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CHAPTER TWO SOLUTIONS (a) 12 µs (b) 750 mJ (c) 1.13 kΩ (d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz Engineering Circuit Analysis, 6th Edition (g) 39 pA (h) 49 kΩ (i) 11.73 pA Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS (a) MW (b) 12.35 mm (c) 47 kW (d) 5.46 mA (e) 33 µJ (f) 5.33 nW (g) ns (h) 5.555 MW Engineering Circuit Analysis, 6th Edition (i) 32 mm Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS Motor power = 175 Hp (a) With 100% efficient mechanical to electrical power conversion, (175 Hp)[1 W/ (1/745.7 Hp)] = 130.5 kW (b) Running for hours, Energy = (130.5×103 W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ (c) A single battery has 430 kW-hr capacity We require (130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS The 400-mJ pulse lasts 20 ns (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 400 t (ns) 20 Then P = 400×10-3/20×10-9 = 20 MW (b) At 20 pulses per second, the average power is Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = W Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS The 1-mJ pulse lasts 75 fs (c) To compute the peak power, we assume the pulse shape is square: Energy (mJ) t (fs) 75 Then P = 1×10-3/75×10-15 = 13.33 GW (d) At 100 pulses per second, the average power is Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS The power drawn from the battery is (not quite drawn to scale): P (W) 10 t (min) 17 24 (a) Total energy (in J) expended is [6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ (b) The average power in Btu/hr is (6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS Total charge q = 18t2 – 2t4 C (a) q(2 s) = 40 C (b) To find the maximum charge within ≤ t ≤ s, we need to take the first and second derivitives: dq/dt = 36t – 8t3 = 0, leading to roots at 0, ± 2.121 s d2q/dt2 = 36 – 24t2 substituting t = 2.121 s into the expression for d2q/dt2, we obtain a value of –14.9, so that this root represents a maximum Thus, we find a maximum charge q = 40.5 C at t = 2.121 s (c) The rate of charge accumulation at t = s is dq/dt|t = 0.8 = 36(0.8) – 8(0.8)3 = 24.7 C/s (d) See Fig (a) and (b) (b) (a) Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS Referring to Fig 2.6c, - + 3e −5t A, i1 (t ) = 3t - + 3e A, t0 Thus, (a) i1(-0.2) = 6.155 A (b) i1 (0.2) = 3.466 A (c) To determine the instants at which i1 = 0, we must consider t < and t > separately: for t < 0, - + 3e-5t = leads to t = -0.2 ln (2/3) = +2.027 s (impossible) for t > 0, -2 + 3e3t = leads to t = (1/3) ln (2/3) = -0.135 s (impossible) Therefore, the current is never negative (d) The total charge passed left to right in the interval 0.08 < t < 0.1 s is q(t) 0.1 = ∫ = ∫ [− + 3e ]dt i (t )dt − 0.08 −5t − 0.08 = − + 3e −5t -0.08 + ∫ [− + 3e ]dt 0.1 3t + − + 3e3t 0.1 = 0.1351 + 0.1499 = 285 mC Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS Referring to Fig 2.28, (a) The average current over one period (10 s) is iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10 = 800 mA (b) The total charge transferred over the interval < t < 12 s is qtotal = 12 ∫ i (t )dt = -4(2) + 2(2) + 6(2) + 0(4) –4(2) = C (d) See Fig below q (C) 16 -8 10 12 t(s) 14 16 -16 Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 10 (a) Pabs = (+3.2 V)(-2 mA) = -6.4 mW (b) Pabs = (+6 V)(-20 A) = -120 W (or +6.4 mW supplied) (or +120 W supplied) (d) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60 W (e) Pabs = (4 sin 1000t V)(-8 cos 1000t mA)| t = ms Engineering Circuit Analysis, 6th Edition = +12.11 W Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 15 v01 1 0 vC1 −3 10 v 0 −1 2 02 q = vC , a = −1 , f = ; w = , b= ,d= iR1 0 −1 −2 iL −2 1 0 iR w = bq + d ∴ q′ = aq + f , w = bq + d , w′ = bq′ = baq + bf 1 1 0 1 10 0 −1 −3 −1 −4 0 −1 10 −1 = ba = −4 , b f = 0 −1 = 0 −1 −2 1 3 0 1 10 1 0 10 vC1 + vC + 10 −1 −4 vC1 −v − 4v + 2i C2 L vC + = C1 ∴ w′ = −4 9vC − 4iL iL 3vC1 + 10 3 0 10 ′ = vC1 + vC + 10, v′02 = −vC1 − 4vC + 2iL , i′R1 = 9v02 − 4iL , i′R = 3vC1 + 10 ∴ v01 Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 16 y1 q1 −3 cos 2π t 3 y q = q2 , a = −2 −2 , f = sin 2π t , y = , b = −1 1 y3 q3 −1 −1 −1 3 y4 10 2 −10 d = 1 , q / = a q + f , q = b y + d , y (0) = −5 3 y1 1 3 2 y1 + y2 + y4 + 7 y ∴ q = −1 1 + 1 = − y2 + y3 + y4 + ∴ q (0) = 11 y3 53 −1 −1 3 y 3 y1 − y2 − y3 + y4 + 3 −3 1 −21 + 11 + 106 + 1 97 q′(0) = −2 −2 11 + = −14 − 22 + 53 = 17 −7 + 33 −1 53 26 Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 17 v q = C iL v 4v′C = − C − iL − iR , 2i′L = vR − 3iR − 3iL + vs + vC , 18iR = 2i′L ∴ iR = i′L 9 vR v = −iR − iL − is ∴ vR = −6iR − 6iL − 6is = − i′L − 6iL − 6is , ∴ 4v′C = − C − iL − i ′L 9 Also, 2i′L = − i′L − 6iL − 6is − i′L − 3iL + vC + vs ∴ 3i′L = vC − 9iL + vs − 6is 3 1 1 1 ∴ i′L = vC − 3iL + vs − 2is (2) 4v′C = − vC − iL − vC + iL − vs + is 3 27 27 2 1 1 ∴ 4v′C = − vC − iL − vs + is ∴ v′C = − vC − iL − vs + is (1) 27 27 27 108 18 vs + is − −1/ 27 −1/ 108 18 ∴[ a ] = [f ]= 1/ − v − 2is s Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 18 (a) (b) 30 = 10i + 2i′ ∴ i′ = −5i + 15 i (0) = 0.5 A, a = −5 ∴ i = e −5 t 0.5 + e −5 t t ∫ 15e 5Z dz ∴ i = 0.5e −5t + e−5t 3(e5t − 1) = − 2.5e−5t A, t > (c) izero state = 3(1 − e −5t ) A, izero input = 0.5e −5t A (d) i f = 3, in = Ae −5t ∴ i = + Ae −5t , i (0) = 0.5 ∴ i = − 2.5e −5t A ∴ in = −2.5e−5t A, i f = A Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 19 ∴ × 10−5 v′C = 0.04t u (t ) − 0.025vC ∴ v′C = −500vC + 800tu (t ) ∴ a = −500 t ∴ vC = e −500t ∫ e500 Z 800Z dz = 800e −500 t t ∫ Ze 500 Z dz t ∴ vC = 800e −500 t 500 Z Z − e 500 500 t ∴ vC = 800e −500t e500t − + 2 500 500 500 1.6 (−1 + e −500t ) ∴ i2 = vC = × 10−3 t − 16 × 10 −6 (1 − e −500t ) A, t > ∴ vC = 1.6t + 500 200 Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 20 (a) 1000 500 100 20 A + u (t ), iL (0) = = 3 15 20 −50t 50t 500 dZ = (e − 1) e + 500e−50t 50 0.3i′L = −15iL + 100 + 50u (t ), i′L = −50iL + ∴ iL = e −50 t t 20 × + e−50t ∫ e50 Z 20 −50t e + 10(1 − e−50t ) t > 20 10 ∴ iL = u (−t ) + 10 − e−50t u (t ) A 3 ∴ iL = (b) 20 −50 t e A, iL , zero state = 10(1 − e −50t ) A 10 iLf = 10 A, iLn = − e −50t A iL , zero input = Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 21 vs = 100[u (t ) − u (t − 0.5)]cos π t V (0.8vs − vC ) ∴ v′C = −2.5vC + 200[u (t ) − u (t − 0.5)]cos π t 40 t < vC = 0.01v′C = ≤ t ≤ 0.5 vC = e −2.5 t t t × 200 ∫ e 2.5 Z cos π Z dz = 200e −2.5 t e 2.5 Z 6.25 + π (2.5cos π Z + π sin π Z) 0 200 200 (2.5cos π t + π sin π t ) − 2.5e −2.5t 6.25 + π 6.25 + π ∴ vC = 31.02 cos π t + 38.98sin π t − 31.02e −2.5t ∴ vC = 0.5 t > 0.5 vC = 200e −2.5 t e 2.5 Z 6.25 + π (2.5cos π Z + π sin π Z) 0 e1.25 ∴ vC = 200e −2.5t π− × 2.5 = 105.03 e−2.5t V 2 6.25 + π 6.25 + π Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 22 ∴ 1 v′ = − v + + 10u (t ) 40 (a) ∴ v′ = −10v + 200 + 400u (t ) (b) t < ∴ v = 20 V Z t > v = 20e−10t + 600e−10t ∫ e10 Z dz = 20e−10t + 60e −10t (e10t − 1) ∴ v = 20e (c) −10 t + 60 − 60e −10 t ∴ v = 60 − 40e−10t V v forced = 60V vnat = −40e−10t V v zero state = 60(1 − e−10t ) V vzero input = 20e−10t V Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 23 15′L = −30iL + vs ∴ i′L = −2iL + 4e −t [u (t ) − u (t − 0.5)] t ∴ iL = 4e−2t ∫ e Z e− Z [u (Z) − u (Z − 0.5)] dz t ≤ t ≤ 0.5 iL = 4e−2t ∫ e Z dZ = 4e−2t (et − 1) = 4e − t − 4e −2t A t ≥ 0.5 iL = 4e −2t 0.5 ∫ e Z dZ + = 4e−2t (e0.5 − 1) = 2.595− e −2t A Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 24 −8 [a] = , t = 0.01 10 −10 (a) −0.08 0.05 −0.08 0.05 −0.08 0.05 e− ta = I − + − 0.1 −0.1 0.1 −0.1 0.1 −0.1 1 −0.08 0.05 0.0114 −0.0090 0.0114 −0.009 −0.08 0.05 = − + − + 0 0.1 −0.1 −0.0180 0.0150 −0.018 0.015 0.1 −0.1 1.08 −0.05 0.0057 −0.0045 −0.00181 0.00147 1.0860 −0.0547 = + − = −0.1 1.1 −0.009 0.0075 0.00294 −0.00240 −0.1095 1.1079 (b) 1 −0.08 0.05 0.0057 −0.0045 −0.0003 0.0002 0.9254 0.0457 eta = + + + = 0 0.1 −0.1 −0.009 0.0075 0.0005 −0.0004 0.0915 0.9071 (c) 1.0860 −0.0547 0.9254 0.0457 1.0000 = e− ta eta = 1.0000 −0.1095 1.1079 0.0915 0.9071 Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 25 (a) (b) (c) x u (t ) −1 q = y , f = cos t , a = −1 z −u (t ) −2 −3 −1 ∴ x′ = x + y − z + u (t ), y′ = − y + z + cos t , z ′ = −2 x − y − z − u (t ) 2 −1 −4 q (0) = −3 , ∴ q /(0) = −1 −3 + = ∆t = 0.1 −2 −3 −1 −1 −0.4 1.6 q (0.1) = q (0) + 0.1q′(0) = −3 + 0.7 = −2.3 0.3 1.3 2 −4 1.8 ∆t = 0.05 ∴ q (0.5) = −3 + 0.05 = −2.65 1.15 −1 1.8 −3.65 q′(0.05) = −1 −2.65 + cos 0.05 = 7.0988 −2 −3 −1 1.15 −1 2.2 1.8 −3.65 1.6175 q (0.1) = q (0.05) + 0.05q′(0.05) = −2.65 + 0.05 7.0988 = −2.2951 1.15 2.2 1.26 Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 26 −1 −1 − s a = −1 Now, a − s I = −1 − s −1 −1 − s det[ ] = (−1 − s )( s + 2s + − 2) + 3(4 + + 3s ) = − s − 3s + 8s + 22 = By trial and error, Solve, or high school algebra (Hormer’s method) we find s = -3.48361 Now divide polynomial by s + 3.48361 Get quadratic, s2 – 0.48361s – 6.31592 = The remaining two roots are: s = -2.28282 and s = + 2.76641 Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 27 −3 −3 − s a= det( a − s I) = det = s + s + 12 − = −4 − s −4 Roots are s1 = −2, s2 = −5 Now, e s1t = uo + u1s1 , es2t = uo + u1s2 1 ∴ e −2t = uo − 2u1 , e −5t = uo − 5u1 ∴ e −2 t − e −5t = 3u1 , u1 = e −2t − e −5t 3 2 ∴ uo = e−2t + e−2t − e−5t , or uo = e−2 t − e −5t 3 3 −2 t −5t e − e − t − t −3 3 eta = uo I + u1a = + e − e −2t −5t 3 −4 e − e 3 −2 t −5 t −2 t −5 t −2 t −5 t e − e −e +e e − e 3 3 eta = −2t −5t −2t −5t −2t −5t e − e e − e − e + e 3 3 3 −2 t −5t −2t −5t e − e 3 e + e 3 ta ∴e = 1 −2 t −5 t e −2 t − e −5 t e + e 3 Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 28 (a) 10i′ = −36 − v i q= 1 v v′ = i − v 90 ∴ i′ = −0.1v − 3.6, v′ = 90i − 10v −0.1 −3.6 ∴a = , f = 90 −10 (b) −0.1 0 − s a − sI = , det(a − s I ) = s + 10 s + ∴ s1 = −1, s2 = −9 90 −10 − s (c) e − t = uo − u1 , e −9 t = uo − 9u1 ∴ e − t − e −9 t = 8u1 1 1 ∴ u1 = e − t − e −9t , uo = e − t − e −9t + e − t = e − t − e −9t 8 8 8 (d) 1 0 −t −9t −0.1 eta = uo I + u1a = (9e −t − e −9t ) + (e − e ) 90 −10 0 1 −t −9 t −t 9e − e 0 −0.1e + 0.1e −9t ta ∴ 8e = + 9e − t − e −9t 90e − t − 90e −9t −10e − t + 10e −9t 9e− t − e−9t −0.1e− t + 0.1e−9t 8e = − t −9 t −e − t + 9e −9t 90e − 90e 1 − t −9t e − e − e − t + e −9t 8 80 80 ∴ eta = − t −9t 90 e − t − 90 e −9t − e + e 8 ta t (e) q = e q (0) + e ta ta ∫e − Za q= 9e − e 90e − t − 90e −9t −t −9 t i (0) q (0) = = v(0) 36 −0.1e− t + 0.1e−9t 1 −e − t + 9e −9t 9 f (Z)dz 9e Z − e9 Z −0.1e Z + 0.1e9 Z −3.6 ∫0 90e Z − 90e9 Z −e Z + 9e9Z dz t 9e Z − e9 Z 9e −t − e −9t − 0.9e − t + 0.9e −9t ta ∴ q = −t − 0.45 e ∫0 90eZ − 90e9Z dZ 90e − 90e −9t − 9e −t + 81e −9t + eta t Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS 9t t 4.05e − t − 0.05e9t e −9+ ta 9e − = − 0.45e 9 −t −9 t 40.5 e 4.5 e − t t 90e − 10e − 90 + 10 4.05e − t − 0.05e −9t 0.45 9e − t − e −9t q= − −t −9 t 90e − t − 90e −9t 40.5e − 4.5e 80 0.1e − t + 0.1e −9t 9et − e9t − 9 −e − t + 9e −9t t 9t 90e − 10e − 80 4.05e − t −0.05e −9t = −t −4.5e −9t 4.05e 80 81 − e8t − 80e −t − 9e −8t + + e −9t − + e8t + 8e −t + 9e−8t − − 8e −9t 0.45 − 9 −t −8 t −9 t −t −8 t −9 t 8t 8t 810 − 10e − 800e − 810e + 10 + 800e − 90 + 10e + 80e + 810e − 90 − 720e 4.05e − t − 0.05e −9 t 0.45 71 + − 72e − t + e −9 t = − 9 −t −9 t −t −9 t 40.5e − 4.5e 640 − 720e + 80e 4.05e − t − 0.05e −9 t − 4.05e − t + 0.05e −9t = − −t −9 t −t −9 t 40.5e − 4.5e 36 − 40.5e + 4.5e −4 + 8.1e− t − 0.1e−9t ∴q = (t > 0) −t −9 t −36 + 81e − 9e Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc All Rights Reserved [...]... 1015 cm-3 L = 8 cm Y = 100 µm 100 µm 250 µm 8 cm Engineering Circuit Analysis, 6th Edition contact Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER THREE SOLUTIONS 1 Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER THREE SOLUTIONS 2 (a) six nodes; (b) nine branches Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill,... (c) path, yes – loop, no Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER THREE SOLUTIONS 4 (a) Five nodes; (b) seven branches; (c) path, yes – loop, no Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER THREE SOLUTIONS 5 (a) 3 A; (b) –3 A; (c) 0 Engineering Circuit Analysis, 6th Edition Copyright... approximately 470 Ω (c) Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 24 Top Left Circuit: I = (5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW Top Right Circuit: I = -(5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW Bottom Left Circuit: I = (-5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW Bottom Right Circuit: I = -(-5/10) mA... assume that it will Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 19 (a) Pabs = i2R = [20e-12t] 2 (1200) µW = [20e-1.2] 2 (1200) µW = 43.54 mW (b) Pabs = v2/R = [40 cos 20t] 2 / 1200 W = [40 cos 2] 2 / 1200 W keep in mind we are using radians = 230.9 mW (c) Pabs = v i = 8t 1.5 W = 253.0 mW Engineering Circuit Analysis, 6th Edition... Circuit: I = -(-5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 25 The voltage vout is given by vout = -10-3 vπ (1000) = - vπ Since vπ = vs = 0.01 cos 1000t V, we find that vout = - vπ = -0.001 cos 1000t Engineering Circuit Analysis, 6th Edition V Copyright ©2002 McGraw-Hill, Inc All Rights... Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 12 (a) Pabs = (40i)(3e-100t)| t = 8 ms = [ di (b) Pabs = 0.2 i = - 180 e −100t dt ( t (c) Pabs = 30∫ idt + 20 3e −100t 0 ) [ 360 e −100t ] 2 t = 8 ms 2 t = 8 ms = 72.68 W = - 36.34 W t = 8 ms t = 90e −100t ∫ 3e −100t ′ dt ′ + 60e −100t 0 Engineering Circuit. .. maximum power allowed, this fuse will provide adequate protection for the application circuitry If a fault occurs and the application circuitry attempts to draw too much power, 1000 W for example, the fuse will blow, no current will flow, and the application circuitry will be protected However, if the application circuitry tries to draw its maximum rated power (500 W), the fuse will also blow In practice,... error, Pmax is roughly (375 mV)(2.5 A) = 938 mW, or just under 1 W Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 14 Note that in the table below, only the –4-A source and the –3-A source are actually “absorbing” power; the remaining sources are supplying power to the circuit Source 2-V source 8-V source -4-A source 10-V source -3-A... –4 – 16 + 40 – 50 + 30 = 0, as demanded from conservation of energy Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 15 We are told that Vx = 1 V, and from Fig 2.33 we see that the current flowing through the dependent source (and hence through each element of the circuit) is 5Vx = 5 A We will compute absorbed power by using the current... vavg = [(+10)(20×10-3) + (-10)(20×10-3)]/(40×10-3) = 0 (c) iavg = vavg /R = 0 (d) pabs (e) pabs max avg = = 2 vmax = (10)2 / 50 = 2 W R 1 (+10) 2 (−10) 2 ⋅ + ⋅ 20 = 2 W 20 40 R R Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 21 We are given that the conductivity σ of copper is 5.8×107 S/m (a) 50 ft of #18 (18 AWG) copper ... 100 µm 100 µm 250 µm cm Engineering Circuit Analysis, 6th Edition contact Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER THREE SOLUTIONS Engineering Circuit Analysis, 6th Edition... Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER THREE SOLUTIONS (a) Five nodes; (b) seven branches; (c) path, yes – loop, no Engineering Circuit Analysis, ... + 60e −100t Engineering Circuit Analysis, 6th Edition ] = 27.63 W t = ms Copyright ©2002 McGraw-Hill, Inc All Rights Reserved CHAPTER TWO SOLUTIONS 13 (a) The short -circuit current is
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