240 × 10−3 = 48 mA 5 1 106 = = 40 2RC 2(2500)(5) 1 106 = = = 2500 LC 400 α2 = 1600; α2 < ωo2 ; .·. underdamped √ s1,2 = −40 ± j 2500 − 1600 = −40 ± j30 rad/s iL = If + B10 e−αt cos ωd t + B20 e−αt sin ωd t = 48 + B10 e−40t cos 30t + B20 e−40t sin 30t iL (0) = 48 + B10 ; B10 = 60 − 48 = 12 mA diL 120 (0) = 30B20 − 40B10 = = 1.5 = 1500 × 10−3 dt 80 .·. 30B20 = 40(12) × 10−3 + 1500 × 10−3 ; B20 = 66 mA .·. iL = 48 + 12e−40t cos 30t + 66e−40t sin 30t mA, t≥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. Problems P 8.34 α= 8–27 1 1 = = 1000 2RC 2(400)(1.25 × 10−6 ) 1 1 = = 64 × 104 LC (1.25 × 10−6 )(1.25) ωo2 = s1,2 = −1000 ± √ 10002 − 64 × 104 = −1000 ± 600 rad/s s1 = −400 rad/s; s2 = −1600 rad/s vo (∞) = 0 = Vf .·. vo = A01e−400t + A02e−1600t vo (0) = 12 = A01 + A02 Note: iC (0+ ) = 0 dvo .·. (0) = 0 = −400A01 − 1600A02 dt A01 = 16 V, Solving, A02 = −4 V vo (t) = 16e−400t − 4e−1600t V, P 8.35 t≥0 [a] io = If + A01 e−400t + A02e−1600t If = 12 = 30mA; 400 io (0) = 0 0 = 30 × 10−3 + A01 + A02 , .·. A01 + A02 = −30 × 10−3 dio 12 (0) = = −400A01 − 1600A02 dt 1.25 Solving, A01 = −32 mA; A02 = 2 mA io = 30 − 32e−400t + 2e−1600t mA, [b] t≥0 dio = [12.8e−400t − 3.2e−1600t] dt vo = L dio = 16e−400t − 4e−1600t V, dt t≥0 This agrees with the solution to Problem 8.34. © 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. 8–28 P 8.36 CHAPTER 8. Natural and Step Responses of RLC Circuits iL (0− ) = iL (0+ ) = 7.5 = 30 mA 250 For t > 0 iL (0− ) = iL (0+ ) = 30 mA α= 1 = 80 rad/s; 2RC ωd = √ ωo2 = 1 = 104 LC so ωo = 100 rad/s 1002 − 802 = 60 rad/s vo (∞) = 0 = Vf ; B10 = v(0) = 0 vo = e−80tB20 sin 60t iC (0+ ) = −30 + 30 + 0 = 0 dvo .·. =0 dt dvo (0) = −αB10 + ωd B20 = 0 + 60B20 = 0 dt .·. B10 = 0; B20 = 0 .·. vo = 0 for t ≥ 0 Note: vo(0) = 0; vo (∞) = 0; dvo (0) =0 dt Hence, the 30 mA current circulates between the current source and the ideal inductor in the equivalent circuit. In the original circuit, the 7.5 V source sustains a current of 30 mA in the inductor. This is an example of a circuit going directly into steady state when the switch is closed. There is no transient period, or interval. © 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. Problems P 8.37 For α= 8–29 t>0 1 = 1000; 2RC 1 = 64 × 104 LC s1,2 = −1000 ± 600 rad/s s1 = −400 rad/s; s2 = −1600 rad/s vo = Vf + A01 e−400t + A02e−1600t Vf = 0; vo (0+ ) = 0; iC (0+ ) = 30 mA .·. A01 + A02 = 0 dvo (0+ ) iC (0+ ) = = 24,000 V/s dt 1.25 × 10−6 dvo (0+ ) = −400A01 − 1600A02 = 24,000 dt Solving, A01 = 20 V; A02 = −20 V vo = 20e−400t − 20e−1600t V, P 8.38 t≥0 [a] From the solution to Prob. 8.37 s1 = −400 rad/s and s2 = −1600 rad/s, therefore io = If + A01e−400t + A02e−1600t If = 30 mA; io (0+ ) = 0; .·. 0 = 30 × 10−3 + A01 + A02; dio (0+ ) =0 dt −400A01 − 1600A02 = 0 Solving A01 = −40 mA; A02 = 10 mA .·. io = 30 − 40e−400t + 10e−1600t mA, t≥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. 8–30 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] dio = 16e−400t − 16e−1600t dt vo = L dio = 20e−400t − 20e−1600t V, dt t≥0 This agrees with the solution to Problem 8.27. P 8.39 q α2 [a] −α + − ω02 = −4000; .·. α = 10,000 rad/s, α= R = 10,000; 2L ωo2 = −α − ω02 = 64 × 106 q α2 − ω02 = −16,000 R = 20,000L 1 = 64 × 106 ; LC L= 109 = 0.5 H 64 × 106 (31.25) R = 10,000 Ω [b] i(0) = 0 L di(0) = vc (0); dt .·. vc2(0) = 576; 1 (31.25) × 10−9 vc2(0) = 9 × 10−6 2 vc (0) = 24 V di(0) 24 = = 48 A/s dt 0.5 [c] i(t) = A1e−4000t + A2e−16,000t i(0) = A1 + A2 = 0 di(0) = −4000A1 − 16,000A2 = 48 dt Solving, .·. A1 = 4 mA; A2 = −4 mA i(t) = 4e−4000t − 4e−16,000t mA, [d] t≥0 di(t) = −16e−4000t + 64e−16,000t dt di = 0 when 64e−16,000t = 16e−4000t dt or e12,000t = 4 .·. t = ln 4 = 115.52 µs 12,000 © 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. Problems [e] imax = 4e−0.4621 − 4e−1.8484 = 1.89 mA di [f] vL(t) = 0.5 = [−8e−1000t + 32e−4000t] V, dt P 8.40 [a] 8–31 t ≥ 0+ 1 = 20,0002 LC There are many possible solutions. This one begins by choosing L = 1 mH. Then, C= 1 (1 × 10−3 )(20,000)2 = 2.5 µF We can achieve this capacitor value using components from Appendix H by combining four 10 µF capacitors in series. Critically damped: α = ω0 = 20,000 so R = 20,000 2L .·. R = 2(10−3 )(20,000) = 40 Ω We can create this resistor value using components from Appendix H by combining a 10 Ω resistor and two 15 Ω resistors in series. The final circuit: q [b] s1,2 = −α ± α2 − ω02 = −20,000 ± 0 Therefore there are two repeated real roots at −20,000 rad/s. P 8.41 [a] Underdamped response: α < ω0 so α < 20,000 Therefore we choose a larger resistor value than the one used in Problem 8.40 to give a smaller value of α. For convenience, pick α = 16,000 rad/s: R = 16,000 so R = 2(16,000)(10−3 ) = 32 Ω 2L We can create a 32 Ω resistance by combining a 10 Ω resistor and a 22 Ω resistor in series. α= s1,2 = −16,000 ± q 16,0002 − 20,0002 = −16,000 ± j12,000 rad/s © 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. 8–32 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] Overdamped response: α > ω0 so α > 20,000 Therefore we choose a smaller resistor value than the one used in Problem 8.40. Choose R = 50 Ω, which can be created by combining two 100 Ω resistors in parallel: α= R = 25,000 2L s1,2 = −25,000 ± q 25,0002 − 20,0002 = −25,000 ± 15,000 = −10,000 rad/s; P 8.42 α = 2000 rad/s; R = 2000; 2L − 40,000 rad/s ωd = 1500 rad/s ωo2 − α2 = 225 × 104 ; α= and ωo2 = 625 × 104 ; wo = 25,000 rad/s R = 4000L 1 = 625 × 104 ; LC L= 1 (625 × 104 )(80 × 10−9 ) = 2H .·. R = 8 kΩ i(0+ ) = B1 = 7.5 mA; at t = 0+ 60 + vL(0+ ) − 30 = 0; . ·. vL (0+ ) = −30 V di(0+ ) −30 = = −15 A/s dt 2 di(0+ ) .·. = 1500B2 − 2000B1 = −15 dt .·. 1500B2 = 2000(7.5 × 10−3 ) − 15; .·. i = 7.5e−2000t sin 1500t mA, .·. B2 = 0 A t≥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. Problems P 8.43 8–33 From Prob. 8.42 we know vc will be of the form vc = B3 e−2000t cos 1500t + B4 e−2000t sin 1500t From Prob. 8.42 we have vc (0) = −30 V = B3 and dvc (0) iC (0) 7.5 × 10−3 = = = 93.75 × 103 −9 dt C 80 × 10 dvc (0) = 1500B4 − 2000B3 = 93,750 dt .·. 1500B4 = 2000(−30) + 93,750; B4 = 22.5 V vc (t) = −30e−2000t cos 1500t + 22.5e−2000t sin 1500t V P 8.44 [a] ωo2 = α= t≥0 1 109 = = 25 × 106 LC (125)(0.32) R = ωo = 5000 rad/s 2L .·. R = (5000)(2)L = 1250 Ω [b] i(0) = iL (0) = 6 mA vL (0) = 15 − (0.006)(1250) = 7.5 V di 7.5 (0) = = 60 A/s dt 0.125 [c] vC = D1 te−5000t + D2 e−5000t vC (0) = D2 = 15 V dvC iC (0) −iL (0) (0) = D1 − 5000D2 = = = −18,750 dt C C .·. D1 = 56,250 V/s vC = 56,250te−5000t + 15e−5000t V, t≥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. 8–34 P 8.45 CHAPTER 8. Natural and Step Responses of RLC Circuits ωo2 = α= 1 1 = = 25 LC (10)(4 × 10−3 ) R 80 = = 4; 2L 2(10) α2 < ωo2 .·. α2 = 16 underdamped √ s1,2 = −4 ± j 9 = −4 ± j3 rad/s i = B1 e−4t cos 3t + B2 e−4t sin 3t i(0) = B1 = −240/100 = −2.4 A di (0) = 3B2 − 4B1 = 0 dt .·. B2 = −3.2 A i = −2.4e−4t cos 3t − 3.2 sin 3t A, P 8.46 t≥0 [a] For t > 0: Since i(0− ) = i(0+ ) = 0 va (0+ ) = 75 V 7 [b] va = 2000i + 10 Z 0 t i dx + 75 dva di = 2000 + 107 i dt dt dva (0+ ) di(0+ ) di(0+ ) = 2000 + 107 i(0+ ) = 2000 dt dt dt −L di(0+ ) = 75 dt di(0+ ) = −2.5(75) = −187.5 A/s dt . ·. dva(0+ ) = −375,000 V/s dt © 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. Problems [c] α = 8–35 R 5000 = = 6250 rad/s 2L 0.8 1 106 = = 25 × 106 LC (0.4)(0.1) √ = −6250 ± 62502 − 25 × 106 = −6250 ± 3750 rad/s ωo2 = s1,2 .·. s1 = −2500 rad/s; s2 = −10,000 rad/s Overdamped: va = A1 e−2500t + A2e−10,000t va (0) = A1 + A2 = 75 V dva (0) = −2500A1 − 10,000A2 = −375,000; dt A2 = 25 V t ≥ 0+ va = 50e−2500t + 25e−10,000t V, P 8.47 .·. A1 = 50 V, [a] t < 0: 80 = 100 mA; vo = 500io = (500)(0.01) = 50 V 800 t > 0: R 500 α= = = 105 rad/s 2L 2(2.5 × 10−3 ) io = ωo2 = 1 1 = = 100 × 108 −3 −9 LC (2.5 × 10 )(40 × 10 ) α2 = ωo2 . ·. critically damped .·. io(t) = D1 te−10 t + D2 e−10 5 5t io (0) = D2 = 100 mA dio (0) = −αD2 + D1 = 0 dt . ·. D1 = 105 (100 × 10−3 ) = 10,000 5 5 io (t) = 10,000te−10 t + 0.1e−10 t A, 5 [b] vo(t) = D3 te−10 t + D4 e−10 t ≥ 0+ 5t vo (0) = D4 = 50 C dvo (0) = −0.1 dt © 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. 8–36 CHAPTER 8. Natural and Step Responses of RLC Circuits dvo −0.1 (0) = = −25 × 105 V/s = −αD4 + D3 dt 40 × 10−9 . ·. D3 = 105 (50) − 25 × 105 = 25 × 105 5 5 vo (t) = 25 × 105 te−10 t + 50e−10 t V, P 8.48 t ≥ 0+ t < 0: i(0) = 240 240 = = 6A 8 + 30k70 + 11 40 vo (0) = 240 − 8(6) − 70 (6)(20) = 108 V 100 t > 0: α= R 20 = = 10, 2L 2(1) ωo2 = α2 = 100 1 1 = = 200 LC (1)(5 × 10−3 ) ωo2 > α2 underdamped s1,2 = −100 ± √ 100 − 200 = −10 ± j10 rad/s vo = B1 e−10t cos 10t + B2e−10t sin 10t © 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. Problems 8–37 vo (0) = B1 = 108 V C dvo (0) = −6, dt dvo −6 = = −1200 V/s dt 5 × 10−3 dvo (0) = −10B1 + 10B2 = −1200 dt 10B2 = −1200 + 10B1 = −1200 + 1080; .·. vo = 108e−10t cos 10t − 12e−10t sin 10t V, P 8.49 iC (0) = 0; α= B2 = −120/10 = −12 V t≥0 vo (0) = 50 V R 8000 = = 25,000 rad/s 2L 2(160 × 10−3 ) ωo2 = 1 1 = = 625 × 106 −3 −9 LC (160 × 10 )(10 × 10 ) .·. α2 = ωo2 ; critical damping vo (t) = Vf + D10 te−25,000t + D20 e−25,000t Vf = 250 V vo (0) = 250 + D20 = 50; D20 = −200 V dvo (0) = −25,000D20 + D10 = 0 dt D10 = 25,000D20 = −5 × 106 V/s vo = 250 − 5 × 106 te−25,000t − 200e−25,000t V, P 8.50 α= t≥0 R = 2000 rad/s 2L ωo2 = 1 1 = = 256 × 104 −3 LC (62.5 × 10 )(6.25 × 10−6 ) s1,2 = −2000 ± √ 4 × 106 − 256 × 104 = −2000 ± j1200 rad/s © 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. 8–38 CHAPTER 8. Natural and Step Responses of RLC Circuits vo = Vf + A01 e−800t + A02e−3200t vo (0) = 0 = Vf + A01 + A02 vo (∞) = 60 V; .·. A01 + A02 = −60 dvo (0) = 0 = −800A01 − 3200A02 dt .·. A01 = −80 V; A02 = 20 V vo = 60 − 80e−800t + 20e−3200t V, P 8.51 α= t≥0 R = 2000 rad/s 2L ωo2 = 1 1 = = 4 × 106 LC (62.5 × 10−3 )(4 × 10−6 ) .·. ωo = 2000 rad/s The response is therefore critically damped vo = Vf + D10 te−2000t + D20 e−2000t vo (0) = 0 = Vf + D20 vo (∞) = 60 V; .·. D20 = −60 V dvo (0) = 0 = D10 − αD20 dt .·. D10 = (2000)(−60) = −120,000 V/s vo = 60 − 120,000te−2000t − 60e−2000t V, P 8.52 α= t≥0 R = 2000 rad/s 2L ωo2 = 1 1 = = 625 × 104 −3 LC (62.5 × 10 )(2.56 × 10−6 ) .·. ωo = 2500 rad/s The response is therefore underdamped. ωd = √ 25002 − 20002 = 1500 rad/s © 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. Problems 8–39 vo = Vf + B10 e−2000t cos 1500t + B20 e−2000t sin 1500t vo (0) = 0 = Vf + B10 .·. B10 = −60 V vo (∞) = 60 V; dvo (0) = 0 = −2000B10 + 1500B20 dt .·. B20 = −80 V vo = V, P 8.53 t≥0 [a] t < 0: io (0− ) = 60 = 6 mA 10,000 vC (0− ) = 20 − (6000)(0.006) = −16 V t = 0+ : 3 kΩk6 kΩ = 2 kΩ .·. vo(0+ ) = (0.006)(2000) − 16 = 12 − 16 = −4 V and vL (0+ ) = 20 − (−4) = 24 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. 8–40 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] vo(t) = 2000io + vC dvo dio dvC (t) = 2000 + dt dt dt dvo + dio dvC + (0 ) = 2000 (0+ ) + (0 ) dt dt dt vL (0+ ) = L dio + (0 ) dt dio + vL(0+ ) 24 (0 ) = = = 48 A/s dt L 0.5 C dvc + (0 ) = io (0+ ) dt . ·. dvc + 6 × 10−3 (0 ) = = 7680 dt 781.25 × 10−9 . ·. dvo + (0 ) = 2000(48) + 7680 = 103,680 V/s dt [c] ωo2 = 1 = 2.56 × 106 ; LC α= R = 2000 rad/s 2L α2 > ωo2 ωo = 1600 rad/s overdamped s1,2 = −2000 ± j1200 rad/s vo (t) = Vf + A01 e−800t + A02 e−3200t Vf = vo (∞) = 20 V 20 + A01 + A02 = −4; Solving A01 = 11.2; −800A01 − 3200A02 = 103,680 A02 = −35.2 .·. vo(t) = 20 + 11.2e−800t − 35.2e−3200t V, P 8.54 t ≥ 0+ [a] Let i be the current in the direction of the voltage drop vo (t). Then by hypothesis i = if + B10 e−αt cos ωd t + B20 e−αt sin ωd t if = i(∞) = 0, i(0) = Vg = B10 R Therefore i = B10 e−αt cos ωd t + B20 e−αt sin ωd t © 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. Problems L di(0) = 0, dt therefore 8–41 di(0) =0 dt di = [(ωd B20 − αB10 ) cos ωd t − (αB20 + ωd B10 ) sin ωd t] e−αt dt Therefore ωd B20 − αB10 = 0; B20 = α 0 α Vg B1 = ωd ωd R Therefore α2 Vg ωd Vg di vo = L = − L + dt ωd R R ( ! LVg =− R α2 + ωd2 −αt e sin ωd t ωd Vg L =− R ωo2 −αt e sin ωd t ωd vo = − Vg L Rωd sin ωd t e−αt ) Vg L =− R =− ) α2 + ωd sin ωd t e−αt ωd ( [b] ! ! ! 1 e−αt sin ωd t LC Vg −αt e sin ωd t V, RCωd t≥0 dvo Vg =− {ωd cos ωd t − α sin ωd t}e−αt dt ωd RC dvo = 0 when dt tan ωd t = ωd α Therefore ωd t = tan−1 (ωd /α) (smallest t) t= P 8.55 1 ωd tan−1 ωd α [a] From Problem 8.54 we have vo = −Vg −αt e sin ωd t RCωd α= R 4800 = = 37,500 rad/s 2L 2(64 × 10−3 ) ωo2 = 1 1 = = 3906.25 × 106 −3 −9 LC (64 × 10 )(4 × 10 ) ωd = q ωo2 − α2 = 50 krad/s © 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. 8–42 CHAPTER 8. Natural and Step Responses of RLC Circuits −Vg −(−72) = = 75 RCωd (4800)(4 × 10−9 )(50 × 103 ) .·. vo = 75e−37,500t sin 50,000t V [b] From Problem 8.54 1 ωd td = tan−1 ωd α 1 50,000 = tan−1 50,000 37,500 ! td = 18.55 µs [c] vmax = 75e−0.0375(18.55) sin[(0.05)(18.55)] = 29.93 V [d] R = 480 Ω; α = 3750 rad/s ωd = 62,387.4 rad/s vo = 601.08e−3750t sin 62,387.4t V, t≥0 td = 24.22 µs vmax = 547.92 V P 8.56 t < 0: iL (0) = −150 = −5 A 30 vC (0) = 18iL (0) = −90 V t > 0: α= R 10 = = 50 rad/s 2L 2(0.1) ωo2 = 1 1 = = 5000 LC (0.1)(2 × 10−3 ) © 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. Problems ωo > α2 .·. s1,2 = −50 ± 8–43 underdamped √ 502 − 5000 = −50 ± j50 vc = 60 + B10 e−50t cos 50t + B20 e−50t sin 50t . ·. vc (0) = −90 = 60 + B10 C dvc (0) = −5; dt B10 = −150 dvc −5 (0) = = −2500 dt 2 × 10−3 dvc (0) = −50B10 + 50B2 = −2500 dt .·. B20 = −200 vc = 60 − 150e−50t cos 50t − 200e−50t sin 50t V, P 8.57 t≥0 [a] vc = Vf + [B10 cos ωd t + B20 sin ωd t] e−αt dvc = [(ωd B20 − αB10 ) cos ωd t − (αB20 + ωd B10 ) sin ωd t]e−αt dt Since the initial stored energy is zero, dvc (0+ ) =0 dt vc (0+ ) = 0 and It follows that B10 = −Vf and B20 = αB10 ωd When these values are substituted into the expression for [dvc/dt], we get dvc = dt α2 + ωd Vf e−αt sin ωd t ωd ! But Vf = V Therefore [b] and dvc = dt dvc = 0 when dt α2 α2 + ωd2 ω2 + ωd = = o ωd ωd ωd ωo2 V e−αt sin ωd t ωd ! sin ωd t = 0, or ωd t = nπ where n = 0, 1, 2, 3, . . . Therefore t = nπ ωd © 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. 8–44 CHAPTER 8. Natural and Step Responses of RLC Circuits nπ , ωd [c] When tn = and cos ωd tn = cos nπ = (−1)n sin ωd tn = sin nπ = 0 Therefore vc (tn ) = V [1 − (−1)n e−αnπ/ωd ] [d] It follows from [c] that v(t1) = V + V e−(απ/ωd ) P 8.58 vc (t3) = V + V e−(3απ/ωd) vc (t1) − V e−(απ/ωd ) = −(3απ/ω ) = e(2απ/ωd) d vc (t3) − V e Therefore But and 2π = t3 − t1 = Td , ωd ( 1 vc (t1) − V ln Td vc (t3) − V ) ; Td = t3 − t1 = 7000 63.84 ln = 1000; α= 2π 26.02 thus α = ωd = 1 [vc(t1 ) − V ] ln Td [vc(t3 ) − V ] 3π π 2π − = ms 7 7 7 2π = 7000 rad/s Td ωo2 = ωd2 + α2 = 49 × 106 + 106 = 50 × 106 L= 1 (50 × 106 )(0.1 × 10−6 ) = 200 mH; R = 2αL = 400 Ω P 8.59 At t = 0 the voltage across each capacitor is zero. It follows that since the operational amplifiers are ideal, the current in the 500 kΩ is zero. Therefore there cannot be an instantaneous change in the current in the 1 µF capacitor. Since the capacitor current equals C(dvo/dt), the derivative must be zero. P 8.60 [a] From Example 8.13 therefore dg(t) = 2, dt g(t) − g(0) = 2t; iR = d2 vo =2 dt2 g(t) = dvo dt g(t) = 2t + g(0); g(0) = dvo (0) dt 5 dvo (0) × 10−3 = 10 µA = iC = −C 500 dt © 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. Problems 8–45 dvo (0) −10 × 10−6 = = −10 = g(0) dt 1 × 10−6 dvo = 2t − 10 dt dvo = 2t dt − 10 dt vo − vo (0) = t2 − 10t; vo = t2 − 10t + 8, vo (0) = 8 V 0 ≤ t ≤ tsat [b] t2 − 10t + 8 = −9 t2 − 10t + 17 = 0 t∼ = 2.17 s P 8.61 Part (1) — Example 8.14, with R1 and R2 removed: [a] Ra = 100 kΩ; d2 vo 1 = dt2 Ra C1 C1 = 0.1 µF; 1 vg ; Rb C2 vg = 250 × 10−3 ; therefore Rb = 25 kΩ; 1 = 100 Ra C1 C2 = 1 µF 1 = 40 Rb C2 d2 vo = 1000 dt2 dvo (0) , our solution is vo = 500t2 dt The second op-amp will saturate when [b] Since vo (0) = 0 = vo = 6 V, or tsat = q 6/500 ∼ = 0.1095 s dvo1 1 =− vg = −25 dt RaC1 [d] Since vo1 (0) = 0, vo1 = −25t V [c] At t = 0.1095 s, vo1 ∼ = −2.74 V Therefore the second amplifier saturates before the first amplifier saturates. Our expressions are valid for 0 ≤ t ≤ 0.1095 s. Once the second op-amp saturates, our linear model is no longer valid. Part (2) — Example 8.14 with vo1(0) = −2 V and vo (0) = 4 V: [a] Initial conditions will not change the differential equation; hence the equation is the same as Example 8.14. © 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. 8–46 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] vo = 5 + A01 e−10t + A02e−20t (from Example 8.14) vo (0) = 4 = 5 + A01 + A02 4 2 + iC (0+ ) − =0 100 25 iC (0+ ) = 4 dvo (0+ ) mA = C 100 dt dvo (0+ ) 0.04 × 10−3 = = 40 V/s dt 10−6 dvo = −10A01 e−10t − 20A02 e−20t dt dvo + (0 ) = −10A01 − 20A02 = 40 dt Therefore −A01 − 2A02 = 4 and Thus, A01 = 2 and A02 = −3 A01 + A02 = −1 vo = 5 + 2e−10t − 3e−20t V [c] Same as Example 8.14: dvo1 + 20vo1 = −25 dt [d] From Example 8.14: vo1 (∞) = −1.25 V; v1(0) = −2 V (given) Therefore vo1 = −1.25 + (−2 + 1.25)e−20t = −1.25 − 0.75e−20t 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. Problems P 8.62 8–47 [a] 2C dva va − vg va + + =0 dt R R (1) Therefore dva va vg + = dt RC 2RC 0 − va d(0 − vb) +C =0 R dt (2) Therefore dvb va + = 0, dt RC va = −RC dvb dt 2vb dvb d(vb − vo ) +C +C =0 R dt dt (3) Therefore From (2) we have dvb vb 1 dvo + = dt RC 2 dt dva d2 vb = −RC 2 dt dt and va = −RC dvb dt When these are substituted into (1) we get (4) − RC d2 vb dvb vg − = 2 dt dt 2RC Now differentiate (3) to get (5) d2 vb 1 dvb 1 d2 vo + = dt2 RC dt 2 dt2 But from (4) we have (6) d2 vb 1 dvb vg + = − dt2 RC dt 2R2 C 2 Now substitute (6) into (5) d2 vo vg =− 2 2 2 dt RC © 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. 8–48 CHAPTER 8. Natural and Step Responses of RLC Circuits d2 vo vg = 2 2 2 dt RC The two equations are the same except for a reversal in algebraic sign. [b] When R1 C1 = R2 C2 = RC : [c] Two integrations of the input signal with one operational amplifier. P 8.63 [a] d2 vo 1 = vg 2 dt R1 C1R2 C2 1 10−6 = = 250 R1 C1 R2 C2 (100)(400)(0.5)(0.2) × 10−6 × 10−6 d2 vo = 250vg dt2 . ·. 0 ≤ t ≤ 0.5− : vg = 80 mV d2 vo = 20 dt2 Let g(t) = Z g(t) g(0) dvo , dt dx = 20 Z t 0 then dy g(t) − g(0) = 20t, g(t) = dg = 20 or dg = 20 dt dt g(0) = dvo (0) = 0 dt dvo = 20t dt dvo = 20t dt Z vo (t) vo (0) dx = 20 Z t 0 vo (t) = 10t2 V, x dx; vo(t) − vo (0) = 10t2 , vo (0) = 0 0 ≤ t ≤ 0.5− dvo1 1 =− vg = −20vg = −1.6 dt R1C1 dvo1 = −1.6 dt Z vo1 (t) vo1 (0) dx = −1.6 Z 0 t dy vo1 (t) − vo1(0) = −1.6t, vo1 (t) = −1.6t V, vo1 (0) = 0 0 ≤ t ≤ 0.5− © 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. Problems 0.5+ ≤ t ≤ tsat: d2 vo = −10, dt2 Z dg(t) = −10 dt g(t) g(0.5+) dvo dt let g(t) = dg(t) = −10; dt 8–49 dx = −10 Z t 0.5 dy g(t) − g(0.5+ ) = −10(t − 0.5) = −10t + 5 dvo (0.5+ ) g(0.5 ) = dt + C dvo 0 − vo1 (0.5+ ) (0.5+ ) = dt 400 × 103 vo1 (0.5+ ) = vo (0.5− ) = −1.6(0.5) = −0.80 V . ·. C dvo1 (0.5+ ) 0.80 = = 2 µA dt 0.4 × 103 dvo1 2 × 10−6 + (0.5 ) = = 10 V/s dt 0.2 × 10−6 .·. g(t) = −10t + 5 + 10 = −10t + 15 = dvo dt .·. dvo = −10t dt + 15 dt Z vo (t) vo (0.5+) dx = Z t 0.5+ −10y dy + vo (t) − vo (0.5+ ) = −5y 2 Z t 0.5+ 15 dy t t + 15y 0.5 0.5 vo (t) = vo (0.5+ ) − 5t2 + 1.25 + 15t − 7.5 vo (0.5+ ) = vo (0.5− ) = 2.5 V .·. vo(t) = −5t2 + 15t − 3.75 V, dvo1 = −20(−0.04) = 0.8, dt dvo1 = 0.8 dt; Z vo1 (t) vo1 (0.5+ ) 0.5+ ≤ t ≤ tsat dx = 0.8 vo1 (t) − vo1(0.5+ ) = 0.8t − 0.4; .·. vo1(t) = 0.8t − 1.2 V, 0.5+ ≤ t ≤ tsat Z t 0.5+ dy vo1 (0.5+ ) = vo1(0.5− ) = −0.8 V 0.5+ ≤ t ≤ tsat © 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. 8–50 CHAPTER 8. Natural and Step Responses of RLC Circuits Summary: 0 ≤ t ≤ 0.5− s : vo1 = −1.6t V, 0.5+ s ≤ t ≤ tsat : vo = 10t2 V vo1 = 0.8t − 1.2 V, vo = −5t2 + 15t − 3.75 V [b] −12.5 = −5t2sat + 15tsat − 3.75 .·. 5t2sat − 15tsat − 8.75 = 0 Solving, tsat = 3.5 sec vo1 (tsat) = 0.8(3.5) − 1.2 = 1.6 V P 8.64 τ1 = (106 )(0.5 × 10−6 ) = 0.50 s 1 = 2; τ1 τ2 = (5 × 106 )(0.2 × 10−6 ) = 1 s; . ·. 1 =1 τ2 d2 vo dvo +3 + 2vo = 20 .·. 2 dt dt s2 + 3s + 2 = 0 (s + 1)(s + 2) = 0; s1 = −1, vo = Vf + A01 e−t + A02e−2t; s2 = −2 Vf = 20 = 10 V 2 vo = 10 + A01e−t + A02 e−2t vo (0) = 0 = 10 + A01 + A02; .·. A01 = −20, dvo (0) = 0 = −A01 − 2A02 dt A02 = 10 V vo (t) = 10 − 20e−t + 10e−2t V, dvo1 + 2vo1 = −1.6; dt 0 ≤ t ≤ 0.5 s .·. vo1 = −0.8 + 0.8e−2t V, 0 ≤ t ≤ 0.5 s vo (0.5) = 10 − 20e−0.5 + 10e−1 = 1.55 V vo1(0.5) = −0.8 + 0.8e−1 = −0.51 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. Problems 8–51 At t = 0.5 s iC = C 0 + 0.51 = 1.26 µA 400 × 103 dvo = 1.26 µA; dt dvo 1.26 = = 6.32 V/s dt 0.2 0.5 s ≤ t ≤ ∞: d2 vo dvo + 3 + 2 = −10 dt2 dt vo (∞) = −5 .·. vo = −5 + A01e−(t−0.5) + A02e−2(t−0.5) 1.55 = −5 + A01 + A02 dvo (0.5) = 6.32 = −A01 − 2A02 dt .·. A01 + A02 = 6.55; −A01 − 2A02 = 6.32 Solving, A01 = 19.42; A02 = −12.87 .·. vo = −5 + 19.42e−(t−0.5) − 12.87e−2(t−0.5) V, 0.5 ≤ t ≤ ∞ dvo1 + 2vo1 = 0.8 dt .·. vo1 = 0.4 + (−0.51 − 0.4)e−2(t−0.5) = 0.4 − 0.91e−2(t−0.5) V, P 8.65 [a] f(t) = = 0.5 ≤ t ≤ ∞ inertial force + frictional force + spring force M[d2 x/dt2 ] + D[dx/dt] + Kx © 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. 8–52 CHAPTER 8. Natural and Step Responses of RLC Circuits d2 x f D [b] = − 2 dt M M ! dx K − x dt M Given vA = d2 x , dt2 1 vB = − R1C1 Z t 1 vC = − R2 C2 Z t vD = − 0 then d2 x dy 2 0 ! vB dy = dy = − 1 dx R1 C1 dt 1 x R1 R2C1 C2 R3 R3 dx · vB = R4 R4 R1 C1 dt R5 + R6 1 R5 + R6 vC = · ·x vE = R6 R6 R1 R2 C1 C2 −R8 vF = f(t), R7 Therefore vA = −(vD + vE + vF ) d2 x R8 R3 dx R5 + R6 = f(t) − − x 2 dt R7 R4R1 C1 dt R6 R1 R2C1 C2 Therefore M = R7 , R8 D= R3 R7 R8 R4 R1 C1 and K= R7 (R5 + R6 ) R8 R6 R1 R2 C1C2 Box Number Function P 8.66 1 inverting and scaling 2 summing and inverting 3 integrating and scaling 4 integrating and scaling 5 inverting and scaling 6 noninverting and scaling [a] Given that the current response is underdamped, we know i will be of the form i = If + [B10 cos ωd t + B20 sin ωd t]e−αt where and α= ωd = R 2L q ωo2 − α2 = s 1 − α2 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. Problems 8–53 The capacitor will force the final value of i to be zero, therefore If = 0. By hypothesis i(0+ ) = Vdc /R; therefore B10 = Vdc /R. At t = 0+ the voltage across the primary winding is approximately zero; hence di(0+ )/dt = 0. From our equation for i we have di = [(ωd B20 − αB10 ) cos ωd t − (ωd B10 + αB20 ) sin ωd t]e−αt dt Hence di(0+ ) = ωd B20 − αB10 = 0 dt Thus α αVdc B20 = B10 = ωd ωd R It follows directly that i= α Vdc cos ωd t + sin ωd t e−αt R ωd [b] Since ωd B20 − αB10 = 0, it follows that di = −(ωd B10 + αB20 )e−αt sin ωd t dt α2 Vdc = ωd R αB20 But and ωd B10 = ωd Vdc R Therefore ωd B10 + αB20 ωd Vdc α2 Vdc Vdc ωd2 + α2 = + = R ωd R R ωd " But ωd2 + α2 = ωo2 = # 1 LC Hence ωd B10 + αB20 = Vdc ωd RLC Now since v1 = L v1 = −L we get Vdc Vdc −αt e−αt sin ωd t = − e sin ωd t ωd RLC ωd RC [c] vc = Vdc − iR − L iR = Vdc di dt di dt α cos ωd t + sin ωd t e−αt ωd © 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. 8–54 CHAPTER 8. Natural and Step Responses of RLC Circuits vc = Vdc − Vdc = Vdc − Vdc e α Vdc −αt cos ωd t + sin ωd t e−αt + e sin ωd t ωd ωd RC −αt Vdc αVdc −αt cos ωd t + − e sin ωd t ωd RC ωd 1 ωd = Vdc 1 − e−αt cos ωd t + 1 − α e−αt sin ωd t RC = Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t] P 8.67 vsp = Vdc a 1− e−αt sin ωd t ωd RC −aVdc d −αt dvsp = [e sin ωd t] dt ωd RC dt = −aVdc [−αe−αt sin ωd t + ωd e−αt cos ωd t] ωd RC = aVdc e−αt [α sin ωd t − ωd cos ωd t] ωd RC dvsp = 0 when α sin ωd t = ωd cos ωd t dt or ωd tan ωd t = ; α −1 ωd t = tan 1 ωd .·. tmax = tan−1 ωd α ωd α Note that because tan θ is periodic, i.e., tan θ = tan(θ ± nπ), where n is an integer, there are an infinite number of solutions for t where dvsp /dt = 0, that is t= tan−1 (ωd /α) ± nπ ωd Because of e−αt in the expression for vsp and knowing t ≥ 0 we know vsp will be maximum when t has its smallest positive value. Hence tmax = tan−1 (ωd /α) . ωd © 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. Problems P 8.68 8–55 [a] vc = Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t] dvc d = Vdc [1 + e−αt(K sin ωd t − cos ωd t)] dt dt = Vdc {(−αe−αt)(K sin ωd t − cos ωd t)+ e−αt[ωd K cos ωd t + ωd sin ωd t]} = Vdc e−αt[(ωd − αK) sin ωd t + (α + ωd K) cos ωd t] dvc = 0 when (ωd − αK) sin ωd t = −(α + ωd K) cos ωd t dt or α + ωd K tan ωd t = αK − ωd .·. ωd t ± nπ = tan−1 α + ωd K αK − ωd α + ωd K 1 tan−1 ± nπ tc = ωd αK − ωd α= R 4 × 103 = = 666.67 rad/s 2L 6 ωd = s 109 − (666.67)2 = 28,859.81 rad/s 1.2 1 K= ωd tc = 1 − α = 21.63 RC o 1 n −1 1 tan (−43.29) + nπ = {−1.55 + nπ} ωd ωd The smallest positive value of t occurs when n = 1, therefore tc max = 55.23 µs [b] vc (tc max ) = 12[1 − e−αtc max cos ωd tc max + Ke−αtc max sin ωd tc max ] = 262.42 V [c] From the text example the voltage across the spark plug reaches its maximum value in 53.63 µs. If the spark plug does not fire the capacitor voltage peaks in 55.23 µs. When vsp is maximum the voltage across the capacitor is 262.15 V. If the spark plug does not fire the capacitor voltage reaches 262.42 V. P 8.69 1 1 [a] w = L[i(0+ )]2 = (5)(16) × 10−3 = 40 mJ 2 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. 8–56 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] α = R 3 × 103 = = 300 rad/s 2L 10 ωd = s 109 − (300)2 = 28,282.68 rad/s 1.25 106 4 × 106 1 = = RC 0.75 3 tmax ωd 1 = tan−1 ωd α vsp (tmax ) = 12 − = 55.16 µs 12(50)(4 × 106 ) −αtmax e sin ωd tmax = −27,808.04 V 3(28,282.68) [c] vc (tmax) = 12[1 − e−αtmax cos ωd tmax + Ke−αtmax sin ωd tmax] 1 1 − α = 47.13 ωd RC K= vc (tmax ) = 568.15 V P 8.70 [a] vc = Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t] dvc d = Vdc [1 + e−αt(K sin ωd t − cos ωd t)] dt dt = Vdc {(−αe−αt)(K sin ωd t − cos ωd t)+ e−αt[ωd K cos ωd t + ωd sin ωd t]} = Vdc e−αt[(ωd − αK) sin ωd t + (α + ωd K) cos ωd t] dvc = 0 when (ωd − αK) sin ωd t = −(α + ωd K) cos ωd t dt or α + ωd K tan ωd t = αK − ωd .·. ωd t ± nπ = tan−1 α + ωd K αK − ωd 1 α + ωd K tc = tan−1 ± nπ ωd αK − ωd α= R 3 = = 300 rad/s 2L 2(5 × 10−3 ) ωd = s 109 − (300)2 = 28,282.68 rad/s 1.25 © 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. Problems 1 K= ωd tc = 8–57 1 − α = 47.13 RC 1 {−1.56 + nπ} ωd The smallest positive value of t occurs when n = 1, therefore tc max = 55.91 µs [b] vc (tc max ) = 12[1 − e−αtc max cos ωd tc max + Ke−αtc max sin ωd tc max ] = 568.28 V [c] From Problem 8.69, the voltage across the spark plug reaches its maximum value in 55.16 µs. If the spark plug does not fire the capacitor voltage peaks in 55.91 µs. When vsp is maximum the voltage across the capacitor is 568.15 V. If the spark plug does not fire the capacitor voltage reaches 568.28 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. 9 Sinusoidal Steady State Analysis Assessment Problems AP 9.1 [a] V = 170/−40◦ V [b] 10 sin(1000t + 20◦ ) = 10 cos(1000t − 70◦ ) . ·. I = 10/−70◦ A [c] I = 5/36.87◦ + 10/−53.13◦ = 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57◦ A [d] sin(20,000πt + 30◦ ) = cos(20,000πt − 60◦ ) Thus, V = 300/45◦ − 100/−60◦ = 212.13 + j212.13 − (50 − j86.60) = 162.13 + j298.73 = 339.90/61.51◦ mV AP 9.2 [a] v = 18.6 cos(ωt − 54◦ ) V [b] I = 20/45◦ − 50/ − 30◦ = 14.14 + j14.14 − 43.3 + j25 = −29.16 + j39.14 = 48.81/126.68◦ Therefore i = 48.81 cos(ωt + 126.68◦ ) mA [c] V = 20 + j80 − 30/15◦ = 20 + j80 − 28.98 − j7.76 = −8.98 + j72.24 = 72.79/97.08◦ v = 72.79 cos(ωt + 97.08◦ ) V AP 9.3 [a] ωL = (104 )(20 × 10−3 ) = 200 Ω [b] ZL = jωL = j200 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 9–1 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. 9–2 CHAPTER 9. Sinusoidal Steady State Analysis [c] VL = IZL = (10/30◦ )(200/90◦ ) × 10−3 = 2/120◦ V [d] vL = 2 cos(10,000t + 120◦ ) V −1 −1 = = −50 Ω ωC 4000(5 × 10−6 ) [b] ZC = jXC = −j50 Ω 30/25◦ V [c] I = = = 0.6/115◦ A ZC 50/−90◦ [d] i = 0.6 cos(4000t + 115◦ ) A AP 9.4 [a] XC = AP 9.5 I1 = 100/25◦ = 90.63 + j42.26 I2 = 100/145◦ = −81.92 + j57.36 I3 = 100/−95◦ = −8.72 − j99.62 I4 = −(I1 + I2 + I3 ) = (0 + j0) A, AP 9.6 [a] I = therefore i4 = 0 A 125/−60◦ 125 /(−60 − θZ )◦ = |Z|/θz |Z| But −60 − θZ = −105◦ .·. θZ = 45◦ Z = 90 + j160 + jXC .·. XC = −70 Ω; . ·. C = [b] I = XC = − 1 = −70 ωC 1 = 2.86 µF (70)(5000) Vs 125/−60◦ = = 0.982/−105◦ A; Z (90 + j90) .·. |I| = 0.982 A AP 9.7 [a] ω = 2000 rad/s ωL = 10 Ω, −1 = −20 Ω ωC © 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. Problems Zxy = 20kj10 + 5 + j20 = 9–3 20(j10) + 5 − j20 (20 + j10) = 4 + j8 + 5 − j20 = (9 − j12) Ω [b] ωL = 40 Ω, Zxy −1 = −5 Ω ωC " (20)(j40) = 5 − j5 + 20kj40 = 5 − j5 + 20 + j40 # = 5 − j5 + 16 + j8 = (21 + j3) Ω [c] Zxy 20(jωL) j106 = + 5− 20 + jωL 25ω " # ! j400ωL j106 20ω 2 L2 + + 5 − 400 + ω 2 L2 400 + ω 2 L2 25ω The impedance will be purely resistive when the j terms cancel, i.e., = 400ωL 106 = 400 + ω 2 L2 25ω Solving for ω yields ω = 4000 rad/s. 20ω 2 L2 [d] Zxy = + 5 = 10 + 5 = 15 Ω 400 + ω 2 L2 AP 9.8 The frequency 4000 rad/s was found to give Zxy = 15 Ω in Assessment Problem 9.7. Thus, V = 150/0◦ , Is = V 150/0◦ = = 10/0◦ A Zxy 15 Using current division, IL = 20 (10) = 5 − j5 = 7.07/−45◦ A 20 + j20 iL = 7.07 cos(4000t − 45◦ ) A, Im = 7.07 A AP 9.9 After replacing the delta made up of the 50 Ω, 40 Ω, and 10 Ω resistors with its equivalent wye, the circuit becomes © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 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Sinusoidal Steady State Analysis The circuit is further simplified by combining the parallel branches, (20 + j40)k(5 − j15) = (12 − j16) Ω Therefore I = 136/0◦ = 4/28.07◦ A 14 + 12 − j16 + 4 AP 9.10 V1 = 240/53.13◦ = 144 + j192 V V2 = 96/−90◦ = −j96 V jωL = j(4000)(15 × 10−3 ) = j60 Ω 1 6 × 106 = −j = −j60 Ω jωC (4000)(25) Perform a source transformation: V1 144 + j192 = = 3.2 − j2.4 A j60 j60 96 V2 = −j = −j4.8 A 20 20 Combine the parallel impedances: Y = 1 1 1 j5 1 1 + + + = = j60 30 −j60 20 j60 12 Z= 1 = 12 Ω Y Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87◦ V vo = 48 cos(4000t + 36.87◦ ) 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. Problems 9–5 AP 9.11 Use the lower node as the reference node. Let V1 = node voltage across the 20 Ω resistor and VTh = node voltage across the capacitor. Writing the node voltage equations gives us V1 V1 − 10Ix − 2/45◦ + = 0 and 20 j10 VTh = −j10 (10Ix ) 10 − j10 We also have Ix = V1 20 Solving these equations for VTh gives VTh = 10/45◦ V. To find the Thévenin impedance, we remove the independent current source and apply a test voltage source at the terminals a, b. Thus It follows from the circuit that 10Ix = (20 + j10)Ix Therefore Ix = 0 and IT = ZTh = VT , IT VT VT + −j10 10 therefore ZTh = (5 − j5) Ω AP 9.12 The phasor domain circuit is as shown in the following diagram: The node voltage equation is −10 + V V V V − 100/−90◦ + + + =0 5 −j(20/9) j5 20 © 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. 9–6 CHAPTER 9. Sinusoidal Steady State Analysis Therefore V = 10 − j30 = 31.62/−71.57◦ Therefore v = 31.62 cos(50,000t − 71.57◦ ) V AP 9.13 Let Ia , Ib , and Ic be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2)Ia + (3 − j5)(Ia − Ib) and 0 = (3 − j5)(Ib − Ia) + 2(Ib − Ic ). But Vx = −j5(Ia − Ib ), therefore Ic = −0.75[−j5(Ia − Ib)]. Solving for I = Ia = 29 + j2 = 29.07/3.95◦ A. √ AP 9.14 [a] M = 0.4 0.0625 = 0.1 H, ωM = 80 Ω Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω Therefore |Z22| = 500 Ω, 80 Zτ = 500 [b] I1 = 2 ∗ Z22 = (400 − j300) Ω (400 − j300) = (10.24 − j7.68) Ω 245.20 = 0.50/ − 53.13◦ A 184 + 100 + j400 + Zτ i1 = 0.5 cos(800t − 53.13◦ ) A [c] I2 = jωM j80 (0.5/ − 53.13◦ ) = 0.08/0◦ A I1 = Z22 500/36.87◦ i2 = 80 cos 800t mA © 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. Problems AP 9.15 I1 = 9–7 Vs 25 × 103 /0◦ = Z1 + 2s2 Z2 1500 + j6000 + (25)2 (4 − j14.4) = 4 + j3 = 5/36.87◦ A V1 = Vs − Z1 I1 = 25,000/0◦ − (4 + j3)(1500 + j6000) = 37,000 − j28,500 V2 = − I2 = 1 V1 = −1480 + j1140 = 1868.15/142.39◦ V 25 1868.15/142.39◦ V2 = = 125/216.87◦ A Z2 4 − j14.4 © 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. 9–8 CHAPTER 9. Sinusoidal Steady State Analysis Problems P 9.1 [a] 80 V [b] 2πf = 1000π; f = 500 Hz [c] ω = 1000π = 3141.59 rad/s −π [d] θ(rad) = = −0.5236 rad 6 [e] θ = −30◦ 1 1 [f] T = = = 2 ms f 500 1 π [g] 1000πt − = 0; .·. t = = 166.67 µs 6 6000 0.002 π − [h] v = 80 cos 1000π t + 3 6 = 80 cos[1000πt + (2π/3) − (π/6)] = 80 cos[1000πt + (π/2)] = −80 sin 1000πt V [i] 1000π(t − to ) − (π/6) = 1000πt − (π/2) .·. 1000πto = π ; 3 to = 1 = 333.33 µs 3000 [j] 1000π(t + to ) − (π/6) = 1000πt .·. 1000πto = P 9.2 [a] π ; 6 to = T = 8 + 2 = 10 ms; 2 f= 1 = 166.67 µs 6000 T = 20 ms 1 1 = = 50 Hz T 20 × 10−3 [b] v = Vm sin(ωt + θ) ω = 2πf = 100π rad/s 100π(−2 × 10−3 ) + θ = 0; . ·. θ = π rad = 36◦ 5 v = Vm sin[100πt + 36◦ ] 80.9 = Vm sin 36◦ ; Vm = 137.64 V v = 137.64 sin[100πt + 36◦ ] = 137.64 cos[100πt − 54◦ ] 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. Problems P 9.3 9–9 [a] By hypothesis i = 20 cos(ωt + θ) di = −20ω sin(ωt + θ) dt .·. 20ω = 8000π; ω = 400π rad/s [b] f = ω = 200 Hz; 2π T = 1 = 5 ms = 5000 µs f 625 1 1 = , .·. θ = − (360) = −45◦ 5000 8 8 .·. i = 20 cos(400πt − 45◦ ) A P 9.4 [a] ω = 2πf = 3769.91 rad/s, f= ω = 600 Hz 2π [b] T = 1/f = 1.67 ms [c] Vm = 10 V [d] v(0) = 10 cos(−53.13◦ ) = 6 V −53.13◦ (2π) [e] φ = −53.13◦ ; φ= = −0.9273 rad 360◦ [f] V = 0 when 3769.91t − 53.13◦ = 90◦ . Now resolve the units: (3769.91 rad/s)t = 143.13◦ = 2.498 rad, 57.3◦ /rad t = 662.64 µs [g] (dv/dt) = (−10)3769.91 sin(3769.91t − 53.13◦ ) (dv/dt) = 0 when 3769.91t − 53.13◦ = 0◦ or 3769.91t = 53.13◦ = 0.9273 rad 57.3◦ /rad Therefore t = 245.97 µs P 9.5 © 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. 9–10 CHAPTER 9. Sinusoidal Steady State Analysis [a] Left as φ becomes more positive [b] Left P 9.6 Z to +T to Vm2 2 cos (ωt + φ) dt = = Vm2 Z Vm2 2 Vm2 = 2 Vm2 = 2 P 9.7 P 9.8 Vm = √ Vrms = Z 0 T /2 2Vrms = s Vm2 1 T Z T /2 0 2 sin √ t 1 1 + cos(2ωt + 2φ) dt 2 2 (o R to +T to dt + Z to +T cos(2ωt + 2φ) dt to ) i 1 h T+ sin(2ωt + 2φ) |ttoo +T 2ω 1 T+ [sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)] 2ω 1 2 T 2 T = Vm + (0) = Vm 2 2ω 2 2(240) = 339.41 V Vm2 sin2 2π t dt T 2π V2 t dt = m T 2 Therefore Vrms = P 9.9 to +T s Z 0 4π V 2T 1 − cos t dt = m T 4 T /2 1 Vm2 T Vm = T 4 2 [a] The numerical values of the terms in Eq. 9.8 are Vm = 20, R/L = 1066.67, √ R2 + ω 2 L2 = 100 ωL = 60 φ = 25◦ , θ = 36.87◦ θ = tan−1 60/80, Substitute these values into Equation 9.9: h i i = −195.72e−1066.67t + 200 cos(800t − 11.87◦ ) mA, t≥0 [b] Transient component = −195.72e−1066.67t mA Steady-state component = 200 cos(800t − 11.87◦ ) mA [c] By direct substitution into Eq 9.9 in part (a), i(1.875 ms) = 28.39 mA [d] 200 mA, 800 rad/s, −11.87◦ [e] The current lags the voltage by 36.87◦ . © 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. Problems P 9.10 9–11 [a] From Eq. 9.9 we have L di Vm R cos(φ − θ) −(R/L)t ωLVm sin(ωt + φ − θ) √ e − = √ 2 dt R + ω 2 L2 R2 + ω 2L2 Ri = −Vm R cos(φ − θ)e−(R/L)t Vm R cos(ωt + φ − θ) √ √ + R2 + ω 2 L2 R2 + ω 2 L2 " di R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ) √ L + Ri = Vm dt R2 + ω 2 L2 # But √ R = cos θ R2 + ω 2 L2 and √ ωL = sin θ R2 + ω 2 L2 Therefore the right-hand side reduces to Vm cos(ωt + φ) At t = 0, Eq. 9.9 reduces to i(0) = −Vm cos(φ − θ) Vm cos(φ − θ) √ + √ 2 =0 R2 + ω 2 L2 R + ω 2 L2 Vm [b] iss = √ 2 cos(ωt + φ − θ) R + ω 2 L2 Therefore diss −ωLVm L =√ 2 sin(ωt + φ − θ) dt R + ω 2L2 and Vm R Riss = √ 2 cos(ωt + φ − θ) R + ω 2L2 " diss R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ) √ L + Riss = Vm dt R2 + ω 2 L2 # = Vm cos(ωt + φ) P 9.11 [a] Y = 50/60◦ + 100/ − 30◦ = 111.8/ − 3.43◦ y = 111.8 cos(500t − 3.43◦ ) [b] Y = 200/50◦ − 100/60◦ = 102.99/40.29◦ y = 102.99 cos(377t + 40.29◦ ) [c] Y = 80/30◦ − 100/ − 225◦ + 50/ − 90◦ = 161.59/ − 29.96◦ y = 161.59 cos(100t − 29.96◦ ) © 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. 9–12 CHAPTER 9. Sinusoidal Steady State Analysis [d] Y = 250/0◦ + 250/120◦ + 250/ − 120◦ = 0 y=0 P 9.12 [a] Vg = 300/78◦ ; . ·. Z = Ig = 6/33◦ 300/78◦ Vg = = 50/45◦ Ω Ig 6/33◦ [b] ig lags vg by 45◦ : 2πf = 5000π; .·. ig lags vg by P 9.13 f = 2500 Hz; T = 1/f = 400 µs 45◦ (400 µs) = 50 µs 360◦ [a] ω = 2πf = 160π × 103 = 502.65 krad/s = 502,654.82 rad/s 25 × 10−3 /0◦ [b] I = = jωC(25 × 10−3 )/0◦ = 25 × 10−3 ωC /90◦ 1/jωC .·. θi = 90◦ [c] 628.32 × 10−6 = 25 × 10−3 ωC 1 25 × 10−3 = = 39.79 Ω, ωC 628.32 × 10−6 [d] C = .·. XC = −39.79 Ω 1 1 = 39.79(ω) (39.79)(160π × 103 ) C = 0.05 × 10−6 = 0.05 µF −1 [e] Zc = j ωC P 9.14 = −j39.79 Ω [a] 400 Hz [b] θv = 0◦ I= 100/0◦ 100 / − 90◦ ; = jωL ωL θi = −90◦ 100 = 20; ωL = 5 Ω ωL 5 [d] L = = 1.99 mH 800π [e] ZL = jωL = j5 Ω [c] © 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. Problems P 9.15 9–13 [a] ZL = j(8000)(5 × 10−3 ) = j40 Ω ZC = −j = −j100 Ω (8000)(1.25 × 10−6 ) 600/20◦ = 8.32/76.31◦ A 40 + j40 − j100 [c] i = 8.32 cos(8000t + 76.31◦ ) A [b] I = P 9.16 [a] jωL = j(2 × 104 )(300 × 10−6 ) = j6 Ω 1 1 = −j = −j10 Ω; 4 jωC (2 × 10 )(5 × 10−6 ) Ig = 922/30◦ A [b] Vo = 922/30◦ Ze Ze = 1 ; Ye Ye = 1 1 1 +j + 10 10 8 + j6 Ye = 0.18 + j0.04 S Ze = 1 = 5.42/ − 12.53◦ Ω 0.18 + j0.04 Vo = (922/30◦ )(5.42/ − 12.53◦ ) = 5000.25/17.47◦ V [c] vo = 5000.25 cos(2 × 104 t + 17.47◦ ) V P 9.17 [a] Z1 = R1 + jωL1 Z2 = R2 (jωL2 ) ω 2 L22 R2 + jωL2 R22 = R2 + jωL2 R22 + ω 2 L22 Z1 = Z2 when R1 = ω 2 L22 R2 R22 + ω 2 L22 and L1 = R22 L2 R22 + ω 2 L22 © 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. 9–14 CHAPTER 9. Sinusoidal Steady State Analysis (4000)2 (1.25)2 (5000) = 2500 Ω 50002 + 40002 (1.25)2 [b] R1 = (5000)2 (1.25) = 625 mH 50002 + 40002 (1.25)2 L1 = P 9.18 [a] Y2 = 1 j − R2 ωL2 R1 − jωL1 1 = 2 R1 + jωL1 R1 + ω 2L21 Y1 = Therefore R2 = R21 + ω 2 L21 R1 and L2 = [a] Z1 = R1 − j 1 ωC1 R2 /jωC2 R2 R2 − jωR22 C2 = = R2 + (1/jωC2 ) 1 + jωR2 C2 1 + ω 2 R22 C22 Z2 = Z1 = Z2 when R1 = 1 ωR22 C2 = ωC1 1 + ω 2 R22 C22 R2 1 + ω 2R22 C22 or C1 = and 1 + ω 2 R22 C22 ω 2 R22 C2 1000 [b] R1 = 1 + (40 × C1 = [a] Y2 = R21 + ω 2 L21 ω 2 L1 80002 + 10002 (4)2 = 20 H 10002 (4) L2 = P 9.20 when 80002 + 10002 (4)2 = 10 kΩ 8000 [b] R2 = P 9.19 Y2 = Y1 103 )2(1000)2 (50 × 10−4 )2 = 200 Ω 1 + (40 × 103 )2 (1000)2 (50 × 10−9 )2 = 62.5 nF (40 × 103 )2(1000)2 (50 × 10−9 ) 1 + jωC2 R2 Y1 = 1 jωC1 ω 2 R1 C12 + jωC1 = = R1 + (1/jωC1 ) 1 + jωR1 C1 1 + ω 2 R21 C12 Therefore R2 = Y1 = Y2 1 + ω 2 R21 C12 ω 2 R1 C12 and when C2 = C1 1 + ω 2 R21 C12 © 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. Problems [b] R2 = 1 + (50 × 103 )2(1000)2 (40 × 10−9 )2 = 1250 Ω (50 × 103 )2 (1000)(40 × 10−9 )2 C2 = P 9.21 9–15 40 × 10−9 = 8 nF 1 + (50 × 103 )2 (1000)2 (40 × 10−9 )2 [a] R = 300 Ω = 120 Ω + 180 Ω ωL − 1 1 = −400 so 10,000L − = −400 ωC 10,000C Choose L = 10 mH. Then, 1 1 = 100 + 400 so C = = 0.2 µF 10,000C 10,000(500) We can achieve the desired capacitance by combining two 0.1 µF capacitors in parallel. The final circuit is shown here: [b] 0.01ω = 1 ω(0.2 × 10−6 ) so ω 2 = 1 = 5 × 108 −6 0.01(0.2 × 10 ) .·. ω = 22,360.7 rad/s P 9.22 [a] Using the notation and results from Problem 9.18: RkL = 40 + j20 so R1 = 40, R2 = L1 = 20 = 4 mH 5000 402 + 50002 (0.004)2 = 50 Ω 40 402 + 50002 (0.004)2 L2 = = 20 mH 50002 (0.004) R2 kjωL2 = 50kj100 = 40 + j20 Ω (checks) The circuit, using combinations of components from Appendix H, is shown here: © 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. 9–16 CHAPTER 9. Sinusoidal Steady State Analysis [b] Using the notation and results from Problem 9.22: RkC = 40 − j20 so R1 = 40, R2 = C2 = C1 = 10 µF 1 + 50002 (40)2 (10 µ)2 = 50 Ω 50002 (40)(10 µ)2 10 µ 1+ 50002 (40)2 (10 µ)2 = 2 µF R2 k(−j/ωC2 ) = 50k(−j100) = 40 − j20 Ω (checks) The circuit, using combinations of components from Appendix H, is shown here: P 9.23 [a] (40 + j20)k(−j/ωC) = 50kj100k(−j/ωC) To cancel out the j100 Ω impedance, the capacitive impedance must be −j100 Ω: −j 1 = −j100 so C = = 2 µF 5000C (100)(5000) Check: RkjωLk(−j/ωC) = 50kj100k(−j100) = 50 Ω Create the equivalent of a 2 µF capacitor from components in Appendix H by combining two 1 µF capacitors in parallel. [b] (40 − j20)k(jωL) = 50k(−j100)k(jωL) To cancel out the −j100 Ω impedance, the inductive impedance must be j100 Ω: j5000L = j100 so L = 100 = 20 mH 5000 Check: RkjωLk(−j/ωC) = 50kj100k(−j100) = 50 Ω Create the equivalent of a 20 mH inductor from components in Appendix H by combining two 10 mH inductors in series. P 9.24 [a] Y = 1 1 1 + + 3 + j4 16 − j12 −j4 = 0.12 − j0.16 + 0.04 + j0.03 + j0.25 = 0.16 + j0.12 = 200/36.87◦ mS © 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. Problems 9–17 [b] G = 160 mS [c] B = 120 mS [d] I = 8/0◦ A, IC = V= 8 I = = 40/−36.87◦ V Y 0.2/36.87◦ 40/−36.87◦ V = = 10/53.13◦ A ZC 4/−90◦ iC = 10 cos(ωt + 53.13◦ ) A, P 9.25 [a] jωL = Rk(−j/ωC) = jωL + jωL + −jR ωCR − j jωL + −jR(ωCR + j) ω 2 C 2 R2 + 1 Im(Zab) = ωL − Im = 10 A −jR/ωC R − j/ωC ωCR2 =0 ω 2 C 2R2 + 1 CR2 ω 2 C 2R2 + 1 . ·. L= . ·. ω 2 C 2 R2 + 1 = . ·. (25×10 )(100) −1 (CR2 /L) − 1 160×10−6 = = 900 × 108 ω = 2 2 −9 2 C R (25 × 10 ) (100)2 CR2 L −9 2 2 ω = 300 krad/s [b] Zab(300 × 103 ) = j48 + P 9.26 (100)(−j133.33) = 64 Ω 100 − j133.33 First find the admittance of the parallel branches Yp = 1 1 1 1 + + + = 0.375 − j0.125 S 6 − j2 4 + j12 5 j10 Zp = 1 1 = = 2.4 + j0.8 Ω Yp 0.375 − j0.125 Zab = −j12.8 + 2.4 + j0.8 + 13.6 = 16 − j12 Ω Yab = 1 1 = = 0.04 + j0.03 S Zab 16 − j12 = 40 + j30 mS = 50/36.87◦ mS © 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. 9–18 P 9.27 CHAPTER 9. Sinusoidal Steady State Analysis Zab = 1 − j8 + (2 + j4)k(10 − j20) + (40kj20) = 1 − j8 + 3 + j4 + 8 + j16 = 12 + j12 Ω = 16.97/45◦ Ω P 9.28 Vg = 40/ − 15◦ V; Z= Ig = 40/ − 68.13◦ mA Vg = 1000/53.13◦ Ω = 600 + j800 Ω Ig 0.4 × 106 Z = 600 + j 3.2ω − ω ! 0.4 × 106 .·. 3.2ω − = 800 ω .·. ω 2 − 250ω − 125,000 = 0 Solving, ω = 500 rad/s P 9.29 1 1 = = −j20 Ω jωC (1 × 10−6 )(50 × 103 ) jωL = j50 × 103 (1.2 × 10−3 ) = j60 Ω Vg = 40/0◦ V Ze = −j20 + 30kj60 = 24 − j8 Ω Ig = 40/0◦ = 1.5 + j0.5 mA 24 − j8 Vo = (30kj60)Ig = 30(j60) (1.5 + j0.5) = 30 + j30 = 42.43/45◦ V 30 + j60 vo = 42.43 cos(50,000t + 45◦ ) 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. Problems P 9.30 [a] 9–19 1 = −j50 Ω jωC jωL = j120 Ω Ze = 100k − j50 = 20 − j40 Ω Ig = 2/0◦ Vg = Ig Ze = 2(20 − j40) = 40 − j80 V Vo = j120 (40 − j80) = 90 − j30 = 94.87/ − 18.43◦ V 80 + j80 vo = 94.87 cos(16 × 105 t − 18.435◦ ) V [b] ω = 2πf = 16 × 105 ; T = . ·. f= 8 × 105 π 1 π = = 1.25π µs f 8 × 105 18.435 (1.25π µs) = 201.09 ns 360 .·. vo lags ig by 201.09 ns. P 9.31 Z = 4 + j(50)(0.24) − j 1 = 5.66/45◦ Ω (50)(0.0025) V 0.1/ − 90◦ = = 17.67/ − 135◦ mA Io = ◦ / Z 5.66 45 io (t) = 17.67 cos(50t − 135◦ ) mA P 9.32 ZL = j(2000)(60 × 10−3 ) = j120 Ω ZC = −j = −j40 Ω (2000)(12.5 × 10−6 ) © 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. 9–20 CHAPTER 9. Sinusoidal Steady State Analysis Construct the phasor domain equivalent circuit: Using current division: I= (120 − j40) (0.5) = 0.25 − j0.25 A 120 − j40 + 40 + j120 Vo = j120I = 30 + j30 = 42.43/45◦ vo = 42.43 cos(2000t + 45◦ ) V P 9.33 [a] Va = (50 + j150)(2/0◦ ) = 100 + j300 V Ib = 100 + j300 = j2.5 A 120 − j40 Ic = 2/0◦ + j2.5 + 6 + j3.5 = 8 + j6 A Vg = 5Ic + Va = 5(8 + j6) + 100 + j300 = 140 + j330 V [b] ib = 2.5 cos(800t + 90◦ ) A ic = 10 cos(800t + 36.87◦ ) A vg = 358.47 cos(800t + 67.01◦ ) 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. Problems P 9.34 9–21 Is = 3/0◦ mA 1 = −j0.4 Ω jωC jωL = j0.4 Ω After source transformation we have Vo = −j0.4kj0.4k5 (66 × 10−3 ) = 10 mV 28 + −j0.4kj0.4k5 vo = 10 cos 200t mV P 9.35 Va − (100 − j50) Va Va − (140 + j30) + + =0 20 j5 12 + j16 Solving, Va = 40 + j30 V IZ + (30 + j20) − 140 + j30 (40 + j30) − (140 + j30) + =0 −j10 12 + j16 Solving, IZ = −30 − j10 A Z= (100 − j50) − (140 + j30) = 2 + j2 Ω −30 − j10 © 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. 9–22 CHAPTER 9. Sinusoidal Steady State Analysis P 9.36 V1 = j5(−j2) = 10 V −25 + 10 + (4 − j3)I1 = 0 . ·. I1 = 15 = 2.4 + j1.8 A 4 − j3 Ib = I1 − j5 = (2.4 + j1.8) − j5 = 2.4 − j3.2 A VZ = −j5I2 + (4 − j3)I1 = −j5(2.4 − j3.2) + (4 − j3)(2.4 + j1.8) = −1 − j12 V −25 + (1 + j3)I3 + (−1 − j12) = 0 . ·. I3 = 6.2 − j6.6 A IZ = I3 − I2 = (6.2 − j6.6) − (2.4 − j3.2) = 3.8 − j3.4 A Z= P 9.37 VZ −1 − j12 = = 1.42 − j1.88 Ω IZ 3.8 − j3.4 Simplify the top triangle using series and parallel combinations: (1 + j1)k(1 − j1) = 1 Ω Convert the lower left delta to a wye: Z1 = (j1)(1) = j1 Ω 1 + j1 − 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. Problems Z2 = (−j1)(1) = −j1 Ω 1 + j1 − j1 Z3 = (j1)(−j1) = 1Ω 1 + j1 − j1 9–23 Convert the lower right delta to a wye: Z4 = (−j1)(1) = −j1 Ω 1 + j1 − j1 Z5 = (−j1)(j1) = 1Ω 1 + j1 − j1 Z6 = (j1)(1) = j1 Ω 1 + j1 − j1 The resulting circuit is shown below: Simplify the middle portion of the circuit by making series and parallel combinations: (1 + j1 − j1)k(1 + 1) = 1k2 = 2/3 Ω Zab = −j1 + 2/3 + j1 = 2/3 Ω P 9.38 [a] Zg = 500 − j 106 103 (j0.5ω) + 3 ω 10 + j0.5ω = 500 − j 106 500jω(1000 − j0.5ω) + ω 106 + 0.25ω 2 = 500 − j 106 250ω 2 5 × 105 ω + 6 + j ω 10 + 0.25ω 2 106 + 0.25ω 2 .·. If Zg is purely real, 106 5 × 105 ω = 6 ω 10 + 0.25ω 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. 9–24 CHAPTER 9. Sinusoidal Steady State Analysis .·. 2(106 + 0.25ω 2 ) = ω 2 . ·. 4 × 106 = ω 2 ω = 2000 rad/s [b] When ω = 2000 rad/s Zg = 500 − j500 + (j1000k1000) = 1000 Ω 20/0◦ = 20/0◦ mA 1000 Vo = Vg − Ig Z1 . ·. Ig = Z1 = 500 − j500 Ω Vo = 20/0◦ − (0.02/0◦ )(500 − j500) = 10 + j10 = 14.14/45◦ V vo = 14.14 cos(2000t + 45◦ ) V P 9.39 50,000 −j20 × 106 + k(1200 + j0.2ω) [a] Zeq = 3 ω = = 50,000 −j20 × 106 (1200 + j0.2ω) + 6 3 ω 1200 + j[0.2ω − 20×10 ] ω 50,000 + 3 −j20×106 (1200 ω h + j0.2ω) 1200 − j 0.2ω − 12002 + 0.2ω − 20×106 ω 20×106 ω 2 i 20 × 106 20 × 106 20 × 106 (1200)2 − 0.2ω 0.2ω − Im(Zeq ) = − ω ω ω 6 2 6 −20 × 10 (1200) − 20 × 10 20 × 106 0.2ω 0.2ω − ω " 20 × 106 −(1200) = 0.2ω 0.2ω − ω 2 " !# !# =0 =0 ! 0.22 ω 2 − 0.2(20 × 106 ) − 12002 = 0 ω 2 = 64 × 106 . ·. ω = 8000 rad/s f = 1273.24 Hz [b] Zeq = 50,000 + −j2500k(1200 + j1600) 3 = Ig = . ·. 50,000 (−j2500)(1200 + j1600) + = 20,000 Ω 3 1200 − j900 30/0◦ = 1.5/0◦ mA 20,000 ig (t) = 1.5 cos 8000t mA © 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. Problems P 9.40 9–25 R R jωC [a] Zp = = R + (1/jωC) 1 + jωRC = 10,000 10,000 = 1 + j(5000)(10,000)C 1 + j50 × 106 C = 10,000(1 − j50 × 106 C) 1 + 25 × 1014 C 2 = 10,000 5 × 1011 C − j 1 + 25 × 1014 C 2 1 + 25 × 1014 C 2 jωL = j5000(0.8) = j4000 .·. 4000 = 5 × 1011 C 1 + 25 × 1014 C 2 .·. 1014 C 2 − 125 × 106 C + 1 = 0 .·. C 2 − 5 × 10−8 C + 4 × 10−16 = 0 Solving, C1 = 40 nF C2 = 10 nF 10,000 [b] Re = 1 + 25 × 1014 C 2 When C = 40 nF Re = 2000 Ω; 80/0◦ = 40/0◦ mA; ig = 40 cos 5000t mA Ig = 2000 When C = 10 nF Re = 8000 Ω; 80/0◦ Ig = = 10/0◦ mA; 8000 P 9.41 [a] ZC = ig = 10 cos 5000t mA 109 = −j4000 Ω j(50,000)(5) Z1 = 10,000kj50,000L = 10,000(j50,000L) 250,000L2 + j50,000L = 10,000 + j50,000L 1 + 25L2 250,000L2 + j50,000L ZT = Z1 + ZR + ZC = − j4000 + 2000 1 + 25L2 ZT is resistive when 50,000L = 4000 or 1 + 25L2 L2 − 0.5L + 0.04 = 0 Solving, L1 = 0.4 H and L2 = 0.1 H. © 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. 9–26 CHAPTER 9. Sinusoidal Steady State Analysis [b] When L = 0.4 H: ZT = 2000 + 250,000(0.16) = 10,000 Ω 1 + 25(0.16) 50/0◦ = 5/0◦ mA 10,000 Ig = ig = 5 cos 50,000t mA When L = 0.1 H: 250,000(0.01) ZT = 2000 + = 4000 Ω 1 + 25(0.01) 50/0◦ = 12.5/0◦ mA 4000 Ig = ig = 12.5 cos 50,000t mA P 9.42 [a] Y1 = 1 = 0.2 × 10−3 S 5000 Y2 = = 1 1200 + j0.2ω 1200 0.2ω − j 1.44 × 106 + 0.04ω 2 1.44 × 106 + 0.04ω 2 Y3 = jω50 × 10−9 YT = Y1 + Y2 + Y3 For ig and vo to be in phase the j component of YT must be zero; thus, ω50 × 10−9 = 0.2ω 1.44 × 106 + 0.04ω 2 or 0.04ω 2 + 1.44 × 106 = 0.2 × 109 = 4 × 106 50 .·. 0.04ω 2 = 2.56 × 106 [b] YT = 0.2 × 10−3 + .·. ω = 8000 rad/s = 8 krad/s 1200 = 0.5 × 10−3 S 1.44 × 106 + 0.04(64) × 106 .·. ZT = 2000 Ω Vo = (2.5 × 10−3 /0◦ )(2000) = 5/0◦ vo = 5 cos 8000t 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. Problems 9–27 P 9.43 IN = 5 − j15 + (1 − j3) mA, ZN IN = −18 − j13.5 + 4.5 − j6 mA, ZN ZN in kΩ ZN in kΩ 5 − j15 −18 − j13.5 + 1 − j3 = + (4.5 − j6) ZN ZN 23 − j15 = 3.5 − j3 ZN IN = P 9.44 .·. ZN = 4 + j3 kΩ 5 − j15 + 1 − j3 = −j6 mA 4 + j3 [a] jωL = j(1000)(100) × 10−3 = j100 Ω 1 106 = −j = −j100 Ω jωC (1000)(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. 9–28 CHAPTER 9. Sinusoidal Steady State Analysis Using voltage division, Vab = (100 + j100)k(−j100) (247.49/45◦ ) = 350/0◦ j100 + (100 + j100)k(−j100) VTh = Vab = 350/0◦ V [b] Remove the voltage source and combine impedances in parallel to find ZTh = Zab : Yab = 1 1 1 + + = 5 − j5 mS j100 100 + j100 −j100 ZTh = Zab = 1 = 100 + j100 Ω Yab [c] P 9.45 Step 1 to Step 2: 240/0◦ = −j20 = 20/ − 90◦ A j12 Step 2 to Step 3: (j12)k36 = 3.6 + j10.8 Ω Step 3 to Step 4: (20/ − 90◦ )(3.6 + j10.8) = 216 − j72 = 227.68/ − 18.43◦ 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. Problems P 9.46 9–29 Step 1 to Step 2: (4/0◦ )(50) = 200/0◦ V Step 2 to Step 3: 50 + 30 + j60 = (80 + j60) Ω Step 3 to Step 4: 200/0◦ = 2/ − 36.87◦ A (80 + j60) Step 4 to Step 5: (80 + j60k − j100 = 100 − j50 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. 9–30 P 9.47 CHAPTER 9. Sinusoidal Steady State Analysis Open circuit voltage: (9 + j4)Ia − Ib = −60/0◦ −Ia + (9 − j4)Ib = 60/0◦ Solving, Ia = −5 + j2.5 A; Ib = 5 + j2.5 A VTh = 4Ia + (4 − j4)Ib = 10/0◦ V Short circuit current: (9 + j4)Ia − 1Ib − 4Isc = −60 −1Ia + (9 − j4)Ib − (4 − j4)Isc = 60 −4Ia − (4 − j4)Ib + (8 − j4)Isc = 0 Solving, Isc = 2.07/0◦ © 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. Problems ZTh = 9–31 VTh 10/0◦ = = 4.83 Ω Isc 2.07/0◦ Alternate calculation for ZTh : X Z = 4 + 1 + 4 − j4 = 9 − j4 Z1 = 4 9 − j4 Z2 = 4 − j4 9 − j4 Z3 = 16 − j16 9 − j4 Za = 4 + j4 + 4 56 + j20 = 9 − j4 9 − j4 Zb = 4 + 4 − j4 40 − j20 = 9 − j4 9 − j4 Za kZb = 2640 − j320 884 − j384 Z3 + Za kZb = 16 − j16 2640 − j320 + = 4.83 Ω 9 − j4 884 − j384 © 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. 9–32 P 9.48 CHAPTER 9. Sinusoidal Steady State Analysis Open circuit voltage: V1 − 250 V1 − 0.03Vo + =0 20 + j10 50 − j100 .·. Vo = −j100 V1 50 − j100 V1 j3V1 V1 250 + + = 20 + j10 50 − j100 50 − j100 20 + j10 V1 = 500 − j250 V; Vo = 300 − j400 V = VTh Short circuit current: 250/0◦ Isc = = 3.5 − j0.5 A 70 + j10 ZTh = VTh 300 − j400 = = 100 − j100 Ω Isc 3.5 − j0.5 The Thévenin equivalent circuit: © 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. Problems P 9.49 9–33 Short circuit current Iβ = −6Iβ 2 2Iβ = −6Iβ ; I1 = 0; . ·. Iβ = 0 .·. Isc = 10/−45◦ A = IN The Norton impedance is the same as the Thévenin impedance. Find it using a test source VT = 6Iβ + 2Iβ = 8Iβ , ZTh = P 9.50 Iβ = j1 IT 2 + j1 VT 8Iβ j8 = = = 1.6 + j3.2 Ω IT [(2 + j1)/j1]Iβ 2 + j1 jωL = j100 × 103 (0.6 × 10−3 ) = j60 Ω 1 −j = = −j25 Ω 3 jωC (100 × 10 )(0.4 × 10−6 ) VT = −j25IT + 5I∆ − 30I∆ © 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. 9–34 CHAPTER 9. Sinusoidal Steady State Analysis I∆ = −j60 IT 30 + j60 VT = −j25IT + 25 j60 IT 30 + j60 VT = Zab = 20 − j15 = 25/ − 36.87◦ Ω IT P 9.51 1 109 = = 8 kΩ ωC1 50,000(2.5) 1 109 = = 4 kΩ ωC2 50,000(5) VT = (2400 − j8000)IT + 40IT (90) ZTh = P 9.52 VT = 6000 − j8000 Ω IT Open circuit voltage: V2 − 15 V2 V2 + 88Iφ + =0 10 −j50 Iφ = 5 − (V2 /5) 200 Solving, V2 = −66 + j88 = 110/126.87◦ V = VTh © 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. Problems 9–35 Find the Thévenin equivalent impedance using a test source: IT = VT 0.8Vt + 88Iφ + 10 −j50 Iφ = −VT /5 200 IT = VT .·. IN = 1 VT /5 0.8 − 88 + 10 200 −j50 ! VT = 30 − j40 = ZTh IT −66 + j88 VTh = = −2.2 + j0 A ZTh 30 − j40 The Norton equivalent circuit: P 9.53 [a] IT = VT VT − αVT + 1000 −j1000 IT 1 (1 − α) j −1+α = − = VT 1000 j1000 j1000 © 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. 9–36 CHAPTER 9. Sinusoidal Steady State Analysis .·. ZTh = VT j1000 = IT α−1+j ZTh is real when α = 1. [b] ZTh = 1000 Ω [c] ZTh = 500 − j500 = = j1000 α−1+j 1000(α − 1) 1000 +j 2 (α − 1) + 1 (α − 1)2 + 1 Equate the real parts: 1000 = 500 (α − 1)2 + 1 . ·. . ·. (α − 1)2 + 1 = 2 (α − 1)2 = 1 so α = 0 Check the imaginary parts: (α − 1)1000 (α − 1)2 + 1 = −500 α=1 Thus, α = 0. 1000 1000(α − 1) [d] ZTh = +j 2 (α − 1) + 1 (α − 1)2 + 1 For Im(ZTh ) > 0, α must be greater than 1. So ZTh is inductive for 1 < α ≤ 10. P 9.54 jωL = j(2000)(1 × 10−3 ) = j2 Ω 1 106 = −j = −j5 Ω jωC (2000)(100) Vg1 = 20/ − 36.87◦ = 16 − j12 V Vg2 = 50/−106.26◦ = −14 − j48 V Vo − (16 − j12) Vo Vo − (−14 − j48) + + =0 j2 10 −j5 © 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. Problems 9–37 Solving, Vo = 36/0◦ vo (t) = 36 cos 2000t V P 9.55 V1 − 240 V1 V1 + + =0 j10 50 30 + j10 Solving for V1 yields V1 = 198.63/ − 24.44◦ V Vo = P 9.56 30 (V1 ) = 188.43/ − 42.88◦ V 30 + j10 Set up the frequency domain circuit to use the node voltage method: V1 − V2 V1 − 20/90◦ + =0 −j8 −j4 At V1: − 5/0◦ + At V2: V2 − V1 V2 V2 − 20/90◦ + + =0 −j8 j4 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. 9–38 CHAPTER 9. Sinusoidal Steady State Analysis In standard form: ! ! V1 1 1 1 + + V2 − −j8 −j4 −j8 V1 1 1 1 1 − + V2 + + −j8 −j8 j4 12 = 5/0◦ + ! ! = 20/90◦ −j4 20/90◦ 12 Solving on a calculator: 8 4 V1 = − + j 3 3 V2 = −8 + j4 Thus 8 56 Vg = V1 − 20/90◦ = − − j V 3 3 P 9.57 jωL = j106 (10 × 10−6 ) = j10 Ω 1 −j = 6 = −j10 Ω jωC 10 (100 × 10−9 ) Va = 50/ − 90◦ = −j50 V Vb = 25/90◦ = j25 V V1 V1 + j25 V1 + j50 + + =0 10 10 −j10 Solving, V1 = 25/ − 53.13◦ V = 15 − j20 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. Problems Ia = 9–39 V1 + j50 −j25 + j50 + −j10 j10 = −0.5 + j1.5 = 1.58/108.43◦ A ia = 1.58 cos(106 t + 108.43◦ ) A Ib = −j25 − V1 −j25 + j50 + 10 j10 = 1 − j0.5 = 1.12/ − 26.57◦ A ib = 1.12 cos(106 t − 26.57◦ ) A Ic = −j50 + j25 j10 = −2.5 A ic = 2.5 cos(106 t + 180◦ ) A P 9.58 Vo Vo + + 20Io = 0 50 −j25 (2 + j4)Vo = −2000Io Vo = (−200 + j400)Io Io = V1 − (Vo /10) j25 .·. V1 = (−20 + j65)Io 0.006 + j0.013 = V1 + Io = (−0.4 + j1.3)Io + Io = (0.6 + j1.3)Io 50 0.6 + j1.3(10 × 10−3 ) .·. Io = = 10/0◦ mA (0.6 + j1.3) Vo = (−200 + j400)Io = −2 + j4 = 4.47/116.57◦ 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. 9–40 P 9.59 CHAPTER 9. Sinusoidal Steady State Analysis Write a KCL equation at the top node: Vo Vo − 2.4I∆ Vo + + − (10 + j10) = 0 −j8 j4 5 The constraint equation is: I∆ = Vo −j8 Solving, Vo = j80 = 80/90◦ V P 9.60 The circuit with the mesh currents identified is shown below: The mesh current equations are: −20/ − 36.87◦ + j2I1 + 10(I1 − I2) = 0 50/ − 106.26◦ + 10(I2 − I1 ) − j5I2 = 0 In standard form: I1 (10 + j2) + I2(−10) = 20/ − 36.87◦ I1 (−10) + I2 (10 − j5) = 50/ − 106.26◦ Solving on a calculator yields: I1 = −6 + j10 A; I2 = −9.6 + j10 A Thus, Vo = 10(I1 − I2 ) = 36 V and vo (t) = 36 cos 2000t 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. Problems P 9.61 Va = 60/0◦ V; 9–41 Vb = 90/90◦ V jωL = j(4 × 104 )(125 × 10−6 ) = j5Ω −j −j106 = = −j20 Ω ωC 40,000(1.25) 60 = (20 + j5)Ia − j5Ib j90 = −j5Ia − j15Ib Solving, Ia = 2.25 − j2.25 A; Ib = −6.75 + j0.75 A Io = Ia − Ib = 9 − j3 = 9.49/ − 18.43◦ A io (t) = 9.49 cos(40,000t − 18.43◦ ) A P 9.62 (12 − j12)Ia − 12Ig − 5(−j8) = 0 −12Ia + (12 + j4)Ig + j20 − 5(j4) = 0 Solving, Ig = 4 − j2 = 4.47/ − 26.57◦ 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. 9–42 CHAPTER 9. Sinusoidal Steady State Analysis P 9.63 10/0◦ = (1 − j1)I1 − 1I2 + j1I3 −5/0◦ = −1I1 + (1 + j1)I2 − j1I3 1 = j1I1 − j1I2 + I3 Solving, I1 = 11 + j10 A; I2 = 11 + j5 A; I3 = 6 A Ia = I3 − 1 = 5 A Ib = I1 − I3 = 5 + j10 A Ic = I2 − I3 = 5 + j5 A Id = I1 − I2 = j5 A P 9.64 jωL = j10,000(5 × 10−3 ) = j50 Ω 1 −j = = −j50 Ω jωC (10,000)(2 × 10−6 ) 130/0◦ = (40 + j50)Ia − 40I∆ + 30I∆ © 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. Problems 9–43 0 = −40Ia + 30I∆ + (140 − j50)I∆ Solving, I∆ = (400 − j400) mA Vo = 100I∆ = 40 − j40 = 56.57/ − 45◦ vo = 56.57 cos(10,000t − 45◦ ) V P 9.65 1 109 = −j = −j100 Ω jωC (12,500)(800) jωL = j(12,500)(0.04) = j500 Ω Let Z1 = 50 − j100 Ω; Z2 = 250 + j500 Ω Ig = 125/0◦ mA Io = −125/0◦ (250 + j500) −Ig Z2 = Z1 + Z2 (300 + j400) = −137.5 − j25 mA = 139.75/ − 169.7◦ mA io = 139.75 cos(12,500t − 169.7◦ ) mA P 9.66 Zo = 12,000 − j 109 = 12,000 − j16,000 Ω (20,000)(3.125) ZT = 6000 + j40,000 + 12,000 − j16,000 = 18,000 + j24,000 Ω = 30,000/53.13◦ Ω Vo = Vg Zo (75/0◦ )(20,000/ − 53.13◦ ) = = 50/ − 106.26◦ V ◦ / ZT 30,000 53.13 vo = 50 cos(20,000t − 106.26◦ ) V P 9.67 1 = −j10 kΩ jωC1 1 = −j100 kΩ jωC2 © 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. 9–44 CHAPTER 9. Sinusoidal Steady State Analysis Va − 2 Va Va Va − Vo + + + =0 5000 −j10,000 20,000 100,000 20Va − 40 + j10V a + 5Va + Va − Vo = 0 .·. (26 + j10)Va − Vo = 40 0 − Va 0 − Vo + =0 20,000 −j100,000 j5Va − Vo = 0 Solving, Vo = 1.43 + j7.42 = 7.56/79.09◦ V vo (t) = 7.56 cos(106 t + 79.09◦ ) V P 9.68 [a] Vg = 25/0◦ V Vp = 20 Vg = 5/0◦ ; 100 Vn = Vp = 5/0◦ V 5 5 − Vo + =0 80,000 Zp Zp = −j80,000k40,000 = 32,000 − j16,000 Ω Vo = 5Zp + 5 = 7 − j = 7.07/ − 8.13◦ 80,000 vo = 7.07 cos(50,000t − 8.13◦ ) 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. Problems [b] Vp = 0.2Vm /0◦ ; 9–45 Vn = Vp = 0.2Vm /0◦ 0.2Vm 0.2Vm − Vo + =0 80,000 32,000 − j16,000 .·. Vo = 0.2Vm + 32,000 − j16,000 Vm (0.2) = Vm (0.28 − j0.04) 80,000 .·. |Vm (0.28 − j0.04)| ≤ 10 .·. Vm ≤ 35.36 V P 9.69 1 = −j20 kΩ jωC Let Va = voltage across the capacitor, positive at upper terminal Then: Vg = 4/0◦ V; Va − 4/0◦ Va Va + + = 0; 20,000 −j20,000 20,000 0 − Va 0 − Vo + = 0; 20,000 10,000 Vo = − .·. Va = (1.6 − j0.8) V Va 2 .·. Vo = −0.8 + j0.4 = 0.89/153.43◦ V vo = 0.89 cos(200t + 153.43◦ ) V P 9.70 [a] Va − 4/0◦ Va + jωCo Va + =0 20,000 20,000 Va = 4 2 + j20,000ωCo Vo = − Va 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. 9–46 CHAPTER 9. Sinusoidal Steady State Analysis Vo = −2 2/180◦ = 2 + j4 × 106 Co 2 + j4 × 106 Co .·. denominator angle = 45◦ so 4 × 106 Co = 2 [b] Vo = .·. C = 0.5 µF 2/180◦ = 0.707/135◦ V 2 + j2 vo = 0.707 cos(200t + 135◦ ) V P 9.71 [a] 1 −j109 = = −j100 Ω jωC (106 )(10) Vg = 30/0◦ V Vp = Vg (1/jωCo ) 30/0◦ = = Vn 25 + (1/jωCo ) 1 + j25ωCo Vn Vn − Vo + =0 100 −j100 1 + j1 30(1 − j1) Vn = (1 − j1)Vn = j 1 + j25ωCo √ 30 2 |Vo | = q =6 1 + 625ω 2 Co2 Vo = Solving, Co = 280 nF [b] Vo = 30(1 − j1) = 6/ − 126.87◦ 1 + j7 vo = 6 cos(106 t − 126.87◦ ) V P 9.72 [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. Problems 9–47 Because the op-amps are ideal Iin = Io , thus Zab = Vab Vab = ; Iin Io Vo1 = Vab ; Vo2 Io = Vab − Vo Z R2 =− Vo1 = −KVo1 = −KVab R1 Vo = Vo2 = −KVab . ·. Io = Vab − (−KVab) (1 + K)Vab = Z Z .·. Zab = [b] Z = P 9.73 1 ; jωC Vab Z Z= (1 + K)Vab (1 + K) Zab = 1 ; jωC(1 + K) .·. Cab = C(1 + K) [a] Superposition must be used because the frequencies of the two sources are different. [b] For ω = 2000 rad/s: 10k − j5 = 2 − j4 Ω so Vo1 = 2 − j4 (20/ − 36.87◦ = 31.62/ − 55.3◦ V 2 − j4 + j2 For ω = 5000 rad/s: j5k10 = 2 + j4 Ω Vo2 = 2 + j4 (10/16.26◦ ) = 15.81/34.69◦ V 2 + j4 − j2 Thus, vo (t) = [31.62 cos(2000t − 55.3◦ ) + 15.81 cos(5000t + 34.69◦ )] V, t≥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. 9–48 P 9.74 CHAPTER 9. Sinusoidal Steady State Analysis [a] Superposition must be used because the frequencies of the two sources are different. [b] For ω = 80,000 rad/s: V0o V0o − 5 V0o + + =0 20 j10 −j10 V0o 1 1 1 + + 20 j10 −j10 ! = 5 20 .·. V0o = 5/0◦ V I0o = V0o = −j0.5 = 500/ − 90◦ mA j10 For ω = 320,000 rad/s: 20kj40 = 16 + j8 Ω V00 = 16 + j8 (2.5/0◦ ) = 2.643/7.59◦ V 16 + j8 − j2.5 V00 00 · . . Io = = 66.08/ − 82.4◦ mA j40 Thus, io (t) = [500 sin 80,000t + 66.08 cos(320,000t − 82.4◦ )] mA, P 9.75 t≥0 [a] jωLL = j100 Ω jωL2 = j500 Ω Z22 = 300 + 500 + j100 + j500 = 800 + j600 Ω ∗ Z22 = 800 − j600 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. Problems 9–49 ωM = 270 Ω Zr = 2 270 1000 [800 − j600] = 58.32 − j43.74 Ω [b] Zab = R1 + jωL1 + Zr = 41.68 + j180 + 58.32 − j43.74 = 100 + j136.26 Ω P 9.76 [a] jωL1 = j(200 × 103 )(10−3 ) = j200 Ω jωL2 = j(200 × 103 )(4 × 10−3 ) = j800 Ω 1 −j = = −j400 Ω 3 jωC (200 × 10 )(12.5 × 10−9 ) .·. Z22 = 100 + 200 + j800 − j400 = 300 + j400 Ω ∗ .·. Z22 = 300 − j400 Ω q M = k L1 L2 = 2k × 10−3 ωM = (200 × 103 )(2k × 10−3 ) = 400k " 400k Zr = 500 #2 (300 − j400) = k 2(192 − j256) Ω Zin = 200 + j200 + 192k 2 − j256k 2 1 |Zin | = [(200 + 192k)2 + (200 − 256k)2 ] 2 d|Zin | 1 1 = [(200 + 192k)2 + (200 − 256k)2 ]− 2 × dk 2 [2(200 + 192k 2 )384k + 2(200 − 256k 2 )(−512k)] d|Zin | = 0 when dk 768k(200 + 192k 2 ) − 1024k(200 − 256k 2 ) = 0 √ .·. k 2 = 0.125; .·. k = 0.125 = 0.3536 [b] Zin (min) = 200 + 192(0.125) + j[200 − 0.125(256)] = 224 + j168 = 280/36.87◦ Ω I1 (max) = 560/0◦ = 2/ − 36.87◦ A 224 + j168 .·. i1 (peak) = 2 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. 9–50 CHAPTER 9. Sinusoidal Steady State Analysis Note — You can test that the k value obtained from setting d|Zin |/dt = 0 leads to a minimum by noting 0 ≤ k ≤ 1. If k = 1, Zin = 392 − j56 = 395.98/ − 8.13◦ Ω Thus, |Zin |k=1 > |Zin |k=√0.125 If k = 0, Zin = 200 + j200 = 282.84/45◦ Ω Thus, |Zin |k=0 > |Zin |k=√0.125 P 9.77 [a] jωL1 = j(5000)(2 × 10−3 ) = j10 Ω jωL2 = j(5000)(8 × 10−3 ) = j40 Ω jωM = j10 Ω 70 = (10 + j10)Ig + j10IL 0 = j10Ig + (30 + j40)IL Solving, Ig = 4 − j3 A; IL = −1 A ig = 5 cos(5000t − 36.87◦ ) A iL = 1 cos(5000t − 180◦ ) A M 2 = √ = 0.5 L1 L2 16 [c] When t = 100π µs, [b] k = √ 5000t = (5000)(100π) × 10−6 = 0.5π = π/2 rad = 90◦ ig (100πµs) = 5 cos(53.13◦ ) = 3 A iL (100πµs) = 1 cos(−90◦ ) = 0 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. Problems 9–51 1 1 1 w = L1 i21 + L2 i22 + Mi1 i2 = (2 × 10−3 )(9) + 0 + 0 = 9 mJ 2 2 2 When t = 200π µs, 5000t = π rad = 180◦ ig (200πµs) = 5 cos(180 − 53.13) = −4 A iL (200πµs) = 1 cos(180 − 180) = 1 A 1 1 w = (2 × 10−3 )(16) + (8 × 10−3 )(1) + 2 × 10−3 (−4)(1) = 12 mJ 2 2 P 9.78 Remove the voltage source to find the equivalent impedance: ZTh 20 = 45 + j125 + |5 + j5| !2 (5 + j5) = 85 + j85 Ω Using voltage division: VTh = Vcd P 9.79 425 = j20I1 = j20 5 + j5 ! = 850 + j850 V jωL1 = j50 Ω jωL2 = j32 Ω 1 = −j20 Ω jωC q jωM = j(4 × 103 )k (12.5)(8) × 10−3 = j40k Ω Z22 = 5 + j32 − j20 = 5 + j12 Ω ∗ Z22 = 5 − j12 Ω " 40k Zr = |5 + j12| #2 (5 − j12) = 47.337k 2 − j113.609k 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. 9–52 CHAPTER 9. Sinusoidal Steady State Analysis Zab = 20 + j50 + 47.337k 2 − j113.609k 2 = (20 + 47.337k 2 ) + j(50 − 113.609k 2 ) Zab is resistive when 50 − 113.609k 2 = 0 or k 2 = 0.44 so k = 0.66 .·. Zab = 20 + (47.337)(0.44) = 40.83 Ω P 9.80 In Eq. 9.69 replace ω 2 M 2 with k 2ω 2 L1 L2 and then write Xab as Xab = ωL1 − = ωL1 k 2 ω 2 L1 L2 (ωL2 + ωLL ) R222 + (ωL2 + ωLL )2 k 2 ωL2 (ωL2 + ωLL ) 1− 2 R22 + (ωL2 + ωLL )2 ( ) For Xab to be negative requires R222 + (ωL2 + ωLL )2 < k 2 ωL2 (ωL2 + ωLL ) or R222 + (ωL2 + ωLL )2 − k 2ωL2 (ωL2 + ωLL ) < 0 which reduces to R222 + ω 2 L22 (1 − k 2) + ωL2 ωLL (2 − k 2 ) + ω 2 L2L < 0 But k ≤ 1, so it is impossible to satisfy the inequality. Therefore Xab can never be negative if XL is an inductive reactance. P 9.81 [a] Zab = Vab V1 + V2 = I1 I1 © 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. Problems V1 V2 = , N1 N2 V2 = N1 I1 = N2 I2 , N2 V1 N1 N1 I1 N2 I2 = V2 = (I1 + I2 )ZL = I1 1 + V1 + V2 = .·. Zab = Zab 9–53 N1 ZL N2 N1 N1 + 1 V2 = 1 + N2 N2 2 ZL I1 (1 + N1 /N2 )2ZL I1 I1 N1 = 1+ N2 2 ZL Q.E.D. [b] Assume dot on N2 is moved to the lower terminal, then V1 −V2 = , N1 N2 N1 I1 = −N2 I2, V1 = −N1 V2 N2 I2 = −N1 I1 N2 and Zab = As in part [a] V2 = (I2 + I1 )ZL Zab = (1 − N1 /N2 )(1 − N1 /N2 )ZL I1 (1 − N1 /N2 )V2 = I1 I1 2 Zab = [1 − (N1 /N2 )] ZL P 9.82 V1 + V2 I1 Q.E.D. [a] N1 I1 = N2 I2 , Zab = I2 = N1 I1 N2 Vab V2 V2 = = I1 + I2 I1 + I2 (1 + N1/N2 )I1 © 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. 9–54 CHAPTER 9. Sinusoidal Steady State Analysis V1 N1 = , V2 N2 V1 = N1 V2 N2 N1 V1 + V2 = ZL I1 = + 1 V2 N2 Zab = I1 ZL (N1/N2 + 1)(1 + N1/N2 )I1 .·. Zab = ZL [1 + (N1 /N2 )]2 Q.E.D. [b] Assume dot on the N2 coil is moved to the lower terminal. Then N1 N1 V2 and I2 = − I1 N2 N2 As before V2 Zab = and V1 + V2 = ZL I1 I1 + I2 V1 = − .·. Zab = Zab = V2 ZL I1 = (1 − N1 /N2 )I1 [1 − (N1 /N2 )]2I1 ZL [1 − (N1 /N2 )]2 Q.E.D. P 9.83 ZL = V3 I3 V2 V3 = ; 10 1 V1 V2 =− ; 8 1 Zab = 10I2 = 1I3 8I1 = −1I2 V1 I1 Substituting, Zab = V1 −8V2 82 V2 = = I1 −I2 /8 I2 © 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. Problems = P 9.84 9–55 82 (10V3 ) (8)2 (10)2 V3 = = (8)2 (10)2 ZL = (8)2 (10)2 (80/60◦ ) = 512,000/60◦ Ω I3/10 I3 The phasor domain equivalent circuit is Vo = Vm − IRx ; 2 I= Vm Rx − jXC As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase angle increases from 0◦ to −180◦ , as shown in the following phasor diagram: P 9.85 [a] I = 240 240 + = (10 − j7.5) A 24 j32 Vs = 240/0◦ + (0.1 + j0.8)(10 − j7.5) = 247 + j7.25 = 247.11/1.68◦ V [b] Use the capacitor to eliminate the j component of I, therefore Ic = j7.5 A, Zc = 240 = −j32 Ω j7.5 Vs = 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13/1.90◦ V [c] Let Ic denote the magnitude of the current in the capacitor branch. Then I = (10 − j7.5 + jIc) = 10 + j(Ic − 7.5) A Vs = 240/α = 240 + (0.1 + j0.8)[10 + j(Ic − 7.5)] = (247 − 0.8Ic ) + j(7.25 + 0.1Ic) It follows that 240 cos α = (247 − 0.8Ic ) and 240 sin α = (7.25 + 0.1Ic ) © 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. 9–56 CHAPTER 9. Sinusoidal Steady State Analysis Now square each term and then add to generate the quadratic equation Ic2 − 605.77Ic + 5325.48 = 0; Ic = 302.88 ± 293.96 Therefore Ic = 8.92 A (smallest value) and Zc = 240/j8.92 = −j26.90 Ω. Therefore, the capacitive reactance is −26.90 Ω. P 9.86 [a] I` = 240 240 + = 30 − j40 A 8 j6 V` = (0.1 + j0.8)(30 − j40) = 35 + j20 = 40.31/29.74◦ V Vs = 240/0◦ + V` = 275 + j20 = 275.73/4.16◦ V [b] [c] I` = 30 − j40 + 240 = 30 + j8 A −j5 V` = (0.1 + j0.8)(30 + j8) = −3.4 + j24.8 = 25.03/97.81◦ Vs = 240/0◦ + V` = 236.6 + j24.8 = 237.9/5.98◦ P 9.87 [a] I1 = 120 240 + = 23.29 − j13.71 = 27.02/−30.5◦ A 24 8.4 + j6.3 I2 = 120 120 − = 5/0◦ A 12 24 © 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. Problems I3 = 120 240 + = 28.29 − j13.71 = 31.44/−25.87◦ A 12 8.4 + j6.3 I4 = 120 = 5/0◦ A; 24 I6 = 240 = 18.29 − j13.71 = 22.86/−36.87◦ A 8.4 + j6.3 I5 = 9–57 120 = 10/0◦ A 12 [b] When fuse A is interrupted, I1 = 0 I3 = 15 A I5 = 10 A I2 = 10 + 5 = 15 A I4 = −5 A I6 = 5 A [c] The clock and television set were fed from the uninterrupted side of the circuit, that is, the 12 Ω load includes the clock and the TV set. [d] No, the motor current drops to 5 A, well below its normal running value of 22.86 A. [e] After fuse A opens, the current in fuse B is only 15 A. P 9.88 [a] The circuit is redrawn, with mesh currents identified: The mesh current equations are: 120/0◦ = 23Ia − 2Ib − 20Ic 120/0◦ = −2Ia + 43Ib − 40Ic 0 = −20Ia − 40Ib + 70Ic Solving, Ia = 24/0◦ A Ib = 21.96/0◦ A Ic = 19.40/0◦ A The branch currents are: I1 = Ia = 24/0◦ A I2 = Ia − Ib = 2.04/0◦ A I3 = Ib = 21.96/0◦ 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. 9–58 CHAPTER 9. Sinusoidal Steady State Analysis I4 = Ic = 19.40/0◦ A I5 = Ia − Ic = 4.6/0◦ A I6 = Ib − Ic = 2.55/0◦ A [b] Let N1 be the number of turns on the primary winding; because the secondary winding is center-tapped, let 2N2 be the total turns on the secondary. From Fig. 9.58, 240 13,200 = N1 2N2 or N2 1 = N1 110 The ampere turn balance requires N1 Ip = N2 I1 + N2 I3 Therefore, Ip = P 9.89 N2 1 (I1 + I3 ) = (24 + 21.96) = 0.42/0◦ A N1 110 [a] The three mesh current equations are 120/0◦ = 23Ia − 2Ib − 20Ic 120/0◦ = −2Ia + 23Ib − 20Ic 0 = −20Ia − 20Ib + 50Ic Solving, Ia = 24/0◦ A; Ib = 24/0◦ A; Ic = 19.2/0◦ A . ·. I2 = Ia − Ib = 0 A [b] Ip = N2 N2 (I1 + I3 ) = (Ia + Ib ) N1 N1 = 1 (24 + 24) = 0.436/0◦ A 110 © 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. Problems 9–59 [c] Yes; when the two 120 V loads are equal, there is no current in the “neutral” line, so no power is lost to this line. Since you pay for power, the cost is lower when the loads are equal. . P 9.90 [a] 125 = (R + 0.05 + j0.05)I1 − (0.03 + j0.03)I2 − RI3 125 = −(0.03 + j0.03)I1 + (R + 0.05 + j0.05)I2 − RI3 Subtracting the above two equations gives 0 = (R + 0.08 + j0.08)I1 − (R + 0.08 + j0.08)I2 . ·. I1 = I2 so In = I1 − I2 = 0 A [b] V1 = R(I1 − I3 ); V2 = R(I2 − I3) Since I1 = I2 (from part [a]) V1 = V2 [c] 250 = (440.04 + j0.04)Ia − 440Ib 0 = −440Ia + 448Ib Solving, Ia = 31.656207 − j0.160343 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. 9–60 CHAPTER 9. Sinusoidal Steady State Analysis Ib = 31.090917 − j0.157479 A I1 = Ia − Ib = 0.56529 − j0.002864 A V1 = 40I1 = 22.612 − j0.11456 = 22.612/ − 0.290282◦ V V2 = 400I1 = 226.116 − j1.1456 = 226.1189/ − 0.290282◦ V [d] 125 = (40.05 + j0.05)I1 − (0.03 + j0.03)I2 − 40I3 125 = −(0.03 + j0.03)I1 + (400.05 + j0.05)I2 − 400I 3 0 = −40I1 − 400I2 + 448I3 Solving, I1 = 34.19 − j0.182 A I2 = 31.396 − j0.164 A I3 = 31.085 − j0.163 A V1 = 40(I1 − I3 ) = 124.2/ − 0.35◦ V V2 = 400(I2 − I3 ) = 124.4/ − 0.18◦ V [e] Because an open neutral can result in severely unbalanced voltages across the 125 V loads. P 9.91 [a] Let N1 = primary winding turns and 2N2 = secondary winding turns. Then 14,000 250 N2 1 = ; .·. = =a N1 2N2 N1 112 In part c), Ip = 2aIa © 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. Problems . ·. Ip = = 9–61 2N2 Ia 1 = Ia N1 56 1 (31.656 − j0.16) 56 Ip = 565.3 − j2.9 mA In part d), Ip N1 = I1 N2 + I2 N2 . ·. Ip = N2 (I1 + I2 ) N1 = 1 (34.19 − j0.182 + 31.396 − j0.164) 112 = 1 (65.586 − j0.346) 112 Ip = 585.6 − j3.1 mA [b] Yes, because the neutral conductor carries non-zero current whenever the load is not balanced. © 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. 10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/ − 45◦ V, I = 20/15◦ A Therefore 1 P = (100)(20) cos[−45 − (15)] = 500 W, 2 Q = 1000 sin −60◦ = −866.03 VAR, [b] V = 100/ − 45◦ , B→A I = 20/165◦ P = 1000 cos(−210◦ ) = −866.03 W, B→A Q = 1000 sin(−210◦ ) = 500 VAR, [c] V = 100/ − 45◦ , A→B I = 20/ − 105◦ P = 1000 cos(60◦ ) = 500 W, A→B Q = 1000 sin(60◦ ) = 866.03 VAR, [d] V = 100/0◦ , A→B A→B I = 20/120◦ P = 1000 cos(−120◦ ) = −500 W, B→A Q = 1000 sin(−120◦ ) = −866.03 VAR, B→A AP 10.2 pf = cos(θv − θi ) = cos[15 − (75)] = cos(−60◦ ) = 0.5 leading rf = sin(θv − θi) = sin(−60◦ ) = −0.866 © 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 10–1 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. 10–2 AP 10.3 CHAPTER 10. Sinusoidal Steady State Power Calculations Iρ 0.18 From Ex. 9.4 Ieff = √ = √ A 3 3 P = 2 Ieff R 0.0324 = (5000) = 54 W 3 AP 10.4 [a] Z = (39 + j26)k(−j52) = 48 − j20 = 52/ − 22.62◦ Ω Therefore I` = 250/0◦ = 4.85/18.08◦ A (rms) 48 − j20 + 1 + j4 VL = ZI` = (52/ − 22.62◦ )(4.85/18.08◦ ) = 252.20/ − 4.54◦ V (rms) IL = VL = 5.38/ − 38.23◦ A (rms) 39 + j26 [b] SL = VL I∗L = (252.20/ − 4.54◦ )(5.38/ + 38.23◦ ) = 1357/33.69◦ = (1129.09 + j752.73) VA PL = 1129.09 W; QL = 752.73 VAR [c] P` = |I` |2 1 = (4.85)2 · 1 = 23.52 W; Q` = |I` |24 = 94.09 VAR [d] Sg (delivering) = 250I∗` = (1152.62 − j376.36) VA Therefore the source is delivering 1152.62 W and absorbing 376.36 magnetizing VAR. |VL |2 (252.20)2 [e] Qcap = = = −1223.18 VAR −52 −52 Therefore the capacitor is delivering 1223.18 magnetizing VAR. Check: 94.09 + 752.73 + 376.36 = 1223.18 VAR and 1129.09 + 23.52 = 1152.62 W AP 10.5 Series circuit derivation: S = 250I∗ = (40,000 − j30,000) Therefore I∗ = 160 − j120 = 200/ − 36.87◦ A (rms) I = 200/36.87◦ A (rms) Z= V 250 = = 1.25/ − 36.87◦ = (1 − j0.75) Ω I 200/36.87◦ Therefore R = 1 Ω, XC = −0.75 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. Problems 10–3 Parallel circuit derivation P = (250)2 ; R therefore R = Q= (250)2 ; XC therefore XC = (250)2 = 1.5625 Ω 40,000 (250)2 = −2.083 Ω −30,000 AP 10.6 S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA S2 = 6000(0.8) − j6000(0.6) = 4800 − j3600 VA ST = S1 + S2 = 13,800 + j8400 VA ST = 200I∗ ; therefore I∗ = 69 + j42 I = 69 − j42 A Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91◦ V (rms) AP 10.7 [a] The phasor domain equivalent circuit and the Thévenin equivalent are shown below: Phasor domain equivalent circuit: Thévenin equivalent: VTh = 3 −j800 = 48 − j24 = 53.67/ − 26.57◦ V 20 − j40 ZTh = 4 + j18 + −j800 = 20 + j10 = 22.36/26.57◦ Ω 20 − j40 For maximum power transfer, ZL = (20 − j10) 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. 10–4 CHAPTER 10. Sinusoidal Steady State Power Calculations [b] I = 53.67/ − 26.57◦ = 1.34/ − 26.57◦ A 40 Therefore P = 1.34 √ 2 !2 20 = 17.96 W [c] RL = |ZTh | = 22.36 Ω 53.67/ − 26.57◦ [d] I = = 1.23/ − 39.85◦ A 42.36 + j10 Therefore P = 1.23 √ 2 !2 (22.36) = 17 W AP 10.8 Mesh current equations: 660 = (34 + j50)I1 + j100(I1 − I2 ) + j40I1 + j40(I1 − I2) 0 = j100(I2 − I1 ) − j40I1 + 100I2 Solving, I2 = 3.5/0◦ A; 1 .·. P = (3.5)2 (100) = 612.50 W 2 AP 10.9 [a] 248 = j400I1 − j500I2 + 375(I1 − I2 ) 0 = 375(I2 − I1) + j1000I 2 − j500I1 + 400I2 Solving, I1 = 0.80 − j0.62 A; I2 = 0.4 − j0.3 = 0.5/ − 36.87◦ 1 .·. P = (0.25)(400) = 50 W 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. Problems 10–5 [b] I1 − I2 = 0.4 − j0.32 A 1 P375 = |I1 − I2 |2(375) = 49.20 W 2 1 [c] Pg = (248)(0.8) = 99.20 W 2 X Pabs = 50 + 49.2 = 99.20 W AP 10.10 [a] VTh = 210 V; V2 = 14 V1 ; Short circuit equations: (checks) I1 = 14 I2 840 = 80I1 − 20I2 + V1 0 = 20(I2 − I1 ) − V2 .·. I2 = 14 A; [b] Pmax = 210 30 2 RTh = 210 = 15 Ω 14 15 = 735 W AP 10.11 [a] VTh = −4(146/0◦ ) = −584/0◦ V (rms) V2 = 4V1; I1 = −4I2 Short circuit equations: 146/0◦ = 80I1 − 20I2 + V1 0 = 20(I2 − I1 ) − V2 .·. I2 = −146/365 = −0.40 A; [b] P = −584 2920 2 RTh = −584 = 1460 Ω −0.4 1460 = 58.40 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. 10–6 CHAPTER 10. Sinusoidal Steady State Power Calculations Problems P 10.1 1 [a] P = (100)(10) cos(50 − 15) = 500 cos 35◦ = 409.58 W 2 (abs) Q = 500 sin 35◦ = 286.79 VAR (abs) 1 [b] P = (40)(20) cos(−15 − 60) = 400 cos(−75◦ ) = 103.53 W 2 (abs) Q = 400 sin(−75◦ ) = −386.37 VAR (del) 1 [c] P = (400)(10) cos(30 − 150) = 2000 cos(−120◦ ) = −1000 W 2 (del) Q = 2000 sin(−120◦ ) = −1732.05 VAR (del) 1 [d] P = (200)(5) cos(160 − 40) = 500 cos(120◦ ) = −250 W 2 (del) Q = 500 sin(120◦ ) = 433.01 VAR (abs) P 10.2 [a] hair dryer = 600 W vacuum = 630 W sun lamp = 279 W air conditioner = 860 W television = 240 W P P = 2609 W 2609 = 21.74 A 120 Yes, the breaker will trip. Therefore Ieff = [b] P 10.3 1700 = 14.17 A 120 Yes, the breaker will not trip if the current is reduced to 14.17 A. X P = 2609 − 909 = 1700 W; p = P + P cos 2ωt − Q sin 2ωt; dp = 0 when dt cos 2ωt = √ Ieff = dp = −2ωP sin 2ωt − 2ωQ cos 2ωt dt − 2ωP sin 2ωt = 2ωQ cos 2ωt or P ; P 2 + Q2 sin 2ωt = − √ tan 2ωt = − Q P Q P 2 + Q2 © 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. Problems 10–7 Let θ = tan−1 (−Q/P ), then p is maximum when 2ωt = θ and p is minimum when 2ωt = (θ + π). q P Q(−Q) √ Therefore pmax = P + P · √ 2 P 2 + Q2 − = P + P + Q2 P 2 + Q2 q P Q √ and pmin = P − P · √ 2 − Q · = P − P 2 + Q2 P + Q2 P 2 + Q2 P 10.4 [a] P = − 1 (240)2 = 60 W 2 480 1 −9 × 106 = = −360 Ω ωC (5000)(5) Q= 1 (240)2 = −80 VAR 2 (−360) q q pmax = P + P 2 + Q2 = 60 + (60)2 + (80)2 = 160 W (del) √ [b] pmin = 60 − 602 + 802 = −40 W (abs) [c] P = 60 W from (a) [d] Q = −80 VAR from (a) [e] generates, because Q < 0 [f] pf = cos(θv − θi ) I= 240 240 + = 0.5 + j0.67 = 0.83/53.13◦ A 480 −j360 .·. pf = cos(0 − 53.13◦ ) = 0.6 leading [g] rf = sin(−53.13◦ ) = −0.8 P 10.5 Ig = 4/0◦ mA; 1 = −j1250 Ω; jωC jωL = j500 Ω Zeq = 500 + [−j1250k(1000 + j500)] = 1500 − j500 Ω 1 1 Pg = − |I|2Re{Zeq} = − (0.004)2 (1500) = −12 mW 2 2 The source delivers 12 mW of power to the circuit. © 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. 10–8 P 10.6 CHAPTER 10. Sinusoidal Steady State Power Calculations jωL = j20,000(0.5 × 10−3 ) = j10 Ω; −6 + 1 106 = = −j40 Ω jωC j20,000(1.25) Vo − 30(Vo /j10) Vo + =0 j10 30 − j40 .·. Vo " # 1 1 + j3 + =6 j10 30 − j40 .·. Vo = 100/126.87◦ V Vo .·. I∆ = = 10/36.87◦ A j10 Io = 6/0◦ − I∆ = 6 − 8 − j6 = −2 − j6 = 6.32/ − 108.43◦ A 1 P30Ω = |Io |2 30 = 600 W 2 P 10.7 Zf = −j10,000k20,000 = 4000 − j8000 Ω Zi = 2000 − j2000 Ω Zf 4000 − j8000 .·. = = 3 − j1 Zi 2000 − j2000 Vo = − Zf Vg ; Zi Vg = 1/0◦ V Vo = −(3 − j1)(1) = −3 + j1 = 3.16/161.57◦ V P = 1 Vm2 1 (10) = = 5 × 10−3 = 5 mW 2 R 2 1000 © 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. Problems P 10.8 10–9 [a] From the solution to Problem 9.59 we have: Vo = j80 = 80/90◦ V 1 1 Sg = − Vo I∗g = − (j80)(10 − j10) = −400 − j400 VA 2 2 Therefore, the independent current source is delivering 400 W and 400 magnetizing vars. I1 = Vo = j16 A 5 1 P5Ω = (16)2 (5) = 640 W 2 Therefore, the 8 Ω resistor is absorbing 640 W. I∆ = Vo = −10 A −j8 1 Qcap = (10)2 (−8) = −400 VAR 2 Therefore, the −j8 Ω capacitor is developing 400 magnetizing vars. 2.4I∆ = −24 V I2 = Vo − 2.4I∆ −j80 + 24 = j4 j4 = 20 − j6 A = 20.88/ − 16.7◦ A 1 Qj4 = |I2 |2 (4) = 872 VAR 2 Therefore, the j4 Ω inductor is absorbing 872 magnetizing vars. Sd.s. = 12 (2.4I∆ )I∗2 = 12 (−24)(20 + j6) = −240 − j72 VA Thus the dependent source is delivering 240 W and 72 magnetizing vars. [b] X Pgen = 400 + 240 = 640 W = X Pabs © 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. 10–10 CHAPTER 10. Sinusoidal Steady State Power Calculations [c] P 10.9 X Qgen = 400 + 400 + 72 = 872 VAR = X Qabs [a] From the solution to Problem 9.61 we have Ia = 2.25 − j2.25 A; Ib = −6.75 + j0.75 A; Io = 9 − j3 A 1 S60V = − (60)I∗a = −30(2.25 + j2.25) = −67.5 − j67.5 VA 2 Thus, the 60 V source is developing 67.5 W and 67.5 magnetizing vars. S90V = − 21 (j90)I∗b = −j45(−6.75 − j0.75) = −33.75 + j303.75 VA Thus, the 90 V source is delivering 33.75 W and absorbing 303.75 magnetizing vars. 1 P20Ω = |Ia |2(20) = 101.25 W 2 Thus the 20 Ω resistor is absorbing 101.25 W. 1 Q−j20Ω = |Ib |2(−20) = −461.25 VAR 2 Thus the −j20 Ω capacitor is developing 461.25 magnetizing vars. 1 Qj5Ω = |Io |2 (5) = 225 VAR 2 Thus the j5 Ω inductor is absorbing 225 magnetizing vars. [b] [c] X X Pdev = 67.5 + 33.75 = 101.25 W = X Pabs Qdev = 67.5 + 461.25 = 528.75 VAR X Qabs = 225 + 303.75 = 528.75 VAR = P 10.10 [a] line loss = 7500 − 2500 = 5 kW line loss = |Ig |2 20 |Ig | = √ X Qdev .·. |Ig |2 = 250 250 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. Problems |Ig |2 RL = 2500 |Ig |2 XL = −5000 10–11 .·. RL = 10 Ω .·. XL = −20 Ω Thus, |Z| = q (30)2 + (X` − 20)2 500 |Ig | = q 900 + (X` − 20)2 25 × 104 = 1000 250 (X` − 20) = ±10. .·. 900 + (X` − 20)2 = Solving, Thus, X` = 10 Ω or X` = 30 Ω [b] If X` = 30 Ω: 500 Ig = = 15 − j5 A 30 + j10 Sg = −500I∗g = −7500 − j2500 VA Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars. Qj30 = |Ig |2X` = 250(30) = 7500 VAR Therefore the line reactance is absorbing 7500 magnetizing vars. Q−j20 = |Ig |2 XL = 250(−20) = −5000 VAR Therefore the load reactance is generating 5000 magnetizing vars. X Qgen = 7500 VAR = X Qabs If X` = 10 Ω: 500 Ig = = 15 + j5 A 30 − j10 Sg = −500I∗g = −7500 + j2500 VA Thus, the voltage source is delivering 7500 W and absorbing 2500 magnetizing vars. Qj10 = |Ig |2(10) = 250(10) = 2500 VAR Therefore the line reactance is absorbing 2500 magnetizing vars. The load continues to generate 5000 magnetizing vars. X Qgen = 5000 VAR = X Qabs © 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. 10–12 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.11 [a] Ieff = 40/115 ∼ = 0.35 A [b] Ieff = 130/115 ∼ = 1.13 A P 10.12 Wdc = Vdc2 T; R V2 .·. dc T = R Ws = Z to +T to Vdc2 = 1 T Z to +T Vdc = s 1 T Z to Z 0.08 0 to vs2 dt R vs2 dt R vs2 dt = Vrms = Veff to 0 ≤ t ≤ 80 ms i(t) = 100 − 1000t .·. Irms = to +T vs2 dt to +T P 10.13 i(t) = 250t Z s 1 0.1 80 ms ≤ t ≤ 100 ms Z 0.08 0 (250)2 t2 dt + (250)2 t2 dt = (250)2 t3 3 0.08 = 0 Z 0.1 0.08 (100 − 1000t)2 dt 32 3 (100 − 1000t)2 = 104 − 2 × 105 t + 106 t2 Z 0.1 Z 0.1 0.08 0.08 6 10 104 dt = 200 2 × 105 t dt = 105 t2 0.1 = 360 0.08 Z 0.1 0.08 t2 dt = .·. Irms = 2 P 10.14 P = Irms R 106 3 t 3 0.1 = 0.08 488 3 q 10{(32/3) + 225 − 360 + (488/3)} = 11.55 A . ·. R = 1280 = 9.6 Ω (11.55)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. Problems 10–13 P 10.15 [a] Area under one cycle of vg2: A = (100)(25 × 10−6 ) + 400(25 × 10−6 ) + 400(25 × 10−6 ) + 100(25 × 10−6 ) = 1000(25 × 10−6 ) Mean value of vg2: M.V. = .·. Vrms [b] P = A 1000(25 × 10−6 ) = = 250 100 × 10−6 100 × 10−6 √ = 250 = 15.81 V (rms) 2 Vrms 250 = = 62.5 W R 4 P 10.16 [a] Vo Vo − 240 Vo + + =0 −j25 12.5 15 + j20 .·. Vo = 183.53 − j14.12 = 184.07/ − 4.4◦ V Ig = 240 − 183.53 + j14.12 = 4.52 + j1.13 A 12.50 Sg = −Vg I∗g = −(240)(4.52 − j1.13) = −1084.24 + j271.06 VA [b] Source is delivering 1084.24 W. [c] Source is absorbing 271.06 magnetizing VAR. [d] Qcap = (184.07)2 = −1355.29 VAR −25 P12.5Ω = |Ig |2 (12.5) = 271.06 W |Io | = 184.07 = 7.36 A 25 P15Ω = |Io |2 (15) = 813.18 W Qind = |Io |2 (20) = 1084.24 VAR © 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. 10–14 CHAPTER 10. Sinusoidal Steady State Power Calculations [e] X Pdel = 1084.24 W X . ·. [f] X Pdiss = 271.06 + 813.18 = 1084.24 W X Pdel = X Pdiss = 1084.24 W Qabs = 271.06 + 1084.24 = 1355.29 VAR X . ·. Qdev = 1355.29 VAR X mag VAR dev = P 10.17 Ig = 40/0◦ mA jωL = j10,000 Ω; Io = X mag VAR abs = 1355.29 VAR 1 = −j10,000 Ω jωC j10,000 (40/0◦ ) = 80/90◦ mA 5000 1 1 P = |Io |2 (5000) = (0.08)2 (5000) = 16 W 2 2 1 Q = |Io |2 (−10,000) = −32 VAR 2 S = P + jQ = 16 − j32 VA |S| = 35.78 VA P 10.18 [a] 1 = −j40 Ω; jωC jωL = j80 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. Problems 10–15 Zeq = 40k − j40 + j80 + 60 = 80 + j60 Ω 40/0◦ = 0.32 − j0.24 A Ig = 80 + j60 1 1 Sg = − Vg I∗g = − 40(0.32 + j0.24) = −6.4 − j4.8 VA 2 2 P = 6.4 W (del); Q = 4.8 VAR (del) |S| = |Sg | = 8 VA [b] I1 = −j40 Ig = 0.04 − j0.28 A 40 − j40 1 P40Ω = |I1 |2(40) = 1.6 W 2 1 P60Ω = |Ig |2 (60) = 4.8 W 2 X Pdiss = 1.6 + 4.8 = 6.4 W = X Pdev [c] I−j40Ω = Ig − I1 = 0.28 + j0.04 A 1 Q−j40Ω = |I−j40Ω |2 (−40) = −1.6 VAR (del) 2 1 Qj80Ω = |Ig |2(80) = 6.4 VAR (abs) 2 X Qabs = 6.4 − 1.6 = 4.8 VAR = P 10.19 ST = 40,800 + j30,600 VA X Qdev S1 = 20,000(0.96 − j0.28) = 19,200 − j5600 VA S2 = ST − S1 = 21,600 + j36,200 = 42,154.48/59.176◦ VA rf = sin(59.176◦ ) = 0.8587 pf = cos(59.176◦ ) = 0.5124 lagging © 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. 10–16 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.20 [a] Let VL = Vm /0◦ : SL = 2500(0.8 + j0.6) = 2000 + j1500 VA I∗` = 2000 1500 +j ; Vm Vm I` = 2000 1500 −j Vm Vm 1500 2000 250/θ = Vm + −j (1 + j2) Vm Vm 250Vm /θ = Vm2 + (2000 − j1500)(1 + j2) = Vm2 + 5000 + j2500 250Vm cos θ = Vm2 + 5000; 250Vm sin θ = 2500 (250)2 Vm2 = (Vm2 + 5000)2 + 25002 62,500Vm2 = Vm4 + 10,000Vm2 + 31.25 × 106 or Vm4 − 52,500Vm2 + 31.25 × 106 = 0 Solving, Vm2 = 26,250 ± 25,647.86; Vm = 227.81 V and Vm = 24.54 V If Vm = 227.81 V: sin θ = 2500 = 0.044; (227.81)(250) If Vm = 24.54 V: 2500 = 0.4075; sin θ = (24.54)(250) .·. θ = 2.52◦ .·. θ = 24.05◦ [b] © 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. Problems 10–17 P 10.21 [a] S1 = 60,000 − j70,000 VA S2 = |VL |2 (2500)2 = = 240,000 − j70,000 VA Z2∗ 24 − j7 S1 + S2 = 300,000 VA 2500I ∗L = 300,000; .·. IL = 120 A(rms) Vg = VL + IL (0.1 + j1) = 2500 + (120)(0.1 + j1) = 2512 + j120 = 2514.86/2.735◦ Vrms [b] T = 1 1 = = 16.67 ms f 60 t 2.735◦ = ; 360◦ 16.67 ms .·. t = 126.62 µs [c] VL lags Vg by 2.735◦ or 126.62 µs P 10.22 [a] 250I∗1 = 7500 + j2500; .·. I1 = 30 − j10 A(rms) 250I∗2 = 2800 − j9600; .·. I2 = 11.2 + j38.4 A(rms) © 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. 10–18 CHAPTER 10. Sinusoidal Steady State Power Calculations I3 = 500 500 + = 40 − j10 A(rms) 12.5 j50 Ig1 = I1 + I3 = 70 − j20 A Sg1 = 250(70 + j20) = 17,500 + j5000 VA Thus the Vg1 source is delivering 17.5 kW and 5000 magnetizing vars. Ig2 = I2 + I3 = 51.2 + j28.4 A(rms) Sg2 = 250(51.2 − j28.4) = 12,800 − j7100 VA Thus the Vg2 source is delivering 12.8 kW and absorbing 7100 magnetizing vars. [b] X Pgen = 17.5 + 12.8 = 30.3 kW X (500)2 = 7500 + 2800 + = 30.3 kW = Pgen 12.5 X Pabs X Qdel = 9600 + 5000 = 14.6 kVAR X Qabs = 2500 + 7100 + X (500)2 = 14.6 kVAR = Qdel 50 P 10.23 S1 = 1200 + 1196 = 2396 + j0 VA 2396 = 19.967 A .·. I1 = 120 S2 = 860 + 600 + 240 = 1700 + j0 VA 1700 .·. I2 = = 14.167 A 120 S3 = 4474 + 12,200 = 16,674 + j0 VA 16,674 .·. I3 = = 69.475 A 240 Ig1 = I1 + I3 = 89.44 A Ig2 = I2 + I3 = 83.64 A Breakers will not trip since both feeder currents are less than 100 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. Problems 10–19 P 10.24 [a] I1 = 5000 − j1250 = 40 − j10 A (rms) 125 I2 = 6250 − j2500 = 50 − j20 A (rms) 125 I3 = 8000 + j0 = 32 + j0 A (rms) 250 .·. Ig1 = 72 − j10 A (rms) In = I1 − I2 = −10 + j10 A (rms) Ig2 = 82 − j20 A Vg1 = 0.05Ig1 + 125 + j0 + 0.15In = 127.1 − j1 V(rms) Vg2 = −0.15In + 125 + j0 + 0.05Ig2 = 130.6 − j2.5 V(rms) Sg1 = −[(127.1 − j1)(72 + j10)] = −[9141.2 + j1343] VA Sg2 = −[(130.6 − j2.5)(82 + j20)] = −[10,759.2 + j2407] VA Note: Both sources are delivering average power and magnetizing VAR to the circuit. [b] P0.05 = |Ig1 |2(0.05) = 264.2 W P0.15 = |In |2 (0.15) = 30 W P0.05 = |Ig2 |2(0.05) = 356.2 W X Pdis = 264.2 + 30 + 356.2 + 5000 + 8000 + 6250 = 19,900.4 W X Qabs = 1250 + 2500 = 3750 VAR X Pdev = 9141.2 + 10,759.2 = 19,900.4 W = X Qdel = 1343 + 2407 = 3750 VAR = X X Pdis Qabs © 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. 10–20 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.25 480I∗1 = 7500 + j9000 .·. I1 = 15.625 − j18.75 A(rms) I∗1 = 15.625 + j18.75; 480I∗2 = 2100 − j1800 I∗2 = 4.375 − j3.75; I3 = .·. I2 = 4.375 + j3.75 A(rms) 480/0◦ = 10 + j0 A; 48 I4 = 480/0◦ = 0 − j25 A j19.2 Ig = I1 + I2 + I3 + I4 = 30 − j40 A Vg = 480 + (30 − j40)(j0.5) = 500 + j15 = 500.22/1.72◦ V (rms) P 10.26 [a] Z1 = 240 + j70 = 250/16.26◦ Ω pf = cos(16.26◦ ) = 0.96 lagging rf = sin(16.26◦ ) = 0.28 Z2 = 160 − j120 = 200/ − 36.87◦ Ω pf = cos(−36.87◦ ) = 0.8 leading rf = sin(−36.87◦ ) = −0.6 Z3 = 30 − j40 = 50/ − 53.13◦ Ω pf = cos(−53.13◦ ) = 0.6 leading rf = sin(−53.13◦ ) = −0.8 © 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. Problems 10–21 [b] Y = Y1 + Y2 + Y3 1 ; 250/16.26◦ Y1 = Y2 = 1 ; 200/ − 36.87◦ Y3 = 1 50/ − 53.13◦ Y = 19.84 + j17.88 mS 1 = 37.44/ − 42.03◦ Ω Y Z= pf = cos(−42.03◦ ) = 0.74 leading rf = sin(−42.03◦ ) = −0.67 P 10.27 [a] S1 = 16 + j18 kVA; S2 = 6 − j8 kVA; S3 = 8 + j0 kVA ST = S1 + S2 + S3 = 30 + j10 kVA 250I∗ = (30 + j10) × 103 ; .·. I = 120 − j40 A 250 = 1.875 + j0.625 Ω = 1.98/18.43◦ Ω 120 − j40 Z= [b] pf = cos(18.43◦ ) = 0.9487 lagging P 10.28 [a] From the solution to Problem 10.26 we have IL = 120 − j40 A (rms) .·. Vs = 250/0◦ + (120 − j40)(0.01 + j0.08) = 254.4 + j9.2 = 254.57/2.07◦ V (rms) [b] |IL | = q 16,000 P` = (16,000)(0.01) = 160 W [c] Ps = 30,000 + 160 = 30.16 kW 30 (100) = 99.47% [d] η = 30.16 P 10.29 [a] I = Q` = (16,000)(0.08) = 1280 VAR Qs = 10,000 + 1280 = 11.28 kVAR 465/0◦ = 2.4 − j1.8 = 3/ − 36.87◦ A(rms) 124 + j93 P = (3)2 (4) = 36 W [b] YL = 1 = 5.33 − j4 mS 120 + j90 .·. XC = 1 = −250 Ω −4 × 10−3 © 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. 10–22 CHAPTER 10. Sinusoidal Steady State Power Calculations 1 = 187.5 Ω 5.33 × 10−3 465/0◦ [d] I = = 2.4279/ − 0.9◦ A 191.5 + j3 [c] ZL = P = (2.4279)2 (4) = 23.58 W 23.58 (100) = 65.5% 36 Thus the power loss after the capacitor is added is 65.5% of the power loss before the capacitor is added. [e] % = P 10.30 IL = 120,000 − j160,000 = 18.75 − j25 A (rms) 6400 IC = 6400 6400 =j = jIC −jXC XC I` = 18.75 − j25 + jIC = 18.75 + j(IC − 25) Vs = 6400 + (4 + j24)[18.75 + j(IC − 25)] = (7075 − 24IC ) + j(350 + 4IC ) |Vs |2 = (7075 − 24IC )2 + (350 + 4IC )2 = (6400)2 .·. 592IC2 − 336,800IC + 9,218,125 = 0 IC = 284.46 ± 255.63 = 28.33 A(rms)∗ *Select the smaller value of IC to minimize the magnitude of I` . 6400 .·. XC = − = −221.99 28.33 .·. C = 1 = 11.95 µF (221.99)(120π) © 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. Problems 10–23 P 10.31 [a] From Problem 9.75, Zab = 100 + j136.26 so I1 = 50 50 = = 160 − j120 mA 100 + j13.74 + 100 + 136.26 200 + j150 I2 = jωM j270 I1 = (0.16 − j0.12) = 51.84 + j15.12 mA Z22 800 + j600 VL = (300 + j100)(0.05184 + j0.01512) = 14.04 + j9.72 |VL | = 17.08 V [b] Pg (ideal) = 50(0.16) = 8 W Pg (practical) = 8 − |I1|2 (100) = 4 W PL = |I2 |2(300) = 0.8748 W % delivered = 0.8748 (100) = 21.87% 4 P 10.32 [a] So = original load = 1600 + j Sf = final load = 1920 + j 1600 (0.6) = 1600 + j1200 kVA 0.8 1920 (0.28) = 1920 + j560 kVA 0.96 .·. Qadded = 560 − 1200 = −640 kVAR [b] deliver [c] Sa = added load = 320 − j640 = 715.54/ − 63.43◦ kVA pf = cos(−63.43) = 0.447 leading [d] I∗L = (1600 + j1200) × 103 = 666.67 + j500 A 2400 IL = 666.67 − j500 = 833.33/ − 36.87◦ A(rms) |IL | = 833.33 A(rms) [e] I∗L = (1920 + j560) × 103 = 800 + j233.33 2400 IL = 800 − j233.33 = 833.33/ − 16.26◦ A(rms) |IL | = 833.33 A(rms) P 10.33 [a] Pbefore = Pafter = (833.33)2 (0.05) = 34,722.22 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. 10–24 CHAPTER 10. Sinusoidal Steady State Power Calculations [b] Vs (before) = 2400 + (666.67 − j500)(0.05 + j0.4) = 2633.33 + j241.67 = 2644.4/5.24◦ V(rms) |Vs (before)| = 2644.4 V(rms) Vs (after) = 2400 + (800 − j233.33)(0.05 + j0.4) = 2533.33 + j308.33 = 2552.028/6.94◦ V(rms) |Vs (after)| = 2552.028 V(rms) P 10.34 [a] SL = 20,000(0.85 + j0.53) = 17,000 + j10,535.65 VA 125I∗L = (17,000 + j10,535.65); I∗L = 136 + j84.29 A(rms) .·. IL = 136 − j84.29 A(rms) Vs = 125 + (136 − j84.29)(0.01 + j0.08) = 133.10 + j10.04 = 133.48/4.31◦ V(rms) |Vs | = 133.48 V(rms) [b] P` = |I` |2 (0.01) = (160)2 (0.01) = 256 W (125)2 [c] = −10,535.65; XC = −1.48306 Ω XC − 1 = −1.48306; ωC C= 1 = 1788.59 µF (1.48306)(120π) [d] I` = 136 + j0 A(rms) Vs = 125 + 136(0.01 + j0.08) = 126.36 + j10.88 = 126.83/4.92◦ V(rms) |Vs | = 126.83 V(rms) [e] P` = (136)2 (0.01) = 184.96 W P 10.35 [a] 10 = j1(I1 − I2 ) + j1(I3 − I2) − j1(I1 − I3 ) © 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. Problems 10–25 0 = 1I2 + j2(I2 − I3 ) + j1(I2 − I1 ) + j1(I2 − I1 ) + j1(I2 − I3 ) 0 = I3 − j1(I3 − I1 ) + j2(I3 − I2 ) + j1(I1 − I2 ) Solving, I1 = 6.25 + j7.5 A(rms); I2 = 5 + j2.5 A(rms); I3 = 5 − j2.5 A(rms) Ia = I1 = 6.25 + j7.5 A Ib = I1 − I2 = 1.25 + j5 A Ic = I2 = 5 + j2.5 A Id = I3 − I2 = −j5 A Ie = I1 − I3 = 1.25 + j10 A If = I3 = 5 − j2.5 A Va = 10 V Vb = j1Ib + j1Id = j1.25 V Vc = 1Ic = 5 + j2.5 V Vd = j2Id − j1Ib = 5 + j1.25 V Ve = −j1Ie = 10 − j1.25 V Vf = 1If = 5 − j2.5 V [b] Sa = −10I∗a = −62.5 + j75 VA Sb = Vb I∗b = 6.25 + j1.5625 VA Sc = Vc I∗c = 31.25 + j0 VA Sd = Vd I∗d = −6.25 + j25 VA Se = Ve I∗e = 0 − j101.5625 VA Sf = Vf I∗f = 31.25 VA [c] X Pdev = 62.5 W X Pabs = 31.25 + 31.25 = 62.5 W Note that the total power absorbed by the coupled coils is zero: 6.25 − 6.25 = 0 = Pb + Pd © 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. 10–26 CHAPTER 10. Sinusoidal Steady State Power Calculations [d] X Qdev = 101.5625 VAR Both the source and the capacitor are developing magnetizing vars. X P 10.36 [a] X Qabs = 75 + 1.5625 + 25 = 101.5625 VAR Q absorbed by the coupled coils is Qb + Qd = 26.5625 272/0◦ = 2Ig + j10Ig + j14(Ig − I2 ) − j6I2 +j14Ig − j8I2 + j20(Ig − I2 ) 0 = j20(I2 − Ig ) − j14Ig + j8I2 + j4I2 +j8(I2 − Ig ) − j6Ig + 8I2 Solving, Ig = 20 − j4 A(rms); I2 = 24/0◦ A(rms) P8Ω = (24)2 (8) = 4608 W [b] Pg (developed) = (272)(20) = 5440 W Vg 272 [c] Zab = −2= − 2 = 11.08 + j2.62 = 11.38/13.28◦ Ω Ig 20 − j4 [d] P2Ω = |Ig |2 (2) = 832 W X Pdiss = 832 + 4608 = 5440 W = P 10.37 [a] Zab = 1 + . ·. I1 = I2 = N1 N2 2 X Pdev (1 − j2) = 25 − j50 Ω 100/0◦ = 2.5/0◦ A 15 + j50 + 25 − j50 N1 I1 = 10/0◦ A N2 .·. IL = I1 + I2 = 12.5/0◦ A(rms) P1Ω = (12.5)2 (1) = 156.25 W P15Ω = (2.5)2 (15) = 93.75 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. Problems 10–27 [b] Pg = −100(2.5/0◦ ) = −250 W X Pabs = 156.25 + 93.75 = 250 W = P 10.38 [a] 25a21 + 4a22 = 500 I25 = a1I; I4 = a2 I; Pdev P25 = a21I2 (25) P4 = a22I2 (4) P4 = 4P25 ; . ·. X a22I2 4 = 100a21 I2 100a21 = 4a22 25a21 + 100a21 = 500; 25(4) + 4a22 = 500; a1 = 2 a2 = 10 2000/0◦ [b] I = = 2/0◦ A(rms) 500 + 500 I25 = a1I = 4 A P25Ω = (16)(25) = 400 W [c] I4 = a2I = 10(2) = 20 A(rms) V4 = (20)(4) = 80/0◦ V(rms) P 10.39 [a] 300 = 60I1 + V1 + 20(I1 − I2) 0 = 20(I2 − I1 ) + V2 + 40I2 1 V2 = V1 ; 4 I2 = −4I1 Solving, V1 = 260 V (rms); V2 = 65 V (rms) © 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. 10–28 CHAPTER 10. Sinusoidal Steady State Power Calculations I1 = 0.25 A (rms); I2 = −1.0 A (rms) V5A = V1 + 20(I1 − I2) = 285 V (rms) .·. P = −(285)(5) = −1425 W Thus 1425 W is delivered by the current source to the circuit. [b] I20Ω = I1 − I2 = 1.25 A(rms) .·. P20Ω = (1.25)2 (20) = 31.25 W P 10.40 ZL = |ZL |/θ◦ = |ZL | cos θ◦ + j|ZL | sin θ◦ Thus |I| = q |VTh | (RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2 Therefore P = 0.5|VTh |2|ZL | cos θ (RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2 Let D = demoninator in the expression for P, then dP (0.5|VTh |2 cos θ)(D · 1 − |ZL |dD/d|ZL |) = d|ZL | D2 dD = 2(RTh + |ZL | cos θ) cos θ + 2(XTh + |ZL | sin θ) sin θ d|ZL | dP dD = 0 when D = |ZL | d|ZL | d|ZL | ! Substituting the expressions for D and (dD/d|ZL |) into this equation gives us 2 the relationship R2Th + XTh = |ZL |2 or |ZTh | = |ZL |. P 10.41 [a] 180 = 3I1 + j4I1 + j3(I2 − I1) + j9(I1 − I2 ) − j3I1 0 = 9I2 + j9(I2 − I1 ) + j3I1 Solving, I1 = 18 − j18 A(rms); I2 = 12/0◦ A(rms) .·. Vo = (12)(9) = 108 V(rms) © 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. Problems 10–29 [b] P = (12)2 (9) = 1296 W [c] Sg = −(180)(18 + j18) = −3240 − j3240 VA % delivered = .·. Pg = −3240 W 1296 (100) = 40% 3240 P 10.42 [a] Open circuit voltage: 180 = 3I1 + j4I1 − j3I1 + j9I1 − j3I1 . ·. I1 = 180 = 9.31 − j21.72 A(rms) 3 + j7 VTh = j9I1 − j3I1 = j6I1 = 130.34 + j55.86 V Short circuit current: 180 = 3I1 + j4I1 + j3(Isc − I1) + j9(I1 − Isc ) − j3I1 0 = −j9(Isc − I1) + j3I1 Solving, Isc = 20 − j20 A ZTh = IL = VTh 130.34 + j55.86 = = 1.86 + j4.66 Ω Isc 20 − j20 130.34 + j55.86 = 35 + j15 = 38.08/23.20◦ A 3.72 © 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. 10–30 CHAPTER 10. Sinusoidal Steady State Power Calculations PL = (38.08)2 (1.86) = 2700 W [b] I1 = Zo + j9 1.86 − j4.66 + j9 I2 = (35 + j15) = 30/0◦ A(rms) j6 j6 Pdev = (180)(30) = 5400 W [c] Begin by choosing the capacitor value from Appendix H that is closest to the required reactive impedance, assuming the frequency of the source is 60 Hz: 1 1 4.66 = so C= = 569.22 µF 2π(60)C 2π(60)(4.66) Choose the capacitor value closest to this capacitance from Appendix H, which is 470 µF. Then, 1 XL = − = −5.6438 Ω 2π(60)(470 × 10−6 ) Now set RL as close as possible to RL = q q R2Th + (XL + XTh )2: 1.8562 + (4.66 − 5.6438)2 = 2.11 Ω The closest single resistor value from Appendix H is 10 Ω. The resulting real power developed by the source is calculated below, using the Thévenin equivalent circuit: 130.34 + j55.86 I= = 11.9157/27.94◦ 1.86 + j4.66 + 10 − j5.6438 P = |130.34 + j55.86|(11.9157) = 1689.7 W (instead of 5400 W) P 10.43 [a] From Problem 9.78, ZTh = 85 + j85 Ω and VTh = 850 + j850 V. Thus, for ∗ maximum power transfer, ZL = ZTh = 85 − j85 Ω: I2 = 850 + j850 = 5 + j5 A 170 425/0◦ = (5 + j5)I1 − j20(5 + j5) © 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. Problems . ·. I1 = 10–31 325 + j100 = 42.5 − j22.5 A 5 + j5 Sg (del) = 425(42.5 + j22.5) = 18,062.5 + j9562.5 VA Pg = 18,062.5 W [b] Ploss = |I1|2 (5) + |I2 |2(45) = 11,562.5 + 2250 = 13,812.5 W % loss in transformer = P 10.44 [a] ZTh = −j40 + 13,812.5 (100) = 76.47% 18,062.5 (40)(j40) = 20 − j20 Ω 40 + j40 ∗ .·. ZL = ZTh = 20 + j20 Ω [b] VTh = √ 80/0◦ (40) = 40(1 − j1) = 40 2/ − 45◦ V 40 + j40 √ 40 2/ − 45◦ √ = 2/ − 45◦ A I= 40 |Irms | = 1 A Pload = (1)2 (20 × 103 ) = 20 W [c] The closest resistor value from Appendix H is 22 Ω. Find the inductor value: (5000)L = 20 so L = 4 mH The closest inductor value is 1 mH. 40/ − 45◦ 40/ − 45◦ I= = = 0.8969/ − 25.35◦ A(rms) 20 − j20 + 22 + j5 42 − j15 Pload = (0.8969)2 (22) = 17.70 W P 10.45 [a] (instead of 20 W) 115.2 − j86.4 − 240 115.2 − j80 + =0 ZTh 90 − j30 .·. ZTh = 240 − 115.2 + j86.4 = 60 + j80 Ω 1.44 − j0.48 .·. ZL = 60 − j80 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. 10–32 CHAPTER 10. Sinusoidal Steady State Power Calculations [b] I = 240/0◦ = 2/0◦ A(rms) 120/0◦ P = (2)2 (60) = 240 W [c] Let R = 15 Ω + 15 Ω + 15 Ω + 15 Ω = 60 Ω 1 = 80 2π(60)C so C= 1 = 33.16 µF 2π(60)(80) Let C = 22 µFk10 µFk1 µF = 33 µF P 10.46 [a] Open circuit voltage: V1 = 5Iφ = 5 100 − 5Iφ 25 + j10 (25 + j10)Iφ = 100 − 5Iφ Iφ = 100 = 3−jA 30 + j10 VTh = j3 (5Iφ ) = 15 V 1 + j3 Short circuit current: V2 = 5Iφ = 100 − 5Iφ 25 + j10 Iφ = 3 − j1 A Isc = 5Iφ = 15 − j5 A 1 ZTh = 15 = 0.9 + j0.3 Ω 15 − j5 © 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. Problems 10–33 ∗ ZL = ZTh = 0.9 − j0.3 Ω IL = 0.3 = 8.33 A(rms) 1.8 P = |IL |2(0.9) = 62.5 W [b] VL = (0.9 − j0.3)(8.33) = 7.5 − j2.5 V(rms) I1 = VL = −0.833 − j2.5 A(rms) j3 I2 = I1 + IL = 7.5 − j2.5 A(rms) 5Iφ = I2 + VL .·. Iφ = 3 − j1 A Id.s. = Iφ − I2 = −4.5 + j1.5 A Sg = −100(3 + j1) = −300 − j100 VA Sd.s. = 5(3 − j1)(−4.5 − j1.5) = −75 + j0 VA Pdev = 300 + 75 = 375 W % developed = 62.5 (100) = 16.67% 375 Checks: P25Ω = (10)(25) = 250 W P1Ω = (67.5)(1) = 67.5 W P0.9Ω = 62.5 W X Pabs = 230 + 62.5 + 67.5 = 375 = X Pdev © 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. 10–34 CHAPTER 10. Sinusoidal Steady State Power Calculations Qj10 = (10)(10) = 100 VAR Qj3 = (6.94)(3) = 20.82 VAR Q−j0.3 = (69.4)(−0.3) = −20.82 VAR Qsource = −100 VAR X Q = 100 + 20.82 − 20.82 − 100 = 0 P 10.47 [a] First find the Thévenin equivalent: jωL = j3000 Ω ZTh = 6000k12,000 + j3000 = 4000 + j3000 Ω VTh = 12,000 (180) = 120 V 6000 + 12,000 −j = −j1000 Ω ωC I= 120 = 18 − j6 mA 6000 + j2000 1 P = |I|2 (2000) = 360 mW 2 [b] Set Co = 0.1 µF so −j/ωC = −j2000 Ω Set Ro as close as possible to Ro = q 40002 + (3000 − 2000)2 = 4123.1 Ω .·. Ro = 4000 Ω [c] I = 120 = 14.77 − j1.85 mA 8000 + j1000 1 P = |I|2 (4000) = 443.1 mW 2 Yes; 443.1 mW > 360 mW © 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. Problems 10–35 120 = 15 mA 8000 1 P = (0.015)2 (4000) = 450 mW 2 [e] Ro = 4000 Ω; Co = 66.67 nF [d] I = [f] Yes; 450 mW > 443.1 mW P 10.48 [a] Set Co = 0.1 µF, so −j/ωC = −j2000 Ω; also set Ro = 4123.1 Ω I= 120 = 14.55 − j1.79 mA 8123.1 + j1000 1 P = |I|2 (4123.1) = 443.18 mW 2 [b] Yes; 443.18 mW > 360 mW [c] Yes; 443.18 mW < 450 mW P 10.49 [a] ZTh = 20 + j60 + (j20)(6 − j18) = 80 + j60 = 100/36.87◦ Ω 6 + j2 .·. R = |ZTh | = 100 Ω [b] VTh = I= j20 (480/0◦ ) = 480 + j1440 V(rms) 6 − j18 + j20 480 + j1440 = 4.8 + j6.4 = 8/53.13◦ A(rms) 180 + j60 P = 82 (100) = 6400 W [c] Pick the 100 Ω resistor from Appendix H to match exactly. P 10.50 [a] Open circuit voltage: Vφ − 100 Vφ + − 0.1Vφ = 0 5 j5 © 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. 10–36 CHAPTER 10. Sinusoidal Steady State Power Calculations .·. Vφ = 40 + j80 V(rms) VTh = Vφ + 0.1Vφ (−j5) = Vφ (1 − j0.5) = 80 + j60 V(rms) Short circuit current: Isc = 0.1Vφ + Vφ = (0.1 + j0.2)Vφ −j5 Vφ − 100 Vφ Vφ + + =0 5 j5 −j5 .·. Vφ = 100 V(rms) Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms) ZTh = VTh 80 + j60 = = 4 − j2 Ω Isc 10 + j20 .·. Ro = |ZTh | = 4.47 Ω [b] 80 + j60 √ = 7.36 + j8.82 A (rms) 4 + 20 − j2 √ P = (11.49)2 ( 20) = 590.17 W I= © 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. Problems 10–37 [c] 80 + j60 = 10 + j7.5 A (rms) 8 P = (102 + 7.52 )(4) = 625 W I= [d] Vφ − 100 Vφ Vo − (25 + j50) + + =0 5 j5 −j5 Vφ = 50 + j25 V (rms) 0.1Vφ = 5 + j2.5 V (rms) 5 + j2.5 + IC = 10 + j7.5 IC = 5 + j5 A (rms) IL = Vφ = 5 − j10 A (rms) j5 IR = IC + IL = 10 − j5 A (rms) Ig = IR + 0.1Vφ = 15 − j2.5 A (rms) Sg = −100I∗g = −1500 − j250 VA 100 = 5(5 + j2.5) + Vcs + 25 + j50 . ·. Vcs = 50 − j62.5 V (rms) Scs = (50 − j62.5)(5 − j2.5) = 93.75 − j437.5 VA Thus, X Pdev = 1500 % delivered to Ro = 625 (100) = 41.67% 1500 © 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. 10–38 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.51 [a] 54 = I1 + j2(I1 − I2 ) + j3I2 0 = 7I2 + j2(I2 − I1 ) − j3I2 + j8I2 + j3(I1 − I2 ) Solving, I1 = 12 − j21 A (rms); I2 = −3 A (rms) Vo = 7I2 = −21/0◦ V(rms) [b] P = |I2 |2(7) = 63 W [c] Pg = (54)(12) = 648 W % delivered = 63 (100) = 9.72% 648 P 10.52 [a] 54 = I1 + j2(I1 − I2 ) + j4kI2 0 = 7I2 + j2(I2 − I1 ) − j4kI2 + j8I2 + j4k(I1 − I2 ) Place the equations in standard form: 54 = (1 + j2)I1 + j(4k − 2)I2 0 = j(4k − 2)I1 + [7 + j(10 − 8k)]I2 I1 = 54 − I2 j(4k − 2) (1 + j2) Substituting, I2 = j54(4k − 2) [7 + j(10 − 8k)](1 + j2) − (4k − 2) For Vo = 0, I2 = 0, so if 4k − 2 = 0, then k = 0.5. © 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. Problems 10–39 [b] When I2 = 0 54 I1 = = 10.8 − j21.6 A(rms) 1 + j2 Pg = (54)(10.8) = 583.2 W Check: Ploss = |I1 |2 (1) = 583.2 W P 10.53 [a] Open circuit: VTh = −j3I1 + j2I1 = −jI1 I1 = 54 = 10.8 − j21.6 1 + j2 VTh = −21.6 − j10.8 V Short circuit: 54 = I1 + j2(I1 − Isc ) + j3Isc 0 = j2(Isc − I1 ) − j3Isc + j8Isc + j3(I1 − Isc ) Solving, Isc = −3.32 + j5.82 ZTh = VTh −21.6 − j10.8 = = 0.2 + j3.6 = 3.6/86.86◦ Ω Isc −3.32 + j5.82 .·. RL = |ZTh | = 3.6 Ω [b] I= −21.6 − j10.8 = 4.614/163.1◦ 3.8 + j3.6 P = |I|2(3.6) = 76.6 W, which is greater than when RL = 7 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. 10–40 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.54 [a] 1 = 100 Ω; ωC C= 1 = 26.53 µF (60)(200π) [b] Vswo = 4000 + (40)(1.25 + j10) = 4050 + j400 = 4069.71/5.64◦ V(rms) Vsw = 4000 + (40 − j40)(1.25 + j10) = 4450 + j350 = 4463.73/4.50◦ V(rms) 4463.73 % increase = − 1 (100) = 9.68% 4069.71 √ [c] P`wo = (40 2)2(1.25) = 4000 W P`w = 402 (1.25) = 2000 W 4000 % increase = − 1 (100) = 100% 2000 P 10.55 Open circuit voltage: I1 = 10/0◦ = 2 − j4 A 1 + j2 VTh = j2I1 + j1.2I1 = j3.2I1 = 12.8 + j6.4 = 14.31/26.57◦ Short circuit current: 10/0◦ = (1 + j2)I1 − j3.2Isc 0 = −j3.2I1 + j5.4Isc Solving, Isc = 5.89/ − 5.92◦ 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. Problems ZTh = 10–41 14.31/26.57◦ = 2.43/32.49◦ = 2.048 + j1.304 Ω 5.89/ − 5.92◦ 14.31/26.57◦ .·. I2 = = 3.49/26.57◦ A 4.096 10/0◦ = (1 + j2)I1 − j3.2I2 10 + j3.2I2 10 + j3.2(3.49/26.57◦ ) = = 5A 1 + j2 1 + j2 .·. I1 = Zg = 10/0◦ = 2 + j0 = 2/0◦ Ω 5 P 10.56 [a] Open circuit: VTh = 120 (j10) = 36 + j48 V 16 + j12 Short circuit: (16 + j12)I1 − j10Isc = 120 −j10I1 + (11 + j23)Isc = 0 Solving, Isc = 2.4 A ZTh = 36 + j48 = 15 + j20 Ω 2.4 ∗ .·. ZL = ZTh = 15 − j20 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. 10–42 CHAPTER 10. Sinusoidal Steady State Power Calculations IL = VTh 36 + j48 = = 1.2 + j1.6 A(rms) ZTh + ZL 30 PL = |IL |2 (15) = 60 W [b] I1 = Z22 I2 26 + j3 = (1.2 + j1.6) = 5.23/ − 30.29◦ A (rms) jωM j10 Ptransformer = (120)(5.23) cos(−30.29◦ ) − (5.23)2 (4) = 432.8 W % delivered = 60 (100) = 13.86% 432.8 P 10.57 [a] jωL1 = j(10,000)(1 × 10−3 ) = j10 Ω jωL2 = j(10,000)(1 × 10−3 ) = j10 Ω jωM = j10 Ω 200 = (5 + j10)Ig + j5IL 0 = j5Ig + (15 + j10)IL Solving, Ig = 10 − j15 A; IL = −5 A Thus, ig = 18.03 cos(10,000t − 56.31◦ ) A iL = 5 cos(10,000t − 180◦ ) A M 0.5 = √ = 0.5 L1 L2 1 [c] When t = 50π µs: [b] k = √ 10,000t = (10,000)(50π) × 10−6 = 0.5π rad = 90◦ ig (50π µs) = 18.03 cos(90◦ − 56.31◦ ) = 15 A iL (50π µs) = 5 cos(90◦ + 180◦ ) = 0 A 1 1 1 w = L1 i21 + L2 i22 + Mi1 i2 = (10−3 )(15)2 + 0 + 0 = 112.5 mJ 2 2 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. Problems 10–43 When t = 100π µs: 10,000t = (104 )(100π) × 10−6 = π = 180◦ ig (100π µs) = 18.03 cos(180 − 56.31◦ ) = −10 A iL (100π µs) = 5 cos(180 − 180◦ ) = 5 A 1 1 w = (10−3 )(10)2 + (10−3 )(5)2 + 0.5 × 10−3 (−10)(5) = 37.5 mJ 2 2 [d] From (a), Im = 5 A, 1 .·. P = (5)2 (15) = 187.5 W 2 [e] Open circuit: 200 (−j5) = −80 − j40 V 5 + j10 VTh = Short circuit: 200 = (5 + j10)I1 + j5Isc 0 = j5I1 + j10Isc Solving, Isc = − ZTh = 80 120 +j 13 13 VTh −80 − j40 = = 1 + j8 Ω Isc −(80/13) + j(120/13) .·. RL = 8.06 Ω [f] I= −80 − j40 = 7.40/165.12◦ A 1 + j8 + 8.06 1 P = (7.40)2 (8.06) = 223.42 W 2 ∗ [g] ZL = ZTh = 1 − j8 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. 10–44 CHAPTER 10. Sinusoidal Steady State Power Calculations [h] I = −80 − j40 = 44.72/ − 153.43◦ 2 1 P = (44.72)2 (1) = 1000 W 2 P 10.58 [a] Replace the circuit to the left of the primary winding with a Thévenin equivalent: VTh = (15)(20kj10) = 60 + j120 V ZTh = 2 + 20kj10 = 6 + j8 Ω Transfer the secondary impedance to the primary side: Zp = 1 XC (100 − jXC ) = 4 − j Ω 25 25 Now maximize I by setting (XC /25) = 8 Ω: . ·. C = [b] I = 1 = 0.25 µF 200(20 × 103 ) 60 + j120 = 6 + j12 A 10 P = |I|2(4) = 720 W Ro = 6 Ω; .·. Ro = 150 Ω 25 60 + j120 = 5 + j10 A [d] I = 12 [c] P = |I|2(6) = 750 W P 10.59 [a] For maximum power transfer, Zab = 90 kΩ Zab N1 = 1− N2 2 ZL © 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. Problems N1 N2 2 . ·. 1− N1 = ±15; N2 1− = 10–45 90,000 = 225 400 N1 = 15 + 1 = 16 N2 2 [b] P = |Ii | (90,000) = 180 180,000 !2 180 [c] V1 = Ri Ii = (90,000) 180,000 (90,000) = 90 mW ! = 90 V [d] Vg = (2.25 × 10−3 )(100,000k80,000) = 100 V Pg (del) = (2.25 × 10−3 )(100) = 225 mW % delivered = 90 (100) = 40% 225 P 10.60 [a] Zab = 50 − j400 = 1 − .·. ZL = N1 N2 2 ZL 1 (50 − j400) = 2 − j16 Ω (1 − 6)2 [b] I1 = 24 = 240/0◦ mA 100 © 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. 10–46 CHAPTER 10. Sinusoidal Steady State Power Calculations N1 I1 = −N2 I2 I2 = −6I1 = −1.44/0◦ A IL = I1 + I2 = −1.68/0◦ A VL = (2 − j16)IL = −3.36 + j26.88 = 27.1/97.13◦ V(rms) P 10.61 [a] ZTh = 720 + j1500 + 200 50 2 (40 − j30) = 1360 + j1020 = 1700/36.87◦ Ω .·. Zab = 1700 Ω Zab = ZL (1 + N1 /N2 )2 (1 + N1/N2 )2 = 6800/1700 = 4 .·. N1/N2 = 1 [b] VTh = IL = or N2 = N1 = 1000 turns 255/0◦ (j200) = 1020/53.13◦ V 40 + j30 1020/53.13◦ = 0.316/34.7◦ A(rms) 3060 + j1020 Since the transformer is ideal, P6800 = P1700. P = |I|2(1700) = 170 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. Problems 10–47 [c] 255/0◦ = (40 + j30)I1 − j200(0.26 + j0.18) .·. I1 = 4.13 − j1.80 A(rms) Pgen = (255)(4.13) = 1053 W Pdiss = 1053 − 170 = 883 W % dissipated = 883 (100) = 83.85% 1053 P 10.62 [a] Open circuit voltage: 500 = 100I1 + V1 V2 = 400I2 V1 V2 = 1 2 . ·. V2 = 2V1 I1 = 2I2 Substitute and solve: 2V1 = 400I1 /2 = 200I1 500 = 100I1 + 100I1 . ·. . ·. .·. V1 = 100I1 I1 = 500/200 = 2.5 A 1 I2 = I1 = 1.25 A 2 V1 = 100(2.5) = 250 V; V2 = 2V1 = 500 V VTh = 20I1 + V1 − V2 + 40I2 = −150 V(rms) © 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. 10–48 CHAPTER 10. Sinusoidal Steady State Power Calculations Short circuit current: 500 = 80(Isc + I1 ) + 360(Isc + 0.5I1 ) 2V1 = 40 I1 + 360(Isc + 0.5I1) 2 500 = 80(I1 + Isc ) + 20I1 + V1 Solving, Isc = −1.47 A RTh = P = VTh −150 = = 102 Ω Isc −1.47 752 = 55.15 W 102 [b] 500 = 80[I1 − (75/102)] − 75 + 360[I2 − (75/102)] 575 + 6000 27,000 + = 80I1 + 180I2 102 102 . ·. I1 = 3.456 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. Problems 10–49 Psource = (500)[3.456 − (75/102)] = 1360.35 W % delivered = 55.15 (100) = 4.05% 1360.35 [c] P80Ω = 80(I1 + IL )2 = 592.13 W P20Ω = 20I21 = 238.86 W P40Ω = 40I22 = 119.43 W P102Ω = 102I2L = 55.15 W P360Ω = 360(I2 + IL )2 = 354.73 W X Pabs = 592.13 + 238.86 + 119.43 + 55.15 + 354.73 = 1360.3 W = P 10.63 [a] Open circuit voltage: X Pdev 40/0◦ = 4(I1 + I3 ) + 12I3 + VTh I1 = −I3 ; 4 I1 = −4I3 Solving, VTh = 40/0◦ V Short circuit current: 40/0◦ = 4I1 + 4I3 + I1 + V1 4V1 = 16(I1 /4) = 4I1 ; .·. V1 = I1 .·. 40/0◦ = 6I1 + 4I3 © 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. 10–50 CHAPTER 10. Sinusoidal Steady State Power Calculations Also, 40/0◦ = 4(I1 + I3 ) + 12I3 Solving, I1 = 6 A; RTh = I= I3 = 1 A; Isc = I1 /4 + I3 = 2.5 A VTh 40 = = 16 Ω Isc 2.5 40/0◦ = 1.25/0◦ A(rms) 32 P = (1.25)2 (16) = 25 W [b] 40 = 4(I1 + I3) + 12I3 + 20 4V1 = 4I1 + 16(I1 /4 + I3 ); .·. V1 = 2I1 + 4I3 40 = 4I1 + 4I3 + I1 + V1 .·. I1 = 6 A; I3 = −0.25 A; I1 + I3 = 5.75/0◦ A P40V (developed) = 40(5.75) = 230 W .·. % delivered = [c] PRL = 25 W; 25 (100) = 10.87% 230 P16Ω = (1.5)2 (16) = 36 W P4Ω = (5.75)2 (4) = 132.25 W; P1Ω = (6)2 (1) = 36 W P12Ω = (−0.25)2 (12) = 0.75 W X Pabs = 25 + 36 + 132.25 + 36 + 0.75 = 230 W = X Pdev © 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. Problems 10–51 P 10.64 30Vo = Va ; Io = Ia ; 30 Vb −Va = ; 1 20 Ib = −20Ia ; therefore Va = 9 kΩ Ia therefore Vb 9000 = = 22.5 Ω Ib 400 Therefore Ib = [50/(2.5 + 22.5)] = 2 A (rms); since the ideal transformers are lossless, P10Ω = P22.5Ω, and the power delivered to the 22.5 Ω resistor is 22 (22.5) or 90 W. a2 10 Vb = = 2.5 Ω; therefore a2 = 100, Ib 400 50 [b] Ib = = 10 A; P = (100)(2.5) = 250 W 5 P 10.65 [a] a = 10 P 10.66 [a] Begin with the MEDIUM setting, as shown in Fig. 10.31, as it involves only the resistor R2. Then, Pmed = 500 W = V2 1202 = R2 R2 Thus, 1202 R2 = = 28.8 Ω 500 [b] Now move to the LOW setting, as shown in Fig. 10.30, which involves the resistors R1 and R2 connected in series: Plow = V2 V2 = = 250 W R1 + R2 R1 + 28.8 Thus, 1202 − 28.8 = 28.8 Ω 250 [c] Note that the HIGH setting has R1 and R2 in parallel: R1 = Phigh = V2 1202 = = 1000 W R1 kR2 28.8k28.8 If the HIGH setting has required power other than 1000 W, this problem could not have been solved. In other words, the HIGH power setting was chosen in such a way that it would be satisfied once the two resistor values were calculated to satisfy the LOW and MEDIUM power settings. © 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. 10–52 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.67 [a] PL = V2 ; R1 + R2 PM = R1 + R2 = V2 ; R2 V2 PL V2 PM R2 = V 2 (R1 + R2) PH = R1 R2 R1 + R2 = PH = V 2 [b] PH = R1 = V 2V 2/PL PL PH = V2 ; PL − V2 PM 2 PM PM − PL V2 PM = V2 V2 − PL PM PM PL PM PL (PM − PL ) (750)2 = 1125 W (750 − 250) P 10.68 First solve the expression derived in P10.67 for PM as a function of PL and PH . Thus PM − PL = 2 PM PH 2 PM − PM + PL = 0 PH or 2 PM − PM PH + PL PH = 0 PH .·. PM = ± 2 s PH 2 s 2 − PL PH PH 1 PL ± PH − = 2 4 PH For the specified values of PL and PH √ PM = 500 ± 1000 0.25 − 0.24 = 500 ± 100 .·. PM 1 = 600 W; PM 2 = 400 W Note in this case we design for two medium power ratings If PM 1 = 600 W R2 = (120)2 = 24 Ω 600 © 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. Problems R1 + R2 = 10–53 (120)2 = 60 Ω 240 R1 = 60 − 24 = 36 Ω (120)2 (60) = 1000 W (36)(24) CHECK: PH = If PM 2 = 400 W R2 = (120)2 = 36 Ω 400 R1 + R2 = 60 Ω (as before) R1 = 24 Ω CHECK: PH = 1000 W P 10.69 R1 + R2 + R3 = (120)2 = 24 Ω 600 (120)2 = 16 Ω R2 + R3 = 900 .·. R1 = 24 − 16 = 8 Ω R3 + R1 kR2 = (120)2 = 12 Ω 1200 8R2 .·. 16 − R2 + = 12 8 + R2 R2 − 8R2 =4 8 + R2 8R2 + R22 − 8R2 = 32 + 4R2 R22 − 4R2 − 32 = 0 R2 = 2 ± √ 4 + 32 = 2 ± 6 .·. R2 = 8 Ω; .·. R3 = 8 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. 10–54 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.70 R2 = (220)2 = 96.8 Ω 500 R1 + R2 = (220)2 = 193.6 Ω 250 .·. R1 = 96.8 Ω CHECK: R1 kR2 = 48.4 Ω PH = (220)2 = 1000 W 48.4 P 10.71 Choose R1 = 22 Ω and R2 = 33 Ω: PL = 1202 = 262 W 22 + 33 PM = 1202 = 436 W 33 PH = 1202 (55) = 1091 W (22)(33) (instead of 240 W) (instead of 400 W) (instead of 1000 W) P 10.72 Choose R1 = R2 = 100 Ω : PL = 2202 = 242 W 100 + 100 PM = 2202 = 484 W 100 PH = 2202 (200) = 968 W (100)(100) (instead of 250 W) (instead of 500 W) (instead of 1000 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. 11 Balanced Three-Phase Circuits Assessment Problems AP 11.1 Make a sketch: We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC . VAB = VAN + VNB = VAN − VBN Since VAN , VBN , and VCN form a balanced set, and VAN = 240/ − 30◦ V, and the phase sequence is positive, VBN = |VAN |//VAN − 120◦ = 240/ − 30◦ − 120◦ = 240/ − 150◦ V Then, VAB = VAN − VBN = (240/ − 30◦ ) − (240/ − 150◦ ) = 415.46/0◦ V Since VAB , VBC , and VCA form a balanced set with a positive phase sequence, we can find VBC from VAB : VBC = |VAB |/(/VAB − 120◦ ) = 415.69/0◦ − 120◦ = 415.69/ − 120◦ 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 11–1 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. 11–2 CHAPTER 11. Balanced Three-Phase Circuits Thus, VBC = 415.69/ − 120◦ V AP 11.2 Make a sketch: We know VCN and wish to find VAB . To do this, write a KVL equation to find VBC , and use the known phase angle relationship between VAB and VBC to find VAB . VBC = VBN + VNC = VBN − VCN Since VAN , VBN , and VCN form a balanced set, and VCN = 450/ − 25◦ V, and the phase sequence is negative, VBN = |VCN |//VCN − 120◦ = 450/ − 23◦ − 120◦ = 450/ − 145◦ V Then, VBC = VBN − VCN = (450/ − 145◦ ) − (450/ − 25◦ ) = 779.42/ − 175◦ V Since VAB , VBC , and VCA form a balanced set with a negative phase sequence, we can find VAB from VBC : VAB = |VBC |//VBC − 120◦ = 779.42/ − 295◦ V But we normally want phase angle values between +180◦ and −180◦ . We add 360◦ to the phase angle computed above. Thus, VAB = 779.42/65◦ V AP 11.3 Sketch the a-phase circuit: © 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. Problems 11–3 [a] We can find the line current using Ohm’s law, since the a-phase line current is the current in the a-phase load. Then we can use the fact that IaA , IbB , and IcC form a balanced set to find the remaining line currents. Note that since we were not given any phase angles in the problem statement, we can assume that the phase voltage given, VAN , has a phase angle of 0◦ . 2400/0◦ = IaA (16 + j12) so IaA = 2400/0◦ = 96 − j72 = 120/ − 36.87◦ A 16 + j12 With an acb phase sequence, /IbB = /IaA + 120◦ and /IcC = /IaA − 120◦ so IaA = 120/ − 36.87◦ A IbB = 120/83.13◦ A IcC = 120/ − 156.87◦ A [b] The line voltages at the source are Vab Vbc, and Vca. They form a balanced set. To find Vab , use the a-phase circuit to find VAN , and use the relationship between phase voltages and line voltages for a y-connection (see Fig. 11.9[b]). From the a-phase circuit, use KVL: Van = VaA + VAN = (0.1 + j0.8)IaA + 2400/0◦ = (0.1 + j0.8)(96 − j72) + 2400/0◦ = 2467.2 + j69.6 2468.18/1.62◦ V From Fig. 11.9(b), √ Vab = Van( 3/ − 30◦ ) = 4275.02/ − 28.38◦ V With an acb phase sequence, /Vbc = /Vab + 120◦ and /Vca = /Vab − 120◦ so Vab = 4275.02/ − 28.38◦ V Vbc = 4275.02/91.62◦ V Vca = 4275.02/ − 148.38◦ 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. 11–4 CHAPTER 11. Balanced Three-Phase Circuits [c] Using KVL on the a-phase circuit Va0n = Va0 a + Van = (0.2 + j0.16)IaA + Van = (0.02 + j0.16)(96 − j72) + (2467.2 + j69.9) = 2480.64 + j83.52 = 2482.05/1.93◦ V With an acb phase sequence, /Vb0 n = /Va0 n + 120◦ and /Vc0 n = /Va0 n − 120◦ so Va0n = 2482.05/1.93◦ V Vb0n = 2482.05/121.93◦ V Vc0n = 2482.05/ − 118.07◦ V AP 11.4 √ √ IcC = ( 3/ − 30◦ )ICA = ( 3/ − 30◦ ) · 8/ − 15◦ = 13.86/ − 45◦ A AP 11.5 IaA = 12/(65◦ − 120◦ ) = 12/ − 55◦ " ! # ! / − 30◦ 1 ◦ √ / − 30 IaA = √ IAB = · 12/ − 55◦ 3 3 = 6.93/ − 85◦ A AP 11.6 [a] IAB = " # ! 1 √ /30◦ [69.28/ − 10◦ ] = 40/20◦ A 3 Therefore Zφ = [b] IAB = " ! 4160/0◦ = 104/ − 20◦ Ω 40/20◦ # 1 √ / − 30◦ [69.28/ − 10◦ ] = 40/ − 40◦ A 3 Therefore Zφ = 104/40◦ Ω AP 11.7 Iφ = 110 110 + = 30 − j40 = 50/ − 53.13◦ A 3.667 j2.75 Therefore |IaA | = AP 11.8 [a] |S| = Q= √ √ 3Iφ = 3(50) = 86.60 A √ 3(208)(73.8) = 26,587.67 VA q (26,587.67)2 − (22,659)2 = 13,909.50 VAR © 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. Problems [b] pf = 11–5 22,659 = 0.8522 lagging 26,587.67 ! 2450 /0◦ V; AP 11.9 [a] VAN = √ VAN I∗aA = Sφ = 144 + j192 kVA 3 Therefore (144 + j192)1000 √ = (101.8 + j135.7) A I∗aA = 2450/ 3 IaA = 101.8 − j135.7 = 169.67/ − 53.13◦ A |IaA | = 169.67 A [b] P = (2450)2 ; R Q= (2450)2 ; X therefore R = (2450)2 = 41.68 Ω 144,000 therefore X = (2450)2 = 31.26 Ω 192,000 √ VAN 2450/ 3 [c] Zφ = = 8.34/53.13◦ = (5 + j6.67) Ω = ◦ / IaA 169.67 − 53.13 .·. R = 5 Ω, X = 6.67 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. 11–6 CHAPTER 11. Balanced Three-Phase Circuits Problems P 11.1 Va = Vm /0◦ = Vm + j0 Vb = Vm / − 120◦ = −Vm (0.5 + j0.866) Vc = Vm /120◦ = Vm (−0.5 + j0.866) Va + Vb + Vc = (Vm )(1 + j0 − 0.5 − j0.866 − 0.5 + j0.866) = Vm (0) = 0 P 11.2 [a] First, convert the cosine waveforms to phasors: Va = 208/27◦ ; Vb = 208/ − 147◦ ; Vc = 208/ − 93◦ Subtract the phase angle of the a-phase from all phase angles: /V0a = 27◦ − 27◦ = 0◦ /V0b = 147◦ − 27◦ = 120◦ /V0c = −93◦ − 27◦ = −120◦ Compare the result to Eqs. 11.1 and 11.2: Therefore acb [b] First, convert the cosine waveforms to phasors: Va = 4160/ − 18◦ ; Vb = 4160/ − 138◦ ; Vc = 4160/ + 102◦ Subtract the phase angle of the a-phase from all phase angles: /V0a = −18◦ + 18◦ = 0◦ /V0b = −138◦ + 18◦ = −120◦ /V0c = 102◦ + 18◦ = 120◦ Compare the result to Eqs. 11.1 and 11.2: Therefore abc P 11.3 [a] Va = 139/0◦ V Vb = 139/120◦ V Vc = 139/ − 120◦ V Balanced, negative phase sequence © 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. Problems 11–7 [b] Va = 381/0◦ V Vb = 381/240◦ V = 622/ − 120◦ V Vc = 381/120◦ V Balanced, positive phase sequence [c] Va = 2771/ − 120◦ V Vb = 2771/0◦ V Vc = 2771/120◦ V Balanced, negative phase sequence [d] Va = 170/ − 60◦ V Vb = 170/180◦ V Vc = 170/60◦ V Balanced, positive phase sequence [e] Unbalanced, due to unequal amplitudes [f] Unbalanced, due to unequal phase angle separation P 11.4 P 11.5 Va + Vb + Vc =0 3(RW + jXW ) √ [a] Van = 1/ 3/30◦ Vab = 240/120◦ V (rms) The a-phase circuit is I= 240/120◦ = 2.4/83.13◦ A (rms) 80 + j60 [c] VAN = (76 + j55)IaA = 225.15/119.02◦ V (rms) √ VAB = 3/ − 30◦ VAN = 389.98/89.02◦ A (rms) [b] IaA = P 11.6 Zga + Zla + ZLa = 60 + j80 Ω Zgb + Zlb + ZLb = 40 + j30Ω Zgc + Zlc + ZLc = 20 + j15Ω VN − 240 VN − 240/120◦ VN − 240/ − 120◦ VN + + + =0 60 + j80 40 + j30 20 + j15 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. 11–8 CHAPTER 11. Balanced Three-Phase Circuits Solving for VN yields VN = 42.94/ − 156.32◦ V (rms) Io = P 11.7 VN = 4.29/ − 156.32◦ A (rms) 10 VAN = 7620/30◦ V VBN = 7620/150◦ V VCN = 7620/ − 90◦ V VAB = VAN − VBN = 13,198.23/0◦ V VBC = VBN − VCN = 13,198.23/120◦ V VCA = VCN − VAN = 13,198.23/ − 120◦ V vAB = 13,198.23 cos ωt V vBC = 13,198.23 cos(ωt + 120◦ ) V vCA = 13,198.23 cos(ωt − 120◦ ) V P 11.8 200 = 8 A (rms) 25 200/ − 120◦ = = 4/ − 66.87◦ A (rms) 30 − j40 [a] IaA = IbB IcC = 200/120◦ = 2/83.13◦ A (rms) 80 + j60 The magnitudes are unequal and the phase angles are not 120◦ apart. b] Io = IaA + IbB + IcC = 9.96/ − 9.79◦ A (rms) P 11.9 [a] 6600 IaA = √ = 15.24/16.26◦ A (rms) 3(240 − j70) |IaA | = |IL | = 15.24 A (rms) © 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. Problems 11–9 [b] Van = (15.24/16.26◦ )(240 − j66) = 3801.24/0.91◦ √ |Vab | = 3(3801.24) = 6583.94 V (rms) P 11.10 [a] IaA 277/0◦ = = 2.77/ − 36.87◦ A (rms) 80 + j60 IbB 277/ − 120◦ = = 2.77/ − 156.87◦ A (rms) 80 + j60 IcC = 277/120◦ = 2.77/83.13◦ A (rms) 80 + j60 Io = IaA + IbB + IcC = 0 [b] VAN = (78 + j54)IaA = 262.79/ − 2.17◦ V (rms) [c] VAB = VAN − VBN VBN = (77 + j56)IbB = 263.73/ − 120.84◦ V (rms) VAB = 262.79/ − 2.17◦ − 263.73/ − 120.84◦ = 452.89/28.55◦ V (rms) [d] Unbalanced — see conditions for a balanced circuit on p. 504 of the text! P 11.11 Make a sketch of the a-phase: [a] Find the a-phase line current from the a-phase circuit: IaA 125/0◦ 125/0◦ = = 0.1 + j0.8 + 19.9 + j14.2 20 + j15 = 4 − j3 = 5/ − 36.87◦ A (rms) Find the other line currents using the acb phase sequence: IbB = 5/ − 36.87◦ + 120◦ = 5/83.13◦ A (rms) IcC = 5/ − 36.87◦ − 120◦ = 5/ − 156.87◦ A (rms) © 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. 11–10 CHAPTER 11. Balanced Three-Phase Circuits [b] The phase voltage at the source is Van = 125/0◦ V. Use Fig. 11.9(b) to find the line voltage, Van, from the phase voltage: √ Vab = Van( 3/ − 30◦ ) = 216.51/ − 30◦ V (rms) Find the other line voltages using the acb phase sequence: Vbc = 216.51/ − 30◦ + 120◦ = 216.51/90◦ V (rms) Vca = 216.51/ − 30◦ − 120◦ = 216.51/ − 150◦ V (rms) [c] The phase voltage at the load in the a-phase is VAN . Calculate its value using IaA and the load impedance: VAN = IaA ZL = (4 − j3)(19.9 + j14.2) = 122.2 − j2.9 = 122.23/ − 1.36◦ V (rms) Find the phase voltage at the load for the b- and c-phases using the acb sequence: VBN = 122.23/ − 1.36◦ + 120◦ = 122.23/118.64◦ V (rms) VCN = 122.23/ − 1.36◦ − 120◦ = 122.23/ − 121.36◦ V (rms) [d] The line voltage at the load in the a-phase is VAB . Find this line voltage from the phase voltage at the load in the a-phase, VAN , using Fig, 11.9(b): √ VAB = VAN ( 3/ − 30◦ ) = 211.72/ − 31.36◦ V (rms) Find the line voltage at the load for the b- and c-phases using the acb sequence: VBC = 211.72/ − 31.36◦ + 120◦ = 211.72/88.64◦ V (rms) VCA = 211.72/ − 31.36◦ − 120◦ = 211.72/ − 151.36◦ V (rms) P 11.12 [a] Van = Vcn − /120◦ = 20/ − 210◦ = 20/150◦ V (rms) Zy = Z∆ /3 = 39 − j33 Ω The a-phase circuit is 20/150◦ [b] IaA = = 0.4/ − 173.13◦ A (rms) 40 − j30 [c] VAN = (39 + j33)IaA = 20.44/146.63◦ V (rms) √ VAB = 3/30◦ VAN = 35.39/176.63◦ A (rms) © 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. Problems 11–11 P 11.13 Zy = Z∆ /3 = 4 + j3 Ω The a-phase circuit is IaA = 240/ − 170◦ = 37.48/151.34◦ A (rms) (1 + j1) + (4 + j3) 1 IAB = √ / − 30◦ IaA = 21.64/121.34◦ A (rms) 3 P 11.14 [a] IAB = 69,000 = 76.67/16.26◦ A (rms) 864 − j252 IBC = 76.67/ − 103.74◦ A (rms) ICA = 76.67/136.26◦ A (rms) √ [b] IaA = 3/ − 30◦ IAB = 132.79/ − 13.74◦ A (rms) IbB = 132.79/ − 133.74◦ A (rms) IcC = 132.79/106.26◦ A (rms) [c] 13,000 √ / − 30◦ + (0.5 + j4)(132.79/ − 13.74◦ ) 3 = 39,755.70/ − 29.24◦ V (rms) Van = Vab = √ 3/30◦ Van = 68,858.88/0.76◦ V (rms) Vbc = 68,858.88/ − 119.24◦ V (rms) Vca = 68,858.88/120.76◦ V (rms) © 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. 11–12 CHAPTER 11. Balanced Three-Phase Circuits P 11.15 [a] IaA = 7650 7650 + = 252.54/ − 6.49◦ A (rms) 72 + j21 50 |IaA | = 252.54 A (rms) √ 7650 3/30◦ = 88.33/30◦ A (rms) [b] IAB = 150 |IAB | = 88.33 A (rms) [c] IAN 7650/0◦ = = 102/ − 16.26◦ A (rms) 72 + j21 |IAN | = 102 A (rms) [d] Van = (252.54/ − 6.49◦ )(j1) + 7650/0◦ = 7682.66/1.87◦ V (rms) √ |Vab | = 3(7682.66) = 13,306.76 V (rms) √ 208 P 11.16 Van = 1/ 3/ − 30◦ Vab = √ /20◦ V (rms) 3 Zy = Z∆ /3 = 1 − j3 Ω The a-phase circuit is Zeq = (4 + j3)k(1 − j3) = 2.6 − j1.8 Ω VAN 2.6 − j1.8 = (1.4 + j0.8) + (2.6 − j1.8) VAB = √ ! 208 √ /20◦ = 92.1/ − 0.66◦ V (rms) 3 3/30◦ VAN = 159.5/29.34◦ V (rms) © 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. Problems P 11.17 [a] IAB = 11–13 13,200/0◦ = 105.6/36.87◦ A (rms) 100 − j75 IBC = 105.6/156.87◦ A (rms) ICA = 105.6/ − 83.13◦ A (rms) √ [b] IaA = 3/ − 30◦ IAB = 182.9/66.87◦ A (rms) IbB = 182.9/ − 173.13◦ A (rms) IcC = 182.9/ − 53.13◦ A (rms) [c] Iba = IAB = 105.6/36.87◦ A (rms) Icb = IBC = 105.6/156.87◦ A (rms) Iac = ICA = 105.6/ − 83.13◦ A (rms) P 11.18 [a] IAB = 480/0◦ = 192/16.26◦ A (rms) 2.4 − j0.7 IBC = 480/120◦ = 48/83.13◦ A (rms) 8 + j6 ICA = 480/ − 120◦ = 24/ − 120◦ A (rms) 20 [b] IaA = IAB − ICA = 210/20.79◦ IbB = IBC − IAB = 178.68/ − 178.04◦ IcC = ICA − IBC = 70.7/ − 104.53◦ P 11.19 [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. 11–14 CHAPTER 11. Balanced Three-Phase Circuits [b] IaA = √ 13,800 = 2917/ − 29.6◦ A (rms) 3(2.375 + j1.349) |IaA | = 2917 A (rms) [c] VAN = (2.352 + j1.139)(2917/ − 29.6◦ ) = 7622.93/ − 3.76◦ V (rms) √ |VAB | = 3|VAN | = 13,203.31 V (rms) [d] Van = (2.372 + j1.319)(2917/ − 29.6◦ ) = 7616.93/ − 0.52◦ V (rms) √ |Vab | = 3|Van | = 13,712.52 V (rms) |IaA | [e] |IAB | = √ = 1684.13 A (rms) 3 [f] |Iab | = |IAB | = 1684.13 A (rms) P 11.20 [a] Since the phase sequence is acb (negative) we have: Van = 2399.47/30◦ V (rms) Vbn = 2399.47/150◦ V (rms) Vcn = 2399.47/ − 90◦ V (rms) 1 ZY = Z∆ = 0.9 + j4.5 Ω/φ 3 √ [b] Vab = 2399.47/30◦ − 2399.47/150◦ = 2399.47 3/0◦ = 4156/0◦ V (rms) Since the phase sequence is negative, it follows that Vbc = 4156/120◦ V (rms) Vca = 4156/ − 120◦ V (rms) © 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. Problems 11–15 [c] Iba = 4156 = 301.87/ − 78.69◦ A (rms) 2.7 + j13.5 Iac = 301.87/ − 198.69◦ A (rms) IaA = Iba − Iac = 522.86/ − 48.69◦ A (rms) Since we have a balanced three-phase circuit and a negative phase sequence we have: IbB = 522.86/71.31◦ A (rms) IcC = 522.86/ − 168.69◦ A (rms) © 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. 11–16 CHAPTER 11. Balanced Three-Phase Circuits [d] IaA = 2399.47/30◦ = 522.86/ − 48.69◦ A (rms) 0.9 + j4.5 Since we have a balanced three-phase circuit and a negative phase sequence we have: IbB = 522.86/71.31◦ A (rms) IcC = 522.86/ − 168.69◦ A (rms) P 11.21 [a] [b] IaA = 2399.47/30◦ = 1.2/46.26◦ A (rms) 1920 − j560 VAN = (1910 − j636)(1.2/46.26◦ ) = 2415.19/27.84◦ V (rms) √ |VAB | = 3(2415.19) = 4183.24 V (rms) 1.2 [c] |Iab| = √ = 0.69 A (rms) 3 © 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. Problems 11–17 [d] Van = (1919.1 − j564.5)(1.2/46.26◦ ) = 2400.48/29.87◦ V (rms) √ |Vab | = 3(2400.48) = 4157.76 V (rms) P 11.22 The a-phase of the circuit is shown below: I1 = 120/20◦ = 12/ − 16.87◦ A (rms) 8 + j6 I∗2 = 600/36◦ = 5/16◦ A (rms) 120/20◦ I = I1 + I2 = 12/ − 16.87◦ + 5/ − 16◦ = 17/ − 16.61◦ A (rms) Sa = VI∗ = (120/20◦ )(17/16.61◦ ) = 2040/36.61◦ VA ST = 3Sa = 6120/36.61◦ VA P 11.23 The complex power of the source per phase is Ss = 20,000/( cos−1 0.6) = 20,000/53.13◦ = 12,000 + j16,000 kVA. This complex power per phase must equal the sum of the per-phase impedances of the two loads: Ss = S1 + S2 so 12,000 + j16,000 = 10,000 + S2 .·. S2 = 2000 + j16,000 VA |Vrms |2 Z2∗ Also, S2 = |Vrms | = |Vload| √ = 120 V (rms) 3 Thus, Z2∗ |Vrms |2 (120)2 = = = 0.11 − j0.89 Ω S2 2000 + j16,000 .·. Z2 = 0.11 + j0.89 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. 11–18 CHAPTER 11. Balanced Three-Phase Circuits P 11.24 [a] ST ∆ = 14,000/41.41◦ − 9000/53.13◦ = 5.5/22◦ kVA S∆ = ST ∆ /3 = 1833.46/22◦ VA 3000/53.13◦ = 300 V (rms) 10/ − 30◦ √ √ |Vline| = |Vab| = 3|Van| = 300 3 = 519.62 V (rms) [b] |Van| = P 11.25 |Iline| = 1600 √ = 11.547 A (rms) 240/ 3 √ |V | 240/ 3 |Zy | = = = 12 |I| 11.547 Zy = 12/ − 50◦ Ω Z∆ = 3Zy = 36/ − 50◦ = 23.14 − j27.58 Ω/φ P 11.26 Let pa , pb , and pc represent the instantaneous power of phases a, b, and c, respectively. Then assuming a positive phase sequence, we have pa = vaniaA = [Vm cos ωt][Im cos(ωt − θφ )] pb = vbnibB = [Vm cos(ωt − 120◦ )][Im cos(ωt − θφ − 120◦ )] pc = vcnicC = [Vm cos(ωt + 120◦ )][Im cos(ωt − θφ + 120◦ )] The total instantaneous power is pT = pa + pb + pc , so pT = Vm Im [cos ωt cos(ωt − θφ ) + cos(ωt − 120◦ ) cos(ωt − θφ − 120◦ ) + cos(ωt + 120◦ ) cos(ωt − θφ + 120◦ )] Now simplify using trigonometric identities. In simplifying, collect the coefficients of cos(ωt − θφ ) and sin(ωt − θφ). We get pT = Vm Im [cos ωt(1 + 2 cos2 120◦ ) cos(ωt − θφ ) +2 sin ωt sin2 120◦ sin(ωt − θφ )] = 1.5Vm Im [cos ωt cos(ωt − θφ ) + sin ωt sin(ωt − θφ )] = 1.5Vm Im cos θφ © 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. Problems 11–19 P 11.27 [a] S1 = (4.864 + j3.775) kVA S2 = 17.636(0.96) + j17.636(0.28) = (16.931 + j4.938) kVA √ 13,853 3VL IL sin θ3 = 13,853; sin θ3 = √ = 0.521 3(208)(73.8) Therefore cos θ3 = 0.854 Therefore 13,853 P3 = × 0.854 = 22,693.584 W 0.521 S3 = 22.694 + j13.853 kVA ST = S1 + S2 + S3 = 44.49 + j22.57 kVA 1 ST /φ = ST = 14.83 + j7.52 kVA 3 208 ∗ √ IaA = (14.83 + j7.52)103 ; I∗aA = 123.49 + j62.64 A 3 IaA = 123.49 − j62.64 = 138.46/ − 26.90◦ A (rms) [b] pf = cos(0◦ − 26.90◦ ) = 0.892 lagging P 11.28 From the solution to Problem 11.18 we have: SAB = (480/0◦ )(192/ − 16.26◦ ) = 88,473.7 − j25,804.5 VA SBC = (480/120◦ )(48/ − 83.13◦ ) = 18,431.98 + j13,824.03 VA SCA = (480/ − 120◦ )(24/120◦ ) = 11,520 + j0 VA P 11.29 [a] S1/φ = 40,000(0.96) − j40,000(0.28) = 38,400 − j11,200 VA S2/φ = 60,000(0.8) + j60,000(0.6) = 48,000 + j36,000 VA S3/φ = 33,600 + j5200 VA ST /φ = S1 + S2 + S3 = 120,000 + j30,000 VA .·. I∗aA = 120,000 + j30,000 = 50 + j12.5 2400 © 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. 11–20 CHAPTER 11. Balanced Three-Phase Circuits .·. IaA = 50 − j12.5 A Van = 2400 + (50 − j12.5)(1 + j8) = 2550 + j387.5 = 2579.27/8.64◦ V (rms) √ |Vab | = 3(2579.27) = 4467.43 V (rms) [b] Sg/φ = (2550 + j387.5)(50 + j12.5) = 122,656.25 + j51,250 VA % efficiency = P 11.30 [a] I∗aA = 120,000 (100) = 97.83% 122,656.25 (160 + j46.67)103 = 133.3 + j38.9 1200 IaA = 133.3 − j38.9 A (rms) Van = 1200 + (133.3 − j38.9)(0.18 + j1.44) = 1280 + j185 V (rms) IC = 1280 + j185 = −3.1 + j21.3 A (rms) −j60 Ina = IaA + IC = 130.25 − j17.556 = 131.4/ − 7.68◦ A (rms) [b] Sg/φ = (1280 + j185)(130.25 + j17.556) = 163,472 + j46,567 VA SgT = 3Sg/φ = −490.4 − j139.7 kVA Therefore, the source is delivering 490.4 kW and 139.7 kvars. [c] Pdel = 490.4 kW Pabs = 3(160,000) + 3|IaA |2(0.18) = 490.4 kW = Pdel [d] Qdel = 3|IC |2(60) + 139.7 × 103 = 223.3 kVAR Qabs = 3(46,667) + 3|IaA |2(1.44) = 223.4 kVAR = Qdel (roundoff) © 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. Problems 11–21 P 11.31 [a] IaA = 1365/0◦ = 27.3/ − 53.13◦ A (rms) 30 + j40 IaA ICA = √ /150◦ = 15.76/96.87◦ A (rms) 3 [b] Sg/φ = −1365I∗aA = −22,358.75 − j29,811.56 VA .·. Pdeveloped/phase = 22.359 kW Pabsorbed/phase = |IaA |228.5 = 21.241 kW % delivered = 21.241 (100) = 95% 22.359 P 11.32 [a] POUT = 746 × 100 = 74,600 W PIN = 74,600/(0.97) = 76,907.22 W √ 3VL IL cos θ = 76,907.22 76,907.22 = 242.58 A (rms) 3(208)(0.88) √ √ [b] Q = 3VL IL sin φ = 3(208)(242.58)(0.475) = 41,511.90 VAR IL = √ P 11.33 [a] √ 24,000 3/0◦ I1 = = 66.5 − j49.9 A (rms) 400 + j300 © 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. 11–22 CHAPTER 11. Balanced Three-Phase Circuits √ 24,000 3/0◦ I2 = = 33.3 + j24.9 A (rms) 800 − j600 I∗3 = 57,600 + j734,400 √ = 1.4 + j17.7 24,000 3 I3 = 1.4 − j17.7 A (rms) IaA = I1 + I2 + I3 = 101.2 − j42.7 A = 109.8/ − 22.9◦ A (rms) √ Van = (2 + j16)(101.2 − j42.7) + 24,000 3 = 42,454.8 + j1533.8 V (rms) Sφ = Van I∗aA = (42,454.8 + j1533.8)(101.2 + j42.7) = 4,230,932.5 + j1,968,040.5 VA ST = 3Sφ = 12,692.8 + j5904.1 kVA √ [b] S1/φ = 24,000 3(66.5 + j49.9) = 2764.4 + j2074.3 kVA √ S2/φ = 24,000 3(33.3 − j24.9) = 1384.3 − j1035.1 kVA S3/φ = 57.6 + j734.4 kVA Sφ (load) = 4206.3 + j1773.6 kVA % delivered = 4206.3 (100) = 99.4% 4230.9 P 11.34 4000I∗1 = (210 + j280)103 I∗1 = 210 280 +j = 52.5 + j70 A (rms) 4 4 I1 = 52.5 − j70 A (rms) © 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. Problems I2 = 11–23 4000/0◦ = 240 + j70 A (rms) 15.36 − j4.48 .·. IaA = I1 + I2 = 292.5 + j0 A (rms) Van = 4000 + j0 + 292.5(0.1 + j0.8) = 4036.04/3.32◦ V (rms) |Vab| = √ 3|Van| = 6990.62 V (rms) P 11.35 Assume a ∆-connect load (series): 1 Sφ = (96 × 103 )(0.8 + j0.6) = 25,600 + j19,200 VA 3 ∗ Z∆φ = |480|2 = 5.76 − j4.32 Ω/φ 25,600 + j19,200 Z∆φ = 5.76 + 4.32 Ω Now assume a Y-connected load (series): 1 ZY φ = Z∆φ = 1.92 + j1.44 Ω/φ 3 Now assume a ∆-connected load (parallel): Pφ = |480|2 R∆ © 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. 11–24 CHAPTER 11. Balanced Three-Phase Circuits R∆φ = Qφ = |480|2 = 9Ω 25,600 |480|2 X∆ X∆ φ = |480|2 = 12 Ω 19,200 Now assume a Y-connected load (parallel): 1 RY φ = R∆φ = 3 Ω 3 1 XY φ = X∆φ = 4 Ω 3 P 11.36 [a] SL/φ 1 720 = 720 + j (0.6) 103 = 240,000 + j180,000 VA 3 0.8 © 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. Problems I∗aA = 11–25 240,000 + j180,000 = 100 + j75 A (rms) 2400 IaA = 100 − j75 A (rms) Van = 2400 + (0.8 + j0.6)(100 − j75) = 2960 + j580 = 3016.29/11.09◦ V (rms) |Vab | = √ 3(3016.29) = 5224.37 V (rms) [b] I1 = 100 − j75 A (from part [a]) 1 S2 = 0 − j (576) × 103 = −j192,000 VAR 3 I∗2 = −j192,000 = −j80 A (rms) 2400 .·. I2 = j80 A (rms) IaA = 100 − j75 + j80 = 100 + j5 A (rms) Van = 2400 + (100 + j5)(0.8 + j6.4) = 2448 + j644 = 2531.29/14.74◦ V (rms) |Vab | = √ 3(2531.29) = 4384.33 V (rms) [c] |IaA | = 125 A (rms) Ploss/φ = (125)2 (0.8) = 12,500 W Pg/φ = 240,000 + 12,500 = 252.5 kW %η = 240 (100) = 95.05% 252.5 © 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. 11–26 CHAPTER 11. Balanced Three-Phase Circuits [d] |IaA | = 100.125 A (rms) P`/φ = (100.125)2 (0.8) = 8020 W %η = 240,000 (100) = 96.77% 248,200 [e] Zcap/Y = −j 24002 = −j30 Ω −192,000 Zcap/∆ = 3Zcap/Y = −j90 Ω . ·. 1 = 90; ωC C= 1 = 29.47 µF (90)(120π) P 11.37 [a] From Assessment Problem 11.9, IaA = (101.8 − j135.7) A (rms) Therefore Icap = j135.7 A (rms) √ 2450/ 3 = −j10.42 Ω Therefore ZCY = j135.7 Therefore CY = 1 = 254.5 µF (10.42)(2π)(60) ZC∆ = (−j10.42)(3) = −j31.26 Ω Therefore C∆ = 254.5 = 84.84 µF 3 [b] CY = 254.5 µF [c] |IaA | = 101.8 A (rms) P 11.38 [a] 1 Sg = (150)(0.8 − j0.6) = 40 − j30 kVA 3 1 S1 = (30 + j30) = 10 + j10 kVA 3 S2 = Sg − S1 = 30 − j40 kVA .·. I∗aA = (30 − j40)103 = 12 − j16 2500 © 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. Problems 11–27 IaA = 12 + j16 A (rms) Z= [b] R = 2500 = 75 − j100 Ω 12 + j16 (2500)2 = 208.33 Ω 30 × 103 XL = (2500)2 = −156.25 Ω −40 × 103 1 P 11.39 [a] Sg/φ = (41.6)(0.707 + j0.707) × 103 = 9803.73 + j9803.73 VA 3 I∗aA = 9803.73 + j9803.73 √ = 70.76 + j70.76 A (rms) 240/ 3 IaA = 70.76 − j70.76 A (rms) 240 VAN = √ − (0.04 + j0.03)(70.76 − j70.76) 3 = 133.61 + j0.71 = 133.61/0.30◦ V (rms) |VAB | = √ 3(133.61) = 231.42 V (rms) © 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. 11–28 CHAPTER 11. Balanced Three-Phase Circuits [b] SL/φ = (133.61 + j0.71)(70.76 + j70.76) = 9404 + j9504.5 VA SL = 3SL/φ = 28,212 + j28,513 VA Check: Sg = 41,600(0.7071 + j0.7071) = 29,415 + j29,415 VA P` = 3|IaA |2 (0.04) = 1202 W Pg = PL + P` = 28,212 + 1202 = 29,414 W (checks) Q` = 3|IaA |2 (0.03) = 901 VAR Qg = QL + Q` = 28,513 + 901 = 29,414 VAR (checks) P 11.40 Zφ = |Z|/θ = VAN IaA θ = /VAN − /IaA θ1 = /VAB − /IaA For a positive phase sequence, /VAB = /VAN + 30◦ Thus, θ1 = /VAN + 30◦ − /IaA = θ + 30◦ Similarly, Zφ = |Z|/θ = VCN IcC θ = /VCN − /IcC θ2 = /VCB − /IcC For a positive phase sequence, /VCB = /VBA − 120◦ = /VAB + 60◦ /IcC = /IaA + 120◦ Thus, θ2 = /VAB + 60◦ − (/IaA + 120◦ ) = θ1 − 60◦ = θ + 30◦ − 60◦ = θ − 30◦ © 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. Problems P 11.41 IaA = 11–29 VAN = |IL |/−θφ A, Zφ Zφ = |Z|/θφ , VBC = |VL |/ − 90◦ V, Wm = |VL | |IL | cos[−90◦ − (−θφ)] = |VL | |IL | cos(θφ − 90◦ ) = |VL | |IL | sin θφ , therefore √ √ 3Wm = 3|VL | |IL | sin θφ = Qtotal P 11.42 [a] Z = 16 − j12 = 20/ − 36.87◦ Ω VAN = 680/0◦ V; .·. IaA = 34/36.87◦ A √ = 680 3/ − 90◦ V VBC = VBN − VCN √ Wm = (680 3)(34) cos(−90 − 36.87◦ ) = −24,027.07 W √ 3Wm = −41,616.1 W [b] Qφ = (342 )(−12) = −13,872 VAR √ QT = 3Qφ = −41,616 VAR = 3Wm P 11.43 [a] W2 − W1 = VL IL [cos(θ − 30◦ ) − cos(θ + 30◦ )] = VL IL [cos θ cos 30◦ + sin θ sin 30◦ − cos θ cos 30◦ + sin θ sin 30◦ ] = 2VL IL sin θ sin 30◦ = VL IL sin θ, therefore √ √ 3(W2 − W1 ) = 3VL IL sin θ = QT [b] Zφ = (8 + j6) Ω √ QT = 3[2476.25 − 979.75] = 2592 VAR, QT = 3(12)2 (6) = 2592 VAR; Zφ = (8 − j6) Ω √ QT = 3[979.75 − 2476.25] = −2592 VAR, QT = 3(12)2 (−6) = −2592 VAR; √ Zφ = 5(1 + j 3) 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. 11–30 CHAPTER 11. Balanced Three-Phase Circuits √ 3[2160 − 0] = 3741.23 VAR, √ QT = 3(12)2 (5 3) = 3741.23 VAR; QT = Zφ = 10/75◦ Ω √ QT = 3[−645.53 − 1763.63] = −4172.80 VAR, QT = 3(12)2 [−10 sin 75◦ ] = −4172.80 VAR P 11.44 Wm1 = |VAB ||IaA | cos(/VAB − /IaA ) = (199.58)(2.4) cos(65.68◦ ) = 197.26 W Wm2 = |VCB ||IcC | cos(/VCB − /IcC ) = (199.58)(2.4) cos(5.68◦ ) = 476.64 W CHECK: W1 + W2 = 673.9 = (2.4)2 (39)(3) = 673.9 W P 11.45 tan φ = √ 3(W2 − W1) = 0.75 W1 + W2 .·. φ = 36.87◦ √ .·. 2400 3|IL | cos 66.87◦ = 40,823.09 |IL | = 25 A |Z| = 2400 = 96 Ω 25 .·. Z = 96/36.87◦ Ω P 11.46 [a] W1 = |VBA ||IbB | cos θ Negative phase sequence: √ VBA = 240 3/150◦ V IaA = 240/0◦ = 18/30◦ A 13.33/ − 30◦ IbB = 18/150◦ A √ W1 = (18)(240) 3 cos 0◦ = 7482.46 W W2 = |VCA ||IcC | cos θ √ VCA = 240 3/ − 150◦ V IcC = 18/ − 90◦ A √ W2 = (18)(240) 3 cos(−60◦ ) = 3741.23 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. Problems 11–31 [b] Pφ = (18)2 (40/3) cos(−30◦ ) = 3741.23 W PT = 3Pφ = 11,223.69 W W1 + W2 = 7482.46 + 3741.23 = 11,223.69 W .·. W1 + W2 = PT (checks) P 11.47 From the solution to Prob. 11.18 we have IaA = 210/20.79◦ A and IbB = 178.68/ − 178.04◦ A [a] W1 = |Vac| |IaA | cos(θac − θaA ) = 480(210) cos(60◦ − 20.79◦ ) = 78,103.2 W [b] W2 = |Vbc| |IbB | cos(θbc − θbB ) = 480(178.68) cos(120◦ + 178.04◦ ) = 40,317.7 W [c] W1 + W2 = 118,421 W PAB = (192)2 (2.4) = 88,473.6 W PBC = (48)2 (8) = 18,432 W PCA = (24)2 (20) = 11,520 W PAB + PBC + PCA = 118,425.7 therefore W1 + W2 ≈ Ptotal (round-off differences) 1 P 11.48 [a] Z = Z∆ = 4.48 + j15.36 = 16/73.74◦ Ω 3 IaA 600/0◦ = = 37.5/ − 73.74◦ A ◦ / 16 73.74 IbB = 37.5/ − 193.74◦ A √ VAC = 600 3/ − 30◦ V √ VBC = 600 3/ − 90◦ V √ W1 = (600 3)(37.5) cos(−30 + 73.74◦ ) = 28,156.15 W √ W2 = (600 3)(37.5) cos(−90 + 193.74◦ ) = −9256.15 W [b] W1 + W2 = 18,900 W PT = 3(37.5)2 (13.44/3) = 18,900 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. 11–32 CHAPTER 11. Balanced Three-Phase Circuits [c] √ 3(W1 − W2 ) = 64,800 VAR QT = 3(37.5)2 (46.08/3) = 64,800 VAR P 11.49 [a] Zφ = 100 − j75 = 125/ − 36.87◦ Ω Sφ = [b] (13,200)2 = 1,115,136 + j836,352 VA 125/36.87◦ 13,200 ◦ ∗ √ /30 IaA = Sφ 3 so IaA = 182.9/66.87◦ Wm1 = (13,200)(182.9) cos(0 − 66.87◦ ) = 948,401.92 W Wm2 = (13,200)(182.9) cos(−60◦ + 53.13◦ ) = 2,397,006.08 W Check: PT = 3(1,115,136) W = Wm1 + Wm2 . P 11.50 [a] Negative phase sequence: √ VAB = 240 3/ − 30◦ V √ VBC = 240 3/90◦ V √ VCA = 240 3/ − 150◦ V √ 240 3/ − 30◦ IAB = = 20.78/ − 60◦ A 20/30◦ √ 240 3/90◦ = 6.93/90◦ A IBC = 60/0◦ √ 240 3/ − 150◦ ICA = = 10.39/ − 120◦ A ◦ / 40 − 30 IaA = IAB + IAC = 18/ − 30◦ A IcC = ICB + ICA = ICA + IBC = 16.75/ − 108.06◦ √ Wm1 = 240 3(18) cos(−30 + 30◦ ) = 7482.46 W √ Wm2 = 240 3(16.75) cos(−90 + 108.07◦ ) = 6621.23 W [b] Wm1 + Wm2 = 14,103.69 W √ PA = (12 3)2 (20 cos 30◦ ) = 7482.46 W √ PB = (4 3)2(60) = 2880 W √ PC = (6 3)2 [40 cos(−30◦ )] = 3741.23 W PA + PB + PC = 14,103.69 = Wm1 + Wm2 © 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. Problems P 11.51 [a] I∗aA = 11–33 144(0.96 − j0.28)103 = 20/ − 16.26◦ A 7200 VBN = 7200/ − 120◦ V; VBC = VBN − VCN VCN = 7200/120◦ V √ = 7200 3/ − 90◦ V IbB = 20/ − 103.74◦ A √ Wm1 = (7200 3)(20) cos(−90◦ + 103.74◦ ) = 242,278.14 W [b] Current coil in line aA, measure IaA . Voltage coil across AC, measure VAC . [c] IaA = 20/16.76◦ A √ VCA = VAN − VCN = 7200 3/ − 30◦ V √ Wm2 = (7200 3)(20) cos(−30◦ − 16.26◦ ) = 172,441.86 W [d] Wm1 + Wm2 = 414.72kW PT = 432,000(0.96) = 414.72 kW = Wm1 + Wm2 P 11.52 [a] [b] [c] © 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. 11–34 CHAPTER 11. Balanced Three-Phase Circuits [d] P 11.53 [a] Q = |V|2 XC .·. |XC | = (13,800)2 = 158.70 Ω 1.2 × 106 1 1 = 158.70; C= = 16.71 µF ωC 2π(60)(158.70) √ (13,800/ 3)2 1 [b] |XC | = = (158.70) 6 1.2 × 10 3 . ·. .·. C = 3(16.71) = 50.14 µF P 11.54 [a] The capacitor from Appendix H whose value is closest to 16.71 µF is 22 µF. |XC | = Q= [b] I∗aA = 1 1 = = 120.57 Ω ωC 2π(60)(22 × 10−6 ) |V |2 (13,800)2 = = 1,579,497 VAR/φ XC 120.57 1,200,000 − j379,497 √ = 50.2 − j15.9 A 13,800/ 3 13,800 ◦ √ /0 + (0.6 + j4.8)(50.2 + j15.9) = 7897.8/1.76◦ 3 √ |Vab | = 3(7897.8) = 13,679.4 V Van = This voltage falls within the allowable range of 13 kV to 14.6 kV. P 11.55 [a] The capacitor from Appendix H whose value is closest to 50.14 µF is 47 µF. |XC | = Q= 1 1 = = 56.4 Ω ωC 2π(60)(47 × 10−6 ) |V |2 (13,800)2 = = 1,124,775.6 VAR 3XC 3(56.4) © 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. Problems [b] I∗aA = 11–35 1,200,000 + j75,224 √ = 150.6 + j9.4 A 13,800/ 3 13,800 ◦ √ /0 + (0.6 + j4.8)(150.6 − j9.4) = 8134.8/5.06◦ 3 √ |Vab | = 3(8134.8) = 14,089.9 V Van = This voltage falls within the allowable range of 13 kV to 14.6 kV. P 11.56 If the capacitors remain connected when the substation drops its load, the expression for the line current becomes 13,800 ∗ √ IaA = −j1.2 × 106 3 or I∗aA = −j150.61 A Hence IaA = j150.61 A Now, Van = 13,800 ◦ √ /0 + (0.6 + j4.8)(j150.61) = 7244.49 + j90.37 = 7245.05/0.71◦ V 3 The magnitude of the line-to-line voltage at the generating plant is |Vab| = √ 3(7245.05) = 12,548.80 V. This is a problem because the voltage is below the acceptable minimum of 13 kV. Thus when the load at the substation drops off, the capacitors must be switched off. P 11.57 Before the capacitors are added the total line loss is PL = 3|150.61 + j150.61|2 (0.6) = 81.66 kW After the capacitors are added the total line loss is PL = 3|150.61|2 (0.6) = 40.83 kW Note that adding the capacitors to control the voltage level also reduces the amount of power loss in the lines, which in this example is cut in half. © 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. 11–36 CHAPTER 11. Balanced Three-Phase Circuits P 11.58 [a] 13,800 ∗ √ IaA = 80 × 103 + j200 × 103 − j1200 × 103 3 √ √ 80 3 − j1000 3 I∗aA = = 10.04 − j125.51 A 13.8 .·. IaA = 10.04 + j125.51 A 13,800 ◦ √ /0 + (0.6 + j4.8)(10.04 + j125.51) 3 = 7371.01 + j123.50 = 7372.04/0.96◦ V Van = .·. |Vab| = √ 3(7372.04) = 12,768.75 V [b] Yes, the magnitude of the line-to-line voltage at the power plant is less than the allowable minimum of 13 kV. P 11.59 [a] 13,800 ∗ √ IaA = (80 + j200) × 103 3 √ √ 80 3 + j200 3 ∗ IaA = = 10.04 + j25.1 A 13.8 .·. IaA = 10.04 − j25.1 A 13,800 ◦ √ /0 + (0.6 + j4.8)(10.04 − j25.1) 3 = 8093.95 + j33.13 = 8094.02/0.23◦ V Van = .·. |Vab| = [b] Yes: √ 3(8094.02) = 14,019.25 V 13 kV < 14,019.25 < 14.6 kV [c] Ploss = 3|10.04 + j125.51|2 (0.6) = 28.54 kW [d] Ploss = 3|10.04 + j25.1|2 (0.6) = 1.32 kW [e] Yes, the voltage at the generating plant is at an acceptable level and the line loss is greatly reduced. © 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. 12 Introduction to the Laplace Transform Assessment Problems eβt + e−βt 2 Therefore, AP 12.1 [a] cosh βt = 1 L{cosh βt} = 2 Z ∞ 0− [e(s−β)t + e−(s−β)t]dt " 1 e−(s−β)t = 2 −(s − β) 1 = 2 ∞ 0− e−(s+β)t + −(s + β) 1 1 + s−β s+β ! = s2 ∞ 0− # s − β2 eβt − e−βt 2 Therefore, [b] sinh βt = L{sinh βt} = 1 2 Z ∞ 0− h i e−(s−β)t − e−(s+β)t dt " 1 e−(s−β)t = 2 −(s − β) = 1 2 #∞ 0− " 1 e−(s+β)t − 2 −(s + β) 1 1 − s−β s+β ! = (s2 #∞ 0− β − β 2) AP 12.2 [a] Let f(t) = te−at : F (s) = L{te−at } = Now, 1 (s + a)2 L{tf(t)} = − dF (s) ds © 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 12–1 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. 12–2 CHAPTER 12. Introduction to the Laplace Transform So, L{t · te −at " # d 1 2 }=− = 2 ds (s + a) (s + a)3 [b] Let f(t) = e−at sinh βt, then L{f(t)} = F (s) = ( df(t) L dt ) β (s + a)2 − β 2 = sF (s) − f(0− ) = s(β) βs − 0 = (s + a)2 − β 2 (s + a)2 − β 2 [c] Let f(t) = cos ωt. Then F (s) = s 2 (s + ω 2 ) and dF (s) −(s2 − ω 2 ) = 2 ds (s + ω 2)2 Therefore L{t cos ωt} = − AP 12.3 F (s) = dF (s) s2 − ω 2 = 2 ds (s + ω 2 )2 6s2 + 26s + 26 K1 K2 K3 = + + (s + 1)(s + 2)(s + 3) s+1 s+2 s+3 K1 = 6 − 26 + 26 = 3; (1)(2) K3 = 54 − 78 + 26 =1 (−2)(−1) K2 = 24 − 52 + 26 =2 (−1)(1) Therefore f(t) = [3e−t + 2e−2t + e−3t] u(t) AP 12.4 7s2 + 63s + 134 K1 K2 K3 F (s) = = + + (s + 3)(s + 4)(s + 5) s+3 s+4 s+5 K1 = 63 − 189 − 134 = 4; 1(2) K3 = 175 − 315 + 134 = −3 (−2)(−1) K2 = 112 − 252 + 134 =6 (−1)(1) f(t) = [4e−3t + 6e−4t − 3e−5t ]u(t) © 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. Problems AP 12.5 F (s) = 10(s2 + 119) (s + 5)(s2 + 10s + 169) s1,2 = −5 ± F (s) = 12–3 √ 25 − 169 = −5 ± j12 K2 K2∗ K1 + + s + 5 s + 5 − j12 s + 5 + j12 K1 = 10(25 + 119) = 10 25 − 50 + 169 K2 = 10[(−5 + j12)2 + 119] = j4.17 = 4.17/90◦ (j12)(j24) Therefore f(t) = [10e−5t + 8.33e−5t cos(12t + 90◦ )] u(t) = [10e−5t − 8.33e−5t sin 12t] u(t) AP 12.6 F (s) = K0 = 4s2 + 7s + 1 K0 K1 K2 = + + 2 2 s(s + 1) s (s + 1) s+1 1 = 1; (1)2 K1 = d 4s2 + 7s + 1 K2 = ds s " = # 4−7+1 =2 −1 = s=−1 s(8s + 7) − (4s2 + 7s + 1) s2 s=−1 1+2 =3 1 Therefore f(t) = [1 + 2te−t + 3e−t ] u(t) AP 12.7 F (s) = = (s2 K1 K2 K1∗ + + (s + 2 − j1)2 (s + 2 − j1) (s + 2 + j1)2 + K1 = 40 40 = 2 + 4s + 5) (s + 2 − j1)2 (s + 2 + j1)2 K2∗ (s + 2 + j1) 40 = −10 = 10/180◦ (j2)2 and K1∗ = −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. 12–4 CHAPTER 12. Introduction to the Laplace Transform " d 40 K2 = ds (s + 2 + j1)2 # = s=−2+j1 −80(j2) = −j10 = 10/ − 90◦ (j2)4 K2∗ = j10 Therefore f(t) = [20te−2t cos(t + 180◦ ) + 20e−2t cos(t − 90◦ )] u(t) = 20e−2t [sin t − t cos t] u(t) AP 12.8 F (s) = 5s2 + 29s + 32 5s2 + 29s + 32 s+8 = =5− 2 (s + 2)(s + 4) s + 6s + 8 (s + 2)(s + 4) s+8 K1 K2 = + (s + 2)(s + 4) s+2 s+4 K1 = −2 + 8 = 3; 2 K2 = −4 + 8 = −2 −2 Therefore, F (s) = 5 − 2 3 + s+2 s+4 f(t) = 5δ(t) + [−3e−2t + 2e−4t ]u(t) AP 12.9 F (s) = 2s3 + 8s2 + 2s − 4 4(s + 1) 4 = 2s − 2 + = 2s − 2 + 2 s + 5s + 4 (s + 1)(s + 4) s+4 f(t) = 2 dδ(t) − 2δ(t) + 4e−4t u(t) dt AP 12.10 7s3 [1 + (9/s) + (134/(7s2 ))] lim sF (s) = lim 3 =7 s→∞ s→∞ s [1 + (3/s)][1 + (4/s)][1 + (5/s)] " # .·. f(0+ ) = 7 7s3 + 63s2 + 134s lim sF (s) = lim =0 s→0 s→0 (s + 3)(s + 4)(s + 5) " # .·. f(∞) = 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. Problems 12–5 s3 [4 + (7/s) + (1/s)2 ] lim sF (s) = lim =4 s→∞ s→∞ s3 [1 + (1/s)]2 " # .·. f(0+ ) = 4 4s2 + 7s + 1 lim sF (s) = lim =1 s→0 s→0 (s + 1)2 " # .·. f(∞) = 1 " # 40s lim sF (s) = lim 4 =0 s→∞ s→∞ s [1 + (4/s) + (5/s2 )]2 .·. f(0+ ) = 0 " # 40s =0 lim sF (s) = lim s→0 s→0 (s2 + 4s + 5)2 .·. f(∞) = 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. 12–6 CHAPTER 12. Introduction to the Laplace Transform Problems P 12.1 P 12.2 [a] (10 + t)[u(t + 10) − u(t)] + (10 − t)[u(t) − u(t − 10)] = (t + 10)u(t + 10) − 2tu(t) + (t − 10)u(t − 10) [b] (−24 − 8t)[u(t + 3) − u(t + 2)] − 8[u(t + 2) − u(t + 1)] + 8t[u(t + 1) − u(t − 1)] +8[u(t − 1) − u(t − 2)] + (24 − 8t)[u(t − 2) − u(t − 3)] = −8(t + 3)u(t + 3) + 8(t + 2)u(t + 2) + 8(t + 1)u(t + 1) − 8(t − 1)u(t − 1) −8(t − 2)u(t − 2) + 8(t − 3)u(t − 3) P 12.3 [a] f(t) = 5t[u(t) − u(t − 2)] + 10[u(t − 2) − u(t − 6)]+ (−5t + 40)[u(t − 6) − u(t − 8)] [b] f(t) = 10 sin πt[u(t) − u(t − 2)] [c] f(t) = 4t[u(t) − u(t − 5)] © 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. Problems P 12.4 12–7 [a] [b] f(t) = −20t[u(t) − u(t − 1)] − 20[u(t − 1) − u(t − 2)] +20 cos( π2 t)[u(t − 2) − u(t − 4)] +(100 − 20t)[u(t − 4) − u(t − 5)] P 12.5 As ε → 0 the amplitude → ∞; the duration → 0; and the area is independent of ε, i.e., A= Z ε 1 dt = 1 π ε2 + t 2 ∞ −∞ P 12.6 F (s) = Z −ε/2 −ε Z ε/2 Z ε 4 −st −4 −st 4 −st e dt + e dt + e dt 3 3 ε ε −ε/2 ε/2 ε3 Therefore F (s) = 4 sε [e − 2esε/2 + 2e−sε/2 − e−sε ] sε3 L{δ 00(t)} = lim F (s) ε→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. 12–8 CHAPTER 12. Introduction to the Laplace Transform After applying L’Hopital’s rule three times, we have 2s s s 2s 3s lim sesε − esε/2 − e−sε/2 + se−sε = ε→0 3 4 4 3 2 Therefore L{δ 00(t)} = s2 P 12.7 P 12.8 1 1 1 [a] A = bh = (2ε) =1 2 2 ε [b] 0; [c] ∞ 1 −st esε − e−sε e dt = 2ε 2εs ε Z F (s) = −ε sesε + se−sε 1 2s 1 lim = · =1 F (s) = 2s ε→0 1 2s 1 " P 12.9 [a] I = 3 Z # 3 (t + 2)δ(t) dt + −1 Z 3 8(t3 + 2)δ(t − 1) dt −1 = (03 + 2) + 8(13 + 2) = 2 + 8(3) = 26 [b] I = 2 Z 2 t δ(t) dt + −2 Z 2 2 t δ(t + 1.5) dt + −2 Z 2 −2 δ(t − 3) dt = 02 + (−1.5)2 + 0 = 2.25 1 P 12.10 f(t) = 2π Z dn f(t) P 12.11 L dtn Therefore ) ( ∞ −∞ (4 + jω) 1 · πδ(ω) · ejtω dω = (9 + jω) 2π ! 4 + j0 −jt0 2 πe = 9 + j0 9 = sn F (s) − sn−1 f(0− ) − sn−2 f 0 (0− ) − · · · , L{δ n (t)} = sn (1) − sn−1 δ(0− ) − sn−2 δ 0(0− ) − sn−3 δ 00(0− ) − · · · = sn P 12.12 [a] Let dv = δ 0(t − a) dt, v = δ(t − a) du = f 0 (t) dt u = f(t), Therefore Z ∞ ∞ f(t)δ (t − a) dt = f(t)δ(t − a) 0 −∞ −∞ − Z ∞ −∞ δ(t − a)f 0 (t) dt = 0 − f 0 (a) 0 [b] L{δ (t)} = Z ∞ 0− 0 δ (t)e −st " d(e−st ) dt = − dt # t=0 h = − −se−st i t=0 =s © 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. Problems P 12.13 L{e−at f(t)} = ∞ Z 0− [e−atf(t)]e−st dt = ( Z 12–9 ∞ 0− f(t)e−(s+a)t dt = F (s + a) ) d sin ωt sω sω P 12.14 [a] L u(t) = 2 − sin(0) = dt s + ω2 s2 + ω 2 s2 s2 −ω 2 d cos ωt [b] L u(t) = 2 − cos(0) = − 1 = dt s + ω2 s2 + ω 2 s2 + ω 2 ( ) d3 (t2) 2 [c] L u(t) = s3 3 − s2(0) − s(0) − 2(0) = 2 3 dt s d sin ωt ωs [d] = (cos ωt) · ω, L{ω cos ωt} = 2 dt s + ω2 ( ) d cos ωt = −ω sin ωt dt L{−ω sin ωt} = − ω2 s2 + ω 2 d3 (t2 u(t)) = 2δ(t); dt3 P 12.15 [a] t Z 0− x dx = t2 L 2 ( ) L{2δ(t)} = 2 t2 2 1 2 = Z ∞ 0− t2e−st dt ∞ " 1 e−st 2 2 (s t + 2st + 2) = 2 −s3 0− # 1 1 (2) = 3 3 2s s Z t 1 . ·. L x dx = 3 s 0− = L{t} 1/s2 1 = = 3 s s s 0− Z t 1 . ·. L x dx = 3 CHECKS s 0− [b] L Z t x dx = P 12.16 L{f(at)} = Z Let u = at, ∞ 0− f(at)e−st dt u = 0− du = a dt, when t = 0− and u = ∞ when t = ∞ Therefore L{f(at)} = Z ∞ 0− f(u)e−(u/a)s du 1 = F (s/a) a 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. 12–10 CHAPTER 12. Introduction to the Laplace Transform P 12.17 [a] L{t} = 1 ; s2 1 (s + a)2 therefore L{te−at } = ejωt − e−jωt j2 Therefore [b] sin ωt = 1 j2 L{sin ωt} = = s2 ! ω + ω2 1 1 − s − jω s + jω ! = 1 j2 ! 2jω s2 + ω 2 [c] sin(ωt + θ) = (sin ωt cos θ + cos ωt sin θ) Therefore L{sin(ωt + θ)} = cos θL{sin ωt} + sin θL{cos ωt} ω cos θ + s sin θ = s2 + ω 2 e−st (−st − 1) s2 0 [e] f(t) = cosh t cosh θ + sinh t sinh θ From Assessment Problem 12.1(a) s L{cosh t} = 2 s −1 [d] L{t} = Z ∞ ∞ te−st dt = 0 =0− 1 1 (0 − 1) = 2 2 s s From Assessment Problem 12.1(b) L{sinh t} = s2 1 −1 .·. L{cosh(t + θ)} = cosh θ = P 12.18 [a] L{f 0 (t)} = Z # s 1 + sinh θ 2 2 (s − 1) s −1 sinh θ + s[cosh θ] (s2 − 1) e−st dt + −ε 2ε ε " Z ∞ ε − ae−a(t−ε) e−st dt 1 sε a = (e − e−sε ) − e−sε = F (s) 2sε s+a a s lim F (s) = 1 − = ε→0 s+a s+a [b] L{e−at} = 1 s+a Therefore L{f 0 (t)} = sF (s) − f(0− ) = s s −0 = s+a s+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. Problems P 12.19 [a] L{40e−8(t−3) u(t − 3)} = 12–11 40e−3s (s + 8) [b] First rewrite f(t) as f(t) = (5t − 10)u(t − 2) + (40 − 10t)u(t − 4) +(10t − 80)u(t − 8) + (50 − 5t)u(t − 10) = 5(t − 2)u(t − 2) − 10(t − 4)u(t − 4) +10(t − 8)u(t − 8) − 5(t − 10)u(t − 10) .·. F (s) = P 12.20 [a] L{te−at } = 5[e−2s − 2e−4s + 2e−8s − e−10s ] s2 Z ∞ 0− te−(s+a)t dt e−(s+a)t = (s + a)2 − (s + a)t − 1 = 0+ 1 (s + a)2 .·. L{te−at} = 1 (s + a)2 ( ∞ 0− ) d −at s (te )u(t) = −0 [b] L dt (s + a)2 ( ) d −at s L (te )u(t) = dt (s + a)2 [c] d −at (te ) = −ate−at + e−at dt −a 1 −a s+a L{−ate−at + e−at } = + = + 2 2 (s + a) (s + a) (s + a) (s + a)2 . ·. L ( ) d −at s (te ) = dt (s + a)2 CHECKS P 12.21 [a] f(t) = 5t[u(t) − u(t − 2)] +(20 − 5t)[u(t − 2) − u(t − 6)] +(5t − 40)[u(t − 6) − u(t − 8)] = 5tu(t) − 10(t − 2)u(t − 2) +10(t − 6)u(t − 6) − 5(t − 8)u(t − 8) .·. F (s) = 5[1 − 2e−2s + 2e−6s − e−8s ] s2 © 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. 12–12 CHAPTER 12. Introduction to the Laplace Transform [b] f 0 (t) = 5[u(t) − u(t − 2)] − 5[u(t − 2) − u(t − 6)] +5[u(t − 6) − u(t − 8)] = 5u(t) − 10u(t − 2) + 10u(t − 6) − 5u(t − 8) L{f 0 (t)} = 5[1 − 2e−2s + 2e−6s − e−8s ] s [c] f 00 (t) = 5δ(t) − 10δ(t − 2) + 10δ(t − 6) − 5δ(t − 8) L{f 00 (t)} = 5[1 − 2e−2s + 2e−6s − e−8s ] P 12.22 [a] L [b] Z Z t 0− t 0− e−ax dx = ( e−ax dx = 1 e−at L − a a F (s) 1 = s s(s + a) 1 e−at − a a ) = 1 1 1 1 − = a s s+a s(s + 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. Problems dF (s) d P 12.23 [a] = ds ds ∞ Z f(t)e 0− −st dt = − Therefore L{tf(t)} = − [b] d2 F (s) = ds2 Z Therefore dn F (s) = (−1)n dsn d4 ds4 d L{t sin βt} = (−1) ds Z ∞ 1 s2 Z ∞ −t3f(t)e−st dt 0− tn f(t)e−st dt = (−1)n L{tn f(t)} 0− β 2 s + β2 1 tf(t)e−st dt 0− d3 F (s) = ds3 t2f(t)e−st dt; [c] L{t5 } = L{t4 t} = (−1)4 ∞ dF (s) ds ∞ 0− Z 12–13 = ! 120 s6 = (s2 2βs + β 2 )2 L{te−t cosh t}: From Assessment Problem 12.1(a), s F (s) = L{cosh t} = 2 s −1 dF (s2 − 1)1 − s(2s) s2 + 1 = = − ds (s2 − 1)2 (s2 − 1)2 Therefore − dF s2 + 1 = 2 ds (s − 1)2 Thus L{t cosh t} = s2 + 1 (s2 − 1)2 (s + 1)2 + 1 s2 + 2s + 2 L{e t cosh t} = = 2 [(s + 1)2 − 1]2 s (s + 2)2 −t P 12.24 [a] Z ∞ F (u)du = s = Z Z ∞ Z ∞ s 0− ∞ 0− f(t) f(t)e−ut dt du = Z ∞ e −ut du dt = s " # Z Z ∞ 0− " ( ∞ ( t sin βt therefore L t ) = Z ∞ s " ∞ f(t)e−ut du dt s e−tu f(t) −t 0− ∞ f(t) −e−st = f(t) dt = L − −t t 0 2βs [b] L{t sin βt} = 2 (s + β 2)2 Z Z ∞ s # dt ) # 2βu du 2 (u + β 2)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. 12–14 CHAPTER 12. Introduction to the Laplace Transform Let ω = u2 + β 2, then ω = s2 + β 2 when u = s, and ω = ∞ when u = ∞; also dω = 2u du, thus ( t sin βt L t ) =β Z ∞ s2 +β 2 P 12.25 [a] f1(t) = e−at sin ωt; " F1(s) = F (s) = sF1(s) − f1 (0− ) = [b] f1(t) = e−at cos ωt; F (s) = [c] # dω −1 = β ω2 ω ∞ = s2 +β 2 β s2 + β 2 ω (s + a)2 + ω 2 sω −0 (s + a)2 + ω 2 F1(s) = s+a (s + a)2 + ω 2 F1(s) s+a = s s[(s + a)2 + ω 2 ] d −at [e sin ωt] = ωe−at cos ωt − ae−at sin ωt dt Therefore F (s) = Z t 0− e−ax cos ωx dx = ω(s + a) − ωa ωs = 2 2 (s + a) + ω (s + a)2 + ω 2 −ae−at cos ωt + ωe−at sin ωt + a a2 + ω 2 Therefore 1 −a(s + a) ω2 a + + F (s) = 2 2 2 2 2 2 a + ω (s + a) + ω (s + a) + ω s " = P 12.26 Ig (s) = # s+a s[(s + a)2 + ω 2 ] 1.2s ; s2 + 1 1 = 1.6; RC 1 = 1; LC 1 = 1.6 C V (s) 1 V (s) + + C[sV (s) − v(0− )] = Ig (s) R L s 1 1 V (s) + + sC = Ig (s) R Ls V (s) = = 1 R 1 sIg (s) Ig (s) LsIg (s) C = = 1 R 1 + Ls + sC RLs + 1 + s2 LC s2 + C s + LC (1.6)(1.2)s2 1.92s2 = (s2 + 1.6s + 1)(s2 + 1) (s2 + 1.6s + 1)(s2 + 1) © 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. Problems 1 P 12.27 [a] L Z t 0 v1 dτ + 12–15 v1 − v2 = ig u(t) R and dv2 v2 v1 C + − =0 dt R R V1 − V2 V1 [b] + = Ig sL R V2 − V1 + sCV2 = 0 R or (R + sL)V1 (s) − sLV2 (s) = RLsIg (s) −V1 (s) + (RCs + 1)V2 (s) = 0 Solving, V2 (s) = C[s2 vo − Vdc 1 P 12.28 [a] + R L . ·. [b] Vo + sIg (s) + (R/L)s + (1/LC)] Z R vo + L t vo dx + C 0 Z dvo =0 dt t 0 vo dx + RC dvo = Vdc dt R Vo Vdc + RCsVo = L s s . ·. sLVo + RVo + RCLs2 Vo = LVdc . ·. Vo (s) = [c] io = s2 (1/RC)Vdc + (1/RC)s + (1/LC) 1Zt vo dx L 0 Vo Vdc /RLC = sL s[s2 + (1/RC)s + (1/LC)] Io (s) = P 12.29 [a] For t ≥ 0+ : Rio + L io = C dio + vo = 0 dt dvo dt .·. RC dio d2 vo =C 2 dt dt dvo d2 vo + LC 2 + vo = 0 dt dt © 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. 12–16 CHAPTER 12. Introduction to the Laplace Transform or d2 vo R dvo 1 + + vo = 0 2 dt L dt LC 1 R Vo (s) = 0 [b] s2Vo (s) − sVdc − 0 + [sVo (s) − Vdc ] + L LC Vo (s) s2 + R 1 s+ = Vdc (s + R/L) L LC Vo (s) = Vdc [s + (R/L)] + (R/L)s + (1/LC)] [s2 1 t vo dvo vo dx + +C L 0 R dt Idc Vo (s) Vo (s) [b] = + + sCVo (s) s sL R Z P 12.30 [a] Idc = .·. Vo (s) = [c] io = C s2 Idc/C + (1/RC)s + (1/LC) dvo dt .·. Io(s) = sCVo (s) = s2 sIdc + (1/RC)s + (1/LC) P 12.31 [a] For t ≥ 0+ : vo dvo +C + io = 0 R dt vo = L dio ; dt dvo d2 io =L 2 dt dt . ·. d2 io L dio + LC 2 + io = 0 R dt dt or d2 io 1 dio 1 + + io = 0 dt2 RC dt LC [b] s2Io (s) − sIdc − 0 + 1 1 [sIo(s) − Idc] + Io (s) = 0 RC LC Io (s) s2 + 1 1 s+ = Idc(s + 1/RC) RC LC Io (s) = Idc[s + (1/RC)] + (1/RC)s + (1/LC)] [s2 © 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. Problems P 12.32 [a] 300 = 60i1 + 25 0=5 12–17 di1 d d di1 + 10 (i2 − i1) + 5 (i1 − i2) − 10 dt dt dt dt d di1 (i2 − i1) + 10 + 40i2 dt dt Simplifying the above equations gives: 300 = 60i1 + 10 0 = 40i2 + 5 [b] di2 di1 +5 dt dt di1 di2 +5 dt dt 300 = (10s + 60)I1 (s) + 5sI2 (s) s 0 = 5sI1 (s) + (5s + 40)I2 (s) [c] Solving the equations in (b), I1(s) = 60(s + 8) s(s + 4)(s + 24) I2(s) = −60 (s + 4)(s + 24) P 12.33 From Problem 12.26: V (s) = 1.92s2 (s2 + 1.6s + 1)(s2 + 1) s2 + 1.6s + 1 = (s + 0.8 + j0.6)(s + 0.8 − j0.6); s2 + 100 = (s − j1)(s + j1) Therefore V (s) = 1.92s2 (s + 0.8 + j0.6)(s + 0.8 − j0.6)(s − j1)(s + j1) K1 K1∗ K2 K2∗ = + + + s + 0.8 − j0.6 s + 0.8 + j0.6 s − j1 s + j1 K1 = K2 = 1.92s2 (s + 0.8 + j0.6)(s2 + 1) 1.92s2 (s + j1)(s2 + 1.6s + 1) s=−0.8+j0.6 = 1/ − 126.87◦ = 0.6/0◦ s=−j1 Therefore v(t) = [2e−0.8t cos(0.6t − 126.87◦ ) + 1.2 cos(t)]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. 12–18 P 12.34 CHAPTER 12. Introduction to the Laplace Transform 1 = 2 × 106 ; C V2 (s) = s2 1 = 4 × 106 ; LC R = 5000; L Ig = 0.015 s 30,000 + 5000s + 4 × 106 s1 = −1000; s2 = −4000 V2 (s) = 30,000 (s + 1000)(s + 4000) = 10 10 − s + 1000 s + 4000 v2(t) = [10e−1000t − 10e−4000t]u(t) V P 12.35 [a] 1 1 = = 50 × 106 −3 LC (200 × 10 )(100 × 10−9 ) 1 1 = = 2000 RC (5000)(100 × 10−9 ) Vo (s) = s2 70,000 + 2000s + 50 × 106 s1,2 = −1000 ± j7000 rad/s Vo (s) = 70,000 (s + 1000 − j7000)(s + 1000 + j7000) = K1 K1∗ + s + 1000 − j7000 s + 1000 + j7000 K1 = 70,000 = 5/ − 90◦ j14,000 vo (t) = 10e−1000t cos(7000t − 90◦ )]u(t) V = [10e−1000t sin 7000t]u(t) V [b] Io (s) = = K1 = K2 = 35(10,000) s(s + 1000 − j7000)(s + 1000 + j7000) K1 K2 K2∗ + + s s + 1000 − j7000 s + 1000 + j7000 35(10,000) = 7 mA 50 × 106 35(10,000) = 3.54/171.87◦ mA (−1000 + j7000)(j14,000) io (t) = [7 + 7.07e−1000t cos(7000t + 171.87◦ )]u(t) mA © 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. Problems P 12.36 R = 5000; L Vo (s) = s2 1 = 4 × 106 LC 48(s + 5000) + 5000s + 4 × 106 s1,2 = −2500 ± √ 6.25 × 106 − 4 × 106 s1 = −1000 rad/s; Vo (s) = K1 = s2 = −4000 rad/s 48(s + 5000) K1 K2 = + (s + 1000)(s + 4000) s + 1000 s + 4000 48(4000) = 64 V; 3000 Vo (s) = 12–19 K2 = 48(1000) = −16 V −3000 16 64 − s + 1000 s + 4000 vo (t) = [64e−1000t − 16e−4000t ]u(t) V P 12.37 [a] 1 1 = = 500 3 RC (1 × 10 )(2 × 10−6 ) 1 1 = = 40,000 LC (12.5)(2 × 10−6 ) Vo (s) = 500,000Idc s + 500s + 40,000 = 500,000Idc (s + 100)(s + 400) = 15,000 (s + 100)(s + 400) = K1 K2 + s + 100 s + 400 K1 = 15,000 = 50; 300 Vo (s) = K2 = 50 50 − s + 100 s + 400 15,000 = −50 −300 vo (t) = [50e−100t − 50e−400t ]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. 12–20 CHAPTER 12. Introduction to the Laplace Transform [b] Io (s) = 0.03s (s + 100)(s + 400) = K1 K2 + s + 100 s + 400 K1 = 0.03(−100) = −0.01 300 K2 = 0.03(−400) = 0.04 −300 Io (s) = −0.01 0.04 + s + 100 s + 400 io (t) = (40e−400t − 10e−100t)u(t) mA [c] io (0) = 40 − 10 = 30 mA Yes. The initial inductor current is zero by hypothesis, the initial resistor current is zero because the initial capacitor voltage is zero by hypothesis. Thus at t = 0 the source current appears in the capacitor. P 12.38 1 = 8000; RC Io (s) = s2 1 = 16 × 106 LC 0.005(s + 8000) + 8000s + 16 × 106 s1,2 = −4000 Io (s) = 0.005(s + 8000) K1 K2 = + 2 2 (s + 4000) (s + 4000) s + 4000 K1 = 0.005(s + 8000) = 20 s=−4000 K2 = d [0.005(s + 8000)]s=−4000 = 0.005 ds Io (s) = 20 0.005 + (s + 4000)2 s + 4000 io (t) = [20te−4000t + 0.005e−4000t ]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. Problems P 12.39 [a] I1(s) = 12–21 K1 K2 K3 + + s s + 4 s + 24 K1 = (60)(8) = 5; (4)(24) K3 = (60)(−16) = −2 (−24)(−20) I1(s) = (60)(4) = −3 (−4)(20) K2 = 5 3 2 − − s s + 4 s + 24 i1 (t) = (5 − 3e−4t − 2e−24t)u(t) A I2(s) = K1 = K1 K2 + s + 4 s + 24 −60 = −3; 20 I2(s) = K2 = −3 3 + s + 4 s + 24 −60 =3 −20 i2 (t) = (3e−24t − 3e−4t )u(t) A [b] i1(∞) = 5 A; i2(∞) = 0 A [c] Yes, at t = ∞ 300 = 5A 60 Since i1 is a dc current at t = ∞ there is no voltage induced in the 10 H inductor; hence, i2 = 0. Also note that i1(0) = 0 and i2 (0) = 0. Thus our solutions satisfy the condition of no initial energy stored in the circuit. i1 = P 12.40 [a] F (s) = K1 = K2 = K3 = K1 K2 K3 + + s+1 s+2 s+4 8s2 + 37s + 32 (s + 2)(s + 4) s=−1 8s2 + 37s + 32 (s + 1)(s + 4) s=−2 8s2 + 37s + 32 (s + 1)(s + 2) s=−4 =1 =5 =2 f(t) = [e−t + 5e−2t + 2e−4t ]u(t) © 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. 12–22 CHAPTER 12. Introduction to the Laplace Transform [b] F (s) = K1 = K2 = K3 = K4 = K1 K2 K3 K4 + + + s s+2 s+4 s+6 13s3 + 134s2 + 392s + 288 (s + 2)(s + 4)(s + 6) s=0 13s3 + 134s2 + 392s + 288 s(s + 4)(s + 6) s=−2 13s3 + 134s2 + 392s + 288 s(s + 2)(s + 6) s=−4 13s3 + 134s2 + 392s + 288 s(s + 2)(s + 4) s=−6 =6 =4 =2 =1 f(t) = [6 + 4e−2t + 2e−4t + e−6t]u(t) [c] F (s) = K1 = K2 K2∗ K1 + + s + 1 s + 1 − 2j s + 1 + 2j 20s2 + 16s + 12 s2 + 2s + 5 =4 s=−1 20s2 + 16s + 12 K2 = (s + 1)(s + 1 + 2j) = 8 + j6 = 10/36.87◦ s=−1+2j f(t) = [4e−t + 20e−t cos(2t + 36.87◦ )]u(t) [d] F (s) = K1 = K2 = K1 K2 K2∗ + + s s+7−j s+7+j 250(s + 7)(s + 14) s2 + 14s + 50 250(s + 7)(s + 14) s(s + 7 + j) = 490 s=0 s=−7+j = 125/ − 163.74◦ f(t) = [490 + 250e−7t cos(t − 163.74◦ )]u(t) P 12.41 [a] F (s) = K1 = K1 K2 K3 + + 2 s s s+5 100 s+5 = 20 s=0 d 100 −100 K2 = = ds s + 5 (s + 5)2 K3 = 100 s2 s=0 = −4 =4 s=−5 f(t) = [20t − 4 + 4e−5t ]u(t) © 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. Problems [b] F (s) = K1 = K2 = 12–23 K1 K2 K3 + + 2 s (s + 1) s+1 50(s + 5) (s + 1)2 s=0 50(s + 5) s s=−1 = 250 " = −200 # " d 50(s + 5) 50 50(s + 5) K3 = = − ds s s s2 # s=−1 = −250 f(t) = [250 − 200te−t − 250e−t ]u(t) [c] F (s) = K1 = K1 K2 K3 K3∗ + + + s2 s s+3−j s+3+j 100(s + 3) s2 + 6s + 10 = 30 s=0 " d 100(s + 3) K2 = ds s2 + 6s + 10 # " 100 100(s + 3)(2s + 6) = 2 − s + 6s + 10 (s2 + 6s + 10)2 K3 = 100(s + 3) + 3 + j) s2 (s # s=0 = 10 − 18 = −8 = 4 + j3 = 5/36.87◦ s=−3+j f(t) = [30t − 8 + 10e−3t cos(t + 36.87◦ )]u(t) [d] F (s) = K1 = K2 = K1 K2 K3 K4 + + + s (s + 1)3 (s + 1)2 s + 1 5(s + 2)2 (s + 1)3 s=0 5(s + 2)2 s s=−1 = 20 = −5 d 5(s + 2)2 10(s + 2) 5(s + 2)2 K3 = = − ds s s s2 " # " # s=−1 = −10 − 5 = −15 1 d 10(s + 2) 5(s + 2)2 K4 = − 2 ds s s2 " # 1 10 10(s + 2) 10(s + 2) 10(s + 2)2 = − − + 2 s s2 s2 s3 " # s=−1 © 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. 12–24 CHAPTER 12. Introduction to the Laplace Transform 1 = (−10 − 10 − 10 − 10) = −20 2 f(t) = [20 − 2.5t2 e−t − 15te−t − 20e−t ]u(t) [e] F (s) = K1 = K2 = K1 K2 K2∗ K3 K3∗ + + + + s (s + 2 − j)2 (s + 2 + j)2 s + 2 − j s + 2 − j 400 (s2 + 4s + 5)2 400 s(s + 2 + j)2 = 16 s=0 = 44.72/26.57◦ s=−2+j " # " 400 400 −800 d K3 = = 2 + 2 2 ds s(s + 2 + j) s (s + 2 + j) s(s + 2 + j)3 # s=−2+j = 12 + j16 − 20 + j40 = −8 + j56 = 56.57/98.13◦ f(t) = [16 + 89.44te−2t cos(t + 26.57◦ ) + 113.14e−2t cos(t + 98.13◦ )]u(t) P 12.42 [a] 5 F (s) = s2 + 6s + 8 5s2 + 38s + 80 5s2 + 30s + 40 8s + 40 F (s) = 5 + K1 = K2 = 8s + 40 K1 K2 = 10 + + + 6s + 8 s+2 s+4 s2 8s + 40 s+4 s=−2 8s + 40 s+2 s=−4 = 12 = −4 f(t) = 5δ(t) + [12e−2t − 4e−4t ]u(t) [b] 10 F (s) = s2 + 48s + 625 10s2 + 512s + 7186 10s2 + 480s + 6250 32s + 936 F (s) = 10 + K1 = 32s + 936 K1 K2∗ = 10 + + s2 + 48s + 625 s + 24 − j7 s + 24 + j7 32s + 936 s + 24 + j7 s=−24+j7 = 16 − j12 = 20/ − 36.87◦ f(t) = 10δ(t) + [40e−24t cos(7t − 36.87◦ )]u(t) © 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. Problems [c] 12–25 s − 10 F (s) = s2 + 15s + 50 s3 + 5s2 − 50s − 100 s3 + 15s2 + 50s −10s2 − 100s − 100 −10s2 − 150s − 500 50s + 400 F (s) = s − 10 + K1 = K2 = K1 K2 + s + 5 s + 10 50s + 400 s + 10 s=−5 50s + 400 s+5 s=−10 = 30 = 20 f(t) = δ 0(t) − 10δ(t) + [30e−5t + 20e−10t ]u(t) K1 K2 K3 K3∗ P 12.43 [a] F (s) = 2 + + + s s s + 1 − j2 s + 1 + j2 K1 = 100(s + 1) s2 + 2s + 5 = 20 s=0 " # " d 100(s + 1) 100 100(s + 1)(2s + 2) K2 = = 2 − 2 ds s + 2s + 5 s + 2s + 5 (s2 + 2s + 5)2 # s=0 = 20 − 8 = 12 K3 = 100(s + 1) + 1 + j2) s2 (s s=−1+j2 = −6 + j8 = 10/126.87◦ f(t) = [20t + 12 + 20e−t cos(2t + 126.87◦ )]u(t) [b] F (s) = K1 = K2 = K1 K2 K3 K4 + + + 3 2 s (s + 5) (s + 5) s+5 500 (s + 5)3 s=0 500 s = −100 =4 s=−5 d 500 −500 K3 = = ds s s2 s=−5 1 d −500 1 1000 K4 = = 2 2 ds s 2 s3 = −20 s=−5 = −4 f(t) = [4 − 50t2 e−5t − 20te−5t − 4e−5t ]u(t) © 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. 12–26 CHAPTER 12. Introduction to the Laplace Transform [c] F (s) = K1 = K2 = K1 K2 K3 K4 + + + 3 2 s (s + 1) (s + 1) s+1 40(s + 2) (s + 1)3 s=0 40(s + 2) s s=−1 = 80 " = −40 # " 40 40(s + 2) d 40(s + 2) = − K3 = ds s s s2 " 1 d 40 40(s + 2) K4 = − 2 ds s s2 # s=−1 = −40 − 40 = −80 # " 1 −40 40 80(s + 2) − 2 + = 2 s2 s s3 # s=−1 1 = (−40 − 40 − 80) = −80 2 f(t) = [80 − 20t2 e−t − 80te−t − 80e−t ]u(t) [d] F (s) = K1 = K2 = K1 K2 K3 K4 K5 + + + + 4 3 2 s (s + 1) (s + 1) (s + 1) s+1 (s + 5)2 (s + 1)4 s=0 (s + 5)2 s s=−1 = 25 = −16 d (s + 5)2 2(s + 5) (s + 5)2 = − K3 = ds s s s2 " # = −8 − 16 = −24 " 1 d 2(s + 5) (s + 5)2 K4 = − 2 ds s s2 " # s=−1 # 1 2 2(s + 5) 2(s + 5) 3(s + 5)2 = − − + 2 s s2 s2 s3 " # s=−1 1 = (−2 − 8 − 8 − 32) = −25 2 1 d 2 2(s + 5) 2(s + 5) 3(s + 5)2 K5 = − − + 6 ds s s2 s2 s3 " # 1 −2 2 4(s + 5) 2 4(s + 5) 4(s + 5) 6(s + 5)2 = − + − + + − 6 s2 s2 s3 s2 s3 s3 s4 " # s=−1 1 = (−2 − 2 − 16 − 2 − 16 − 16 − 96) = −25 6 f(t) = [25 − (8/3)t3 e−t − 12t2 e−t − 25te−t − 25e−t ]u(t) © 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. Problems P 12.44 f(t) = L −1 ( K K∗ + s + α − jβ s + α + jβ 12–27 ) = Ke−αt ejβt + K ∗ e−αte−jβt = |K|e−αt [ejθ ejβt + e−jθ e−jβt ] = |K|e−αt [ej(βt+θ) + e−j(βt+θ)] = 2|K|e−αt cos(βt + θ) n P 12.45 [a] L{t f(t)} = (−1) Let f(t) = 1, n " dn F (s) dsn # 1 then F (s) = , s Therefore L{tn } = (−1)n " It follows that L{t(r−1) } = and L{t(r−1)e−at } = [b] f(t) = L Therefore f(t) = = (−1)n n! n! = (n+1) (n+1) s s # (r − 1)! sr (r − 1)! (s + a)r K K Ktr−1 e−at L{tr−1 e−at} = = L (r − 1)! (s + a)r (r − 1)! ( Therefore −1 dn F (s) (−1)n n! = dsn s(n+1) thus ( K K∗ + (s + α − jβ)r (s + α + jβ)r ) ) Ktr−1 −(α−jβ)t K ∗tr−1 −(α+jβ)t e + e (r − 1)! (r − 1)! i |K|tr−1e−αt h jθ jβt e e + e−jθ e−jβt (r − 1)! " # 2|K|tr−1 e−αt = cos(βt + θ) (r − 1)! 1.92s3 P 12.46 [a] lim sV (s) = lim 4 =0 s→∞ s→∞ s [1 + (1.6/s) + (1/s2 )][1 + (1/s2 )] " # Therefore v(0+ ) = 0 [b] No, V has a pair of poles on the imaginary axis. © 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. 12–28 CHAPTER 12. Introduction to the Laplace Transform P 12.47 sVo (s) = sVdc /RC + (1/RC)s + (1/LC) s2 lim sVo (s) = 0, s→0 .·. vo(∞) = 0 .·. vo (0+ ) = 0 lim sVo (s) = 0, s→∞ sIo (s) = s2 Vdc /RC) + (1/RC)s + (1/LC) Vdc/RLC Vdc = , 1/LC R lim sIo (s) = s→0 .·. io (0+ ) = 0 lim sIo (s) = 0, s→∞ P 12.48 sVo (s) = s2 (Idc/C)s + (1/RC)s + (1/LC) lim sVo (s) = 0, s→0 .·. vo(∞) = 0 .·. vo (0+ ) = 0 lim sVo (s) = 0, s→∞ sIo (s) = s2Idc s2 + (1/RC)s + (1/LC) lim sIo (s) = 0, s→0 .·. io (∞) = 0 .·. io (0+ ) = Idc lim sIo (s) = Idc, s→∞ P 12.49 sIo (s) = Idcs[s + (1/RC)] s2 + (1/RC)s + (1/LC) lim sIo (s) = 0, s→0 .·. io (∞) = 0 lim sIo (s) = Idc, s→∞ P 12.50 [a] sF (s) = .·. io (0+ ) = Idc 8s3 + 37s2 + 32s (s + 1)(s + 2)(s + 4) lim sF (s) = 0, s→0 lim sF (s) = 8, s→∞ Vdc .·. io (∞) = R .·. f(∞) = 0 .·. f(0+ ) = 8 © 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. Problems [b] sF (s) = 12–29 13s3 + 134s2 + 392s + 288 (s + 2)(s2 + 10s + 24) lim sF (s) = 6; s→0 lim sF (s) = 13, s→∞ .·. f(∞) = 6 .·. f(0+ ) = 13 20s3 + 16s2 + 12s [c] sF (s) = (s + 1)(s2 + 2s + 5) lim sF (s) = 0, s→0 lim sF (s) = 20, s→∞ [d] sF (s) = .·. f(∞) = 0 .·. f(0+ ) = 20 250(s + 7)(s + 14) (s2 + 14s + 50) 250(7)(14) = 490, .·. f(∞) = 490 s→0 50 lim sF (s) = 250, .·. f(0+ ) = 250 lim sF (s) = s→∞ 100 s(s + 5) F (s) has a second-order pole at the origin so we cannot use the final value theorem. P 12.51 [a] sF (s) = lim sF (s) = 0, s→∞ [b] sF (s) = .·. f(0+ ) = 0 50(s + 5) (s + 1)2 lim sF (s) = 250, s→0 lim sF (s) = 0, s→∞ .·. f(∞) = 250 .·. f(0+ ) = 0 100(s + 3) s(s2 + 6s + 10) F (s) has a second-order pole at the origin so we cannot use the final value theorem. [c] sF (s) = lim sF (s) = 0, s→∞ [d] sF (s) = .·. f(0+ ) = 0 5(s + 2)2 (s + 1)3 lim sF (s) = 20, .·. f(∞) = 20 lim sF (s) = 0, .·. f(0+ ) = 0 s→0 s→∞ © 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. 12–30 CHAPTER 12. Introduction to the Laplace Transform [e] sF (s) = (s2 400 + 4s + 5)2 lim sF (s) = 16, .·. f(∞) = 16 lim sF (s) = 0, .·. f(0+ ) = 0 s→0 s→∞ P 12.52 All of the F (s) functions referenced in this problem are improper rational functions, and thus the corresponding f(t) functions contain impulses (δ(t)). Thus, neither the initial value theorem nor the final value theorem may be applied to these F (s) functions! 100(s + 1) s(s2 + 2s + 5) F (s) has a second-order pole at the origin, so we cannot use the final value theorem here. P 12.53 [a] sF (s) = lim sF (s) = 0, s→∞ [b] sF (s) = 500 (s + 5)3 lim sF (s) = 4, s→0 lim sF (s) = 0, s→∞ [c] sF (s) = .·. f(0+ ) = 0 .·. f(∞) = 4 .·. f(0+ ) = 0 40(s + 2) (s + 1)3 lim sF (s) = 80, .·. f(∞) = 80 lim sF (s) = 0, .·. f(0+ ) = 0 s→0 s→∞ [d] sF (s) = (s + 5)2 (s + 1)4 lim sF (s) = 25, .·. f(∞) = 25 lim sF (s) = 0, .·. f(0+ ) = 0 s→0 s→∞ © 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. Problems P 12.54 [a] ZL = j120π(0.01) = j3.77 Ω; ZC = The phasor-transformed circuit is IL = 12–31 −j = −j26.526 Ω 120π(100 × 10−6 ) 1 = 36.69/56.61◦ mA 15 + j3.77 − j26.526 .·. iL−ss (t) = 36.69 cos(120πt + 56.61◦ ) mA [b] The steady-state response is the second term in Eq. 12.109, which matches the steady-state response just derived in part (a). P 12.55 The transient and steady-state components are both proportional to the magnitude of the input voltage. Therefore, K= 40 = 0.947 42.26 So if we make the amplitude of the sinusoidal source 0.947 instead of 1, the current will not exceed the 40 mA limit. A plot of the current through the inductor is shown below with the amplitude of the sinusoidal source set at 0.947. P 12.56 We begin by using Eq. 12.105, and changing the right-hand side so it is the Laplace transform of Kte−100t: 15IL (s) + 0.01sIL (s) + 104 IL (s) A = s (s + 100)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. 12–32 CHAPTER 12. Introduction to the Laplace Transform Solving for IL(s), IL (s) = 100Ks K1 K1∗ = + (s2 + 1500s + 106 )(s + 100)2 s + 750 − j661.44 s + 750 + j661.44 + K1 = K2 = K3 K2 + 2 (s + 100) s + 100 100Ks (s + 750 + j661.44)(s + 100)2 (s2 100Ks + 1500s + 106 ) " s=−100 100Ks d K3 = 2 ds (s + 1500s + 106 ) # = 87.9K /139.59◦ µA s=−750+j661.44 = −11.63K mA = 133.86K µA s=−100 Therefore, iL (t) = K[0.176e−750t cos(661.44t + 139.59◦ ) − 11.63te−100t + 0.134e−100t ]u(t) mA Plot the expression above with K = 1: The maximum value of the inductor current is 0.068K mA. Therefore, K= 40 = 588 0.068 So the inductor current rating will not be exceeded if the input to the RLC circuit is 588te−100t 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. 13 The Laplace Transform in Circuit Analysis Assessment Problems AP 13.1 [a] Y = 1 1 C[s2 + (1/RC)s + (1/LC) + + sC = R sL s 1 106 = = 80,000; RC (500)(0.025) 1 = 25 × 108 LC 25 × 10−9 (s2 + 80,000s + 25 × 108 ) s √ [b] z1,2 = −40,000 ± 16 × 108 − 25 × 108 = −40,000 ± j30,000 rad/s Therefore Y = −z1 = −40,000 − j30,000 rad/s −z2 = −40,000 + j30,000 rad/s p1 = 0 rad/s AP 13.2 [a] Z = 2000 + 1 4 × 107 s = 2000 + 2 Y s + 80,000s + 25 × 108 2000(s2 + 105 s + 25 × 108 ) 2000(s + 50,000)2 = s2 + 80,000s + 25 × 108 s2 + 80,000s + 25 × 108 [b] −z1 = −z2 = −50,000 rad/s = −p1 = −40,000 − j30,000 rad/s −p2 = −40,000 + j30,000 rad/s © 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 13–1 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. 13–2 CHAPTER 13. The Laplace Transform in Circuit Analysis AP 13.3 [a] At t = 0− , 0.2v1 = (0.8)v2; v1 = 4v2 ; Therefore v1(0− ) = 80V = v1(0+ ); I= v1 + v2 = 100 V v2(0− ) = 20V = v2(0+ ) (80/s) + (20/s) 20 × 10−3 = 5000 + [(5 × 106 )/s] + (1.25 × 106 /s) s + 1250 80 5 × 106 V1 = − s s 20 × 10−3 s + 1250 20 1.25 × 106 V2 = − s s [b] i = 20e−1250tu(t) mA; ! = 20 × 10−3 s + 1250 ! 80 s + 1250 = 20 s + 1250 v1 = 80e−1250tu(t) V v2 = 20e−1250tu(t) V AP 13.4 [a] I= Vdc /L Vdc/s = 2 R + sL + (1/sC) s + (R/L)s + (1/LC) Vdc = 40; L I= R = 1.2; L 1 = 1.0 LC 40 K1 K1∗ = + (s + 0.6 − j0.8)(s + 0.6 + j0.8) s + 0.6 − j0.8 s + 0.6 + j0.8 K1 = 40 = −j25 = 25/ − 90◦ ; j1.6 K1∗ = 25/90◦ [b] i = 50e−0.6t cos(0.8t − 90◦ ) = [50e−0.6t sin 0.8t]u(t) 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. Problems [c] V = sLI = 160s (s + 0.6 − j0.8)(s + 0.6 + j0.8) = K1 K1∗ + s + 0.6 − j0.8 s + 0.6 + j0.8 K1 = 13–3 160(−0.6 + j0.8) = 100/36.87◦ j1.6 [d] v(t) = [200e−0.6t cos(0.8t + 36.87◦ )]u(t) V AP 13.5 [a] The two node voltage equations are V1 − V2 5 V2 V2 − V1 V2 − (15/s) + V1 s = and + + =0 s s 3 s 15 Solving for V1 and V2 yields V1 = 5(s + 3) , 2 s(s + 2.5s + 1) V2 = 2.5(s2 + 6) s(s2 + 2.5s + 1) [b] The partial fraction expansions of V1 and V2 are 50/3 5/3 15 125/6 25/3 15 − + and V2 = − + s s + 0.5 s + 2 s s + 0.5 s + 2 It follows that 50 5 v1 (t) = 15 − e−0.5t + e−2t u(t) V and 3 3 V1 = 125 −0.5t 25 −2t v2 (t) = 15 − e + e u(t) V 6 3 [c] v1 (0+ ) = 15 − 50 5 + =0 3 3 v2 (0+ ) = 15 − [d] v1(∞) = 15 V; 125 25 + = 2.5 V 6 3 v2 (∞) = 15 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. 13–4 CHAPTER 13. The Laplace Transform in Circuit Analysis AP 13.6 [a] With no load across terminals a − b Vx = 20/s: 1 20 20 − VTh s + 1.2 − VTh = 0 2 s s therefore VTh = Vx = 5IT and 20(s + 2.4) s(s + 2) ZTh = VT IT Solving for IT gives (VT − 5IT )s + VT − 6IT 2 Therefore IT = 14IT = VT s + 5sIT + 2VT ; therefore ZTh = 5(s + 2.8) s+2 [b] I= VTh 20(s + 2.4) = ZTh + 2 + s s(s + 3)(s + 6) © 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. Problems AP 13.7 [a] i2 = 1.25e−t − 1.25e−3t ; Therefore 13–5 di2 = −1.25e−t + 3.75e−3t dt therefore di2 = 0 when dt 1.25e−t = 3.75e−3t or e2t = 3, t = 0.5(ln 3) = 549.31 ms i2 (max) = 1.25[e−0.549 − e−3(0.549)] = 481.13 mA [b] From Eqs. 13.68 and 13.69, we have ∆ = 12(s2 + 4s + 3) = 12(s + 1)(s + 3) and N1 = 60(s + 2) Therefore I1 = N1 5(s + 2) = ∆ (s + 1)(s + 3) A partial fraction expansion leads to the expression 2.5 2.5 + I1 = s+1 s+3 Therefore we get i1 = 2.5[e−t + e−3t ]u(t) A di1 di1 (0.54931) = −2.5[e−t + 3e−3t ]; = −2.89 A/s dt dt [d] When i2 is at its peak value, [c] di2 =0 dt Therefore L2 [e] i2(max) = di2 dt ! M = 0 and i2 = − 12 di1 dt ! −2(−2.89) = 481.13 mA (checks) 12 AP 13.8 [a] The s-domain circuit with the voltage source acting alone is V 0 − (20/s) V0 V 0s + + =0 2 1.25s 20 V0 = 200 100/3 100/3 = − (s + 2)(s + 8) s+2 s+8 v0 = 100 −2t [e − e−8t]u(t) V 3 © 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. 13–6 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] With the current source acting alone, V 00 V 00 V 00s 5 + + = 2 1.25s 20 s V 00 = 100 50/3 50/3 = − (s + 2)(s + 8) s+2 s+8 v 00 = 50 −2t [e − e−8t ]u(t) V 3 [c] v = v 0 + v 00 = [50e−2t − 50e−8t ]u(t) V Vo Vo s + = Ig ; s+2 10 [b] −z1 = −2 rad/s; Vo 10(s + 2) = H(s) = 2 Ig s + 2s + 10 −p1 = −1 + j3 rad/s; −p2 = −1 − j3 rad/s AP 13.9 [a] therefore AP 13.10 [a] Vo = 10(s + 2) 1 Ko K1 K1∗ · = + + s2 + 2s + 10 s s s + 1 − j3 s + 1 + j3 Ko = 2; K1 = 5/3/ − 126.87◦ ; K1∗ = 5/3/126.87◦ vo = [2 + (10/3)e−t cos(3t − 126.87◦ )]u(t) V [b] Vo = 10(s + 2) K2 K2∗ · 1 = + s2 + 2s + 10 s + 1 − j3 s + 1 + j3 K2 = 5.27/ − 18.43◦ ; K2∗ = 5.27/18.43◦ vo = [10.54e−t cos(3t − 18.43◦ )]u(t) V AP 13.11 [a] H(s) = L{h(t)} = L{vo (t)} vo (t) = 10,000 cos θe−70t cos 240t − 10,000 sin θe−70t sin 240t = 9600e−70t cos 240t − 2800e−70t sin 240t Therefore H(s) = = 9600(s + 70) 2800(240) − 2 2 (s + 70) + (240) (s + 70)2 + (240)2 s2 9600s + 140s + 62,500 © 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. Problems [b] Vo (s) = H(s) · 1 9600 = 2 s s + 140s + 62,500 = K1 = 13–7 K1 K1∗ + s + 70 − j240 s + 70 + j240 9600 = −j20 = 20/ − 90◦ j480 Therefore vo (t) = [40e−70t cos(240t − 90◦ )]u(t) V = [40e−70t sin 240t]u(t) V AP 13.12 From Assessment Problem 13.9: H(s) = 10(s + 2) + 2s + 10 s2 Therefore H(j4) = 10(2 + j4) = 4.47/ − 63.43◦ 10 − 16 + j8 Thus, vo = (10)(4.47) cos(4t − 63.43◦ ) = 44.7 cos(4t − 63.43◦ ) V AP 13.13 [a] Let R1 = 10 kΩ, then V1 = V2 = Also R2 = 50 kΩ, C = 400 pF, Vg R2 R2 + (1/sC) R2 C = 2 × 10−5 V1 − Vg V1 − Vo + =0 R1 R1 therefore Vo = 2V1 − Vg Now solving for Vo /Vg , we get H(s) = It follows that H(j50,000) = R2 Cs − 1 R2 Cs + 1 j−1 = j1 = 1/90◦ j+1 Therefore vo = 10 cos(50,000t + 90◦ ) V [b] Replacing R2 by Rx gives us H(s) = Rx Cs − 1 Rx Cs + 1 Therefore H(j50,000) = j20 × 10−6 Rx − 1 Rx + j50,000 = j20 × 10−6 Rx + 1 Rx − j50,000 Thus, 50,000 = tan 60◦ = 1.7321, Rx Rx = 28,867.51 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. 13–8 CHAPTER 13. The Laplace Transform in Circuit Analysis Problems P 13.1 P 13.2 P 13.3 P 13.4 1 i= L Z t 0− vdτ + I0 ; VTh = Vab = CV0 therefore I = 1 sC = V0 ; s 1 L ZTh = V s + I0 V I0 = + s sL s 1 sC −I0 −LI0 = ; ZN = sL sL s Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4. Iscab = IN = [a] Y = 1 1 C[s2 + (1/RC)s + (1/LC)] + + sC = R sL s Z= s/C 8 × 107 s 1 = 2 = 2 Y s + (1/RC)s + (1/LC) s + 40,000s + 256 × 106 [b] zero at z1 = 0 poles at −p1 = −8000 rad/s and −p2 = −32,000 rad/s P 13.5 [a] Z= (R + 1/sC)(sL) (Rs)(s + 1/RC) = 2 R + sL + (1/sC) s + (R/L)s + (1/LC) R = 10,000; L Z= [b] Z = s2 1 = 1600; RC 1000s(s + 1600) + 10,000s + 16 × 106 1000s(s + 1600) (s + 2000)(s + 8000) z1 = 0; −z2 = −1600 rad/s −p1 = −2000 rad/s; P 13.6 1 = 16 × 106 LC −p2 = −8000 rad/s 1 L[s2 + (R/L)s + (1/LC)] [a] Z = R + sL + = sC s = [s2 + 8000s + 25 × 106 ] s © 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. Problems 13–9 [b] s1,2 = −4000 ± j3000 rad/s Zeros at −4000 + j3000 rad/s and −4000 − j3000 rad/s Pole at 0. P 13.7 Zab = 1k[s + (1/sk1)] = 1k[s + (1/(s + 1))] = = s + (1/(s + 1)) 1 + s + (1/(s + 1)) (s + 0.5 + j0.866)(s + 0.5 − j0.866) s2 + s + 1 = 2 s + 2s + 2 (s + 1 + j1)(s + 1 − j1) Zeros at −0.5 + j0.866 rad/s and −0.5 − j0.866 rad/s; poles at −1 + j1 rad/s and −1 − j1 rad/s. P 13.8 Transform the Y-connection of the two resistors and the inductor into the equivalent delta-connection: where Za = (s)(1) + (1)(s) + (1)(1) 2s + 1 = s s Zb = Zc = (s)(1) + (1)(s) + (1)(1) = 2s + 1 1 Then Zab = Za k[(1/skZc ) + (1/skZb )] = Za k2(1/skZb ) 1/skZb = Zab = = 1 (2s + 1) s 1 + 2s + 1 s = 2s + 1 +s+1 2s2 2s + 1 2(2s + 1) k 2 s 2s + s + 1 2(2s + 1)2 2 = 2 (2s + 1)(2s + s + 1) + 2s(2s + 1) s+1 No zeros; one pole at −1 rad/s. © 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. 13–10 P 13.9 CHAPTER 13. The Laplace Transform in Circuit Analysis [a] For t > 0: 2.5s −150 [b] Vo = 5 (16 × 10 )/s + 5000 + 2.5s s = s2 −150s + 2000s + 64 × 104 −150s (s + 400)(s + 1600) K1 K2 [c] Vo = + s + 400 s + 1600 = K1 = −150s s + 1600 K2 = −150s s + 400 Vo = = 50 s=−400 s=−1600 = −200 50 200 − s + 400 s + 1600 vo (t) = (50e−400t − 200e−1600t)u(t) V P 13.10 [a] For t < 0: Vc − 50 Vc Vc − 137.5 + + =0 400 1200 500 Vc 1 1 1 50 137.5 + + = + 400 1200 500 400 500 © 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. Problems 13–11 Vc = 75 V iL (0− ) = 75 − 137.5 = −0.125 A 500 For t > 0: [b] Vo = 75 5 × 105 I+ s 5 0=− 137.5 5 × 105 75 + 100I + I+ − 1.25 × 10−3 + 0.01sI s s s 5 × 105 62.5 I 100 + + 0.01s = + 1.25 × 10−3 s s ! . ·. I = 6250 + 0.125s s2 + 104 s + 5 × 107 5 × 105 Vo = s = [c] Vo = 6250 + 0.125s 75 + 2 4 7 s + 10 s + 5 × 10 s 75s2 + 812,500s + 6875 × 106 s(s2 + 104 s + 5 × 107 ) K1 K2 K2∗ + + s s + 5000 − j5000 s + 5000 + j5000 K1 = K2 = 75s2 + 812,500s + 6875 × 106 s2 + 104 s + 5 × 107 75s2 + 812,500s + 6875 × 106 s(s + 5000 + j5000 = 137.5 s=0 = 40.02/141.34◦ s=−5000+j5000 vo (t) = [137.5 + 80.04e−5000t cos(5000t + 141.34◦ )]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. 13–12 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.11 [a] For t < 0: iL (0− ) = −100 −100 = = −5 A 4 + 10k40 + 8 20 10 (5) = 1 A 50 i1 = vC (0− ) = 10(1) + 4(5) − 100 = −70 V For t > 0: [b] (20 + 2s + 100/s)I = 10 + 5(s + 7) + 10s + 50 . ·. I= Vo = 100 70 I− s s s2 70 s −70s2 − 200s −70(s + 20/7) = = 2 2 s(s + 10s + 50) s + 10s + 50 = K1 = K1 K1∗ + s + 5 − j5 s + 5 + j5 −70(s + 20/7) s + 5 + j5 s=−5+j5 = 38.1/ − 156.8◦ © 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. Problems 13–13 [c] vo (t) = 76.2e−5t cos(5t − 156.8◦ )u(t) V P 13.12 [a] For t < 0: 1 1 1 1 = + + = 0.1875; Re 8 80 20 Re = 5.33 Ω v1 = (9)(5.33) = 48 V iL (0− ) = 48 = 2.4 A 20 vC (0− ) = −v1 = −48 V For t = 0+ : s-domain circuit: where R = 20 Ω; C = 6.25 µF; L = 6.4 mH; [b] and γ = −48 V; ρ = −2.4 A Vo Vo ρ + Vo sC − γC + − =0 R sL s .·. Vo = s2 γ[s + (ρ/γC)] + (1/RC)s + (1/LC) ρ −2.4 = = 8000 γC (−48)(6.25 × 10−6 ) © 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. 13–14 CHAPTER 13. The Laplace Transform in Circuit Analysis 1 1 = = 8000 RC (20)(6.25 × 10−6 ) 1 1 = = 25 × 106 −3 LC (6.4 × 10 )(6.25 × 10−6 ) Vo = [c] IL = s2 −48(s + 8000) + 8000s + 25 × 106 ρ Vo 2.4 Vo − = + sL s 0.0064s s = [d] Vo = = s(s2 −7500(s + 8000) 2.4 2.4(s + 4875) − = 2 6 + 8000s + 25 × 10 ) s (s + 8000s + 25 × 106 ) −48(s + 8000) s2 + 8000s + 25 × 106 K1 K1∗ + s + +4000 − j3000 s + 4000 + j3000 K1 = −48(s + 8000) s + 4000 + j3000 = 40/126.87◦ s=−4000+j3000 vo (t) = [80e−4000t cos(3000t + 126.87◦ )]u(t) V [e] IL = = s2 2.4(s + 4875) + 8000s + 25 × 106 K1 K1∗ + s + 4000 − j3000 s + 4000 + j3000 K1 = 2.4(s + 4875) s + 4000 + j3000 s=−4000+j3000 = 1.25/ − 16.26◦ iL (t) = [2.5e−4000t cos(3000t − 16.26◦ )]u(t) A P 13.13 [a] io(0− ) = Io = 20 = 5 mA 4000 20/s + Lρ sC(20/s + Lρ) = 2 R + sL + 1/sC s LC + RsC + 1 © 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. Problems = s2 13–15 20/L + sρ 40 + s(0.005) = 2 + sR/L + 1/LC s + 8000s + 16 × 106 4000(40 + 0.005s) 0.0025s(s + 8000) − 0.0025 + 2 6 + 8000s + 16 × 10 s + 8000s + 16 × 106 20 40,000 20s + 120,000 = + = (s + 4000)2 (s + 4000)2 s + 4000 Vo = RIo − Lρ + sLIo = s2 vo (t) = [20te−4000t + 40,000e−4000t]u(t) V [b] Io = 0.005(s + 8000) + 8000s + 16 × 106 K1 K2 = + (s + 4000)2 s + 4000 s2 K1 = 20 K2 = 0.005 io (t) = [20te−4000t + 0.005e−4000t]u(t) A P 13.14 For t < 0: vo (0− ) − 500 vo(0− ) vo (0− ) + + =0 5 25 100 25vo (0− ) = 10,000 iL (0− ) = .·. vo (0− ) = 400 V vo (0− ) 400 = = 16 A 25 25 For t > 0 : Vo + 400 Vo Vo − (400/s) + + =0 25 + 25s 100 100/s © 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. 13–16 CHAPTER 13. The Laplace Transform in Circuit Analysis Vo 1 1 s + + 25 + 25s 100 100 .·. Vo = =4− 400 25 + 25s 400(s − 3) s2 + 2s + 5 −20s − 20 Vo − (400/s) = 2 100/s s + 2s + 5 Io = = K1 = K1∗ K1 + s + 1 − j2 s + 1 + j2 −20(s + 1) s + 1 + j2 s=−1+j2 = −10 io (t) = [−20e−t cos 2t]u(t) A P 13.15 Vo = (18/s)(8 × 106 /s) 2800 + 0.2s + (8 × 106 /s) = 720 × 106 s(s2 + 14,000s + 40 × 106 ) = 720 × 106 s(s + 4000)(s + 10,000) = K1 K2 K3 + + s s + 4000 s + 10,000 K1 = 720 × 106 = 18 4 × 107 720 × 106 K2 = = −30 (−4000)(6000) K3 = 720 × 106 = 12 (−6000)(−10,000) Vo = 18 30 12 − + s s + 4000 s + 10,000 vo (t) = [18 − 30e−4000t + 12e−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. Problems 13–17 P 13.16 With a non-zero initial voltage on the capacitor, the s-domain circuit becomes: Vo − 18/s (Vo − 30/s)s + =0 0.2s + 2800 8 × 106 Vo " # 5 s 30 90 + = + s + 14,000 8 × 106 80 × 106 s(s + 14,000) 30s2 + 420,000s + 720 × 106 · . . Vo = s(s + 4000)(s + 10,000) = K1 K2 K3 + + s s + 4000 s + 10,000 720 × 106 K1 = = 18 40 × 106 K2 = 30s2 + 420,000s + 720 × 106 s(s + 10,000) s=−4000 K3 = 30s2 + 420,000s + 720 × 106 s(s + 4000) s=−10,000 Vo = 18 20 8 + − s s + 4000 s + 10,000 = 20 = −8 vo (t) = [18 + 20e−4000t − 8e−10,000t]u(t) V P 13.17 [a] For t < 0: V2 = 10 (450) = 90 V 10 + 40 © 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. 13–18 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] V1 = 25(450/s) (125,000/s) + 25 + 1.25 × 10−3 s = 9 × 106 9 × 106 = s2 + 20, 000s + 108 (s + 10,000)2 v1 (t) = (9 × 106 te−10,000t)u(t) V [c] V2 = 90 (25,000/s)(450/s) − s (125,000/s) + 1.25 × 10−3 s + 25 = 90(s + 20,000) s2 + 20,000s + 108 = 900,000 90 + 2 (s + 10,000) s + 10,000 v2 (t) = [9 × 105 te−10,000t + 90e−10,000t]u(t) V P 13.18 [a] iL(0− ) = iL (0+ ) = " 24 = 8A 3 # directed upward 20(10/s) 25IT (10/s) 200 VT = 25Iφ + IT = + IT 20 + (10/s) 20 + (10/s) 10 + 20s VT 250 + 200 45 =Z= = IT 20s + 10 2s + 1 © 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. Problems 13–19 Vo Vo (2s + 1) Vo 8 + + = 5 45 5.625s s [9s + (2s + 1)s + 8]Vo 8 = 45s s Vo [2s2 + 10s + 8] = 360 Vo = [b] Vo = 2s2 360 180 = 2 + 10s + 8 s + 5s + 4 180 K1 K2 = + (s + 1)(s + 4) s+1 s+4 K1 = 180 = 60; 3 Vo = 60 60 − s+1 s+4 K2 = 180 = −60 −3 vo (t) = [60e−t − 60e−4t ]u(t) V P 13.19 vC (0− ) = vC (0+ ) = 0 0.01 Vo Vo Vo s 6Vo = + + − 6 s 50,000 5000 50 × 10 50,000 500 × 103 = (1000 + 10,000 + s − 6000)Vo s Vo = = 500 × 103 K1 K2 = + s(s + 5000) s s + 5000 100 100 − s s + 5000 vo (t) = [100 − 100e−5000t ]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. 13–20 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.20 5 × 10−3 Vo Vo = + 3.75 × 10−3 Vφ + 6 s 200 + 4 × 10 /s 0.04s Vφ = 4 × 106 /s 4 × 106 Vo V = o 200 + 4 × 106 /s 200s + 4 × 106 .·. 5 × 10−3 Vo s 15,000Vo 25Vo = + + 6 6 s 200s + 4 × 10 200s + 4 × 10 s .·. Vo = s2 s + 20,000 K1 K2 = + 8 2 + 20,000s + 10 (s + 10,000) s + 10,000 K1 = 10,000; Vo = K2 = 1 10,000 1 + 2 (s + 10,000) s + 10,000 vo (t) = [10,000te−10,000t + e−10,000t]u(t) V P 13.21 [a] Vo − 8Iφ Vo − 35/s + 0.4V∆ + =0 2 s + (250/s) " # Vo − 8Iφ V∆ = s; s + (250/s) Iφ = (35/s) − Vo 2 Solving for Vo yields: Vo = 29.4s2 + 56s + 1750 29.4s2 + 56s + 1750 = s(s2 + 2s + 50) s(s + 1 − j7)(s + 1 + j7) © 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. Problems Vo = K1 K2 K2∗ + + s s + 1 − j7 s + 1 + j7 K1 = 29.4s2 + 56s + 1750 s2 + 2s + 50 K2 = 29.4s2 + 56s + 1750 s(s + 1 + j7) 13–21 = 35 s=0 s=−1+j7 = −2.8 + j0.6 = 2.86/167.91◦ .·. vo(t) = [35 + 5.73e−t cos(7t + 167.91◦ )]u(t) V [b] At t = 0+ vo = 35 + 5.73 cos(167.91◦ ) = 29.4 V vo − 35 + 0.4v∆ = 0; 2 vo − 35 + 0.8v∆ = 0 vo = v∆ + 8iφ = v∆ + 8(0.4v∆ ) = 4.2 V vo + (0.8) vo = 35; 4.2 .·. vo (0+ ) = 29.4 V(checks) At t = ∞, the circuit is v∆ = 0, .·. vo = 35 V(checks) iφ = 0 P 13.22 [a] I1 + s(I1 − I2) = 10 s and 1 I2 + I2 + s(I2 − I1 ) = 0 s © 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. 13–22 CHAPTER 13. The Laplace Transform in Circuit Analysis Solving the second equation for I1 : s2 + s + 1 I2 s2 Substituting into the first equation and solving for I2: I1 = 10 s2 + s + 1 (s + 1) − s I2 = 2 s s " # . ·. I2 = . ·. I1 = 2s2 s2 + s + 1 10s 10(s2 + s + 1) · = s2 2s2 + 2s + 1 s(2s2 + 2s + 1) Io = I1 − I2 = = 10s 5(s + 1) 10(s2 + s + 1) − 2 = 2 2 s(2s + 2s + 1) 2s + 2s + 1 s(s + s + 0.5) K1 K2 K2∗ + + s s + 0.5 − j0.5 s + 0.5 + j0.5 K1 = 10; . ·. 10s + 2s + 1 K2 = 5/ − 180◦ io (t) = [10 − 10e−0.5t cos 0.5t]u(t) A [b] Vo = sIo = 5(s + 1) K1 K1∗ = + s2 + s + 0.5 s + 0.5 − j0.5 s + 0.5 + j0.5 K1 = 3.54/ − 45◦ . ·. vo (t) = 7.07e−0.5t cos(0.5t − 45◦ )u(t) V [c] At t = 0+ the circuit is .·. vo(0+ ) = 5 V = 7.07 cos(−45◦ ); Io(0+ ) = 0 Both values agree with our solutions for vo and io . At t = ∞ the circuit is .·. vo(∞) = 0; io(∞) = 10 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. Problems 13–23 Both values agree with our solutions for vo and io . P 13.23 [a] Vo = (1/sC)(sL)(Ig /s) Ig /C = 2 R + sL + (1/sC) s + (R/L)s + (1/LC) Ig 0.015 = = 0.15 C 0.1 R = 7; L 1 = 10 LC Vo = 0.15 s2 + 7s + 10 [b] sVo = 0.15s s2 + 7s + 10 lim sVo = 0; s→0 lim sVo = 0; s→∞ [c] Vo = .·. vo (∞) = 0 .·. vo (0+ ) = 0 0.15 0.05 −0.05 = + (s + 2)(s + 5) s+2 s+5 vo = [50e−2t − 50e−5t ]u(t) mV P 13.24 IL = IL = Ig Vo Ig + = − sCVo s 1/sC s 15 15s 15 −10 25 − = − + s (s + 2)(s + 5) s s+2 s+5 iL (t) = [15 + 10e−2t − 25e−5t ]u(t) mA Check: iL (0+ ) = 0 (ok); iL(∞) = 15 mA (ok) © 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. 13–24 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.25 [a] [b] Zeq = 50,000 + 107 20 × 1012 /s2 + 3s 12 × 106 /s = 50,000 + = 107 20 × 1012 + 3s 12 × 106 s 100,000s + 107 2s I1 = 20/s 0.4 × 10−3 = Zeq s + 100 V1 = 107 4000/3 I1 = 3s s(s + 100) V2 = 107 0.4 × 10−4 2000/3 · = 6s s + 100 s(s + 100) [c] i1 (t) = 0.4e−100tu(t) mA V1 = 40/3 40/3 − ; s s + 100 v1(t) = (40/3)(1 − 1e−100t)u(t) V V2 = 20/3 20/3 − ; s s + 100 v2(t) = (20/3)(1 − 1e−100t)u(t) V [d] i1(0+ ) = 0.4 mA i1 (0+ ) = 20 × 10−3 = 0.44 mA(checks) 50 v1 (0+ ) = 0; v2(0+ ) = 0(checks) v1 (∞) = 40/3 V; v2(∞) = 20/3 V(checks) v1 (∞) + v2 (∞) = 20 V(checks) (0.3 × 10−6 )v1 (∞) = 4 µC (0.6 × 10−6 )v2 (∞) = 4 µC(checks) © 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. Problems 13–25 P 13.26 [a] V1 − 75/s V1 V1 − V2 + + =0 10 20 10 V2 V2 − V1 (V2 − 75/s)s + + =0 5s 10 250 Thus, 5V1 − 2V2 = 150 s −25sV1 + (s2 + 25s + 50)V2 = 75s ∆= N2 = 5 −2 −25s s2 + 25s + 50 5 150/s = 5(s + 5)(s + 10) = 375(s + 10) −25s 75s V2 = N2 375(s + 10) 75 = = ∆ 5(s + 5)(s + 10) s+5 Vo = 75 75 375 − = s s+5 s(s + 5) Io = V2 15 3 3 = = − 5s s(s + 5) s s+5 [b] vo(t) = (75 − 75e−5t )u(t) V io (t) = (3 − 3e−5t )u(t) 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. 13–26 CHAPTER 13. The Laplace Transform in Circuit Analysis [c] At t = 0+ the circuit is vo (0+ ) = 0; io (0+ ) = 0 Checks At t = ∞ the circuit is vo (∞) = 75 V; io (∞) = 75 20 · = 3 A Checks 10 + (200/30) 30 P 13.27 [a] 10 10 I1 + (I1 − I2) + 10(I1 − 9/s) = 0 s s 10 10 (I2 − 9/s) + (I2 − I1) + 10I2 = 0 s s Simplifying, (s + 2)I1 − I2 = 9 −I1 + (s + 2)I2 = 9 s © 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. Problems ∆= N1 = (s + 2) −1 −1 (s + 2) 9 −1 13–27 = s2 + 4s + 3 = (s + 1)(s + 3) = 9/s (s + 2) 9s2 + 18s + 9 9 = (s + 1)2 s s 9 (s + 1)2 9(s + 1) N1 I1 = = = ∆ s (s + 1)(s + 3) s(s + 3) " N2 = # (s + 2) 9 −1 = 9/s 18 (s + 1) s N2 18(s + 1) 18 = = ∆ s(s + 1)(s + 3) s(s + 3) I2 = Ia = I1 = 9(s + 1) 3 6 = + s(s + 3) s s+3 9 9 9(s + 1) 6 6 − I1 = − = − s s s(s + 3) s s+3 Ib = [b] ia(t) = 3(1 + 2e−3t )u(t) A ib (t) = 6(1 − e−3t )u(t) A 10 3 6 10 + [c] Va = Ib = s s s s+3 = 30 60 30 20 20 + = 2 + − 2 s s(s + 3) s s s+3 10 10 Vb = (I2 − I1) = s s 6 6 3 6 − − + s s+3 s s+3 = 10 3 12 30 40 40 − = 2 − + s s s+3 s s s+3 Vc = 10 10 9 6 6 (9/s − I2) = − + s s s s s+3 = 30 20 20 + − s2 s s+3 [d] va(t) = [30t + 20 − 20e−3t ]u(t) V vb (t) = [30t − 40 + 40e−3t ]u(t) V vc (t) = [30t + 20 − 20e−3t ]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. 13–28 CHAPTER 13. The Laplace Transform in Circuit Analysis [e] Calculating the time when the capacitor voltage drop first reaches 1000 V: 30t + 20 − 20e−3t = 1000 or 30t − 40 + 40e−3t = 1000 Note that in either of these expressions the exponential tem is negligible when compared to the other terms. Thus, 30t + 20 = 1000 or 30t − 40 = 1000 Thus, 980 1040 = 32.67 s or t = = 34.67 s 30 30 Therefore, the breakdown will occur at t = 32.67 s. t= P 13.28 [a] V1 V1 − 50/s V1 − Vo + + =0 10 25/s 4s −5 Vo − V1 Vo − 50/s + + =0 s 4s 30 Simplfying, (4s2 + 10s + 25)V1 − 25Vo = 200s −15V1 + (2s + 15)Vo = 400 ∆= No = (4s2 + 10s + 25) −25 −15 (2s + 15) (4s2 + 10s + 25) 200s −15 = 8s(s + 5)2 = 200(8s2 + 35s + 50) 400 Vo = No 200(8s2 + 35s + 50) 25(8s2 + 35s + 50) K1 K2 K3 = = = + + 2 2 2 ∆ 8s(s + 5) s(s + 5) s (s + 5) s+5 K1 = (25)(50) = 50; 25 K2 = 25(200 − 175 + 50) = −375 −5 © 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. Problems d 8s2 + 35s + 50 K3 = 25 ds s " # s=−5 13–29 s(16s + 35) − (8s2 + 35s + 50) = 25 s2 " # s=−5 = −5(−45) − 75 = 150 .·. Vo = 50 375 150 − + 2 s (s + 5) s+5 [b] vo(t) = [50 − 375te−5t + 150e−5t ]u(t) V [c] At t = 0+ : vo (0+ ) = 50 + 150 = 200 V(checks) At t = ∞: vo (∞) vo (∞) − 50 −5+ =0 10 30 .·. 3vo (∞) − 150 + vo(∞) − 50 = 0; . ·. .·. 4vo (∞) = 200 vo (∞) = 50 V(checks) © 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. 13–30 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.29 [a] 5 0 = 2.5s(I1 − 6/s) + (I1 − I2) + 10I1 s −75 5 = (I2 − I1) + 5(I2 − 6/s) s s or (s2 + 4s + 2)I1 − 2I2 = 6s −I1 + (s + 1)I2 = −9 ∆= −1 (s + 1) −2 9 (s + 1) = 5(s + 2)(s + 3) = 6(s2 + s − 3) N1 6(s2 + s − 3) = ∆ s(s + 2)(s + 3) (s2 + 4s + 2) 6s N2 = I2 = −2 6s N1 = I1 = (s2 + 4s + 2) −1 9 = −9s2 − 30s − 18 N2 −9s2 − 30s − 18 = ∆ s(s + 2)(s + 3) 6(s2 + s − 3) [b] sI1 = (s + 2)(s + 3) lim sI1 = i1 (0+ ) = 6 A; s→∞ sI2 = lim sI1 = i1(∞) = −3 A s→0 −9s2 − 30s − 18 (s + 2)(s + 3) © 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. Problems lim sI2 = i2 (0+ ) = −9 A; lim sI2 = i2(∞) = −3 A s→∞ [c] I1 = 13–31 s→0 6(s2 + s − 3) K1 K2 K3 = + + s(s + 2)(s + 3) s s+2 s+3 K1 = 6(−3) = −3; 6 K3 = 6(9 − 3 − 3) =6 (−3)(−1) K2 = 6(4 − 2 − 3) =3 (−2)(1) i1 (t) = [−3 + 3e−2t + 6e−3t ]u(t) A I2 = −9s2 − 30s − 18 K1 K2 K3 = + + s(s + 2)(s + 3) s s+2 s+3 K1 = −18 = −3; 6 K3 = −81 + 90 − 18 = −3 (−3)(−1) K2 = −36 + 60 − 18 = −3 (−2)(1) i2 (t) = [−3 − 3e−2t − 3e−3t ]u(t) A P 13.30 [a] At Vo : Vo − V1 50 Vo − V2 − + =0 10s s 5 . ·. Vo (2s + 1) − 2sV2 − V1 = 500 Supernode: V1 s V1 − Vo V2 V2 − V1 + + + =0 5 10s 1 5 . ·. −Vo (2s + 1) + 12sV2 + (2s2 + 1)V1 = 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. 13–32 CHAPTER 13. The Laplace Transform in Circuit Analysis Constraint: V1 s V1 − V2 = 4I∆ = 4 − 5 . ·. V2 = (0.8s + 1)V1 Simplifying: Vo (2s + 1) − V1 (1.6s2 + 2s + 1) = 500 −Vo (2s + 1) − V1 (11.6s2 + 12s + 1) = 0 ∆= No = Vo = 2s + 1 −(1.6s2 + 2s + 1) −(2s + 1) (11.6s2 + 12s + 1) 500 −(1.6s2 + 2s + 1) 2 = 20(s2 + 1.5s + 0.5) = 500(11.6s2 + 12s + 1) 0 (11.6s + 12s + 1) No 25(11.6s2 + 12s + 1) = ∆ s(s + 0.5)(s + 1) [b] vo(0+ ) = lim sVo = 25(11.6) = 290 V s→∞ vo (∞) = lim sVo = s→0 25 = 50 V 0.5 [c] At t = 0+ the circuit is 4I∆ + 1I1 = 0; I1 − I∆ = 50 .·. 4Iφ + 50 + I∆ = 0; 5I∆ = −50 .·. I∆ = Io (0+ ) = −10 A Also I1 = 50 − 10 = 40 A Vo (0+ ) = 5(I1 − I∆ ) + 1I1 = 6I1 − 5I∆ = 240 − 5(−10) = 290 V (checks) © 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. Problems 13–33 At t = ∞ the circuit is Vo (∞) = 50(1) = 50 V(checks) [d] Vo = 25(11.6s2 + 12s + 1) K1 K2 K3 = + + s(s + 0.5)(s + 1) s s + 0.5 s + 1 K1 = 25 = 50; (0.5)(1) K3 = 15 = 30 (−1)(−0.5) Vo = 50 210 30 + + s s + 0.5 s + 1 K2 = −52.5 = 210 (−0.5)(0.5) vo (t) = (50 + 210e−0.5t + 30e−t )u(t) V vo (∞) = 50 V(checks) vo (0+ ) = 50 + 210 + 30 = 290 V(checks) P 13.31 [a] 120 250 = 50(I1 − 0.05Vφ ) + (I1 − I2 ) s s 250 250 250 250 = 50I1 − 2.5 (I2 − I1) + I1 − I2 s s s s Simplfying, (50s + 875)I1 − 875I2 = 120 © 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. 13–34 CHAPTER 13. The Laplace Transform in Circuit Analysis 250(s − 1)I1 + (20s2 + 450s + 250)I2 = 0 ∆= (50s + 875) 2 −875 = 1000s(s2 + 40s + 625) 250(s − 1) (20s + 450s + 250) N1 = N2 = 120 −875 = 1200(2s2 + 45s + 25) 2 0 (20s + 450s + 250) (50s + 875) 120 250(s − 1) 0 = −30,000(s − 1) I1 = N1 1200(2s2 + 45s + 25) = ∆ s(s2 + 40s + 625) I2 = N2 −30,000(s − 1) = ∆ s(s2 + 40s + 625) Io = I2 − 0.05Vφ = I2 − 0.05 I2 − I1 = 250 (I2 − I1) s −2400(s + 35) s(s2 + 40s + 625) 250 −600,000(s + 35) (I2 − I1 ) = s s(s2 + 40s + 625) . ·. [b] sIo = Io = (s2 30,000(s + 35) 1080 −30,000(s − 1) + = 2 2 2 s(s + 40s + 625) s(s + 40s + 625) s(s + 40s + 625) 1080 + 40s + 625) io (0+ ) = lim sIo = 0 s→∞ io (∞) = lim sVo = s→0 1080 = 1728 mA 625 [c] At t = 0+ the circuit is i(0+ ) = 0 (checks) © 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. Problems 13–35 At t = ∞ the circuit is 120 = 50(ia − i1) + 700ia = 50(ia − 0.05vφ ) + 700ia = 750ia − 2.5vφ . ·. vφ = −700ia 120 = 750ia + 1750ia = 2500ia 120 = 48 mA 2500 vφ = −700ia = −33.60 V ia = io (∞) = 48 × 10−3 − 0.05(−33.60) = 48 × 10−3 + 1.68 = 1728 mA (checks) [d] Io = 1080 K1 K2 K2∗ = + + s(s2 + 40s + 625) s s + 20 − j15 s + 20 + j15 1080 = 1.728 625 1080 = 1.44/126.87◦ K2 = (−20 + j15)(j30) K1 = io (t) = [1728 + 2880e−20t cos(15t + 126.87◦ )]u(t) mA Check: io (0+ ) = 0 mA; 100k5s = 500s 100s = 5s + 100 s + 20 io (∞) = 1728 mA P 13.32 [a] " # 100s 50 5000s Vo = = 2 s + 20 (s + 25) (s + 20)(s + 25)2 Io = Vo 50s = 100 (s + 20)(s + 25)2 IL = Vo 1000 = 5s (s + 20)(s + 25)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. 13–36 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] Vo = K1 K2 K3 + + 2 s + 20 (s + 25) s + 25 K1 = K2 = 5000s (s + 25)2 5000s (s + 20) s=−20 = −4000 = 25,000 s=−25 d 5000s K3 = ds s + 20 s=−25 " 5000 5000s = − s + 20 (s + 20)2 # = 4000 s=−25 vo (t) = [−4000e−20t + 25,000te−25t + 4000e−25t ]u(t) V Io = K1 K2 K3 + + s + 20 (s + 25)2 s + 25 K1 = 50s (s + 25)2 K2 = 50s (s + 20) s=−20 = −40 = 250 s=−25 d 50s K3 = ds s + 20 s=−25 " 50 50s = − s + 20 (s + 20)2 # = 40 s=−25 io (t) = [−40e−20t + 250te−25t + 40e−25t ]u(t) V IL = K1 K2 K3 + + 2 s + 20 (s + 25) s + 25 K1 = 1000 (s + 25)2 K2 = 1000 (s + 20) = 40 s=−20 s=−25 d 1000 K3 = ds s + 20 = −200 s=−25 " 1000 = − (s + 20)2 # s=−25 = −40 iL (t) = [40e−20t − 200te−25t − 40e−25t ]u(t) V P 13.33 vC = 12 × 105 te−5000t V, dvC iC = C dt ! C = 5 µF; therefore = 6e−5000t(1 − 5000t) A iC > 0 when 1 > 5000t or iC > 0 when 0 < t < 200 µs © 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. Problems 13–37 and iC < 0 when t > 200 µs iC = 0 when 1 − 5000t = 0, or t = 200 µs dvC = 12 × 105 e−5000t[1 − 5000t] dt .·. iC = 0 when dvC =0 dt P 13.34 10s 40 400 40 · = = 10s + 1000 s 10s + 1000 s + 100 VTh = ZTh = 1000 + 1000k10s = 1000 + I= (5 × = K1 = 105 )/s 10,000s 2000(s + 50) = 10s + 1000 s + 100 40/(s + 100) 40s = 2 + 2000(s + 50)/(s + 100) 2000s + 600,000s + 5 × 107 0.02s K1 K1∗ = + s2 + 300s + 25,000 s + 150 − j50 s + 150 + j50 0.02s s + 150 + j50 s=−150+j50 = 31.62 × 10−3 /71.57◦ i(t) = 63.25e−150t cos(50t + 71.57◦ )u(t) mA © 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. 13–38 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.35 [a] The s-domain equivalent circuit is I= Vg /L Vg = , R + sL s + (R/L) I= K0 K1 K1∗ + + s + R/L s − jω s + jω K0 = Vg = Vm (ωL cos φ − R sin φ) , R2 + ω 2 L2 Vm (ω cos φ + s sin φ) s2 + ω 2 K1 = Vm /φ − 90◦ − θ(ω) √ 2 R2 + ω 2 L2 where tan θ(ω) = ωL/R. Therefore, we have i(t) = Vm (ωL cos φ − R sin φ) −(R/L)t Vm sin[ωt + φ − θ(ω)] √ e + R2 + ω 2 L2 R2 + ω 2 L2 Vm sin[ωt + φ − θ(ω)] + ω 2 L2 Vm (ωL cos φ − R sin φ) −(R/L)t [c] itr = e R2 + ω 2 L2 Vg , Vg = Vm /φ − 90◦ [d] I = R + jωL [b] iss (t) = √ R2 Therefore I = √ Vm /φ − 90◦ Vm √ /φ − θ(ω) − 90◦ = 2 2 2 2 2 2 / R + ω L θ(ω) R +ω L Vm Therefore iss = √ 2 sin[ωt + φ − θ(ω)] R + ω 2 L2 [e] The transient component vanishes when ωL cos φ = R sin φ or tan φ = ωL R or φ = θ(ω) 1 1 P 13.36 [a] W = L1 i21 + L2 i22 + Mi1i2 2 2 W = 4(15)2 + 9(100) + 150(6) = 2700 J [b] 120i1 + 8 di1 di2 −6 =0 dt dt 270i2 + 18 di2 di1 −6 =0 dt dt © 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. Problems 13–39 Laplace transform the equations to get 120I1 + 8(sI1 − 15) − 6(sI2 + 10) = 0 270I2 + 18(sI2 + 10) − 6(sI1 − 15) = 0 In standard form, (8s + 120)I1 − 6sI2 = 180 −6sI1 + (18s + 270)I2 = −270 ∆= N1 = N2 = 8s + 120 −6s −6s 18s + 270 180 = 108(s + 10)(s + 30) −6s = 1620(s + 30) −270 18s + 270 8s + 120 180 −6s −270 = −1080(s + 30) I1 = N1 1620(s + 30) 15 = = ∆ 108(s + 10)(s + 30) s + 10 I2 = N2 −1080(s + 30) −10 = = ∆ 108(s + 10)(s + 30) s + 10 [c] i1 (t) = 15e−10t u(t) A; i2 (t) = −10e−10t u(t) A Z ∞ e−20t ∞ [d] W120Ω = (225e−20t )(120) dt = 27,000 = 1350 J −20 0 0 W270Ω = Z 0 ∞ (100e−20t )(270) dt = 27,000 e−20t −20 ∞ = 1350 J 0 W120Ω + W270Ω = 2700 J 1 1 [e] W = L1 i21 + L2 i22 + Mi1 i2 = 900 + 900 − 900 = 900 J 2 2 With the dot reversed the s-domain equations are (8s + 120)I1 + 6sI2 = 60 6sI1 + (18s + 270)I2 = −90 As before, N1 = N2 = ∆ = 108(s + 10)(s + 30). Now, 60 −6s = 1620(s + 10) −90 18s + 270 8s + 120 60 −6s −90 = −1080(s + 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. 13–40 CHAPTER 13. The Laplace Transform in Circuit Analysis I1 = N1 15 = ; ∆ s + 30 I2 = i1 (t) = 15e−30t u(t) A; Z ∞ W270Ω = Z ∞ W120Ω = i2 (t) = −10e−30tu(t) A (100e−60t )(270) dt = 450 J 0 0 N2 −10 = ∆ s + 30 (225e−60t )(120) dt = 450 J W120Ω + W270Ω = 900 J P 13.37 The s-domain equivalent circuit is V1 − 48/s V1 + 9.6 V1 + + =0 4 + (100/s) 0.8s 0.8s + 20 V1 = Vo = s2 20 −30,000 V1 = 0.8s + 20 (s + 25)(s + 5 − j10)(s + 5 + j10) = K1 = K2 = −1200 + 10s + 125 K1 K2 K2∗ + + s + 25 s + 5 − j10 s + 5 + j10 s2 −30,000 + 10s + 125 s=−25 −30,000 (s + 25)(s + 5 + j10) = −60 = 67.08/63.43◦ s=−5+j10 vo (t) = [−60e−25t + 134.16e−5t cos(10t + 63.43◦ )]u(t) V P 13.38 For t < 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. Problems 13–41 For t > 0+ : Note that because of the dot locations on the coils, the sign of the mutual inductance is negative! (See Example C.1 in Appendix C.) L1 − M = 3 + 1 = 4 H; 18 × 4 = 72; L2 − M = 2 + 1 = 3 H 18 × 3 = 54 V − 72 V V + 54 + + =0 4s + 20 −s + 10 3s V 1 1 1 + + 4s + 20 −s + 10 3s = 72 54 − 4s + 20 3s " # 3s(−s + 10) + 3s(4s + 20) + (4s + 20)(−s + 10) 72(3s) − 54(4s + 20) V = 3s(−s + 10)(4s + 20) 3s(4s + 20) V = [72(3s) − 54(4s + 20)](−s + 10) 5s2 + 110s + 200 Io = V −108 −1.2 1.2 = = + −s + 10 (s + 2)(s + 20) s + 2 s + 20 io (t) = 1.2[e−20t − e−2t]u(t) 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. 13–42 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.39 [a] 150 = (25 + 0.9375s)I1 + 0.625sI2 s 0 = 0.625sI2 + (50 − 1.25s)I1 ∆= 0.625s 0.625s 1.25s + 50 150 N1 = I1 = 0.9375s + 25 0.625s = 0 1.25s + 50 = 0.78125(s2 + 100s + 1600) 187.5(s + 40) s N1 240(s + 40) = ∆ s(s + 20)(s + 80) 240(s + 40) (s + 20)(s + 80) [b] sI1 = lim sI1 = i1(∞) = 6 A s→0 lim sI1 = i1 (0) = 0 s→∞ [c] I1 = K1 K2 K3 + + s s + 20 s + 80 K1 = 6; K2 = −4; K3 = −2 i1 (t) = (6 − 4e−20t − 2e−80t )u(t) A P 13.40 [a] From the solution to Problem 13.39 we have 0.9375s + 25 150 N2 = I2 = 0.625s 0 = −93.75 −120 K1 K2 = + (s + 20)(s + 80) s + 20 s + 80 K1 = −120 = −2; 60 K2 = −120 =2 −60 i2 (t) = (−2e−20t + 2e−80t )u(t) 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. Problems 13–43 [b] Reversing the dot on the 1.25 H coil will reverse the sign of M, thus the circuit becomes The two simulanteous equations are 150 = (25 + 0.9375s)I1 − 0.625sI2 s 0 = −0.625sI1 + (1.25s + 50)I2 When these equations are compared to those derived in Problem 13.39 we see the only difference is the algebraic sign of the 0.625s term. Thus reversing the dot will have no effect on I1 and will reverse the sign of I2 . Hence, i2 (t) = (2e−20t − 2e−80t )u(t) A P 13.41 [a] s-domain equivalent circuit is i2(0+ ) = − Note: [b] 20 = −2 A 10 24 = (120 + 3s)I1 + 3sI2 + 6 s 0 = −6 + 3sI1 + (360 + 15s)I2 + 36 In standard form, (s + 40)I1 + sI2 = (8/s) − 2 sI1 + (5s + 120)I2 = −10 ∆= N1 = s + 40 s s 5s + 120 = 4(s + 20)(s + 60) (8/s) − 2 s −10 5s + 120 = −200(s − 4.8) s © 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. 13–44 CHAPTER 13. The Laplace Transform in Circuit Analysis I1 = [c] sI1 = N1 −50(s − 4.8) = ∆ s(s + 20)(s + 60) −50(s − 4.8) (s + 20)(s + 60) lim sI1 = i1 (0+ ) = 0 A s→∞ lim sI1 = i1(∞) = s→0 [d] I1 = (−50)(−4.8) = 0.2 A (20)(60) K1 K2 K3 + + s s + 20 s + 60 K1 = 240 = 0.2; 1200 K3 = −50(−60) + 240 = 1.35 (−60)(−40) K2 = −50(−20) + 240 = −1.55 (−20)(40) i1 (t) = [0.2 − 1.55e−20t + 1.35e−60t ]u(t) A P 13.42 [a] Voltage source acting alone: Vo1 − 60/s V01 s V01 + + =0 10 80 20 + 10s .·. V01 = 480(s + 2) s(s + 4)(s + 6) Vo2 V02 s V02 − 30/s + + =0 10 80 10(s + 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. Problems .·. V02 = 240 s(s + 4)(s + 6) Vo = Vo1 + Vo2 = [b] Vo = 13–45 480(s + 2) + 240 480(s + 2.5) = s(s + 4)(s + 6) s(s + 4)(s + 6) K2 K3 K1 + + s s+4 s+6 (480)(2.5) = 50; (4)(6) K1 = K2 = 480(−1.5) = 90; (−4)(2) K3 = 480(−3.5) = −140 (−6)(−2) vo (t) = [50 + 90e−4t − 140e−6t ]u(t) V Y11 Y12 P 13.43 ∆ = Y12 Y22 N2 = V2 = = Y11Y22 − Y122 Y11 [(Vg /R1 ) + γC − (ρ/s)] Y12 (Ig − γC) N2 ∆ Substitution and simplification lead directly to Eq. 13.90. P 13.44 [a] Vo = − Zf = Zf Vg Zi 104 (80 × 106 /s) 80 × 106 = 104 + 80 × 106 /s s + 8000 Zi = 4000 + Vg = 16,000 s2 .·. Vo = [b] Vo = 109 4000(s + 4000) = 62.5s s −320 × 106 s(s + 4000)(s + 8000) K1 K2 K3 + + s s + 4000 s + 8000 K1 = −20,000(16,000) = −10 (4000)(8000) K2 = −320 × 106 = 20 (−4000)(4000) © 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. 13–46 CHAPTER 13. The Laplace Transform in Circuit Analysis K3 = −320 × 106 = −10 (−8000)(−4000) .·. vo(t) = (−10 + 20e−4000t − 10e−8000t)u(t) V [c] −10 + 20e−4000ts − 10e−8000ts = −5 .·. 20e−4000ts − 10e−8000ts = 5 Let x = e−4000ts . Then 20x − 10x2 = 5; Solving, x = 1± √ 0.5 or x2 − 2x + 0.5 = 0 so x = 0.2929 .·. e−4000ts = 0.2929; [d] vg = m tu(t); Vg = .·. ts = 306.99 µs m s2 Vo = −20,000m s(s + 4000)(s + 8000) K1 = −20,000m −20,000m = (4000)(8000) 32 × 106 .·. −5 = −20,000m 32 × 106 .·. m = 8000 V/s Thus, m must be less than or equal to 8000 V/s to avoid saturation. P 13.45 [a] Let va be the voltage across the 0.5 µF capacitor, positive at the upper terminal. Let vb be the voltage across the 100 kΩ resistor, positive at the upper terminal. Also note 106 2 × 106 106 4 × 106 0.5 = and = ; Vg = 0.5s s 0.25s s s sVa Va − (0.5/s) Va + + =0 6 s × 10 200,000 200,000 sVa + 10Va − Va = 5 + 10Va = 0 s 5 s(s + 20) 0 − Va (0 − Vb )s + =0 200,000 4 × 106 © 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. Problems . ·. Vb = − 13–47 20 −100 Va = 2 s s (s + 20) Vb (Vb − 0)s (Vb − Vo )s + + =0 100,000 4 × 106 4 × 106 40Vb + sVb + sVb = sVo . ·. 2(s + 20)Vb Vo = ; s −100 −200 Vo = 2 = 3 s s3 [b] vo(t) = −100t2 u(t) V [c] −100t2 = −4; t = 0.2 s = 200 ms P 13.46 Va − 0.016/s Va s (Va − Vo )s + + =0 6 2000 50 × 10 50 × 106 (0 − Va )s (0 − Vo ) + =0 50 × 106 10,000 Va = −5000Vo s −5000Vo 0.016 .·. (2s + 25,000) − sVo = 25,000 s s Vo = −4000 (s + 5000 − j10,000)(s + 5000 + j10,000) K1 = −400 = j0.02 = 0.02/90◦ j10,000 vo (t) = 40e−5000t cos(10,000t + 90◦ ) = −40e−5000t sin(10,000t)u(t) mV © 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. 13–48 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.47 [a] Vp = 50/s 50 Vg2 = Vg2 5 + 50/s 5s + 50 Vp − 40/s Vp − Vo Vp − Vo + + =0 20 5 100/s Vp 1 1 s 1 s + + − Vo + 20 5 100 5 100 = 2 s s + 25 50 16 2 1 s − = Vo + 100 5s + 50 s s 5 100 " = Vo s + 20 100 # 100 16(s + 25) 2 −40s + 2000 − = Vo = s + 20 10(s + 10)(s) s s(s + 10)(s + 20) = K1 K2 K3 + + s s + 10 s + 20 K1 = 10; . ·. K2 = −24; K3 = 14 vo (t) = [10 − 24e−10t + 14e−20t]u(t) V [b] 10 − 24x + 14x2 = 5 14x2 − 24x + 5 = 0 x = 0 or 0.242691 e−10t = 0.242691 .·. t = 141.60 ms P 13.48 Let vo1 equal the output voltage of the first op amp. Then Vo1 = −Zf 1 Vg ZA1 ZA1 = 25,000 + where Zf 1 = 25 × 103 Ω 25,000(20 × 104 /s) 25,000 + (20 × 104 /s) © 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. Problems = .·. 13–49 25,000(s + 16) Ω (s + 8) Vo1 = −(s + 8) Vg (s + 16) .·. vg (t) = 16u(t) mV; Vg = 16 × 10−3 s Vo1 = −16 × 10−3 (s + 8) −0.008 −0.008 = + s(s + 16) s s + 16 .·. vo1 = −0.008(1 + e−16t ) V The op amp will saturate when vo1 = ±6 V. Hence, saturation will occur when −0.008(1 + e−16t) = −6 Thus t= e−16t = 749 so ln 749 = −0.414 s −16 Thus, the first op amp never saturates. We must investigate the output of the second op amp: 2 × 108 Ω and ZA2 = 25,000 Ω s Vo = −Zf 2 Vo1 ZA2 .·. −8000 −8000 −(s + 8) Vo = Vo1 = Vg s s (s + 16) where Zf 2 = " = # 8000(s + 8) Vg s(s + 16) .·. vg (t) = 16u(t) mV; Vg = 16 × 10−3 s Vo = 128(s + 8) K1 K2 K3 = 2 + + 2 s (s + 16) s s s + 16 K1 = 128(8) = 64 16 d s+8 K2 = 128 ds s + 16 =4 s=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. 13–50 CHAPTER 13. The Laplace Transform in Circuit Analysis K3 = 128(−8) = −4 256 vo (t) = [64t + 4 − 4e−16t ]u(t) V The op amp will saturate when vo = ±6 V. Hence, saturation will occur when 64t + 4 − 4e−16t = 6 or 16t − 0.5 = e−16t This equation can be solved by trial and error. First note that t > 0.5/16 or t > 31.25 ms. Try 40 ms: 0.64 − 0.5 = 0.14; e−0.64 = 0.53 Try 50 ms: 0.80 − 0.5 = 0.30; e−0.80 = 0.45 Try 60 ms: 0.96 − 0.5 = 0.46; e−0.96 = 0.38 Further trial and error gives tsat ∼ = 56.5 ms P 13.49 [a] 4 100,000Vi Vi = Vo + 20 100,000 + (4 × 108 /s) 0.2Vi − . ·. sVi = Vo s + 4000 Vo −0.8(s − 1000) = H(s) = Vi (s + 4000) © 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. Problems 13–51 [b] −z1 = 1000 rad/s −p1 = −4000 rad/s P 13.50 [a] Vo 1/sC 1 = = Vi R + 1/sC RCs + 1 H(s) = [b] (1/RC) 250 = ; s + (1/RC) s + 250 −p1 = −250 rad/s Vo R RCs s = = = Vi R + 1/sC RCs + 1 s + (1/RC) s ; z1 = 0, −p1 = −250 rad/s s + 250 sL s s Vo = = = [c] Vi R + sL s + R/L s + 8000 = z1 = 0; [d] −p1 = −8000 rad/s Vo R R/L 8000 = = = Vi R + sL s + (R/L) s + 8000 −p1 = −8000 rad/s [e] Vo s Vo Vo − Vi + + =0 6 4 × 10 10,000 40,000 sVo + 400Vo + 100Vo = 100Vi H(s) = Vo 100 = Vi s + 500 −p1 = −500 rad/s P 13.51 [a] [b] 1/sC 1 1/RC = = R + 1/sC RsC + 1 s + 1/RC There are no zeros, and a single pole at −1/RC rad/sec. R R/L = R + sL s + R/L There are no zeros, and a single pole at −R/L rad/sec. © 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. 13–52 CHAPTER 13. The Laplace Transform in Circuit Analysis [c] There are several possible solutions. One is R = 10 Ω; L = 10 mH; C = 100 µF R RsC s = = R + 1/sC RsC + 1 s + 1/RC There is a single zero at 0 rad/sec, and a single pole at −1/RC rad/sec. sL s [b] = R + sL s + R/L There is a single zero at 0 rad/sec, and a single pole at −R/L rad/sec. P 13.52 [a] [c] There are several possible solutions. One is R = 100 Ω; P 13.53 [a] L = 10 mH; C = 1 µF R (R/L)s = 2 1/sC + sL + R s + (R/L)s + 1/LC There is a single zero at 0 rad/sec, and two poles: p1 = −(R/2L) + q (R/2L)2 − (1/LC); p2 = −(R/2L) − [b] There are several possible solutions. One is R = 250 Ω; L = 10 mH; q (R/2L)2 − (1/LC) C = 1 µF These component values yield the following poles: −p1 = −5000 rad/sec and − p2 = −20,000 rad/sec [c] There are several possible solutions. One is R = 200 Ω; L = 10 mH; C = 1 µF These component values yield the following poles: −p1 = −10,000 rad/sec and − p2 = −10,000 rad/sec [d] There are several possible solutions. One is R = 120 Ω; L = 10 mH; C = 1 µF These component values yield the following poles: −p1 = −6000 + j8000 rad/sec P 13.54 [a] Zi = 1000 + Zf = and − p2 = −6000 − j8000 rad/sec 5 × 106 1000(s + 5000) = s s 40 × 106 40 × 106 k40,000 = s s + 1000 H(s) = − Zf −40 × 106 /(s + 1000) −40,000s = = Zi 1000(s + 5000)/s (s + 1000)(s + 5000) © 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. Problems [b] Zero at z1 = 0; 13–53 Poles at −p1 = −1000 rad/s and −p2 = −5000 rad/s P 13.55 [a] Let R1 = 250 kΩ; R2 = 125 kΩ; C2 = 1.6 nF; and Cf = 0.4 nF. Then (R2 + 1/sC2 )1/sCf (s + 1/R2 C2 ) = Zf = C +C 1 1 R2 + sC2 + sCf Cf s s + C22Cf Rf2 1 = 2.5 × 109 Cf 1 62.5 × 107 = = 5000 rad/s R2 C2 125 × 103 C2 + Cf 2 × 10−9 = = 25,000 rad/s C2Cf R2 (0.64 × 10−18 )(125 × 103 ) .·. Zf = 2.5 × 109 (s + 5000) Ω s(s + 25,000) Zi = R1 = 250 × 103 Ω H(s) = Vo −Zf −104 (s + 5000) = = Vg Zi s(s + 25,000) [b] −z1 = −5000 rad/s −p1 = 0; −p2 = −25,000 rad/s P 13.56 [a] Va − Vg sVa (Va − Vo )s + + =0 6 1000 5 × 10 5 × 106 5000Va − 5000Vg + 2sVa − sVo = 0 (5000 + 2s)Va − sVo = 5000Vg (0 − Va )s 0 − Vo + =0 5 × 106 5000 © 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. 13–54 CHAPTER 13. The Laplace Transform in Circuit Analysis .·. −sVa − 1000Vo = 0; Va − −1000 Vo s −1000 (2s + 5000) Vo − sVo = 5000Vg s 1000Vo (2s + 5000) + s2 Vo = −5000sVg Vo (s2 + 2000s + 5 × 106 ) = −5000sVg −5000s Vo = 2 Vg s + 2000s + 5 × 106 √ s1,2 = −1000 ± 106 − 5 × 106 = −1000 ± j2000 Vo −5000s = Vg (s + 1000 − j2000)(s + 1000 + j2000) [b] z1 = 0; −p1 = −1000 + j2000; −p2 = −1000 − j2000 P 13.57 [a] Vo Vo + + Vo (10−7 )s = Ig 5000 0.2s .·. Vo = Ig = 10 × 106 s · Ig s2 + 2000s + 50 × 106 0.1s ; s2 + 108 .·. H(s) = Io = 10−7 sVo s2 s2 + 2000s + 50 × 106 (s2 )(0.1s) [b] Io = (s + 1000 − j7000)(s + 1000 + j7000)(s2 + 108 ) Io = 0.1s3 (s + 1000 − j7000)(s + 1000 + j7000)(s + j104 )(s − j104 ) [c] Damped sinusoid of the form Me−1000t cos(7000t + θ1) [d] Steady-state sinusoid of the form N cos(104 t + θ2 ) © 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. Problems [e] Io = 13–55 K1 K1∗ K2 K2∗ + + + s + 1000 − j7000 s + 1000 + j7000 s − j104 s + j104 K1 = 0.1(−1000 + j7000)3 = 46.9 × 10−3 / − 140.54◦ (j14,000)(−1000 − j5000)(−1000 + j17,000) K2 = 0.1(j104 )3 = 92.85 × 10−3 /21.8◦ (j20,000)(1000 + j3000)(1000 + j17,000) io (t) = [93.8e−1000t cos(7000t − 140.54◦ ) + 185.7 cos(104 t + 21.8◦ )] mA Test: Z= 1 ; Y . ·. Z = Y = 1 1 1 2 + j5 + + = 5000 j2000 −j1000 10,000 10,000 = 1856.95/ − 68.2◦ Ω 2 + j5 Vo = Ig Z = (0.1/0◦ )(1856.95/ − 68.2◦ ) = 185.695/ − 68.2◦ V Io = (10−7 )(j104 )Vo = 185.7/21.8◦ mA ioss = 185.7 cos(104 t + 21.8◦ ) mA(checks) P 13.58 Vg = 25sI1 − 35sI2 16 × 106 0 = −35sI1 + 50s + 10,000 + I2 s ! ∆= N2 = 25s −35s −35s 50s + 10,000 + 16 × 106 /s 25s Vg −35s 0 = 25(s + 2000)(s + 8000) = 35sVg © 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. 13–56 CHAPTER 13. The Laplace Transform in Circuit Analysis I2 = N2 35sVg = ∆ 25(s + 2000)(s + 8000) H(s) = .·. I2 1.4s = Vg (s + 2000)(s + 8000) z1 = 0; −p1 = −2000 rad/s; −p2 = −8000 rad/s P 13.59 [a] 2000(Io − Ig ) + 8000Io + µ(Ig − Io )(2000) + 2sIo = 0 . ·. I o = 1000(1 − µ) Ig s + 1000(5 − µ) .·. H(s) = 1000(1 − µ) s + 1000(5 − µ) [b] µ < 5 [c] µ H(s) −3 4000/(s + 8000) Io 20,000/s(s + 8000) 0 1000/(s + 5000) 5000/s(s + 5000) 4 −3000/(s + 1000) −15,000/s(s + 1000) 5 −4000/s −20,000/s2 6 −5000/(s − 1000) µ = −3: Io = 2.5 2.5 − ; s (s + 8000) −25,000/s(s − 1000) io = [2.5 − 2.5e−8000t]u(t) A µ = 0: 1 1 − ; s s + 5000 µ = 4: Io = Io = −15 15 − ; s s + 1000 io = [1 − e−5000t]u(t) A io = [−15 + 15e−1000t ]u(t) 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. Problems 13–57 µ = 5: −20,000 ; s2 µ = 6: Io = Io = io = −20,000t u(t) A 25 25 − ; s s − 1000 io = 25[1 − e1000t]u(t) A P 13.60 [a] y(t) = 0 t<0 0 ≤ t ≤ 10 : y(t) = Z t 625 dλ = 625t 0 Z 10 ≤ t ≤ 20 : y(t) = 20 ≤ t < ∞ : y(t) = 0 10 t−10 625 dλ = 625(10 − t + 10) = 625(20 − t) © 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. 13–58 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] y(t) = 0 t<0 0 ≤ t ≤ 10 : t Z y(t) = 312.5 dλ = 312.5t 0 y(t) = Z 10 10 20 ≤ t ≤ 30 : y(t) = Z 30 ≤ t < ∞ : y(t) = 0 10 ≤ t ≤ 20 : 0 312.5 dλ = 3125 t−20 312.5 dλ = 312.5(30 − t) [c] y(t) = 0 0≤t≤1: t<0 y(t) = Z 0 t 625 dλ = 625t © 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. Problems 1 ≤ t ≤ 10 : y(t) = Z 1 0 Z 10 ≤ t ≤ 11 : y(t) = 11 ≤ t < ∞ : y(t) = 0 13–59 625 dλ = 625 1 t−10 625 dλ = 625(11 − t) P 13.61 [a] 0 ≤ t ≤ 40: y(t) = Z 0 t t (10)(1)(dλ) = 10λ = 10t 0 40 ≤ t ≤ 80: y(t) = Z t ≥ 80 : 40 40 t−40 (10)(1)(dλ) = 10λ t−40 = 10(80 − t) y(t) = 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. 13–60 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] 0 ≤ t ≤ 10: y(t) = Z t t 40 dλ = 40λ = 40t 0 0 10 ≤ t ≤ 40: y(t) = Z t t t−10 40 dλ = 40λ = 400 t−10 40 ≤ t ≤ 50: y(t) = Z t ≥ 50 : 40 40 t−10 40 dλ = 40λ t−10 = 40(50 − t) y(t) = 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. Problems 13–61 [c] The expressions are 0≤t≤1: y(t) = 1 ≤ t ≤ 40 : y(t) = t t Z 400 dλ = 400λ = 400t 0 0 Z t t t−1 Z 40 ≤ t ≤ 41 : y(t) = 41 ≤ t < ∞ : y(t) = 0 400 dλ = 400λ 40 40 t−1 = 400 t−1 400 dλ = 400λ t−1 = 400(41 − t) [d] [e] Yes, note that h(t) is approaching 40δ(t), therefore y(t) must approach 40x(t), i.e. y(t) = Z 0 t h(t − λ)x(λ) dλ → Z 0 t 40δ(t − λ)x(λ) dλ → 40x(t) This can be seen in the plot, e.g., in part (c), y(t) ∼ = 40x(t). Vo 1 = ; Vi s+1 For 0 ≤ t ≤ 1: P 13.62 H(s) = vo = Z 0 h(t) = e−t t e−λ dλ = (1 − e−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. 13–62 CHAPTER 13. The Laplace Transform in Circuit Analysis For 1 ≤ t ≤ ∞: vo = Z t t−1 P 13.63 H(s) = e−λ dλ = (e − 1)e−t V Vo s 1 = = 1− ; Vi s+1 s+1 h(t) = δ(t) − e−t h(λ) = δ(λ) − e−λ For 0 ≤ t ≤ 1: vo = Z t 0 [δ(λ) − e−λ ] dλ = [1 + e−λ ] |t0= e−t V For 1 ≤ t ≤ ∞: vo = Z t t t−1 (−e−λ ) dλ = e−λ t−1 = (1 − e)e−t V P 13.64 [a] From Problem 13.50(a) H(s) = 250 s + 250 h(λ) = 250e−250λ 0 ≤ t ≤ 4 ms: vo = Z t 16(250)e−250λ dλ = 16(1 − e−250t) V 0 4 ms ≤ t ≤ ∞: vo = Z t t−0.004 16(250)e−250λ dλ = 16(e − 1)e−250t V [b] © 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. Problems 2500 s + 2500 0 ≤ t ≤ 4 ms: .·. h(λ) = 2500e−2500λ P 13.65 [a] H(s) = vo = 13–63 t Z 16(2500)e−2500λ dλ = 16(1 − e−2500t) V 0 4 ms ≤ t ≤ ∞: vo = t Z t−0.004 16(2500)e−2500λ dλ = 16(e10 − 1)e−2500t V [b] decrease [c] The circuit with R = 10 kΩ. P 13.66 [a] vo = Z 0 t 10(10e−4λ ) dλ = 100 e−4λ −4 t 0 = −25[e−4t − 1] = 25(1 − e−4t ) V, 0≤t≤∞ © 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. 13–64 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] 0 ≤ t ≤ 0.5: vo = Z t 0 t 100(1 − 2λ) dλ = 100(λ − λ2 ) = 100t(1 − t) 0 0.5 ≤ t ≤ ∞: vo = Z 0.5 0 100(1 − 2λ) dλ = 100(λ − λ2 ) 0.5 = 25 0 [c] P 13.67 [a] −1 ≤ t ≤ 4: vo = Z t+1 0 10λ dλ = 5λ2 Z = 5t2 + 10t + 5 V 0 4 ≤ t ≤ 9: vo = t+1 t+1 t−4 10λ dλ = 5λ2 t−4 9 ≤ t ≤ 14: vo = 10 Z t+1 10 t−4 λ dλ + 10 Z = 50t − 75 V t+1 10 10 dλ © 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. Problems 10 = 5λ2 t+1 +100λ 10 t−4 13–65 = −5t2 + 140t − 480 V 14 ≤ t ≤ 19: vo = 100 Z t+1 t−4 dλ = 500 V 19 ≤ t ≤ 24: vo = Z 20 t−4 100λ dλ + Z t+2 10(30 − λ) dλ 20 20 t+1 = 100λ +300λ 20 t−2 t+2 −5λ2 20 = −5t2 + 190t − 1305 V 24 ≤ t ≤ 29: vo = 10 Z t+1 t+1 (30 − λ) dλ = 300λ t−4 t−4 −5λ2 t+1 t−4 = 1575 − 50t V 29 ≤ t ≤ 34: vo = 10 Z 30 30 t−4 (30 − λ) dλ = 300λ t−4 −5λ2 30 t−2 = 5t2 − 340t + 5780 V Summary: vo = 0 − ∞ ≤ t ≤ −1 vo = 5t2 + 10t + 5 V vo = 50t − 75 V −1≤t≤4 4≤t≤9 vo = −5t2 + 140t − 480 V vo = 500 V 9 ≤ t ≤ 14 14 ≤ t ≤ 19 vo = −5t2 + 190t − 1305 V vo = 1575 − 50t V 19 ≤ t ≤ 24 24 ≤ t ≤ 29 vo = 5t2 − 340t + 5780 V vo = 0 29 ≤ t ≤ 34 34 ≤ t ≤ ∞ © 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. 13–66 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] 2 P 13.68 [a] h(λ) = λ 5 0≤λ≤5 2 h(λ) = 4 − λ 5 5 ≤ λ ≤ 10 0 ≤ t ≤ 5: 2 λ dλ = 2t2 0 5 5 ≤ t ≤ 10: Z t vo = 10 vo = 10 Z 5 = 0 4λ2 2 2 λ dλ + 10 5 5 t 5 t +40λ 0 Z 5 − 2 4 − λ dλ 5 4λ2 2 t 5 = −100 + 40t − 2t2 10 ≤ t ≤ ∞: vo = 10 Z 5 0 2 λ dλ + 10 5 Z 5 10 2 4 − λ dλ 5 © 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. Problems = 4λ2 2 5 10 +40λ 0 5 − 4λ2 2 13–67 10 5 = 50 + 200 − 150 = 100 vo = 2t2 V 0≤t≤5 vo = 40t − 100 − 2t2 V vo = 100 V 5 ≤ t ≤ 10 10 ≤ t ≤ ∞ [b] 1 [c] Area = (10)(2) = 10 2 5 h(λ) = λ 2 0 ≤ t ≤ 2: vo = 10 Z t 0 1 (4)h = 10 so h = 5 2 0≤λ≤2 5 h(λ) = 10 − λ 2 . ·. 2≤λ≤4 5 λ dλ = 12.5t2 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. 13–68 CHAPTER 13. The Laplace Transform in Circuit Analysis 2 ≤ t ≤ 4: vo = 10 = Z 2 0 5 λ dλ + 10 2 25λ2 2 2 Z 5 10 − λ dλ 2 t 2 t +100λ 0 2 − 25λ2 2 t 2 = −100 + 100t − 12.5t2 4 ≤ t ≤ ∞: vo = 10 Z 2 0 Z 4 5 5 λ dλ + 10 10 − λ dλ 2 2 2 25λ2 = 2 2 4 +100λ 0 2 25λ2 − 2 4 2 = 50 + 200 − 150 = 100 vo = 12.5t2 V 0≤t≤2 vo = 100t − 100 − 12.5t2 V vo = 100 V 2≤t≤4 4≤t≤∞ [d] The waveform in part (c) is closer to replicating the input waveform because in part (c) h(λ) is closer to being an ideal impulse response. That is, the area was preserved as the base was shortened. P 13.69 [a] Vo = . ·. 16 Vg 20 H(s) = Vo 4 = Vg 5 h(λ) = 0.8δ(λ) [b] 0 < t < 0.5 s : vo = Z 0 t 75[0.8δ(λ)] dλ = 60 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. Problems 13–69 0.5 s ≤ t ≤ 1.0 s: vo = Z t−0.5 0 −75[0.8δ(λ)] dλ = −60 V 1s < t < ∞ : vo = 0 [c] Yes, because the circuit has no memory. P 13.70 [a] Vo − Vg Vo s Vo + + =0 5 4 20 (5s + 5)Vo = 4Vg H(s) = 0.8 Vo = ; Vg s+1 h(λ) = 0.8e−λ u(λ) © 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. 13–70 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] 0 ≤ t ≤ 0.5 s; vo = Z 0 t 75(0.8e−λ ) dλ = 60 vo = 60 − 60e−t V, e−λ −1 t 0 0 ≤ t ≤ 0.5 s 0.5 s ≤ t ≤ 1 s: vo = Z t−0.5 (−75)(0.8e 0 = −60 e−λ −1 −λ t−0.5 +60 0 ) dλ + e−λ −1 Z t t−0.5 t t−0.5 = 120e−(t−0.5) − 60e−t − 60 V, 1 s ≤ t ≤ ∞; vo = Z t−0.5 t−1 = −60 (−75)(0.8e−λ ) dλ + e−λ −1 t−0.5 +60 t−1 e−λ −1 75(0.8e−λ ) dλ Z 0.5 s ≤ t ≤ 1 s t t−0.5 75(0.8e−λ ) dλ t t−0.5 = 120e−(t−0.5) − 60e−(t−1) − 60e−t V, 1s ≤ t ≤ ∞ © 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. Problems 13–71 [c] [d] No, the circuit has memory because of the capacitive storage element. P 13.71 vi = 25 sin 10λ [u(λ) − u(λ − π/10)] H(s) = 32 s + 32 h(λ) = 32e−32λ h(t − λ) = 32e−32(t − λ) = 32e−32t e32λ vo = 800e −32t = 800e −32t Z t " e32λ (32 sin 10λ − 10 cos 10λ 322 + 102 0 e32λ sin 10λ dλ = 800e−32t 32t [e (32 sin 10t − 10 cos 10t) + 10] 1124 = 800 [32 sin 10t − 10 cos 10t + 10e−32t ] 1124 t 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. 13–72 CHAPTER 13. The Laplace Transform in Circuit Analysis vo (0.075) = 10.96 V 16s 0.8s 2 1.6 P 13.72 H(s) = = = 0.8 1 − = 0.8 − 40 + 4s + 16s s+2 s+2 s+2 h(λ) = 0.8δ(λ) − 1.6e−2λ u(λ) Z vo = t 0 75[0.8δ(λ) − 1.6e = 60 − 120 e−2λ −2 −2λ ] dλ = Z 0 t 60δ(λ) dλ − 120 Z t 0 e−2λ dλ t 0 = 60 + 60(e−2t − 1) = 60e−2t u(t) V P 13.73 Vo = 5 × 103 Ig (20 × 103 ) 25 × 103 + 2.5 × 106 /s Vo 4000s = H(s) = Ig s + 100 H(s) = 4000 1 − 100 4 × 105 = 4000 − s + 100 s + 100 h(t) = 4000δ(t) − 4 × 105 e−100t vo = Z + 10−3 0 Z (−20 × 10−3 )[4000δ(λ) − 4 × 105 e−100λ] dλ 5×10−3 10−3 (10 × 10−3 )[−4 × 105 e−100λ] dλ Z = −80 + 8000 10−3 0 Z e−100λ dλ − 4000 5×10−3 10−3 e−100λ dλ = −80 − 80(e−0.1 − 1) + 40(e−0.5 − e−0.1) = 40e−0.5 − 120e−0.1 = −84.32 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. Problems 13–73 Alternate: Ig = Z 4×10−3 (10 × 10−3 )e−st dt + 0 Z 6×10−3 4×10−3 (−20 × 10−3 )e−st dt 10 30 −4×10−3 s 20 −6×10−3 s = − e + e × 10−3 s s s 40 −3 −3 [1 − 3e−4×10 s + 2e−6×10 s ] s + 100 Vo = Ig H(s) = −3 s 40 120e−4×10 = − s + 100 s + 100 80e−6×10 s + ] s + 100 vo (t) = 40e−100t − 120e−100(t−4×10 +80e−100(t−6×10 −3 ) −3 −3 ) u(t − 4 × 10−3 ) u(t − 6 × 10−3 ) vo (5 × 10−3 ) = 40e−0.5 − 120e−0.1 + 80(0) = −84.32 V P 13.74 [a] H(s) = = (checks) Vo 1/LC = 2 Vi s + (R/L)s + (1/LC) s2 100 100 = + 20s + 100 (s + 10)2 h(λ) = 100λe−10λ u(λ) 0 ≤ t ≤ 0.5: Z vo = 500 t 0 λe−10λ dλ ( e−10λ = 500 (−10λ − 1) 100 t 0 ) = 5[1 − e−10t(10t + 1)] © 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. 13–74 CHAPTER 13. The Laplace Transform in Circuit Analysis 0.5 ≤ t ≤ ∞: Z vo = 500 t t−0.5 λe−10λ dλ ( e−10λ = 500 (−10λ − 1) 100 t t−0.5 ) = 5e−10t [e5(10t − 4) − 10t − 1] [b] P 13.75 [a] Io = Vo Vo s Vo (s + 50) + = 5 6 10 5 × 10 5 × 106 Vo 5 × 106 = H(s) = Ig s + 50 h(λ) = 5 × 106 e−50λu(λ) © 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. Problems 0 ≤ t ≤ 0.1 s: vo = t Z 0 (50 × 10−6 )(5 × 106 )e−50λ dλ = 250 e−50λ −50 13–75 t 0 = 5(1 − e−50t) V 0.1 s ≤ t ≤ 0.2 s: vo = t−0.1 Z 0 + Z t t−0.1 = −250 h (−50 × 10−6 )(5 × 106 e−50λ dλ) (50 × 10−6 )(5 × 106 e−50λ dλ) e−50λ −50 t−0.1 +250 0 i h e−50λ −50 t t−0.1 = 5 e−50(t−0.1) − 1 − 5 e−50t − e−50(t−0.1) vo = [10e−50(t−0.1) − 5e−50t − 5] V i © 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. 13–76 CHAPTER 13. The Laplace Transform in Circuit Analysis 0.2 s ≤ t ≤ ∞: vo = Z t−0.1 t−0.2 " − 250e−50λ dλ + t−0.1 = 5e −50λ t−0.2 Z t t−0.1 t −5e −50λ t−0.1 250e−50λ dλ # vo = [10e−50(t−0.1) − 5e−50(t−0.2) − 5e−50t ] V [b] Io = Vo s s 5 × 106 Ig = · 5 × 106 5 × 106 s + 50 Io s 50 = H(s) = = 1− Ig s + 50 s + 50 h(λ) = δ(λ) − 50e−50λ 0 < t < 0.1 s: io = Z t 0 (50 × 10−6 )[δ(λ) − 50e−50λ ] dλ = 50 × 10−6 − 25 × 10−3 e−50λ −50 t 0 = 50 × 10−6 + 50 × 10−6 [e−50t − 1] = 50e−50t µA © 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. Problems 13–77 0.1 s < t < 0.2 s: io = Z + t−0.1 (−50 × 10−6 )[δ(λ) − 50e−50λ ] dλ 0 Z t t−0.1 (50 × 10−6 )(−50e−50λ) dλ = −50 × 10−6 + 2.5 × 10−3 e−50λ −50 t−0.1 0 −2.5 × 10−3 e−50λ −50 t t−0.1 = −50 × 10−6 − 50 × 10−6 e−50(t−0.1) + 50 × 10−6 +50 × 10−6 e−50t − 50 × 10−6 e−50(t−0.1) = 50e−50t − 100e−50(t−0.1) µA © 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. 13–78 CHAPTER 13. The Laplace Transform in Circuit Analysis 0.2 s < t < ∞: io = Z + t−0.1 t−0.2 Z (−50 × 10−6 )(−50e−50λ) dλ t t−0.1 (50 × 10−6 )(−50e−50λ) dλ = 50e−50t − 100e−50(t−0.1) + 50e−50(t−0.2) µA [c] At t = 0.1− : vo = 5(1 − e−5 ) = 4.97 V; i100kΩ = 4.97 = 49.66 µA 0.1 .·. io = 50 − 49.66 = 0.34 µA From the solution for io we have io (0.1− ) = 50e−5 = 0.34 µA (checks) At t = 0.1+ : vo (0.1+ ) = vo (0.1− ) = 4.97 V i100kΩ = 49.66 µA .·. io(0.1+ ) = −(50 + 49.66) = −99.66 µA From the solution for io we have io (0.1+ ) = 50e−5 − 100 = 99.66 µA (checks) At t = 0.2− : vo = 10e−5 − 5e−10 − 5 = −4.93 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. Problems 13–79 i100kΩ = 49.33 µA io = −50 + 49.33 = −0.67 µA From the solution for io , vo (0.2− ) = 50e−10 − 100e−5 = −0.67 µA (checks) At t = 0.2+ : vo (0.2+ ) = vo (0.2− ) = −4.93 V; i100kΩ = −49.33 µA io = 0 + 49.33 = 49.33 µA From the solution for io , io (0.2+ ) = 50e−10 − 100e−5 + 50 = 49.33 µA(checks) P 13.76 [a] Y (s) = Z ∞ 0 y(t)e−st dt Z ∞ Y (s) = = Z ∞Z ∞ = Z 0 0 0 e−st ∞ Z h(λ)x(t − λ) dλ dt 0 ∞ 0 e−st h(λ)x(t − λ) dλ dt h(λ) Z ∞ e−st x(t − λ) dt dλ 0 But x(t − λ) = 0 when t < λ. Therefore Y (s) = Let u = t − λ; Z ∞ Y (s) = Z ∞ = Z ∞ = 0 0 0 h(λ) Z ∞ h(λ) 0 du = dt; Z Z ∞ λ e−st x(t − λ) dt dλ u = 0 when t = λ; u = ∞ when t = ∞. ∞ 0 e−s(u+λ) x(u) du dλ h(λ)e−sλ Z ∞ 0 e−su x(u) du dλ h(λ)e−sλ X(s) dλ = H(s) X(s) Note on x(t − λ) = 0, t<λ We are using one-sided Laplace transforms; therefore h(t) and x(t) are assumed zero for t < 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. 13–80 CHAPTER 13. The Laplace Transform in Circuit Analysis a 1 a = · = H(s)X(s) 2 s(s + a) s (s + a)2 [b] F (s) = .·. h(t) = u(t), .·. f(t) = = Z x(t) = at e−atu(t) t 0 (1)aλe −aλ t " e−aλ dλ = a (−aλ − 1) a2 0 1 −at 1 [e (−at − 1) − 1(−1)] = [1 − e−at − ate−at] a a 1 1 −at = − e − te−at u(t) a a Check: F (s) = K0 = 1 ; a f(t) = P 13.77 H(j3) = K0 K1 K2 a = + + 2 2 s(s + a) s (s + a) s+a K1 = −1; K2 = d a ds s s=−a =− 1 a 1 1 − te−at − e−at u(t) a a 4(3 + j3) = 0.42/8.13◦ −9 + j24 + 41 .·. vo (t) = 16.97 cos(3t + 8.13◦ ) 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. Problems P 13.78 Vo = 13–81 50 20 30(s + 3000) − = s + 8000 s + 5000 (s + 5000)(s + 8000) 30 Vo = H(s)Vg = H(s) s .·. H(s) = s(s + 3000) (s + 5000)(s + 8000) H(j6000) = (j6000)(3000 + j6000) = 0.52/66.37◦ (5000 + j6000)(8000 + j6000) .·. vo (t) = 61.84 cos(6000t + 66.37◦ ) V P 13.79 [a] Vp = 0.01s s Vg = Vg 80 + 0.01s s + 8000 Vn = Vp Vn Vn − Vo + + (Vn − Vo )8 × 10−9 s = 0 5000 25,000 5Vn + Vn − Vo + (Vn − Vo )2 × 10−4 s = 0 6Vn + 2 × 10−4 sVn = Vo + 2 × 10−4 sVo 2 × 10−4 Vn (s + 30,000) = 2 × 10−4 Vo (s + 5000) s + 30,000 Vo = Vi = s + 5000 H(s) = s + 30,000 s + 5000 sVg s + 8000 Vo s(s + 30,000) = Vg (s + 5000)(s + 8000) © 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. 13–82 CHAPTER 13. The Laplace Transform in Circuit Analysis [b] vg = 0.6u(t); Vg = 0.6 s Vo = 0.6(s + 30,000) K1 K2 = + (s + 5000)(s + 8000) s + 5000 s + 8000 K1 = 0.6(25,000) = 5; 3000 K2 = 0.6(22,000) = −4.4 −3000 .·. vo(t) = (5e−5000t − 4.4e−8000t)u(t) V [c] Vg = 2 cos 10,000t V H(jω) = . ·. j10,000(30,000 + j10,000) = 2.21/ − 6.34◦ (5000 + j10,000)(8000 + j10,000) vo = 4.42 cos(10,000t − 6.34◦ ) V P 13.80 [a] H(s) = −Zf Zi Zf = (1/Cf ) 108 = s + (1/Rf Cf ) s + 1000 Zi = Ri [s + (1/Ri Ci )] 10,000(s + 400) = s s H(s) = −104 s (s + 400)(s + 1000) [b] H(j400) = −104 (j400) = 6.565/ − 156.8◦ (400 + j400)(1000 + j400) vo (t) = 13.13 cos(400t − 156.8◦ ) V P 13.81 Original charge on C1 ; q1 = V0 C1 The charge transferred to C2 ; q2 = V0 Ce = V0 C1 C2 C1 + C2 The charge remaining on C1; q10 = q1 − q2 = Therefore V2 = q2 V0 C1 = C2 C1 + C2 and V1 = V0 C12 C1 + C2 q10 V0 C1 = C1 C1 + C2 © 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. Problems 13–83 P 13.82 [a] The s-domain circuit is The node-voltage equation is Therefore V = ρR s + (R/Le ) V V ρ V + + = sL1 R sL2 s where Le = L1 L2 L1 + L2 Therefore v = ρRe−(R/Le)t u(t) V [b] I1 = V V ρ[s + (R/L2 )] K0 K1 + = = + R sL2 s[s + (R/Le )] s s + (R/Le ) K0 = ρL1 ; L1 + L2 K1 = Thus we have i1 = [c] I2 = ρL2 L1 + L2 ρ [L1 + L2 e−(R/Le)t]u(t) A L1 + L2 V (ρR/L2 ) K2 K3 = = + sL2 s[s + (R/Le )] s s + (R/Le ) K2 = ρL1 ; L1 + L2 Therefore i2 = K3 = −ρL1 L1 + L2 ρL1 [1 − e−(R/Le)t ]u(t) L1 + L2 [d] λ(t) = L1 i1 + L2i2 = ρL1 P 13.83 [a] As R → ∞, v(t) → ρLe δ(t) since the area under the impulse generating function is ρLe . ρL1 i1 (t) → u(t) A as R → ∞ L1 + L2 i2 (t) → ρL1 u(t) A as R → ∞ L1 + L2 [b] The s-domain circuit is V V ρ + = ; sL1 sL2 s therefore V = ρL1 L2 = ρLe L1 + L2 © 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. 13–84 CHAPTER 13. The Laplace Transform in Circuit Analysis Therefore v(t) = ρLe δ(t) I1 = I2 = V ρL1 = sL2 L1 + L2 Therefore i1 = i2 = 1 s ρL1 u(t) A L1 + L2 P 13.84 [a] Vo = = 5 · 16 × 10−3 s 200 + 20 × 10−3 s 4s 40,000 =4− s + 10,000 s + 10,000 vo (t) = 4δ(t) − 40,000e−10,000tu(t) V [b] At t = 0 the voltage impulse establishes a current in the inductors; thus 103 iL (0) = 20 Z 0+ 0− 5δ(t) dt = 250 A It follows that since iL (0− ) = 0 that diL (0) = 250δ(t) dt .·. vo(0) = (16 × 10−3 )(250δ(t)) = 4δ(t) This agrees with our solution. At t = 0+ our circuit is .·. iL(t) = 250e−t/τ A, t ≥ 0+ τ = L/R = 0.1 ms .·. iL(t) = 250e−10,000t A, t ≥ 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. Problems vo (t) = 16 × 10−3 diL = −40,000e−10,000t V, dt 13–85 t ≥ 0+ which agrees with our solution. 1/C1 25 × 1010 P 13.85 [a] Z1 = = Ω s + 1/R1 C1 s + 20 × 104 1/C2 6.25 × 1010 Z2 = = Ω s + 1/R2 C2 s + 12,500 V0 V0 − 10/s + =0 Z2 Z1 10 (s + 20 × 104 ) V0 (s + 12,500) V0 (s + 20 × 104 ) + = 6.25 × 1010 25 × 1010 s 25 × 1010 V0 = 2(s + 200,000) K1 K2 = + s(s + 50,000) s s + 50,000 K1 = 2(200,000) =8 50,000 K2 = 2(150,000) = −6 −50,000 .·. vo = [8 − 6e−50,000t]u(t) V [b] I0 = V0 2(s + 200,000)(s + 12,500) = Z2 s(s + 50,000)6.25 × 1010 162,500s + 25 × 108 1+ s(s + 50,000) = 32 × 10 " = 32 × 10−12 " −12 K1 = 50,000; K1 K2 + 1+ s s + 50,000 # # K2 = 112,500 io = 32δ(t) + [1.6 × 106 + 3.6 × 106 e−50,000t]u(t) pA [c] When Z1 = C1 = 64 pF 156.25 × 108 Ω s + 12,500 V0 (s + 12,500) V0 (s + 12,500) 10 (s + 12,500) + = 8 8 625 × 10 156.25 × 10 s 156.25 × 108 .·. V0 + 4V0 = 40 s © 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. 13–86 CHAPTER 13. The Laplace Transform in Circuit Analysis V0 = 8 s vo = 8u(t) V I0 = V0 8 (s + 12,500) 12,500 −12 = = 128 × 10 1 + Z2 s 6.25 × 1010 s io (t) = 128δ(t) + 1.6 × 106 u(t) pA P 13.86 Let a = 1 1 = R1C1 R2C2 Then Z1 = 1 C1 (s + a) and Z2 = 1 C2(s + a) Vo 10/s Vo + = Z2 Z1 Z1 Vo C2(s + a) + V0 C1 (s + a) = (10/s)C1 (s + a) 10 C1 Vo = s C1 + C2 Thus, vo is the input scaled by the factor C1 . C1 + C2 P 13.87 [a] For t < 0: Req = 0.8 kΩk4 kΩk16 kΩ = 0.64 kΩ; i1 (0− ) = 3200 = 0.8 A; 4000 i2(0− ) = v = 5(640) = 3200 V 3200 = 0.2 A 1600 [b] For t > 0: i1 + i2 = 0 8(∆i1 ) = 2(∆i2 ) i1 (0− ) + ∆i1 + i2(0− ) + ∆i2 = 0; ∆i2 = −0.8 A; therefore ∆i1 = −0.2 A i1 (0+ ) = 0.8 − 0.2 = 0.6 A [c] i2 (0− ) = 0.2 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. Problems 13–87 [d] i2(0+ ) = 0.2 − 0.8 = −0.6 A [e] The s-domain equivalent circuit for t > 0 is I1 = 0.006 0.6 = 0.01s + 20,000 s + 2 × 106 6 i1 (t) = 0.6e−2×10 t u(t) A 6 [f] i2(t) = −i1(t) = −0.6e−2×10 t u(t) A [g] V = −0.0064 + (0.008s + 4000)I1 = = −1.6 × 10−3 − 7200 s + 2 × 106 −0.0016(s + 6.5 × 106 ) s + 2 × 106 6 v(t) = [−1.6 × 10−3 δ(t)] − [7200e−2×10 t u(t)] V P 13.88 [a] For t < 0, 0.5v1 = 2v2 ; v1 + v2 = 100; therefore v1 = 4v2 therefore v1(0− ) = 80 V [b] v2(0− ) = 20 V [c] v3 (0− ) = 0 V [d] For t > 0: I= 100/s × 10−6 = 32 × 10−6 3.125/s i(t) = 32δ(t) µA © 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. 13–88 CHAPTER 13. The Laplace Transform in Circuit Analysis 106 0+ 32 × 10−6 δ(t) dt + 80 = −64 + 80 = 16 V 0.5 0− 106 Z 0+ [f] v2(0+ ) = − 32 × 10−6 δ(t) dt + 20 = −16 + 20 = 4 V 2 0− 20 0.625 × 106 · 32 × 10−6 = [g] V3 = s s Z [e] v1(0+ ) = − v3(0+ ) = 20 V v3 (t) = 20u(t) V; Check: v1(0+ ) + v2(0+ ) = v3(0+ ) P 13.89 [a] 0.5 106 · 50,000 + 5 × 106 /s s Vo = 500,000 10 = 50,000s + 5 × 106 s + 100 vo = 10e−100tu(t) V [b] At t = 0 the current in the 1 µF capacitor is 10δ(t) µA . ·. + 6 vo (0 ) = 10 Z 0+ 0− 10 × 10−6 δ(t) dt = 10 V After the impulsive current has charged the 1 µF capacitor to 10 V it discharges through the 50 kΩ resistor. Ce = C1 C2 0.25 = = 0.2 µF C1 + C2 1.25 τ = (50,000)(0.2 × 10−6 ) = 10−2 1 = 100 (checks) τ Note – after the impulsive current passes the circuit becomes © 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. Problems 13–89 The solution for vo in this circuit is also vo = 10e−100tu(t) V P 13.90 [a] After making a source transformation, the circuit is as shown. The impulse current will pass through the capacitive branch since it appears as a short circuit to the impulsive current, + 6 Therefore vo(0 ) = 10 Z 0+ 0− " # δ(t) dt = 1000 V 1000 Therefore wC = (0.5)Cv 2 = 0.5 J [b] iL(0+ ) = 0; therefore wL = 0 J Vo Vo [c] Vo (10−6 )s + + = 10−3 250 + 0.05s 1000 Therefore 1000(s + 5000) Vo = 2 s + 6000s + 25 × 106 = K1 K1∗ + s + 3000 − j4000 s + 3000 + j4000 K1 = 559.02/ − 26.57◦ ; K1∗ = 559.02/26.57◦ vo = [1118.03e−3000t cos(4000t − 26.57◦ )]u(t) V [d] The s-domain circuit is Vo s Vo Vo + + = 10−3 106 250 + 0.05s 1000 © 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. 13–90 CHAPTER 13. The Laplace Transform in Circuit Analysis Note that this equation is identical to that derived in part [c], therefore the solution for Vo will be the same. P 13.91 [a] 20 = sI1 − 0.5sI2 0 = −0.5sI1 + s + s ∆= 3 I2 s −0.5s −0.5s (s + 3/s) 20 −0.5s N1 = I1 = = = s2 + 3 − 0.25s2 = 0.75(s2 + 4) = 20s + 0 (s + 3/s) 60 20s2 + 60 20(s2 + 3) = = s s s N1 20(s2 + 3) 80 s2 + 3 = = · ∆ s(0.75)(s2 + 4) 3 s(s2 + 4) K0 K1 K1∗ + + s s − j2 s + j2 " 80 3 K0 = = 20; 3 4 20 cos 2t u(t) A 3 .·. i1 = 20 + s [b] N2 = # 80 −4 + 3 10 K1 = = /0◦ 3 (j2)(j4) 3 20 = 10s −0.5s 0 N2 10s 40 s K1 K1∗ I2 = = = = + ∆ 0.75(s2 + 4) 3 s2 + 4 s − j2 s + j2 40 K1 = 3 j2 j4 ! = 20 ◦ /0 3 40 i2 = cos 2t u(t) A 3 © 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. Problems 13–91 3 3 40 s 40 K1 K1∗ [c] V0 = I2 = = = + s s 3 s2 + 4 s2 + 4 s − j2 s + j2 K1 = 40 = −j10 = 10/90◦ j4 vo = 20 cos(2t − 90◦ ) = 20 sin 2t vo = [20 sin 2t]u(t) V [d] Let us begin by noting i1 jumps from 0 to (80/3) A between 0− and 0+ and in this same interval i2 jumps from 0 to (40/3) A. Therefore in the derivatives of i1 and i2 there will be impulses of (80/3)δ(t) and (40/3)δ(t), respectively. Thus di1 80 40 = δ(t) − sin 2t A/s dt 3 3 di2 40 80 = δ(t) − sin 2t A/s dt 3 3 From the circuit diagram we have 20δ(t) = 1 = di1 di2 − 0.5 dt dt 40 20δ(t) 40 80 δ(t) − sin 2t − + sin 2t 3 3 3 3 = 20δ(t) Thus our solutions for i1 and i2 are in agreement with known circuit behavior. Let us also note the impulsive voltage will impart energy into the circuit. Since there is no resistance in the circuit, the energy will not dissipate. Thus the fact that i1, i2, and vo exist for all time is consistent with known circuit behavior. Also note that although i1 has a dc component, i2 does not. This follows from known transformer behavior. Finally we note the flux linkage prior to the appearance of the impulsive voltage is zero. Now since v = dλ/dt, the impulsive voltage source must be matched to an instantaneous change in flux linkage at t = 0+ of 20. For the given polarity dots and reference directions of i1 and i2 we have λ(0+ ) = L1 i1(0+ ) + Mi1 (0+ ) − L2 i2(0+ ) − Mi2 (0+ ) 80 80 40 40 λ(0 ) = 1 + 0.5 −1 − 0.5 3 3 3 3 + = 120 60 − = 20 (checks) 3 3 © 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. 13–92 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.92 [a] The circuit parameters are 1202 1202 288 1202 = 12 Ω Rb = = 8Ω Xa = = Ω 1200 1800 350 7 The branch currents are 120/0◦ 120/0◦ 35 35 I1 = I2 = = 10/0◦ A(rms) = −j = / − 90◦ A(rms) 12 j1440/35 12 12 Ra = I3 = 120/0◦ = 15/0◦ A(rms) 8 .·. Io = I1 + I2 + I3 = 25 − j 35 = 25.17/ − 6.65◦ A(rms) 12 Therefore, i2 = 35 √ 2 cos(ωt − 90◦ ) A 12 Thus, i2 (0− ) = i2(0+ ) = 0 A and and √ iL = 25.17 2 cos(ωt − 6.65◦ ) A √ iL (0− ) = iL (0+ ) = 25 2 A [b] Begin by using the s-domain circuit in Fig. 13.60 to solve for V0 symbolically. Write a single node voltage equation: V0 − (Vg + L` Io ) V0 V0 + + =0 sL` Ra sLa .·. V0 = (Ra /L` )Vg + Io Ra s + [Ra (La + L` )]/La L` √ where L` = 1/120π H, La = 12/35π H, Ra = 12 Ω, and I0Ra = 300 2 V. Thus, √ √ √ 1440π(122.92 2s − 3000π 2) 300 2 V0 = + (s + 1475π)(s2 + 14,400π 2 ) s + 1475π √ K1 K2 K2∗ 300 2 = + + + s + 1475π s − j120π s + j120π s + 1475π The coefficients are √ K1 = −121.18 2 V √ K2 = 61.03 2/6.85◦ V √ K2∗ = 61.03 2/ − 6.85◦ © 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. Problems 13–93 √ √ Note that K1 + 300 2 = 178.82 2 V. Thus, the inverse transform of V0 is √ √ v0 = 178.82 2e−1475πt + 122.06 2 cos(120πt + 6.85◦ ) V Initially, √ √ √ v0 (0+ ) = 178.82 2 + 122.06 2 cos 6.85◦ = 300 2 V √ Note that at t = 0+ the initial value of iL , which is √ 25 2 A, exists√in the 12 Ω resistor Ra . Thus, the initial value of V0 is (25 2)(12) = 300 2 V. [c] The phasor domain equivalent circuit has a j1 Ω inductive impedance in series with the parallel combination of a 12 Ω resistive impedance and a j1440/35 Ω inductive impedance (remember that ω = 120π rad/s). Note that Vg = 120/0◦ + (25.17/ − 6.65◦ )(j1) = 125.43/11.50◦ V(rms). The node voltage equation in the phasor domain circuit is V0 − 125.43/11.50◦ V0 35V0 + + =0 j1 12 1440 .·. V0 = 122.06/6.85◦ V(rms) √ Therefore, v0 = 122.06 2 cos(120πt + 6.85◦ ) V, agreeing with the steady-state component of the result in part (b). [d] A plot of v0 , generated in Excel, is shown below. P 13.93 [a] At t = 0− the phasor domain equivalent circuit is I1 = −j120 = −j10 = 10/ − 90◦ A (rms) 12 I2 = 35 35 −j120(35) = − = /180◦ A (rms) j1440 12 12 I3 = −j120 = −j15 = 15/ − 90◦ A (rms) 8 © 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. 13–94 CHAPTER 13. The Laplace Transform in Circuit Analysis IL = I1 + I2 + I3 = − 35 − j25 = 25.17/ − 96.65◦ A (rms) 12 √ iL = 25.17 2 cos(120πt − 96.65◦ )A √ iL (0− ) = iL (0+ ) = −2.92 2A 35 √ 2 cos(120πt + 180◦ )A 12 √ 35 √ i2 (0− ) = i2(0+ ) = − 2 = −2.92 2A 12 i2 = Vg = Vo + j1IL 35 12 25 − j122.92 = 125.43/ − 78.50◦ V (rms) √ 125.43 2 cos(120πt − 78.50◦ )V √ 125.43 2[cos 120πt cos 78.50◦ + sin 120πt sin 78.50◦ ] √ √ 25 2 cos 120πt + 122.92 2 sin 120πt Vg = −j120 + 25 − j = vg = = = √ √ 25 2s + 122.92 2(120π) · . . Vg = s2 + (120π)2 s-domain circuit: where Ll = 1 H; 120π La = √ iL (0) = −2.92 2 A; 12 H; 35π Ra = 12 Ω √ i2(0) = −2.92 2 A The node voltage equation is 0= Vo − (Vg + iL (0)Ll ) Vo Vo + i2(0)La + + sLl Ra sLa Solving for Vo yields Vo = Vg Ra /Ll Ra [iL(0) − i2 (0)] + [s + Ra (Ll + La )/La Ll ] [s + Ra (Ll + La )/Ll La ] © 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. Problems 13–95 Ra = 1440π Ll 1 12 + 35π ) 12( 120π Ra (Ll + La ) = = 1475π 12 1 Ll La ( 35π )( 120π ) √ √ iL (0) − i2 (0) = −2.92 2 + 2.92 2 = 0 √ √ 1440π[25 2s + 122.92 2(120π)] · . . Vo = (s + 1475π)[s2 + (120π)2 ] K1 K2 K2∗ = + + s + 1475π s − j120π s + j120π √ √ K1 = −14.55 2 K2 = 61.03 2/ − 83.15◦ √ √ .·. vo(t) = −14.55 2e−1475πt + 122.06 2 cos(120πt − 83.15◦ )V Check: √ vo (0) = (−14.55 + 14.55) 2 = 0 [b] © 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. 13–96 CHAPTER 13. The Laplace Transform in Circuit Analysis © 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. Problems 13–97 √ [c] In Problem 13.92, the line-to-neutral voltage spikes at 300 2 V. Here the line-to-neutral voltage has no spike. Thus the amount of voltage disturbance depends on what part of the cycle the sinusoidal steady-state voltage is switched. © 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. 13–98 CHAPTER 13. The Laplace Transform in Circuit Analysis P 13.94 [a] First find Vg before Rb is disconnected. The phasor domain circuit is 120/θ◦ 120/θ◦ 120/θ◦ + + Ra Rb jXa 120/θ◦ = [(Ra + Rb )Xa = jRa Rb ] Ra Rb Xa IL = Since Xl = 1 Ω we have Vg = 120/θ◦ + 120/θ◦ [Ra Rb + j(Ra + Rb )Xa ] Ra Rb Xa Ra = 12 Ω; Rb = 8 Ω; Vg = Xa = 1440 Ω 35 120/θ◦ (1475 + j300) 1400 25 ◦ /θ (59 + j12) = 125.43/(θ + 11.50)◦ 12 √ vg = 125.43 2 cos(120πt + θ + 11.50◦ )V = Let β = θ + 11.50◦ . Then √ vg = 125.43 2(cos 120πt cos β − sin 120πt sin β)V Therefore √ 125.43 2(s cos β − 120π sin β) Vg = s2 + (120π)2 The s-domain circuit becomes where ρ1 = iL(0+ ) and ρ2 = i2(0+ ). The s-domain node voltage equation is Vo − (Vg + ρ1 Ll ) Vo Vo + ρ2 La + + =0 sLl Ra sLa © 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. Problems 13–99 Solving for Vo yields Vo = Vg Ra /Ll + (ρ1 − ρ2 )Ra [s + (La +Ll )Ra ] La Ll Substituting the numerical values Ll = 1 H; 120π La = 12 H; 35π Ra = 12 Ω; Rb = 8 Ω; gives Vo = 1440πVg + 12(ρ1 − ρ2 ) (s + 1475π) Now determine the values of ρ1 and ρ2 . ρ1 = iL(0+ ) IL = ρ2 = i2 (0+ ) and 120/θ◦ [(Ra + Rb )Xa − jRa Rb ] Ra Rb Xa " # 120/θ◦ (20)(1440) = − j96 96(1440/35) 35 = 25.17/(θ − 6.65)◦ A(rms) √ .·. iL = 25.17 2 cos(120πt + θ − 6.65◦ )A √ iL (0+ ) = ρ1 = 25.17 2 cos(θ − 6.65◦ )A √ √ .·. ρ1 = 25 2 cos θ + 2.92 2 sin θA 120/θ◦ 35 I2 = = /(θ − 90)◦ j(1440/35) 12 35 √ 2 cos(120πt + θ − 90◦ )A 12 √ 35 √ ρ2 = i2(0+ ) = 2 sin θ = 2.92 2 sin θA 12 √ .·. ρ1 = ρ2 = 25 2 cos θ √ (ρ1 − ρ2 )Ra = 300 2 cos θ √ 1440π 300 2 cos θ . ·. V o = · Vg + s + 1475π s + 1475π √ √ " # 1440π 125.43 2(s cos β − 120π sin β) 300 2 cos θ = + s + 1475π s2 + 14,400π 2 s + 1475π √ K1 + 300 2 cos θ K2 K2∗ = + + s + 1475π s − j120π s + j120π i2 = © 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. 13–100 CHAPTER 13. The Laplace Transform in Circuit Analysis Now √ (1440π)(125.43 2)[−1475π cos β − 120π sin β] K1 = 14752 π 2 + 14,400π 2 √ −1440(125.43 2)[1475 cos β + 120 sin β] = 14752 + 14,000 Since β = θ + 11.50◦ , K1 reduces to √ √ K1 = −121.18 2 cos θ + 14.55 2 sin θ From the partial fraction expansion√for Vo we see vo (t) will go directly into steady state when K1 = −300 2 cos θ. It follow that √ √ 14.55 2 sin θ = −178.82 2 cos θ or tan θ = −12.29 Therefore, θ = −85.35◦ [b] When θ = −85.35◦ , β = −73.85◦ √ 1440π(125.43 2)[−120π sin(−73.85◦ ) + j120π cos(−73.85◦ ) .·. K2 = (1475π + j120π)(j240π) √ 720 2(120.48 + j34.88) = −120 + j1475 √ = 61.03 2/ − 78.50◦ √ .·. vo = 122.06 2 cos(120πt − 78.50◦ ) V t > 0 = 172.61 cos(120πt − 78.50◦ ) V t > 0 [c] vo1 = 169.71 cos(120πt − 85.35◦ )V vo2 = 172.61 cos(120πt − 78.50◦ )V t<0 t>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. Problems 13–101 © 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. 13–102 CHAPTER 13. The Laplace Transform in Circuit Analysis © 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. 14 Introduction to Frequency-Selective Circuits Assessment Problems AP 14.1 fc = 8 kHz, ωc = 1 ; RC ωc = 2πfc = 16π krad/s R = 10 kΩ; 1 1 .·. C = = = 1.99 nF ωc R (16π × 103 )(104 ) AP 14.2 [a] ωc = 2πfc = 2π(2000) = 4π krad/s L= R 5000 = = 0.40 H ωc 4000π [b] H(jω) = ωc 4000π = ωc + jω 4000π + jω When ω = 2πf = 2π(50,000) = 100,000π rad/s H(j100,000π) = 4000π 1 = = 0.04/87.71◦ 4000π + j100,000π 1 + j25 .·. |H(j100,000π)| = 0.04 [c] .·. θ(100,000π) = −87.71◦ AP 14.3 ωc = R 5000 = = 1.43 Mrad/s L 3.5 × 10−3 © 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 14–1 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. 14–2 CHAPTER 14. Introduction to Frequency-Selective Circuits 1 106 106 = = = 10 krad/s RC R 100 106 [b] ωc = = 200 rad/s 5000 106 [c] ωc = = 33.33 rad/s 3 × 104 AP 14.4 [a] ωc = AP 14.5 Let Z represent the parallel combination of (1/SC) and RL . Then RL (RL Cs + 1) Z= Thus H(s) = = where K = AP 14.6 Z RL = R+Z R(RL Cs + 1) + RL (1/RC) s+ R+RL RL 1 RC = (1/RC) s+ 1 K 1 RC RL R + RL ωo2 = 1 LC Q= ωo ωo = β R/L so L = 1 ωo2 C = so R = 1 (24π × 103 )2 (0.1 × 10−6 ) = 1.76 mH ωo L (24π × 103 )(1.76 × 10−3 ) = = 22.10 Ω Q 6 AP 14.7 ωo = 2π(2000) = 4000π rad/s; β = 2π(500) = 1000π rad/s; β= 1 RC so C = ωo2 = 1 LC so L = ωo2 = 1 LC so L = AP 14.8 β= 1 RC so R = R = 250 Ω 1 1 = = 1.27 µF βR (1000π)(250) 1 106 = = 4.97 mH ωo2 C (4000π)2 (1.27) 1 ωo2 C = 1 (104 π)2 (0.2 × 10−6 ) = 5.07 mH 1 1 = = 3.98 kΩ βC 400π(0.2 × 10−6 ) © 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. Problems AP 14.9 ωo2 = 1 LC Q= fo 5 × 103 = = 25 = ωo RC β 200 so L = 1 ωo2 C = 1 (400π)2 (0.2 × 10−6 ) 14–3 = 31.66 mH 25 Q .·. R = = = 9.95 kΩ ωo C (400π)(0.2 × 10−6 ) AP 14.10 ωo = 8000π rad/s C = 500 nF ωo2 = 1 LC Q= ωo ωo L 1 = = β R ωo CR .·. R = so L = 1 ωo2 C = 3.17 mH 1 1 = = 15.92 Ω ωo CQ (8000π)(500)(5 × 10−9 ) AP 14.11 ωo = 2πfo = 2π(20,000) = 40π krad/s; Q= ωo ωo = β (R/L) ωo2 = 1 LC so L = so C = R = 100 Ω; Q=5 QR 5(100) = = 3.98 mH ωo (40π × 103 ) 1 1 = = 15.92 nF ωo2 L (40π × 103 )2 (3.98 × 10−3 ) © 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. 14–4 CHAPTER 14. Introduction to Frequency-Selective Circuits Problems P 14.1 [a] ωc = R 127 = = 12.7 krad/s L 10 × 10−3 .·. fc = [b] H(s) = ωc 12,700 = = 2021.27 Hz 2π 2π 12,700 ωc = s + ωc s + 12,700 12,700 12,700 + jω H(jω) = H(jωc ) = 12,700 = 0.7071/ − 45◦ 12,700 + j12,700 H(j0.2ωc ) = H(j5ωc ) = 12,700 = 0.981/ − 11.31◦ 12,700 + j2540 12,700 = 0.196/ − 78.69◦ 12,700 + j63,500 [c] vo (t)|ωc = 7.07 cos(12,700t − 45◦ ) V vo (t)|0.2ωc = 9.81 cos(2540t − 11.31◦ ) V vo (t)|5ωc = 1.96 cos(63,500t − 78.69◦ ) V P 14.2 [a] R = 10,000π rad/s L R = (0.001)(10,000)(π) = 31.42 Ω [b] Re = 31.42k68 = 21.49 Ω ωloaded = Re = 21,488.34 rad/s L .·. floaded = 3419.98 Hz [c] The 33 Ω resistor in Appendix H is closest to the desired value of 31.42 Ω. Therefore, ωc = 33 krad/s so fc = 5252.11 Hz P 14.3 [a] H(s) = Vo R (R/L) = = Vi sL + R + Rl s + (R + Rl )/L © 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. Problems 14–5 (R/L) [b] H(jω) = R+R l + jω L |H(jω)| = r (R/L) R+Rl 2 L + ω2 |H(jω)|max occurs when ω = 0 R R + Rl R R/L [d] |H(jωc )| = √ = r 2(R + Rl ) R+Rl 2 [c] |H(jω)|max = L + ωc2 R + Rl 2 ; .·. ωc = (R + Rl )/L L 127 + 75 = 20,200 rad/s [e] ωc = 0.01 12,700 H(jω) = 20,200 + jω .·. ωc2 = H(j0) = 0.6287 0.6287 √ / − 45◦ = 0.4446/ − 45◦ 2 12,700 H(j6060) = = 0.6022/ − 16.70◦ 20,200 + j6060 H(j20,200) = H(j60,600) = P 14.4 12,700 = 0.1988/ − 71.57◦ 20,200 + j60,600 1 1 = = 10 krad/s 3 RC (10 )(100 × 10− 9) ωc fc = = 1591.55 Hz 2π ωc 10,000 [b] H(jω) = = s + ωc s + 10,000 [a] ωc = H(jω) = H(jωc ) = 10,000 10,000 + jω 10,000 = 0.7071/ − 45◦ 10,000 + j10,000 H(j0.1ωc ) = 10,000 = 0.9950/ − 5.71◦ 10,000 + j1000 H(j10ωc ) = 10,000 = 0.0995/ − 84.29◦ 10,000 + j100,000 © 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. 14–6 CHAPTER 14. Introduction to Frequency-Selective Circuits [c] vo (t)|ωc = 200(0.7071) cos(10,000t − 45◦ ) = 141.42 cos(10,000t − 45◦ ) mV vo (t)|0.1ωc = 200(0.9950) cos(1000t − 5.71◦ ) = 199.01 cos(1000t − 5.71◦ ) mV vo (t)|10ωc = 200(0.0995) cos(100,000t − 84.29◦ ) = 19.90 cos(100,000t − 84.29◦ ) mV P 14.5 [a] fc = [b] ωc 50,000 50 = = × 103 = 7957.75 Hz 2π 2π 2π 1 = 50 × 103 RC R= 1 (50 × 103 )(0.5 × 10−6 ) = 40 Ω [c] With a load resistor added in parallel with the capacitor the transfer function becomes RL k(1/sC) RL /sC H(s) = = R + RL k(1/sC) R[RL + (1/sC)] + RL /sC = 1/RC RL = RRL sC + R + RL s + [(R + RL )/RRL C] This transfer function is in the form of a low-pass filter, with a cutoff frequency equal to the quantity added to s in the denominator. Therefore, R + RL 1 R ωc = = 1+ RRL C RC RL . ·. R = 0.05 RL [d] H(j0) = P 14.6 .·. RL = 20R = 800 Ω RL 800 = = 0.9524 R + RL 840 [a] ωc = 2π(100) = 628.32 rad/s 1 1 1 [b] ωc = so R = = = 338.63 Ω RC ωc C (628.32)(4.7 × 10−6 ) [c] © 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. Problems 14–7 Vo 1/sC 1/RC 628.32 = = = Vi R + 1/sC s + 1/RC s + 628.32 Vo (1/sC)kRL 1/RC 628.32 [e] H(s) = = = = R + RL Vi R + (1/sC)kRL s + 2(628.32) s+ 1/RC RL [f] ωc = 2(628.32) = 1256.64 rad/s [d] H(s) = [g] H(0) = 1/2 P 14.7 [a] Let Z = RL RL (1/SC) = RL + 1/SC RL Cs + 1 Then H(s) = = = [b] |H(jω)| = q Z Z +R RL RRL Cs + R + RL (1/RC) R + RL s+ RRL C (1/RC) ω 2 + [(R + RL )/RRL C]2 |H(jω)| is maximum at ω = 0. RL R + RL RL (1/RC) [d] |H(jωc )| = √ =q 2(R + RL ) ωC2 + [(R + RL )/RRL C]2 [c] |H(jω)|max = .·. ωc = [e] ωc = R + RL 1 = (1 + (R/RL )) RRL C RC 1 (103 )(10−7 ) H(j0) = [1 + (103 /104 )] = 10,000(1 + 0.1) = 11,000 rad/s 10,000 = 0.9091/0◦ 11,000 H(jωc ) = 10,000 = 0.6428/ − 45◦ 11,000 + j11,000 H(j0.1ωc ) = 10,000 = 0.9046/ − 5.71◦ 11,000 + j1100 H(j10ωc ) = 10,000 = 0.0905/ − 84.29◦ 11,000 + j110,000 © 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. 14–8 P 14.8 CHAPTER 14. Introduction to Frequency-Selective Circuits [a] ZL = jωL = j0L = 0 At ω = 0, so it is a short circuit. Vo = Vi [b] ZL = jωL = j∞L = ∞ At ω = ∞, so it is an open circuit. Vo = 0 [c] This is a low pass filter, with a gain of 1 at low frequencies and a gain of 0 at high frequencies. Vo R R/L [d] H(s) = = = Vi R + sL s + R/L R 330 [e] ωc = = = 33 krad/s L 0.01 P 14.9 R RL Vo RkRL L R + RL [a] H(s) = = = R RL Vi RkRL + sL s+ L R + RL R R RL [b] ωc(U L) = ; ωc(L) = so the cutoff frequencies are different. L L R + RL H(0)(U L) = 1; H(0)(L) = 1 so the passband gains are the same. [c] ωc(U L) = 33,000 rad/s ωc(L) = 33,000 − 0.05(33,000) = 31,350 rad/s 31,350 = 330 RL 0.01 330 + RL so RL = 0.95 330 + RL .·. 0.05RL = 313.5 so RL ≥ 6270 Ω P 14.10 [a] 1 1 = = 4000 rad/s 3 RC (50 × 10 )(5 × 10−9 ) fc = 4000 = 636.62 Hz 2π [b] H(s) = s s + ωc .·. H(jωc ) = H(j4000) = H(jω) = jω 4000 + jω j4000 = 0.7071/45◦ 4000 + j4000 H(j0.2ωc ) = H(j800) = H(j5ωc ) = H(j20ωc ) = j800 = 0.1961/78.69◦ 4000 + j800 j20,000 = 0.9806/11.31◦ 4000 + j20,000 © 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. Problems 14–9 [c] vo (t)|ωc = (0.7071)(500) cos(4000t + 45◦ ) = 353.55 cos(4000t + 45◦ ) mV vo (t)|0.2ωc = (0.1961)(500) cos(800t + 78.60◦ ) = 98.06 cos(800t + 78.69◦ ) mV vo (t)|5ωc = (0.9806)(500) cos(20,000t + 11.31◦ ) = 490.29 cos(20,000t + 11.31◦ ) mV P 14.11 [a] H(s) = Vo R = Vi R + Rc + (1/sC) R s · R + Rc [s + (1/(R + Rc )C)] jω R [b] H(jω) = · R + Rc jω + (1/(R + Rc )C) = |H(jω)| = R ω ·q R + Rc ω 2 + (R+R1c )2 C 2 The magnitude will be maximum when ω = ∞. R [c] |H(jω)|max = R + Rc Rωc q [d] |H(jωc )| = (R + Rc ) ωc2 + [1/(R + Rc )C]2 .·. |H(jω)| = √ .·. ωc2 = or ωc = [e] ωc = R 2(R + Rc ) when 1 (R + Rc )2 C 2 1 (R + Rc )C 1 (62.5 × 103 )(5 × 10−9 ) = 3200 rad/s R 50 = = 0.8 R + Rc 62.5 . ·. H(jω) = H(jωc ) = 0.8jω 3200 + jω (0.8)j3200 = 0.5657/45◦ 3200 + j3200 © 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. 14–10 CHAPTER 14. Introduction to Frequency-Selective Circuits H(j0.2ωc ) = H(j5ωc ) = (0.8)j640 = 0.1569/78.69◦ 3200 + j640 (0.8)j16,000 = 0.7845/11.31◦ 3200 + j16,000 P 14.12 [a] ωc = 2π(500) = 3141.59 rad/s 1 1 1 [b] ωc = so R = = = 1.45 MΩ RC ωc C (3141.59)(220 × 10−12 ) [c] Vo R s s = = = Vi R + 1/sC s + 1/RC s + 3141.59 Vo RkRL s s [e] H(s) = = = = R + RL Vi RkRL + (1/sC) s + 2(3141.59) 1/RC s+ RL [f] ωc = 2(3141.59) = 6283.19 rad/s [d] H(s) = [g] H(∞) = 1 1 = 2π(300) = 600π rad/s RC 1 1 . ·. R = = = 5305.16 Ω = 5.305 kΩ ωc C (600π)(100 × 10−9 ) P 14.13 [a] ωc = [b] Re = 5305.16k47,000 = 4767.08 Ω ωc = fc = 1 1 = = 2097.7 rad/s Re C (4767.08)(100 × 10−9 ) ωc 2097.7 = = 333.86 Hz 2π 2π P 14.14 [a] R = ωc L = (1500 × 103 )(100 × 10−6 ) = 150 Ω (a value from Appendix H) [b] With a load resistor in parallel with the inductor, the transfer function becomes sLkRL sLRL s[RL /(R + RL )] H(s) = = = R + sLkRL R(sL + RL ) + sLRL s + [RRL /(R + RL )] This transfer function is in the form of a high-pass filter whose cutoff frequency is the quantity added to s in the denominator. Thus, ωc = RRL L(R + RL ) © 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. Problems 14–11 Substituting in the values of R and L from part (a), we can solve for the value of load resistance that gives a cutoff frequency of 1200 krad/s: 150RL = 1200 × 103 100 × 10−6 (150 + RL ) so RL = 600 Ω The smallest resistor from Appendix H that is larger than 600 Ω is 680 Ω. P 14.15 [a] For ω = 0, the inductor behaves as a short circuit, so Vo = 0. For ω = ∞, the inductor behaves as an open circuit, so Vo = Vi . Thus, the circuit is a high-pass filter. sL s s [b] H(s) = = = R + sL s + R/L s + 15,000 R = 15,000 rad/s [c] ωc = L jR/L j 1 [d] |H(jR/L)| = = =√ jR/L + R/L j +1 2 RL s Vo RL ksL R + RL P 14.16 [a] H(s) = = = R RL Vi R + RL ksL s+ L R + RL = 1 s 2 1 s + (15,000) 2 R RL 1 [b] ωc = = (15,000) = 7500 rad/s L R + RL 2 1 [c] ωc(L) = ωc(U L) 2 [d] The gain in the passband is also reduced by a factor of 1/2 for the loaded filter. P 14.17 By definition Q = ωo /β therefore β = ωo /Q. Substituting this expression into Eqs. 14.34 and 14.35 yields ωo ωc1 = − + 2Q v u u t v u ωo 2Q ωo u ωo ωc2 = +t 2Q 2Q !2 !2 + ωo2 + ωo2 © 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. 14–12 CHAPTER 14. Introduction to Frequency-Selective Circuits Now factor ωo out to get ωc1 = ωo − 1 + 2Q + v u u t ωc1 ωc2 = q 1 ωc2 = ωo √ P 14.18 ωo = v u u t 2Q 1 2Q 1+ 1 2Q 1+ !2 !2 (121)(100) = 110 krad/s ωo = 17.51 kHz 2π fo = β = 121 − 100 = 21 krad/s or 2.79 kHz Q= P 14.19 β = 110 ωo = = 5.24 β 21 ωo 50,000 = = 12.5 krad/s; Q 4 1 ωc2 = 50,000 + 8 fc2 = s 1 1+ 8 = 56.64 krad/s 56.64 k = 9.01 kHz 2π 1 ωc1 = 50,000 − + 8 fc1 = 2 12,500 = 1.99 kHz 2π s 2 1 1+ 8 = 44.14 krad/s 44.14 k = 7.02 kHz 2π P 14.20 [a] ωo2 = R= 1 LC 1 = 79.16 mH [8000(2π)]2 (5 × 10−9 ) so L = ωo L 8000(2π)(79.16 × 10−3 ) = = 1.99 kΩ Q 2 1 [b] fc1 = 8000 − + 4 s 1 1+ = 6.25 kHz 16 © 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. Problems s 1 [c] fc2 = 8000 + 4 14–13 1 1+ = 10.25 kHz 16 [d] β = fc2 − fc1 = 4 kHz or fo 8000 β= = = 4 kHz Q 2 P 14.21 [a] We need ωc close to 2π(8000) = 50,265.48 rad/s. There are several possible approaches – this one starts by choosing L = 10 mH. Then, 1 = 39.58 nF [2π(8000)]2 (0.01) C= Use the closest value from Appendix H, which is 0.047 µF to give ωc = s 1 = 46,126.56 rad/s or fc = 7341.27 Hz (0.01)(47 × 10−9 ) Then, R = ωo L (46,126.56)(0.01) = = 230 Ω Q 2 Use the closest value from Appendix H, which is 220 Ω to give (46,126.56)(0.01) = 2.1 220 7341.27 − 8000 (100) = −8.23% [b] % error in fc = 8000 Q= % error in Q = P 14.22 [a] ωo2 = 2.1 − 2 (100) = 5% 2 1 1 = = 1010 −3 LC (10 × 10 )(10 × 10−9 ) ωo = 105 rad/s = 100 krad/s ωo 105 = = 15.9 kHz 2π 2π [c] Q = ωo RC = (100 × 103 )(8000)(10 × 10−9 ) = 8 [b] fo = [d] ωc1 = ωo − 1 + 2Q v u u t 1+ 1 2Q !2 ωc1 [e] .·. fc1 = = 14.95 kHz 2π 1 [f] ωc2 = ωo + 2Q v u u t 1 1+ 2Q !2 1 5 + = 10 − 16 1 5 + = 10 16 s s 1+ 1 = 93.95 krad/s 256 1 1+ = 106.45 krad/s 256 © 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. 14–14 CHAPTER 14. Introduction to Frequency-Selective Circuits ωc2 [g] .·. fc2 = = 16.94 kHz 2π ωo 105 [h] β = = = 12.5 krad/s or 1.99 kHz Q 8 P 14.23 [a] L = 1 ωo2 C 1 (50 × 10−9 )(20 1 = ωo + 2Q v u u t 1 1+ 2Q . ·. = 22.10 krad/s 1 ωc1 = ωo − + 2Q !2 = 50 mH s 1 1 + 1+ = 20,000 1 1+ 2Q . ·. 10 fc2 = v u u t = 18.10 krad/s [c] β = × 103 )2 5 Q = = 5 kΩ 3 ωo C (20 × 10 )(50 × 10−9 ) R= [b] ωc2 = !2 ωc2 = 3.52 kHz 2π 1 + = 20,000 − 10 fc1 = 100 ωc1 = 2.88 kHz 2π s 1 1+ 100 ωo 20,000 = = 4000 rad/s or 636.62 Hz Q 5 P 14.24 [a] We need ωc = 20,000 rad/s. There are several possible approaches – this one starts by choosing L = 1 mH. Then, C= 1 20,0002 (0.001) = 2.5 µF Use the closest value from Appendix H, which is 2.2 µF to give ωc = s 1 = 21,320 rad/s (0.001)(2.2 × 10−6 ) Then, R = Q 5 = = 106.6 Ω ωo C (21320)(2.2 × 10−6 ) Use the closest value from Appendix H, which is 100 Ω to give Q = 100(21,320)(2.2 × 10−6 ) = 4.69 [b] % error in ωc = % error in Q = 21,320 − 20,000 (100) = 6.6% 20,000 4.69 − 5 (100) = −6.2% 5 © 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. Problems 14–15 1 1 = = 625 × 106 −3 −9 LC (40 × 10 )(40 × 10 ) P 14.25 [a] ωo2 = ωo = 25 × 103 rad/s = 25 krad/s fo = [b] Q = [c] ωc1 25,000 = 3978.87 Hz 2π ωo L (25 × 103 )(40 × 10−3 ) = =5 R + Ri 200 v u u 1 = ωo − + t1 + 1 2Q 2Q !2 s 1 1 + 1+ = 25,000 − 10 100 = 22.62 krad/s or 3.60 kHz [d] wc2 1 = ωo + 2Q v u u t 1 1+ 2Q !2 s 1 1 + 1+ = 25,000 10 100 = 27.62 krad/s or 4.4 kHz [e] β = ωc2 − ωc1 = 27.62 − 22.62 = 5 krad/s or ωo 25,000 β= = = 5 krad/s or 795.77 Hz Q 5 P 14.26 [a] H(s) = (R/L)s (R+Ri ) s L 1 + + LC For the numerical values in Problem 14.25 we have 4500s H(s) = 2 s + 5000s + 625 × 106 s2 .·. H(jω) = H(jωo ) = (625 × 4500jω − ω 2 ) + j5000ω 106 j4500(25 × 103 ) = 0.9/0◦ j5000(25 × 103 ) .·. vo(t) = 500(0.9) cos 25,000 = 450 cos 25,000t mV [b] From the solution to Problem 14.25, ωc1 = 22.62 krad/s H(j22.62 k) = j4500(22.62 × 103 ) = 0.6364/45◦ (113.12 + j113.12) × 106 .·. vo(t) = 500(0.6364) cos(22,620t + 45◦ ) = 318.2 cos(22,620t + 45◦ ) mV © 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. 14–16 CHAPTER 14. Introduction to Frequency-Selective Circuits [c] From the solution to Problem 14.25, ωc2 = 27.62 krad/s H(j27.62 k) = j4500(27.62 × 103 ) = 0.6364/ − 45◦ 6 (−138.12 + j138.12) × 10 .·. vo(t) = 500(0.6364) cos(27,620t − 45◦ ) = 318.2 cos(27,620t − 45◦ ) mV P 14.27 [a] ωo = q 1/LC Q= ωo β β= R L so L = so β = 1 ωo2 C = 1 (50,000)2 (0.01 × 10−6 ) = 40 mH ωo 50,000 = = 6250 rad/s Q 8 so R = Lβ = (40 × 10−3 )(6250) = 250 Ω [b] From part (a), β = 6250 rad/s. Then, ωc1,2 β =± + 2 s β 6250 + ωo2 = ± + 2 2 ωc1 = 46,972.56 rad/s P 14.28 H(jω) = 50,0002 6250 2 2 + 50,0002 = ±3125 + 50,097.56 ωc2 = 53,222.56 rad/s jω(6250) − ω 2 + jω(6250) [a] H(j50,000) = Vo = (1)Vi s 50,0002 j50,000(6250) =1 − 50,0002 + j(50,000)(6250) .·. vo (t) = 50 cos 50,000t mV [b] H(j46,972.56) = 50,0002 1 Vo = √ /45◦ Vi 2 [c] H(j53,222.56) = j46,972.56(6250) 1 = √ /45◦ 2 − 46,972.56 + j(46,972.56)(6250) 2 .·. vo(t) = 35.36 cos(46,972.56t + 45◦ ) mV 50,0002 1 Vo = √ / − 45◦ Vi 2 j53,222.56(6250) 1 = √ / − 45◦ 2 − 53,222.56 + j(53,222.56)(6250) 2 .·. vo(t) = 35.36 cos(53,222.56t − 45◦ ) mV © 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. Problems [d] H(j10,000) = j10,000(6250) = 0.026/88.5◦ − 10,0002 + j(10,000)(6250) 50,0002 .·. vo (t) = 1.3 cos(10,000t + 88.5◦ ) mV Vo = 0.026/88.5◦ Vi [e] H(j250,000) = 50,0002 j250,000(6250) = 0.026/ − 88.5◦ − 250,0002 + j(250,000)(6250) Vo = 0.026/ − 88.5◦ Vi P 14.29 H(s) = 1 − 14–17 .·. vo (t) = 1.3 cos(250,000t − 88.5◦ ) mV s2 + (1/LC) (R/L)s = s2 + (R/L)s + (1/LC) s2 + (R/L)s + (1/LC) 50,0002 − ω 2 H(jω) = 50,0002 − ω 2 + jω(6250) [a] H(j50,000) = Vo = (0)Vi 50,0002 − 50,0002 =0 50,0002 − 50,0002 + j(50,000)(6250) .·. vo (t) = 0 mV [b] H(j46,972.56) = 50,0002 − 46,972.562 1 = √ / − 45◦ 2 2 50,000 − 46,972.56 + j(46,972.56)(6250) 2 1 Vo = √ / − 45◦ Vi 2 [c] H(j53,222.56) = 1 Vo = √ /45◦ Vi 2 [d] H(j10,000) = .·. vo(t) = 35.36 cos(46,972.56t − 45◦ ) mV 50,0002 − 53,222.562 1 = √ /45◦ 2 2 50,000 − 53,222.56 + j(53,222.56)(6250) 2 .·. vo(t) = 35.36 cos(53,222.56t + 45◦ ) mV 50,0002 − 10,0002 = 0.9997/ − 1.49◦ 50,0002 − 10,0002 + j(10,000)(6250) Vo = 0.9997/ − 1.49◦ Vi [e] H(j250,000) = .·. vo (t) = 49.98 cos(10,000t − 1.49◦ ) mV 50,0002 − 250,0002 = 0.9997/1.49◦ 2 2 50,000 − 250,000 + j(250,000)(6250) Vo = 0.9997/1.49◦ Vi .·. vo (t) = 49.98 cos(250,000t + 1.49◦ ) mV © 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. 14–18 CHAPTER 14. Introduction to Frequency-Selective Circuits P 14.30 [a] [b] L = 1 ωo2 C R= 1 = (50 × 103 )2 (20 × 10−4 ) = 20 mH ωo L (50 × 103 )(20 × 10−3 ) = = 160 Ω Q 6.25 [c] Re = 160k480 = 120 Ω Re + Ri = 120 + 80 = 200 Ω Qsystem = [d] βsystem = ωo L (50 × 103 )(20 × 10−3 ) = =5 Re + Ri 200 ωo Qsystem βsystem(Hz) = P 14.31 [a] = 50 × 103 = 10 krad/s 5 10,000 = 1591.55 Hz 2π Vo Z 1 = where Z = Vi Z +R Y and Y = sC + H(s) = = = = RL RLCs2 RL Ls + (R + RL )Ls + RRL (1/RC)s s2 + h R+RL RL 1 RC i s+ 1 LC RL R+RL 1 s R+RL RL RC h i 1 1 L s2 + R+R s + LC RL RC s2 Kβs , + βs + ωo2 R + RL [b] β = RL 1 1 LCRL s2 + sL + RL + = sL RL RL Ls K= RL , R + RL β= 1 (RkRL )C 1 RC © 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. Problems [c] βU = 14–19 1 RC R + RL R . ·. β L = βU = 1 + βU RL RL [d] Q = ωo ωo RC = R+R L β RL [e] QU = ωo RC . ·. Q L = [f] H(jω) = RL 1 QU = QU R + RL [1 + (R/RL )] Kjωβ ωo2 − ω 2 + jωβ H(jωo ) = K Let ωc represent a corner frequency. Then K Kωc β |H(jωc )| = √ = q 2 (ωo2 − ωc2 )2 + ωc2 β 2 1 ωc β . ·. √ = q 2 (ωo2 − ωc2 )2 + ωc2 β 2 Squaring both sides leads to (ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β .·. ωc2 ± ωc β − ωo2 = 0 or β ωc = ∓ ± 2 s β2 + ωo2 4 The two positive roots are β ωc1 = − + 2 s β2 + ωo2 4 β and ωc2 = + 2 s β2 + ωo2 4 where R β = 1+ RL P 14.32 [a] ωo2 = 1 1 and ωo2 = RC LC 1 1 = = 1012 −3 LC (5 × 10 )(200 × 10−12 ) ωo = 1 Mrad/s © 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. 14–20 CHAPTER 14. Introduction to Frequency-Selective Circuits R + RL 1 [b] β = · = RL RC 500 × 103 400 × 103 ωo 106 = = 16 β 62.5 × 103 RL [d] H(jωo ) = = 0.8/0◦ R + RL ! 1 3 (100 × 10 )(200 × 10−12 ) ! = 62.5 krad/s [c] Q = .·. vo(t) = 250(0.8) cos(106 t) = 200 cos 106 t mV R [e] β = 1 + RL 1 100 = 1+ (50 × 103 ) rad/s RC RL ωo = 106 rad/s Q= ωo 20 = β 1 + (100/RL ) where RL is in kilohms [f] P 14.33 ωo2 = 1 1 = = 1016 −6 LC (2 × 10 )(50 × 10−12 ) ωo = 100 Mrad/s QU = ωo RC = (100 × 106 )(2.4 × 103 )(50 × 10−12 ) = 12 .·. RL 12 = 7.5; R + RL 7.5 .·. RL = R = 4 kΩ 4.5 P 14.34 [a] In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage and we seek the steady-state nature of the output voltage vo . At zero frequency the inductor provides a direct connection between the input and the output, hence vo = vi when ω = 0. At infinite frequency the capacitor provides the direct connection, hence vo = vi when 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. Problems 14–21 At the resonant frequency of the parallel combination of L and C the impedance of the combination is infinite and hence the output voltage will be zero when ω = ωo . At frequencies on either side of ωo the amplitude of the output voltage will be nonzero but less than the amplitude of the input voltage. Thus the circuit behaves like a band-reject filter. [b] Let Z represent the impedance of the parallel branches L and C, thus Z= sL(1/sC) sL = 2 sL + 1/sC s LC + 1 Then H(s) = = H(s) = Vo R R(s2 LC + 1) = = Vi Z +R sL + R(s2 LC + 1) [s2 + (1/LC)] s2 + 1 RC s+ s2 + ωo2 s2 + βs + ωo2 1 LC [c] From part (b) we have H(jω) = ωo2 − ω 2 ωo2 − ω 2 + jωβ It follows that H(jω) = 0 when ω = ωo . 1 .·. ωo = √ LC [d] |H(jω)| = q ωo2 − ω 2 (ωo2 − ω 2 )2 + ω 2 β 2 1 |H(jω)| = √ when ω 2 β 2 = (ωo2 − ω 2)2 2 or ± ωβ = ωo2 − ω 2, thus ω 2 ± βω − ωo2 = 0 The two positive roots of this quadratic are ωc1 = −β + 2 v ! u u β 2 t v u β u β ωc2 = + t 2 2 2 !2 + ωo2 + ωo2 © 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. 14–22 CHAPTER 14. Introduction to Frequency-Selective Circuits Also note that since β = ωo /Q −1 ωc1 = ωo 2Q 1 ωc2 = ωo 2Q + v u u t + v u u t 1+ 1+ 1 2Q !2 1 2Q !2 [e] It follows from the equations derived in part (d) that β = ωc2 − ωc1 = 1/RC [f] By definition Q = ωo /β = ωo RC. P 14.35 [a] ωo2 = 1 1 = = 1012 −6 LC (50 × 10 )(20 × 10−9 ) .·. ωo = 1 Mrad/s ωo = 159.15 kHz 2π [c] Q = ωo RC = (106 )(750)(20 × 10−9 ) = 15 [b] fo = 1 [d] ωc1 = ωo − + 2Q v u u t 1 1+ 2Q !2 = 967.22 krad/s [e] fc1 = 1 6 + = 10 − 30 s 1 1+ 900 ωc1 = 153.94 kHz 2π 1 [f] ωc2 = ωo + 2Q v u u t 1 1+ 2Q = 1.03 Mrad/s !2 1 6 + = 10 30 s 1 1+ 900 ωc1 = 164.55 kHz 2π [h] β = fc2 − fc1 = 10.61 kHz [g] fc2 = P 14.36 [a] ωo = 2πfo = 8π krad/s L= R= 1 ωo2 C = 1 (8000π)2 (0.5 × 10−6 ) = 3.17 mH Q 5 = = 397.89 Ω ωo C (8000π)(0.5 × 10−6 ) © 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. Problems [b] fc2 1 = fo + 2Q v u u t 1 1+ 2Q !2 = 4.42 kHz v u 14–23 s 1 1 + 1+ = 4000 u 1 1 fc1 = fo − + t1 + 2Q 2Q 10 !2 = 3.62 kHz 1 + = 4000 − 10 100 s 1 1+ 100 [c] β = fc2 − fc1 = 800 Hz or fo 4000 β= = = 800 Hz Q 5 P 14.37 [a] Re = 397.89k1000 = 284.63 Ω Q = ωo Re C = (8000π)(284.63)(0.5 × 10−6 ) = 3.58 [b] β = fo 4000 = = 1.12 kHz Q 3.58 1 [c] fc2 = 4000 + 7.16 s 1 + [d] fc1 = 4000 − 7.16 1 = 4.60 kHz 1+ 7.162 s 1 = 3.48 kHz 1+ 7.162 P 14.38 [a] We need ωc = 2π(4000) = 25,132.74 rad/s. There are several possible approaches – this one starts by choosing L = 100 µH. Then, 1 C= = 15.83 µF 2 [2π(4000)] (100 × 10−6 ) Use the closest value from Appendix H, which is 22 µF, to give ωc = s 1 100 × Then, R = 10−6 )(22 × 10−6 ) = 21,320.07 rad/s so fc = 3393.19 Hz Q 5 = = 10.66 Ω ωo C (21320.07)(22 × 10−6 ) Use the closest value from Appendix H, which is 10 Ω, to give Q = 10(21,320.07)(22 × 10−6 ) = 4.69 [b] % error in fc = % error in Q = 3393.19 − 4000 (100) = −15.2% 4000 4.69 − 5 (100) = −6.2% 5 © 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. 14–24 CHAPTER 14. Introduction to Frequency-Selective Circuits P 14.39 [a] ωo = q 1/LC Q= ωo β β= R L so L = so β = 1 ωo2 C = 1 (25,000)2 (200 × 10−9 ) = 8 mH ωo 25,000 = = 10,000 rad/s Q 2.5 so R = Lβ = (8 × 10−3 )(10,000) = 80 Ω [b] From part (a), β = 10,000 rad/s. ωc1,2 β =± + 2 s β 10,000 + ωo2 = ± + 2 2 ωc1 = 20,495.1 rad/s P 14.40 H(jω) = s 10,000 2 2 + 25,0002 = ±5000 + 25,495.1 ωc2 = 30,495.1 rad/s ωo2 − ω 2 25,0002 − ω 2 = ωo2 − ω 2 + jωβ 25,0002 − ω 2 + jω(10,000) [a] H(j25,000) = Vo = (0)Vi 25,0002 − 25,0002 =0 25,0002 − 25,0002 + j(25,000)(10,000) .·. vo (t) = 0 mV [b] H(j20,495.1) = 25,0002 − 20,495.12 1 = √ / − 45◦ 2 2 25,000 − 20,495.1 + j(20,495.1)(10,000) 2 1 Vo = √ / − 45◦ Vi 2 [c] H(j30,495.1) = 1 Vo = √ /45◦ Vi 2 [d] H(j5000) = .·. vo(t) = 176.78 cos(20,495.1t − 45◦ ) mV 25,0002 − 30,495.12 1 = √ /45◦ 2 2 25,000 − 30,495.1 + j(30,495.1)(10,000) 2 .·. vo(t) = 176.78 cos(30,495.1t + 45◦ ) mV 25,0002 − 50002 = 0.9965/ − 4.76◦ 25,0002 − 50002 + j(5000)(10,000) Vo = 0.9965/ − 4.76◦ Vi .·. vo (t) = 249.1 cos(5000t − 4.76◦ ) mV © 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. Problems 25,0002 − 125,0002 = 0.9965/4.76◦ 25,0002 − 125,0002 + j(125,000)(10,000) [e] H(j125,000) = .·. vo (t) = 249.1 cos(125,000t + 4.76◦ ) mV Vo = 0.9965/4.76◦ Vi P 14.41 H(jω) = ωo2 jωβ jω(10,000) = 2 2 − ω + jωβ 25,000 − ω 2 + jω(10,000) [a] H(j25,000) = j(25,000)(10,000) =1 − 25,0002 + j(25,000)(10,000) 25,0002 .·. vo (t) = 250 cos 25,000t mV Vo = (1)Vi [b] H(j20,495.1) = 1 Vo = √ /45◦ Vi 2 [c] H(j30,495.1) = j(20,495.1)(10,000) 1 /45◦ √ = 2 2 25,000 − 20,495.1 + j(20,495.1)(10,000) 2 .·. vo(t) = 176.78 cos(20,495.1t + 45◦ ) mV j(30,495.1)(10,000) 1 / − 45◦ √ = 2 2 25,000 − 30,495.1 + j(30,495.1)(10,000) 2 1 Vo = √ / − 45◦ Vi 2 [d] H(j5000) = .·. vo(t) = 176.78 cos(30,495.1t − 45◦ ) mV j(5000)(10,000) = 0.083/85.24◦ 2 2 25,000 − 5000 + j(5000)(10,000) .·. vo (t) = 20.75 cos(5000t + 85.24◦ ) mV Vo = 0.083/85.24◦ Vi [e] H(j125,000) = 25,0002 j(125,000)(10,000) = 0.083/ − 85.24◦ − 125,0002 + j(125,000)(10,000) Vo = 0.083/ − 85.24◦ Vi P 14.42 [a] Let Z = Z= 14–25 .·. vo (t) = 20.75 cos(125,000t − 85.24◦ ) mV RL (sL + (1/sC)) RL + sL + (1/sC) RL (s2LC + 1) s2LC + RL Cs + 1 Then H(s) = Vo s2 RL CL + RL = Vi (R + RL )LCs2 + RRL Cs + R + RL Therefore H(s) = = RL [s2 + (1/LC)] h i · RRL s 1 R + RL s2 + R+R + L LC L K(s2 + ωo2 ) s2 + βs + ωo2 where K = RL ; R + RL ωo2 = 1 ; LC β= RRL R + RL 1 L © 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. 14–26 CHAPTER 14. Introduction to Frequency-Selective Circuits 1 [b] ωo = √ LC RRL 1 [c] β = R + RL L ωo ωo L [d] Q = = β [RRL /(R + RL )] K(ωo2 − ω 2 ) [e] H(jω) = 2 (ωo − ω 2 ) + jβω H(jωo ) = 0 [f] H(j0) = Kωo2 =K ωo2 [g] H(jω) = nh H(j∞) = [h] H(jω) = h i K (ωo /ω)2 − 1 i (ωo /ω)2 − 1 + jβ/ω −K =K −1 o K(ωo2 − ω 2) (ωo2 − ω 2 ) + jβω H(j0) = H(j∞) = K Let ωc represent a corner frequency. Then K |H(jωc )| = √ 2 K K(ωo2 − ωc2 ) . ·. √ = q 2 (ωo2 − ωc2 )2 + ωc2 β 2 Squaring both sides leads to (ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β .·. ωc2 ± ωc β − ωo2 = 0 or s β β2 ωc = ∓ ± + ωo2 2 4 The two positive roots are s β β2 β ωc1 = − + + ωo2 and ωc2 = + 2 4 2 where RRL 1 1 β= · and ωo2 = R + RL L LC s β2 + ωo2 4 © 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. Problems P 14.43 [a] ωo2 = 14–27 1 1 = = 0.25 × 1018 = 25 × 1016 −6 −12 LC (10 )(4 × 10 ) ωo = 5 × 108 = 500 Mrad/s β= RRL 1 (30)(150) 1 · = · −6 = 25 Mrad/s = 3.98 MHz R + RL L 180 10 Q= 500 M ωo = = 20 β 25 M [b] H(j0) = RL 150 = = 0.8333 R + RL 180 RL = 0.8333 R + RL H(j∞) = 250 1 [c] fc2 = + π 40 fc1 = s 1 1+ = 81.59 MHz 1600 250 1 − + π 40 s 1+ 1 = 77.61 MHz 1600 Check: β = fc2 − fc1 = 3.98 MHz. ωo 500 × 106 [d] Q = = RRL 1 β ·L R+R L 50 30 500(R + RL ) = 1+ RRL 3 RL where RL is in ohms. = [e] © 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. 14–28 CHAPTER 14. Introduction to Frequency-Selective Circuits P 14.44 [a] ωo2 = 1 = 625 × 106 LC . ·. L = 1 = 64 mH (625 × 106 )(25 × 10−9 ) RL = 0.9; R + RL . ·. .·. 0.1RL = 0.9R 500 .·. R = = 55.6 Ω 9 RL = 9R RL 1 [b] β = R · = 781.25 rad/s R + RL L ωo 25,000 = = 32 β 781.25 Q= P 14.45 [a] |H(jω)| = q . ·. 1010 (1010 − ω 2 )2 + (50,000ω)2 =1 1020 = (1010 − ω 2 )2 + (50,000ω)2 = −2 × 1010 ω 2 + ω 4 + 25 × 108 ω 2 . ·. ω 2 = 175 × 108 so ω = 132,287.57 rad/s [b] From the equation for |H(jω)| in part (a), the frequency for which the magnitude is maximum is the frequency for which the denominator is minimum. This is the frequency at which √ (1010 − ω 2 )2 = 0 so ω = 1010 = 100,000 rad/s [c] |H(j100,000)| = q 1010 (1010 − 100,0002 )2 + [50,000(100,000)]2 =2 P 14.46 [a] Use the cutoff frequencies to calculate the bandwidth: ωc1 = 2π(697) = 4379.38 rad/s Thus ωc2 = 2π(941) = 5912.48 rad/s β = ωc2 − ωc1 = 1533.10 rad/s Calculate inductance using Eq. (14.32) and capacitance using Eq. (14.31): L= C= R 600 = = 0.39 H β 1533.10 1 Lωc1 ωc2 = 1 = 0.10 µF (0.39)(4379.38)(5912.48) © 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. Problems 14–29 [b] At the outermost two frequencies in the low-frequency group (687 Hz and 941 Hz) the amplitudes are |V697Hz| = |V941Hz | = |Vpeak| √ = 0.707|Vpeak | 2 because these are cutoff frequencies. We calculate the amplitudes at the other two low frequencies using Eq. (14.32): ωβ |V | = (|Vpeak|)(|H(jω)|) = |Vpeak| q (ωo2 − ω 2 )2 + (ωβ)2 Therefore |V770Hz | = |Vpeak| = q (4838.05)(1533.10) (5088.522 − 4838.052 )2 + [(4838.05)(1533.10)]2 = 0.948|Vpeak | and |V852Hz | = |Vpeak| = q (5353.27)(1533.10) (5088.522 − 5353.272 )2 + [(5353.27)(1533.10)]2 = 0.948|Vpeak | It is not a coincidence that these two magnitudes are the same. The frequencies in both bands of the DTMF system were carefully chosen to produce this type of predictable behavior with linear filters. In other words, the frequencies were chosen to be equally far apart with respect to the response produced by a linear filter. Most musical scales consist of tones designed with this dame property – note intervals are selected to place the notes equally far apart. That is why the DTMF tones remind us of musical notes! Unlike musical scales, DTMF frequencies were selected to be harmonically unrelated, to lower the risk of misidentifying a tone’s frequency if the circuit elements are not perfectly linear. [c] The high-band frequency closest to the low-frequency band is 1209 Hz. The amplitude of a tone with this frequency is |V1209Hz| = |Vpeak| = q (7596.37)(1533.10) (5088.522 − 7596.372 )2 + [(7596.37)(1533.10)]2 = 0.344|Vpeak | This is less than one half the amplitude of the signals with the low-band cutoff frequencies, ensuring adequate separation of the bands. © 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. 14–30 CHAPTER 14. Introduction to Frequency-Selective Circuits P 14.47 The cutoff frequencies and bandwidth are ωc1 = 2π(1209) = 7596 rad/s ωc2 = 2π(1633) = 10.26 krad/s β = ωc2 − ωc1 = 2664 rad/s Telephone circuits always have R = 600 Ω. Therefore, the filters inductance and capacitance values are L= R 600 = = 0.225 H β 2664 C= 1 = 0.057 µF ωc1 ωc2 L At the highest of the low-band frequencies, 941 Hz, the amplitude is ωβ |Vω | = |Vpeak| q (ωo2 − ω 2 )2 + ω 2 β 2 where ωo = √ ωc1 ωc2 . Thus, |Vpeak|(5912)(2664) |Vω | = q [(8828)2 − (5912)2 ]2 + [(5912)(2664)]2 = 0.344 |Vpeak| Again it is not coincidental that this result is the same as the response of the low-band filter to the lowest of the high-band frequencies. P 14.48 From Problem 14.46 the response to the largest of the DTMF low-band tones is 0.948|Vpeak |. The response to the 20 Hz tone is |V20Hz | = [(50892 |Vpeak|(125.6)(1533) − 125.62 )2 + [(125.6)(1533)]2 ]1/2 = 0.00744|Vpeak | .·. |V20Hz | |V20Hz | 0.00744|Vpeak | = = = 0.5 |V770Hz | |V852Hz | 0.948|Vpeak | .·. |V20Hz | = 63.7|V770Hz | Thus, the 20Hz signal can be 63.7 times as large as the DTMF tones. © 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. 15 Active Filter Circuits Assessment Problems AP 15.1 H(s) = −(R2 /R1 )s s + (1/R1 C) 1 = 1 rad/s; R1 C R2 = 1, R1 .·. R1 = 1 Ω, .·. R2 = R1 = 1 Ω Hprototype(s) = AP 15.2 H(s) = .·. C = 1 F −s s+1 −(1/R1 C) −20,000 = s + (1/R2 C) s + 5000 1 = 20,000; R1 C .·. R1 = C = 5 µF 1 = 10 Ω (20,000)(5 × 10−6 ) 1 = 5000 R2 C .·. R2 = 1 = 40 Ω (5000)(5 × 10−6 ) © 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 15–1 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. 15–2 CHAPTER 15. Active Filter Circuits AP 15.3 ωc = 2πfc = 2π × 104 = 20,000π rad/s .·. kf = 20,000π = 62,831.85 C0 = C kf km .·. km = .·. 0.5 × 10−6 = 1 kf km 1 = 31.83 (0.5 × 10−6 )(62,831.85) AP 15.4 For a 2nd order Butterworth high pass filter H(s) = s2 √ s2 + 2s + 1 For the circuit in Fig. 15.25 H(s) = s2 s2 + 2 R2 C s+ 1 R1 R2 C 2 Equate the transfer functions. For C = 1F, √ 2 = 2, R2 C .·. R2 = 1 = 1, R1 R2 C 2 √ 2 = 0.707 Ω 1 .·. R1 = √ = 1.414 Ω 2 AP 15.5 Q = 8, K = 5, ωo = 1000 rad/s, C = 1 µF For the circuit in Fig 15.26 1 − s R1 C ! H(s) = R + R 2 2 1 s2 + s+ R3 C R1 R2 R3C 2 Kβs = 2 s + βs + ωo2 β= 2 , R3 C β= ωo 1000 = = 125 rad/s Q 8 .·. R3 = 2 βC © 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. Problems 15–3 2 × 106 .·. R3 = = 16 kΩ (125)(1) Kβ = 1 R1 C .·. R1 = 1 1 = = 1.6 kΩ KβC 5(125)(1 × 10−6 ) R1 + R2 R1 R2 R3 C 2 ωo2 = 106 = (1600 + R2 ) (1600)(R2 )(16,000)(10−6 )2 Solving for R2, R2 = (1600 + R2 )106 , 256 × 105 246R2 = 16,000, R2 = 65.04 Ω AP 15.6 ωo = 1000 rad/s; Q = 4; C = 2 µF s2 + (1/R2 C 2) # 4(1 − σ) 1 2 s + s+ RC R2 C 2 2 2 s + ωo 1 = 2 ; ωo = ; 2 s + βs + ωo RC H(s) = " R= 1 1 = = 500 Ω ωo C (1000)(2 × 10−6 ) β= ωo 1000 = = 250 Q 4 .·. β= 4(1 − σ) RC 4(1 − σ) = 250 RC 4(1 − σ) = 250RC = 250(500)(2 × 10−6 ) = 0.25 1−σ = 0.25 = 0.0625; 4 .·. σ = 0.9375 © 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. 15–4 CHAPTER 15. Active Filter Circuits Problems P 15.1 Summing the currents at the inverting input node yields 0 − Vi 0 − Vo + =0 Zi Zf .·. Vo Vi =− Zf Zi Vo Zf .·. H(s) = =− Vi Zi P 15.2 [a] Zf = R2 (1/sC2 ) R2 = [R2 + (1/sC2 )] R2 C2 s + 1 (1/C2 ) s + (1/R2 C2) Likewise (1/C1 ) Zi = s + (1/R1 C1 ) = .·. H(s) = −(1/C2 )[s + (1/R1 C1)] [s + (1/R2 C2)](1/C1 ) =− C1 [s + (1/R1 C1)] C2 [s + (1/R2 C2)] " −C1 jω + (1/R1 C1 ) [b] H(jω) = C2 jω + (1/R2 C2 ) H(j0) = −C1 C2 R2C2 R1C1 = # −R2 R1 ! C1 j −C1 [c] H(j∞) = − = C2 j C2 [d] As ω → 0 the two capacitor branches become open and the circuit reduces to a resistive inverting amplifier having a gain of −R2/R1 . As ω → ∞ the two capacitor branches approach a short circuit and in this case we encounter an indeterminate situation; namely vn → vi but vn = 0 because of the ideal op amp. At the same time the gain of the ideal op amp is infinite so we have the indeterminate form 0 · ∞. Although ω = ∞ is indeterminate we can reason that for finite large values of ω H(jω) will approach −C1 /C2 in value. In other words, the circuit approaches a purely capacitive inverting amplifier with a gain of (−1/jωC2 )/(1/jωC1 ) or −C1 /C2 . © 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. Problems P 15.3 15–5 (1/C2 ) s + (1/R2 C2 ) [a] Zf = Zi = R1 + 1 R1 = [s + (1/R1 C1 )] sC1 s H(s) = − s (1/C2 ) · [s + (1/R2 C2)] R1 [s + (1/R1 C1)] =− [b] H(jω) = − H(j0) = 0 1 s R1C2 [s + (1/R1 C1 )][s + (1/R2 C2 )] 1 R1C2 jω + jω 1 R1 C1 jω + 1 R2 C2 [c] H(j∞) = 0 [d] As ω → 0 the capacitor C1 disconnects vi from the circuit. Therefore vo = vn = 0. As ω → ∞ the capacitor short circuits the feedback network, thus ZF = 0 and therefore vo = 0. P 15.4 [a] ωc = 1 R2 C K= R2 R1 so R2 = so R1 = 1 1 = = 6366 Ω ωc C 2π(2500)(10 × 10−9 ) R2 6366 = = 1273 Ω K 5 [b] Both the cutoff frequency and the passband gain are changed. P 15.5 [a] 5(2) = 10 V so Vcc ≥ 10 V [b] H(jω) = −5(2π)(2500) jω + 2π(2500) H(j5000π) = −5(5000π) 5 = −2.5 + j2.5 = √ /135◦ 5000π + j5000π 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. 15–6 CHAPTER 15. Active Filter Circuits 10 Vo = √ /135◦ Vi 2 [c] H(j1000π) = −5(5000π) = 4.9/168.7◦ 5000π + j1000π Vo = 4.9/168.7◦ Vi [d] H(j25,000π) = so vo(t) = 9.8 cos(1000πt + 168.7◦ ) V −5(5000π) = 0.98/101.3◦ 5000π + j25,000π Vo = 0.98/101.3◦ Vi P 15.6 so vo (t) = 7.07 cos(5000πt + 135◦ ) V so vo (t) = 1.96 cos(25,000πt + 101.3◦ ) V [a] K = 10(10/20) = 3.16 = R2 R1 R2 = 1 1 = = 212.21 Ω 3 ωc C (2π)(10 )(750 × 10−9 ) R1 = R2 212.21 = = 67.16 Ω K 3.16 [b] P 15.7 [a] 1 = 2π(1000) so RC = 1.5915 × 10−4 RC There are several possible approaches. Here, choose Rf = 150 Ω. Then 1.5915 × 10−4 = 1.06 × 10−6 150 Choose C = 1 µF. This gives C= ωc = 1 = 6.67 × 103 rad/s so fc = 1061 Hz (150)(10−6 ) To get a passband gain of 10 dB, choose Ri = Rf 150 = = 47.47 Ω 3.16 3.16 © 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. Problems 15–7 Choose Ri = 47 Ω to give K = 20 log 10(150/47) = 10.08 dB. The resulting circuit is 1061 − 1000 (100) = 6.1% 1000 10.08 − 10 % error in passband gain = (100) = 0.8% 10 [b] % error in fc = P 15.8 [a] ωc = 1 R1 C K= R2 R1 so R1 = 1 1 = = 159 Ω ωc C 2π(4000)(250 × 10−9 ) so R2 = KR1 = (8)(159) = 1273 Ω [b] The passband gain changes but the cutoff frequency is unchanged. P 15.9 [a] 8(0.25) = 2 V so Vcc ≥ 2 V −8jω [b] H(jω) = jω + 8000π H(j600π) = −8(j8000π) 8 = √ / − 135◦ 8000π + j8000π 2 8 Vo = √ / − 135◦ Vi 2 [c] H(j1600π) = so vo (t) = 1.41 cos(8000πt − 135◦ ) V −8(j1600π) = 1.57/ − 101.3◦ 8000π + j1600π Vo = 1.57/ − 101.3◦ Vi so vo (t) = 392.2 cos(1600πt − 101.3◦ ) mV © 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. 15–8 CHAPTER 15. Active Filter Circuits [d] H(j40,000π) = −8(j40,000π) = 7.84/ − 168.7◦ 8000π + j40,000π Vo = 7.84/ − 168.7◦ Vi P 15.10 [a] R1 = so vo (t) = 1.96 cos(40,000πt − 168.7◦ ) V 1 1 = = 5.10 kΩ 3 ωc C (2π)(8 × 10 )(3.9 × 10−9 ) K = 10(14/20) = 5.01 = R2 R1 .·. R2 = 5.01R1 = 25.55 kΩ [b] P 15.11 [a] 1 = 2π(8000) so RC = 19.89 × 10−6 RC There are several possible approaches. Here, choose C = 0.047 µF. Then 19.89 × 10−6 = 423 0.047 × 10−6 Choose Ri = 390 Ω. This gives Ri = ωc = 1 = 54.56 krad/s so fc = 8.68 kHz (0.047 × 10−6 )(390) To get a passband gain of 14 dB, choose Rf = 5Ri = 5(390) = 1950 Ω Choose Rf = 1.8 kΩ to give a passband gain of 20 log 10(1800/390) = 13.3 dB. The resulting circuit is © 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. Problems [b] % error in fc = 15–9 8683.76 − 8000 (100) = 8.5% 8000 % error in passband gain = 13.3 − 14 (100) = −5.1% 14 P 15.12 For the RC circuit H(s) = s Vo = Vi s + (1/RC ) R0 = km R; C0 = C km kf 1 RC .·. R0 C 0 = = ; kf kf H 0 (s) = 1 = kf R0 C 0 s s (s/kf ) = = s + (1/R0 C 0) s + kf (s/kf ) + 1 For the RL circuit H(s) = s s + (R/L) R0 = km R; L0 = km L kf R0 = kf L0 H 0 (s) = s (s/kf ) = s + kf (s/kf ) + 1 R = kf L P 15.13 For the RC circuit H(s) = Vo (1/RC) = Vi s + (1/RC ) R0 = km R; C0 = C km kf C 1 1 .·. R0 C 0 = km R = RC = km kf kf kf © 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. 15–10 CHAPTER 15. Active Filter Circuits 1 = kf R0 C 0 H 0 (s) = (1/R0 C 0) kf = s + (1/R0 C 0) s + kf H 0 (s) = 1 (s/kf ) + 1 For the RL circuit R0 = km R; L0 = km R R0 = km = kf 0 L L kf R L H(s) = R/L s + R/L so km L kf = kf H 0 (s) = (R0 /L0 ) kf = 0 0 s + (R /L ) s + kf H 0 (s) = 1 (s/kf ) + 1 (R/L)s βs = s2 + (R/L)s + (1/LC) s2 + βs + ωo2 For the prototype circuit ωo = 1 and β = ωo /Q = 1/Q. For the scaled circuit P 15.14 H(s) = (R0 /L0 )s H (s) = 2 s + (R0 /L0 )s + (1/L0 C 0) 0 where R0 = km R; L0 = km C L; and C 0 = kf kf km R0 km R .·. = km = kf 0 L L kf R L = kf β kf2 1 kf km = = = kf2 km L0 C 0 LC LC kf Q0 = ωo0 kf ωo = =Q 0 β kf β © 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. Problems 15–11 therefore the Q of the scaled circuit is the same as the Q of the unscaled circuit. Also note β 0 = kf β. .·. H 0 (s) = s2 + H 0 (s) = 2 s kf P 15.15 [a] L = 1 H; R= kf s Q kf s+ Q 1 Q + 1 Q s kf s kf kf2 +1 C = 1F 1 1 = = 0.05 Ω Q 20 ωo0 = 40,000; ωo Thus, [b] kf = km = R0 5000 = = 100,000 R 0.05 R0 = km R = (0.05)(100,000) = 5 kΩ L0 = km 100,000 L= (1) = 2.5 H kf 40,000 C0 = 1 C = = 250 pF km kf (40,000)(100,000) [c] P 15.16 [a] Since ωo2 = 1/LC and ωo = 1 rad/s, C= 1 1 = F L Q [b] H(s) = s2 H(s) = (R/L)s + (R/L)s + (1/LC) s2 (1/Q)s + (1/Q)s + 1 © 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. 15–12 CHAPTER 15. Active Filter Circuits [c] In the prototype circuit R = 1 Ω; L = 16 H; .·. km = 1 = 0.0625 F L C= R0 = 10,000; R kf = ωo0 = 25,000 ωo Thus R0 = km R = 10 kΩ L0 = km 10,000 L= (16) = 6.4 H kf 25,000 C0 = C 0.0625 = = 250 pF km kf (10,000)(25,000) [d] [e] H 0 (s) = H 0 (s) = s 25,000 s2 1 16 2 + s 25,000 1 16 s 25,000 +1 1562.5s + 1562.5s + 625 × 106 P 15.17 [a] Using the first prototype ωo = 1 rad/s; km = C = 1 F; L = 1 H; R0 40,000 = = 1600; R 25 kf = R = 25 Ω ωo0 = 50,000 ωo Thus, R0 = km R = 40 kΩ; C0 = L0 = km 1600 L= (1) = 32 mH; kf 50,000 C 1 = = 12.5 nF km kf (1600)(50,000) Using the second prototype ωo = 1 rad/s; L= 1 = 40 mH; 25 C = 25 F R = 1Ω © 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. Problems R0 = 40,000; R kf = ωo0 = 50,000 ωo R0 = km R = 40 kΩ; L0 = km 40,000 L= (0.04) = 32 mH; kf 50,000 km = 15–13 Thus, C0 = C 25 = = 12.5 nF km kf (40,000)(50,000) [b] P 15.18 For the scaled circuit H 0 (s) = L0 = .·. s2 + s2 + km L; kf R0 L0 1 L0 C 0 s+ C0 = R L 1 L0 C 0 C km kf kf2 1 = ; L0 C 0 LC R0 .·. = kf L0 R0 = km R It follows then that 2 s + H 0 (s) = s2 + R L = 2 s kf kf s + s kf + kf2 LC 2 + = H(s)|s=s/kf R L kf2 LC 1 LC s kf + 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. 15–14 CHAPTER 15. Active Filter Circuits P 15.19 For the circuit in Fig. 15.31 H(s) = s2 + s RC s2 + 1 LC + It follows that 1 LC s2 + L01C 0 H (s) = 2 s + R0sC 0 + L01C 0 0 where R0 = km R; C0 = .·. km L; kf L0 = C km kf kf2 1 = L0 C 0 LC 1 kf = 0 0 RC RC 2 s + H 0 (s) = s2 + = 2 s kf kf RC + kf2 LC s kf kf2 LC s+ 2 1 RC = H(s)|s=s/kf 1 LC + s kf + 1 LC P 15.20 [a] For the circuit in Fig. P15.20(a) 1 s+ Vo s2 + 1 s H(s) = = = 1 1 2 + 1 s+1 Vi s +s+ Q Q s For the circuit in Fig. P15.18(b) H(s) = Qs + Qs Vo = Vi 1 + Qs + Qs = H(s) = Q(s2 + 1) Qs2 + s + Q s2 + 1 s2 + 1 Q s+1 © 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. Problems 2 s 50,000 2 s + 15 50,000 2 [b] H 0 (s) = 15–15 +1 s 50,000 8 +1 s + 25 × 10 s2 + 10,000s + 25 × 108 = P 15.21 For prototype circuit (a): H(s) = Vo Q Q = 1 = Vi Q + s+ 1 Q + s2s+1 s H(s) = Q(s2 + 1) s2 + 1 = Q(s2 + 1) + s s2 + Q1 s + 1 For prototype circuit (b): H(s) = = Vo 1 = Vi 1 + (s(s/Q) 2 +1) s2 + 1 s2 + P 15.22 [a] km = L0 = [b] 1 Q s+1 R0 1000 = = 1000; R 1 kf = C 1 = = 5000 0 km C (1000)(200 × 10−9 ) km 1000 (L) = (1) = 200 mH kf 5000 V − 10/s V V + + =0 1000 0.2s 1000 + (5 × 106 /s) V 1 5 s + + 1000 s 1000s + 5 × 106 = 1 100s V = 5(s + 5000) 10(s + 5000) = 2s2 + 10,000s + 25 × 106 s2 + 5000s + 12.5 × 106 Io = V 25(s + 5000) = 2 0.2s s(s + 5000s + 12.5 × 106 ) © 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